aQuestion #1A=a

a2	6	8
3	b2	9
4	5	c2

${\displaystyle A=\left[\begin{array}{ccc}{a}^2& 6& 8\\ 3& b^2& 9\\ 4& 5& c^2\end{array}\right]}$ and B=a

2a	3	5
2	2b	6
1	4	2c-3

${\displaystyle B=\left[\begin{array}{ccc}2a& 3& 5\\ 2& 2b& 6\\ 1& 4& 2c-3\end{array}\right]}$If Tr(A)=Tr(B)${\displaystyle T_r(A)=T_r(B)}$, then the value of (a-1+b-1+c-1)${\displaystyle (a^{-1}+b^{-1}+c^{-1})}$ is ____. Source: Howard Anton Linear Algebra Textbook SolutionGiven: Tr(A)=Tr(B)${\displaystyle T_r(A)=T_r(B)}$⟹a2+b2+c2=2a+2b+2c-3${\displaystyle ⟹a^2+b^2+c^2=2a+2b+2c-3}$⟹(a-1)2+(b-1)2+(c-1)2=0${\displaystyle ⟹(a-1{)}^2+(b-1{)}^2+(c-1{)}^2=0}$⟹a=1,b=1,c=1${\displaystyle ⟹a=1,b=1,c=1}$ ∴(a-1+b-1+c-1)=1+1+1=

3

${\displaystyle \therefore (a^{-1}+b^{-1}+c^{-1})=1+1+1=\boxed{3}}$

Question #2The trace of a square matrix is defined to be the sum of its diagonal entries. If A is a 2 x 2 matrix such that the trace of A is 3 and the trace of A3${\displaystyle A^3}$ is -18${\displaystyle -18}$, then the value of the determinant of A is ___ Source: JEE (Advanced) 2020SolutionLet A=a

a	b
c	d

${\displaystyle A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}$ with tr(A)=a+d=3${\displaystyle tr(A)=a+d=3}$.Let det(A)=ad-bc=D${\displaystyle \det (A)=ad-bc=D}$ Now use the Cayley–Hamilton theorem for a 2×2 ${\displaystyle 2\times 2}$matrix: A2-(tr(A))A+(detA)I=0${\displaystyle A^2-(tr(A))A+(\det A)I=0}$,Since tr(A)=a+d=3${\displaystyle tr(A)=a+d=3}$,⟹A2-3A+DI=0${\displaystyle ⟹A^2-3A+DI=0}$⟹(A2-3A+DI)A=(0)A${\displaystyle ⟹(A^2-3A+DI)A=(0)A}$⟹A3-3A2+DA=0${\displaystyle ⟹A^3-3A^2+DA=0}$ From the earlier equation A2-3A+DI=0⟹A2=3A-DI…(1)${\displaystyle A^2-3A+DI=0⟹A^2=3A-DI\dots (1)}$ Using (1) in A3-3A2+DA=0${\displaystyle A^3-3A^2+DA=0}$⟹A3-3(3A-DI)+DA=0${\displaystyle ⟹A^3-3(3A-DI)+DA=0}$⟹A3-9A+3DI+DA=0${\displaystyle ⟹A^3-9A+3DI+DA=0}$ Take trace on both sides⟹A3-9A+3DI+DA=0${\displaystyle ⟹A^3-9A+3DI+DA=0}$⟹tr(A3)-9.tr(A)+3D.tr(I)+D.tr(A)=0${\displaystyle ⟹tr\left(A^3\right)-9.tr(A)+3D.tr(I)+D.tr(A)=0}$ Given: tr(A3)=-18${\displaystyle tr\left(A^3\right)=-18}$, tr(A)=3${\displaystyle tr(A)=3}$, tr(I)=2${\displaystyle tr(I)=2}$So, tr(A3)-9.tr(A)+3D.tr(I)+D.tr(A)=0${\displaystyle tr\left(A^3\right)-9.tr(A)+3D.tr(I)+D.tr(A)=0}$⟹-18-9×3+3D×2+D×3=0${\displaystyle ⟹-18-9\times 3+3D\times 2+D\times 3=0}$⟹9D=45${\displaystyle ⟹9D=45}$⟹D=

5

${\displaystyle ⟹D=\boxed{5}}$ Answer (5)

Question #3Let A=a

a	b
c	d

${\displaystyle A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}$ and B=a

p

q

≠a

0

0

${\displaystyle B=\left[\begin{array}{c}p\\ q\end{array}\right]\ne \left[\begin{array}{c}0\\ 0\end{array}\right]}$ such that AB=B${\displaystyle AB=B}$ and a+d=2021${\displaystyle a+d=2021}$, then the value of ad-bc${\displaystyle ad-bc}$ is equal to __. Source: JEE (Main) 2021 SolutionWe have, A=a

a	b
c	d

${\displaystyle A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}$ and B=a

p

q

≠a

0

0

${\displaystyle B=\left[\begin{array}{c}p\\ q\end{array}\right]\ne \left[\begin{array}{c}0\\ 0\end{array}\right]}$ and Also, AB=B${\displaystyle AB=B}$⟹(A-I)B=0${\displaystyle ⟹(A-I)B=0}$Since B≠0${\displaystyle B\ne 0}$, the homogeneous system (A−I)B=0${\displaystyle (A-I)B=0}$ has a non-trivial solution.∴det(A-I)=0${\displaystyle \therefore \det (A-I)=0}$ So,⟹det(A-I)=0${\displaystyle ⟹\det (A-I)=0}$⟹a

a-1	b
c	d-1

=0${\displaystyle ⟹\left|\begin{array}{cc}a-1& b\\ c& d-1\end{array}\right|=0}$⟹(a-1)(d-1)-bc=0${\displaystyle ⟹(a-1)(d-1)-bc=0}$⟹ad-a-d+1-bc=0${\displaystyle ⟹ad-a-d+1-bc=0}$⟹ad-bc=a+d-1${\displaystyle ⟹ad-bc=a+d-1}$ Given: a+d=2021${\displaystyle a+d=2021}$, we get,⟹ad-bc=2021-1${\displaystyle ⟹ad-bc=2021-1}$⟹ad-bc=

2020

${\displaystyle ⟹ad-bc=\boxed{2020}}$ Answer 2020

Question #4Let A=a

cosa	-sina
sina	cosa

, (a∈R)${\displaystyle A=\left[\begin{array}{cc}\cos a& -\sin a\\ \sin a& \cos a\end{array}\right],(a\in \mathbb{R})}$ such that A32=a

0	-1
1	0

${\displaystyle A^{32}=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]}$. Then, value of a${\displaystyle a}$ is 𝜋

m${\displaystyle \frac{𝜋}{m}}$. Find log2m${\displaystyle {\log }_{2}m}$.Source: JEE (Main) 2019 SolutionA=a

cosa	-sina
sina	cosa

, A2=a

cos2a	-sin2a
sin2a	cos2a

…, An=a

cosna	-sinna
sinna	cosna

${\displaystyle A=\left[\begin{array}{cc}\cos a& -\sin a\\ \sin a& \cos a\end{array}\right],A^2=\left[\begin{array}{cc}\cos 2a& -\sin 2a\\ \sin 2a& \cos 2a\end{array}\right]\dots ,A^n=\left[\begin{array}{cc}\cos na& -\sin na\\ \sin na& \cos na\end{array}\right]}$A32=a

cos(32a)	-sin(32a)
sin(a32)	cos(32a)

=a

0	-1
1	0

${\displaystyle A^{32}=\left[\begin{array}{cc}\cos (32a)& -\sin (32a)\\ \sin (a32)& \cos (32a)\end{array}\right]=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]}$ a

⟹	cos(32a)=0
	-sin(32a)=-1
⟹	32a=𝜋 2+2k𝜋	k∈Z
⟹	a=𝜋 64+k𝜋 16

${\displaystyle \begin{array}{rlr}⟹& \cos (32a)=0& \\ & -\sin (32a)=-1& \\ ⟹& 32a=\frac{𝜋}{2}+2k𝜋& k\in \mathbb{Z}\\ ⟹& a=\frac{𝜋}{64}+\frac{k𝜋}{16}& \end{array}}$ The least positive value⟹32a=𝜋

2⟹a=𝜋

64${\displaystyle ⟹32a=\frac{𝜋}{2}⟹a=\frac{𝜋}{64}}$; ∴m=64${\displaystyle \therefore m=64}$∴log2m=log2(64)=

6

${\displaystyle \therefore {\log }_{2}m={\log }_2(64)=\boxed{6}}$ Answer (6)

Question #5If a

1	1
0	1

.a

1	2
0	1

.a

1	3
0	1

…a

1	n-1
0	1

=a

1	78
0	1

${\displaystyle \left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right].\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right].\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]\dots \left[\begin{array}{cc}1& n-1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 78\\ 0& 1\end{array}\right]}$, then n${\displaystyle n}$ is ___Source: JEE (Main) 2019 Solutionn-1∑i=1a

1	i
0	1

=a

1	78
0	1

${\displaystyle \sum _{i=1}^{n-1}\left[\begin{array}{cc}1& i\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 78\\ 0& 1\end{array}\right]}$⟹a

1	1
0	1

.a

1	2
0	1

.a

1	3
0	1

…a

1	n-1
0	1

=a

1	78
0	1

${\displaystyle ⟹\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right].\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right].\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]\dots \left[\begin{array}{cc}1& n-1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 78\\ 0& 1\end{array}\right]}$⟹a

1	1+2+3+…+(n-1)
0	1

=a

1	78
0	1

${\displaystyle ⟹\left[\begin{array}{cc}1& 1+2+3+\dots +(n-1)\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 78\\ 0& 1\end{array}\right]}$⟹1+2+3+…+(n-1)=78${\displaystyle ⟹1+2+3+\dots +(n-1)=78}$⟹n(n-1)

2=78${\displaystyle ⟹\frac{n(n-1)}{2}=78}$⟹n=

13

${\displaystyle ⟹n=\boxed{13}}$ Answer (13)

Question #6How many 3×3${\displaystyle 3\times 3}$ matrices M with entries from {0,1,2}${\displaystyle \{0,1,2\}}$ are there, for which the sum of the diagonal entires of MTM${\displaystyle M^{T}M}$ is 5?Source: JEE (Advanced) 2017 SolutionM=a

a11	a12	a13
a21	a22	a23
a31	a32	a33

${\displaystyle M=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]}$, MT=a

a11	a21	a31
a12	a22	a32
a13	a23	a33

${\displaystyle M^T=\left[\begin{array}{ccc}{a}_{11}& a_{21}& a_{31}\\ a_{12}& a_{22}& a_{32}\\ a_{13}& a_{23}& a_{33}\end{array}\right]}$MTM=a

a211+a221+a231
	a212+a222+a232
		a213+a223+a233

${\displaystyle M^{T}M=\left[\begin{array}{ccc}{a}_{11}^2+a_{21}^2+a_{31}^2& & \\ & a_{12}^2+a_{22}^2+a_{32}^2& \\ & & a_{13}^2+a_{23}^2+a_{33}^2\end{array}\right]}$ tr(MTM)=a211+a221+a231+a212+a222+a232+a213+a223+a233=5${\displaystyle tr\left(M^{T}M\right)=a_{11}^2+a_{21}^2+a_{31}^2+a_{12}^2+a_{22}^2+a_{32}^2+a_{13}^2+a_{23}^2+a_{33}^2=5}$Case 1: 1+1+1+1+1+0+0+0+0${\displaystyle 1+1+1+1+1+0+0+0+0}$, 9!

5!4!=126${\displaystyle \frac{9!}{5!4!}=126}$Choose 5 positions for 1's Case 2: 4+1+0+0+0+0+0+0+0${\displaystyle 4+1+0+0+0+0+0+0+0}$, P(9,2)=9!

7!=72${\displaystyle P(9,2)=\frac{9!}{7!}=72}$Choose position of 2 (whose square is 4) and position of 1 Adding case 1 and case 2: 126+72=

198

${\displaystyle 126+72=\boxed{198}}$

Question #7If A is a symmetric matrix and B is a skew-symmetric matrix such that A+B=a

2	3
5	-1

${\displaystyle A+B=\left[\begin{array}{cc}2& 3\\ 5& -1\end{array}\right]}$, then -1×det(AB)${\displaystyle -1\times \det (AB)}$ is __Source: JEE Main 2019 SolutionAssume A+B=a

2	3
5	-1

=M${\displaystyle A+B=\left[\begin{array}{cc}2& 3\\ 5& -1\end{array}\right]=M}$Any square matrix can be uniquely written as:A=M+MT

2,B=M-MT

2${\displaystyle A=\frac{M+M^T}{2},B=\frac{M-M^T}{2}}$ First, MT=a

2	5
3	-1

${\displaystyle M^T=\left[\begin{array}{cc}2& 5\\ 3& -1\end{array}\right]}$, M=a

2	3
5	-1

${\displaystyle M=\left[\begin{array}{cc}2& 3\\ 5& -1\end{array}\right]}$ Compute A=1

2a

2+2	3+5
5+3	-1-1

=a

2	4
4	-1

${\displaystyle A=\frac{1}{2}\left[\begin{array}{cc}2+2& 3+5\\ 5+3& -1-1\end{array}\right]=\left[\begin{array}{cc}2& 4\\ 4& -1\end{array}\right]}$; Compute B=1

2${\displaystyle B=\frac{1}{2}}$a

0	-2
2	0

=a

0	-1
1	0

${\displaystyle \left[\begin{array}{cc}0& -2\\ 2& 0\end{array}\right]=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]}$ Now, det(A)=2(-1)-4×4=-18${\displaystyle \det (A)=2(-1)-4\times 4=-18}$;det(B)=0×0-(-1)×(1)=1${\displaystyle \det (B)=0\times 0-(-1)\times (1)=1}$ Now, det(AB)=det(A)det(B)=(-18)(1)=-18${\displaystyle \det (AB)=\det (A)\det (B)=(-18)(1)=-18}$∴-1×det(AB)=-1×-18=

18

${\displaystyle \therefore -1\times \det (AB)=-1\times -18=\boxed{18}}$ Answer (18)

Question #8If A=a

1	2	2
2	1	-2
a	2	b

${\displaystyle A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right]}$ is a matrix satisfying the equation AAT=9I${\displaystyle A{A}^T=9I}$, where I${\displaystyle I}$ is 3×3${\displaystyle 3\times 3}$ identity matrix, then ||a+b||${\displaystyle |a+b|}$ is __ Source: JEE Main 2015 SolutionGiven: A=a

1	2	2
2	1	-2
a	2	b

⟹AT=a

1	2	a
2	1	2
2	-2	b

${\displaystyle A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right]⟹A^T=\left[\begin{array}{ccc}1& 2& a\\ 2& 1& 2\\ 2& -2& b\end{array}\right]}$ and a

AAT=	a 1 2 2 2 1 -2 a 2 b a 1 2 a 2 1 2 2 -2 b
=	a 9 0 a+4+2b 0 9 2a+2-2b a+4+2b 2a+2-2b a2+4+b2

${\displaystyle \begin{array}{rl}A{A}^T=& \left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right]\left[\begin{array}{ccc}1& 2& a\\ 2& 1& 2\\ 2& -2& b\end{array}\right]\\ =& \left[\begin{array}{ccc}9& 0& a+4+2b\\ 0& 9& 2a+2-2b\\ a+4+2b& 2a+2-2b& a^2+4+b^2\end{array}\right]\end{array}}$ It is given that, AAT=9I${\displaystyle A{A}^T=9I}$a

⟹	a 9 0 a+4+2b 0 9 2a+2-2b a+4+2b 2a+2-2b a2+4+b2 =9a 1 0 0 0 1 0 0 0 1
⟹	a 9 0 a+4+2b 0 9 2a+2-2b a+4+2b 2a+2-2b a2+4+b2 =a 9 0 0 0 9 0 0 0 9

${\displaystyle \begin{array}{rl}⟹& \left[\begin{array}{ccc}9& 0& a+4+2b\\ 0& 9& 2a+2-2b\\ a+4+2b& 2a+2-2b& a^2+4+b^2\end{array}\right]=9\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ ⟹& \left[\begin{array}{ccc}9& 0& a+4+2b\\ 0& 9& 2a+2-2b\\ a+4+2b& 2a+2-2b& a^2+4+b^2\end{array}\right]=\left[\begin{array}{ccc}9& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right]\end{array}}$ On comparing we get, a+4+2b=0⟹a+2b=-4${\displaystyle a+4+2b=0⟹a+2b=-4}$2a+2-2b=0⟹a-b=-1${\displaystyle 2a+2-2b=0⟹a-b=-1}$ On solving a+2b=-4${\displaystyle a+2b=-4}$ and a-b=-1${\displaystyle a-b=-1}$, get a=-2,b=-1${\displaystyle a=-2,b=-1}$ ∴||a+b||=||-2-1||=

3

${\displaystyle \therefore |a+b|=|-2-1|=\boxed{3}}$ Answer (3)

Question #9Let A=a

0	2q	r
p	q	-r
p	-q	r

${\displaystyle A=\left(\begin{array}{ccc}0& 2q& r\\ p& q& -r\\ p& -q& r\end{array}\right)}$. If AAT=I3${\displaystyle A{A}^T=I_3}$, then find the value of ||p||-4${\displaystyle |p{|}^{-4}}$ where ||.||${\displaystyle |.|}$ is the absolute value function. Source: JEE Main 2019 Solution Given: AAT=I3${\displaystyle A{A}^T=I_3}$⟹a

0	2q	r
p	q	-r
p	-q	r

a

0	p	p
2q	q	-q
r	-r	r

=a

4q2+r2	2q2-r2	-2q2+r2
2q2-r2	p2+q2+r2	p2-q2-r2
-2q2+r2	p2-q2-r2	p2+q2+r2

=a

1	0	0
0	1	0
0	0	1

${\displaystyle ⟹\left[\begin{array}{ccc}0& 2q& r\\ p& q& -r\\ p& -q& r\end{array}\right]\left[\begin{array}{ccc}0& p& p\\ 2q& q& -q\\ r& -r& r\end{array}\right]=\left[\begin{array}{ccc}4q^2+r^2& 2q^2-r^2& -2q^2+r^2\\ 2q^2-r^2& p^2+q^2+r^2& p^2-q^2-r^2\\ -2q^2+r^2& p^2-q^2-r^2& p^2+q^2+r^2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$ 4q2+r2=p2+q2+r2=1${\displaystyle 4q^2+r^2=p^2+q^2+r^2=1}$2q2-r2=p2-q2-r2=0${\displaystyle 2q^2-r^2=p^2-q^2-r^2=0}$⟹p2=3q2${\displaystyle ⟹p^2=3q^2}$ and r2=2q2${\displaystyle r^2=2q^2}$4q2+r2=1⟹4q2+2q2=1⟹q2=1

6${\displaystyle 4q^2+r^2=1⟹4q^2+2q^2=1⟹q^2=\frac{1}{6}}$⟹p2+q2+r2=1⟹p2+1

6+2(1

6)=1⟹p2=1

2⟹p=±1

2${\displaystyle ⟹p^2+q^2+r^2=1⟹p^2+\frac{1}{6}+2\left(\frac{1}{6}\right)=1⟹p^2=\frac{1}{2}⟹p=\frac{\pm 1}{\sqrt{2}}}$∴||p||-4=|±1

2|-4=

4

${\displaystyle \therefore |p{|}^{-4}=|\frac{\pm 1}{\sqrt{2}}{|}^{-4}=\boxed{4}}$ Answer (4)

Question #10Let P=a

1	0	0
4	1	0
16	4	1

${\displaystyle P=\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]}$ and I${\displaystyle I}$ be the identity matrix of order 3${\displaystyle 3}$. If Q=[qij]${\displaystyle Q=[q_{ij}]}$ is a matrix, such that P50-Q=I${\displaystyle P^{50}-Q=I}$, then q31+q32

q21${\displaystyle \frac{q_{31}+q_{32}}{q_{21}}}$ equals __Source: JEE Advanced 2016 SolutionGiven: P=a

1	0	0
4	1	0
16	4	1

${\displaystyle P=\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]}$∴P2=a

1	0	0
4	1	0
16	4	1

a

1	0	0
4	1	0
16	4	1

=a

1	0	0
4+4	1	0
16+32	4+4	1

=a

1	0	0
4×2	1	0
16(1+2)	4×2	1

${\displaystyle \therefore P^2=\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 4+4& 1& 0\\ 16+32& 4+4& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 4\times 2& 1& 0\\ 16(1+2)& 4\times 2& 1\end{array}\right]}$ and P3=a

1	0	0
4×4	1	0
16(1+2)	4×4	1

a

1	0	0
4	1	0
16	4	1

=a

1	0	0
4×3	1	0
16(1+2+3)	4×3	1

${\displaystyle P^3=\left[\begin{array}{ccc}1& 0& 0\\ 4\times 4& 1& 0\\ 16(1+2)& 4\times 4& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 4\times 3& 1& 0\\ 16(1+2+3)& 4\times 3& 1\end{array}\right]}$ From symmetry, P50=a

1	0	0
4×50	1	0
16(1+2+3+…+50)	4×50	1

${\displaystyle P^{50}=\left[\begin{array}{ccc}1& 0& 0\\ 4\times 50& 1& 0\\ 16(1+2+3+\dots +50)& 4\times 50& 1\end{array}\right]}$ ∵P50-Q=I${\displaystyle \because P^{50}-Q=I}$ (Given)∴a

1	0	0
4×50	1	0
16(1+2+3+…+50)	4×50	1

-Q=I${\displaystyle \therefore \left[\begin{array}{ccc}1& 0& 0\\ 4\times 50& 1& 0\\ 16(1+2+3+\dots +50)& 4\times 50& 1\end{array}\right]-Q=I}$⟹a

1-q11	0-q12	0-q13
4×50-q21	1-q22	0-q23
16(50 2×51)-q31	4×50-q32	1-q33

=a

1	0	0
0	1	0
0	0	1

${\displaystyle ⟹\left[\begin{array}{ccc}1-q_{11}& 0-q_{12}& 0-q_{13}\\ 4\times 50-q_{21}& 1-q_{22}& 0-q_{23}\\ 16(\frac{50}{2}\times 51)-q_{31}& 4\times 50-q_{32}& 1-q_{33}\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$⟹200-q21=0,16(50

2×51)-q31=0,4×50-q32=0${\displaystyle ⟹200-q_{21}=0,16(\frac{50}{2}\times 51)-q_{31}=0,4\times 50-q_{32}=0}$∴q21=200${\displaystyle \therefore q_{21}=200}$, q32=200${\displaystyle q_{32}=200}$, q31=20400${\displaystyle q_{31}=20400}$ Thus, q31+q32

q21=20400+200

200=

103

${\displaystyle \frac{q_{31}+q_{32}}{q_{21}}=\frac{20400+200}{200}=\boxed{103}}$ Answer (103)

Question #11Let P=a

1	0	0
3	1	0
9	3	1

${\displaystyle P=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 9& 3& 1\end{array}\right]}$ and Q=[qij]${\displaystyle Q=[q_{ij}]}$ be two 3×3${\displaystyle 3\times 3}$ matrices such that Q-P5=I3${\displaystyle Q-P^5=I_3}$, Then q21+q31

q32${\displaystyle \frac{q_{21}+q_{31}}{q_{32}}}$ is __ Source: JEE Main 2019 Solution⟹P=a

1	0	0
3	1	0
9	3	1

=a

0	0	0
3	0	0
9	3	0

+a

1	0	0
0	1	0
0	0	1

${\displaystyle ⟹P=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 9& 3& 1\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 9& 3& 0\end{array}\right]+\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$⟹P=X+I${\displaystyle ⟹P=X+I}$ (Assume) Now, P5=(I+X)5${\displaystyle P^5=(I+X{)}^5}$⟹P5=I+C(5,1)(X1)+C(5,2)(X2)+C(5,3)(X3)+…${\displaystyle ⟹P^5=I+C(5,1)\left(X^1\right)+C(5,2)\left(X^2\right)+C(5,3)\left(X^3\right)+\dots }$∵In=I;${\displaystyle \because I^n=I;}$I.A=A${\displaystyle I.A=A}$ X=a

0	0	0
3	0	0
9	3	0

${\displaystyle X=\left[\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 9& 3& 0\end{array}\right]}$⟹X2=a

0	0	0
3	0	0
9	3	0

a

0	0	0
3	0	0
9	3	0

=a

0	0	0
0	0	0
9	0	0

${\displaystyle ⟹X^2=\left[\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 9& 3& 0\end{array}\right]\left[\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 9& 3& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 9& 0& 0\end{array}\right]}$⟹X4=X5=a

0	0	0
0	0	0
0	0	0

${\displaystyle ⟹X^4=X^5=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]}$ So, P5=I+5a

0	0	0
3	0	0
9	3	0

+10a

0	0	0
0	0	0
9	0	0

+0=a

1	0	0
15	1	0
135	15	1

${\displaystyle P^5=I+5\left[\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 9& 3& 0\end{array}\right]+10\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 9& 0& 0\end{array}\right]+0=\left[\begin{array}{ccc}1& 0& 0\\ 15& 1& 0\\ 135& 15& 1\end{array}\right]}$ and Q=I+P5=a

1	0	0
15	1	0
135	15	1

=[qij]${\displaystyle Q=I+P^5=\left[\begin{array}{ccc}1& 0& 0\\ 15& 1& 0\\ 135& 15& 1\end{array}\right]=[q_{ij}]}$∴ q21+q31

q32=15+135

15=

10

${\displaystyle \therefore \frac{q_{21}+q_{31}}{q_{32}}=\frac{15+135}{15}=\boxed{10}}$ Answer (10)

Question #12If P=a

3 2	1 2
-1 2	3 2

, A=a

1	1
0	1

${\displaystyle P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right],A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]}$ and Q=PAPT${\displaystyle Q=PA{P}^T}$ and X=(PTQ2005P)${\displaystyle X=\left(P^T{Q}^{2005}P\right)}$, then the sum of all entries in matrix X${\displaystyle X}$ is __Source: AIEEE 2005 SolutionNow, P=a

3 2	1 2
-1 2	3 2

⟹PT=a

3 2	-1 2
1 2	3 2

${\displaystyle P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]⟹P^T=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ \frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]}$⟹PTP=a

1	0
0	1

${\displaystyle ⟹P^{T}P=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]}$⟹PTP=I${\displaystyle ⟹P^{T}P=I}$⟹PT=P-1${\displaystyle ⟹P^T=P^{-1}}$ X=PTQ2005P${\displaystyle X=P^T{Q}^{2005}P}$=PTQQQ…Q⏠⏣⏣⏣⏡⏣⏣⏣⏢2005 timesP${\displaystyle =P^T\underset{2005\text{times}}{\underbrace{QQQ\dots Q}}P}$=PTPAPTPAPTPAPT…PAPT⏠⏣⏣⏣⏣⏣⏣⏡⏣⏣⏣⏣⏣⏣⏢2005 timesP${\displaystyle =P^T\underset{2005\text{times}}{\underbrace{PA{P}^{T}PA{P}^{T}PA{P}^T\dots PA{P}^T}}P}$=PT(PAPT)2005P${\displaystyle =P^T(PA{P}^T{)}^{2005}P}$=

PTP

A

PTP

A

PTP

A

PTP

…

PTP

A

PTP

${\displaystyle =\boxed{P^{T}P}A\boxed{P^{T}P}A\boxed{P^{T}P}A\boxed{P^{T}P}\dots \boxed{P^{T}P}A\boxed{P^{T}P}}$=IAIAIA…IA=A2005${\displaystyle =IAIAIA\dots IA=A^{2005}}$ A=a

1	1
0	1

⟹An=a

1	n
0	1

⟹A2005=a

1	2005
0	1

${\displaystyle A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]⟹A^n=\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]⟹A^{2005}=\left[\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right]}$ Now, sum=1+2005+0+1=

2007

${\displaystyle =1+2005+0+1=\boxed{2007}}$ Answer (2007)

Question #13If A=a

a	0
1	1

${\displaystyle A=\left[\begin{array}{cc}a& 0\\ 1& 1\end{array}\right]}$ and B=a

1	0
5	1

${\displaystyle B=\left[\begin{array}{cc}1& 0\\ 5& 1\end{array}\right]}$, then the number of all possible values of a${\displaystyle a}$ for which A2=B${\displaystyle A^2=B}$, is __Source: AIEEE 2003 SolutionGiven: A=a

a	0
1	1

${\displaystyle A=\left[\begin{array}{cc}a& 0\\ 1& 1\end{array}\right]}$ and B=a

1	0
5	1

${\displaystyle B=\left[\begin{array}{cc}1& 0\\ 5& 1\end{array}\right]}$⟹A2=a

a	0
1	1

a

a	0
1	1

=a

a2	0
a+1	1

${\displaystyle ⟹A^2=\left[\begin{array}{cc}a& 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}a& 0\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}{a}^2& 0\\ a+1& 1\end{array}\right]}$ Also, given A2=B${\displaystyle A^2=B}$⟹a

a2	0
a+1	1

=a

1	0
5	1

${\displaystyle ⟹\left[\begin{array}{cc}{a}^2& 0\\ a+1& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 5& 1\end{array}\right]}$⟹a2=1${\displaystyle ⟹a^2=1}$ and a+1=5${\displaystyle a+1=5}$ Which is not possible at the same time.∴${\displaystyle \therefore }$No real values of a${\displaystyle a}$ exists. Answer (0)

Question #14Let X${\displaystyle X}$ and Y${\displaystyle Y}$ be two arbitrary, 3×3${\displaystyle 3\times 3}$, non-zero, skew-symmetric matrices and Z${\displaystyle Z}$ be an arbitrary, 3×3${\displaystyle 3\times 3}$, non-zero, symmetric matrix. Then, which of the following matrices is/are skew-symmetric? (A) Y3Z4-Z4Y3${\displaystyle Y^3{Z}^4-Z^4{Y}^3}$(B) X44+Y44${\displaystyle X^{44}+Y^{44}}$(C) X4Z3-Z3X4${\displaystyle X^4{Z}^3-Z^3{X}^4}$(D) X23+Y23${\displaystyle X^{23}+Y^{23}}$ SolutionGiven: XT=-X${\displaystyle X^T=-X}$, YT=-Y${\displaystyle Y^T=-Y}$, ZT=Z${\displaystyle Z^T=Z}$ (A) Y3Z4-Z4Y3${\displaystyle Y^3{Z}^4-Z^4{Y}^3}$Let P=Y3Z4-Z4Y3${\displaystyle P=Y^3{Z}^4-Z^4{Y}^3}$a

PT	=(Y3Z4)T-(Z4Y3)T
	=(ZT)4(YT)3-(YT)3(ZT)4
	=(Z)4(-Y)3-(-Y)3(Z)4
	=-Z4Y3+Y3Z4
	=Y3Z4-Z4Y3
	=P

${\displaystyle \begin{array}{rl}{P}^T& =(Y^3{Z}^4{)}^T-(Z^4{Y}^3{)}^T\\ & =\left(Z^T{)}^4\right(Y^T{)}^3-\left(Y^T{)}^3\right(Z^T{)}^4\\ & =(Z{)}^4(-Y{)}^3-(-Y{)}^3(Z{)}^4\\ & =-Z^4{Y}^3+Y^3{Z}^4\\ & =Y^3{Z}^4-Z^4{Y}^3\\ & =P\end{array}}$ ∵PT=P${\displaystyle \because P^T=P}$. So, P${\displaystyle P}$ is a symmetric matrix. (B) X44+Y44${\displaystyle X^{44}+Y^{44}}$Let P=X44+Y44${\displaystyle P=X^{44}+Y^{44}}$a

PT	=(XT)44+(YT)44
	=(-X)44+(-Y)44
	=X44+Y44
	=P

${\displaystyle \begin{array}{rl}{P}^T& =(X^T{)}^{44}+(Y^T{)}^{44}\\ & =(-X{)}^{44}+(-Y{)}^{44}\\ & =X^{44}+Y^{44}\\ & =P\end{array}}$ ∵PT=P${\displaystyle \because P^T=P}$. So, P${\displaystyle P}$ is a symmetric matrix. (C) X4Z3-Z3X4${\displaystyle X^4{Z}^3-Z^3{X}^4}$Let P=X4Z3-Z3X4${\displaystyle P=X^4{Z}^3-Z^3{X}^4}$a

PT	=(X4Z3)T-(Z3X4)T
	=(ZT)3(XT)4-(XT)4(ZT)3
	=(Z)3(-X)4-(-X)4(Z)3
	=Z3X4-X4Z3
	=-(X4Z3-Z3X4)
	=-P

${\displaystyle \begin{array}{rl}{P}^T& =(X^4{Z}^3{)}^T-(Z^3{X}^4{)}^T\\ & =\left(Z^T{)}^3\right(X^T{)}^4-\left(X^T{)}^4\right(Z^T{)}^3\\ & =(Z{)}^3(-X{)}^4-(-X{)}^4(Z{)}^3\\ & =Z^3{X}^4-X^4{Z}^3\\ & =-(X^4{Z}^3-Z^3{X}^4)\\ & =-P\end{array}}$ ∵PT=-P${\displaystyle \because P^T=-P}$. So, P${\displaystyle P}$ is a skew-symmetric matrix. (D) X23+Y23${\displaystyle X^{23}+Y^{23}}$Let P=X23+Y23${\displaystyle P=X^{23}+Y^{23}}$a

PT	=(XT)23+(YT)23
	=(-X)23+(-Y)23
	=-X23-Y23
	=-P

${\displaystyle \begin{array}{rl}{P}^T& =(X^T{)}^{23}+(Y^T{)}^{23}\\ & =(-X{)}^{23}+(-Y{)}^{23}\\ & =-X^{23}-Y^{23}\\ & =-P\end{array}}$ ∵PT=-P${\displaystyle \because P^T=-P}$. So, P${\displaystyle P}$ is a skew-symmetric matrix. Answer (C,D)Source: JEE (Advanced) 2015

Question #15Let w≠1${\displaystyle w\ne 1}$ be the cube root of unity and S be the set of all non-singular matrices of the form a

1	a	b
w	1	c
w2	w	1

${\displaystyle \left[\begin{array}{ccc}1& a& b\\ w& 1& c\\ w^2& w& 1\end{array}\right]}$ where each of a, b and c is either w${\displaystyle w}$ or w2${\displaystyle w^2}$. Then the number of distinct matrices in the set S is ___ Solution||A||≠0${\displaystyle |A|\ne 0}$ (non-singular matrix)⟹a

1	a	b
w	1	c
w2	w	1

≠0${\displaystyle ⟹\left|\begin{array}{ccc}1& a& b\\ w& 1& c\\ w^2& w& 1\end{array}\right|\ne 0}$⟹1(1-cw)-a(w-cw2)+b(w2-w2)≠0${\displaystyle ⟹1(1-cw)-a(w-c{w}^2)+b(w^2-w^2)\ne 0}$⟹1-cw-aw+acw2≠0${\displaystyle ⟹1-cw-aw+ac{w}^2\ne 0}$⟹(1-cw)(1-aw)≠0${\displaystyle ⟹(1-cw)(1-aw)\ne 0}$⟹a≠1

w,c≠1

w${\displaystyle ⟹a\ne \frac{1}{w},c\ne \frac{1}{w}}$⟹a=w,c=w${\displaystyle ⟹a=w,c=w}$ and b∈{w,w2}${\displaystyle b\in \{w,w^2\}}$Hence, 2 solutions Answer

2

${\displaystyle \boxed{2}}$ Source: IIT JEE 2011https://jeeadv.ac.in/past_qps/2011_2.pdfQuestion 43 (Paper 2)

Question #16For 3×3${\displaystyle 3\times 3}$ matrices M and N, which of the following statement(s) is (are) NOT correct? (A) NTMN${\displaystyle N^{T}MN}$ is symmetric or skew symmetric, according as M is symmetric or skew symmetric(B) MN-NM${\displaystyle MN-NM}$ is skew symmetric for all symmetric matrices M and N(C) MN${\displaystyle MN}$ is symmetric for all symmetric matrices M and N(D) (adjM)(adjN)=adj (MN)${\displaystyle \left(\mathrm{a}\mathrm{d}\mathrm{j}M\right)\left(\mathrm{a}\mathrm{d}\mathrm{j}N\right)=\mathrm{a}\mathrm{d}\mathrm{j}(MN)}$ for all invertible matrices M and N SolutionNTMN${\displaystyle N^{T}MN}$ is symmetric or skew symmetric, according as M is symmetric or skew symmetricIf MT=M${\displaystyle M^T=M}$ (symmetric)Let X=NTMN${\displaystyle X=N^{T}MN}$⟹X=NTMN${\displaystyle ⟹X=N^{T}MN}$⟹X'=(NTMN)T${\displaystyle ⟹X\text{'}=(N^{T}MN{)}^T}$⟹X'=NTMTN=X${\displaystyle ⟹X\text{'}=N^T{M}^{T}N=X}$ If MT=-M${\displaystyle M^T=-M}$ (skew symmetric)XT=-X${\displaystyle X^T=-X}$⟹X=NTMN${\displaystyle ⟹X=N^{T}MN}$⟹X'=(NTMN)T${\displaystyle ⟹X\text{'}=(N^{T}MN{)}^T}$⟹X'=-(NTMN)=-X${\displaystyle ⟹X\text{'}=-\left(N^{T}MN\right)=-X}$Hence, (A) is correct. MN-NM${\displaystyle MN-NM}$ is skew symmetric for all symmetric matrices M and NMT=M${\displaystyle M^T=M}$, NT=N${\displaystyle N^T=N}$X=MN-NM${\displaystyle X=MN-NM}$⟹XT=(MN-NM)T${\displaystyle ⟹X^T=(MN-NM{)}^T}$=(MN)T-(NM)T=NTMT-MTNT=NM-MN=-X${\displaystyle =(MN{)}^T-(NM{)}^T=N^T{M}^T-M^T{N}^T=NM-MN=-X}$(B) is correct MN is symmetric for all symmetric matrices M and NXT=(MN)T=NTMT=NM${\displaystyle X^T=(MN{)}^T=N^T{M}^T=NM}$Since, in general, MN≠NM${\displaystyle MN\ne NM}$XT≠X${\displaystyle X^T\ne X}$Therefore, MN is not necessarily symmetric, so (C) is false. Actually, (adjMN)=adj (N)(adjM)${\displaystyle \left(\mathrm{a}\mathrm{d}\mathrm{j}MN\right)=\mathrm{a}\mathrm{d}\mathrm{j}(N)\left(\mathrm{a}\mathrm{d}\mathrm{j}M\right)}$ is a standard identity.(adjM)(adjN)≠adj (MN)${\displaystyle \left(\mathrm{a}\mathrm{d}\mathrm{j}M\right)\left(\mathrm{a}\mathrm{d}\mathrm{j}N\right)\ne \mathrm{a}\mathrm{d}\mathrm{j}(MN)}$ for all invertible matrices M and N (false)(D) is false Answer (C,D)JEE Advanced 2013

Question #17Let w${\displaystyle w}$ be the complex cube root of unity with w≠1${\displaystyle w\ne 1}$ and P=[pij]${\displaystyle P=[p_{ij}]}$ be a n×n${\displaystyle n\times n}$ matrix with pij=w(i+j)${\displaystyle p_{ij}=w^{(i+j)}}$. Then P2≠0${\displaystyle P^2\ne 0}$, when n=${\displaystyle n=}$__(A) 57 (B) 55 (C) 58 (D) 56 SolutionWe have pij=w(i+j)${\displaystyle p_{ij}=w^{(i+j)}}$, w3=1,w≠1${\displaystyle w^3=1,w\ne 1}$We need to check when P2≠0${\displaystyle P^2\ne 0}$ Compute an entry of P2${\displaystyle P^2}$P2=a

w2	1
1	w

a

w2	1
1	w

=a

w+1	?
?	?

≠0${\displaystyle P^2=\left[\begin{array}{cc}{w}^2& 1\\ 1& w\end{array}\right]\left[\begin{array}{cc}{w}^2& 1\\ 1& w\end{array}\right]=\left[\begin{array}{cc}w+1& ?\\ ?& ?\end{array}\right]\ne 0}$ Formally,(P2)ij=n∑k=1(pik)(pkj)=n∑k=1(w(i+k))(w(k+j))=w(i+j)n∑k=1w2k${\displaystyle (P^2{)}_{ij}=\sum _{k=1}^n(p_{ik})(p_{kj})=\sum _{k=1}^n(w^{(i+k)}\left)\right(w^{(k+j)})=w^{(i+j)}\sum _{k=1}^n{w}^{2k}}$Thus, P2=(n∑k=1w2k)P${\displaystyle P^2=\left(\sum _{k=1}^n{w}^{2k}\right)P}$Hence, (P2≠0)⇔(n∑k=1w2k≠0)${\displaystyle (P^2\ne 0)\iff (\sum _{k=1}^n{w}^{2k}\ne 0)}$ because P≠0${\displaystyle P\ne 0}$ Evaluate the sum now,Since w3=1${\displaystyle w^3=1}$, w2+w+w3=w2+w+1=0${\displaystyle w^2+w+w^3=w^2+w+1=0}$ The sequence repeats every 3 termsw2,w,1,w2,w,1,w2,w,1,w2,w,1,…${\displaystyle w^2,w,1,w^2,w,1,w^2,w,1,w^2,w,1,\dots }$ So, the sum is zero whenever (3∣n)${\displaystyle (3\mid n)}$ Now, check the options57≡0(mod3)⇒ sum=0${\displaystyle 57\equiv 0(\mathrm{mod}3)\Rightarrow \mathrm{s}\mathrm{u}\mathrm{m}=0}$55≡1(mod3)⇒ sum=w2≠0${\displaystyle 55\equiv 1(\mathrm{mod}3)\Rightarrow \mathrm{s}\mathrm{u}\mathrm{m}=w^2\ne 0}$58≡1(mod3)⇒ sum=w2≠0${\displaystyle 58\equiv 1(\mathrm{mod}3)\Rightarrow \mathrm{s}\mathrm{u}\mathrm{m}=w^2\ne 0}$56≡2(mod3)⇒ sum=w2+w=-1≠0${\displaystyle 56\equiv 2(\mathrm{mod}3)\Rightarrow \mathrm{s}\mathrm{u}\mathrm{m}=w^2+w=-1\ne 0}$ Answer (B,C,D)Source: JEE Advanced 2013

Question #18If A=a

a	b	c
b	c	a
c	a	b

${\displaystyle A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]}$ is orthogonal, where a,b,c${\displaystyle a,b,c}$ are positive and abc=1${\displaystyle abc=1}$, then value of (a3+b3+c3)${\displaystyle (a^3+b^3+c^3)}$ is __ SolutionGiven: A${\displaystyle A}$ is orthogonal.⟹ATA=I…(i)${\displaystyle ⟹A^{T}A=I\dots (i)}$⟹a

a	b	c
b	c	a
c	a	b

a

a	b	c
b	c	a
c	a	b

=a

1	0	0
0	1	0
0	0	1

${\displaystyle ⟹\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$⟹a

a2+b2+c2	ab+bc+ca	ab+bc+ca
ab+bc+ca	a2+b2+c2	ab+bc+ca
ab+bc+ca	ab+bc+ca	a2+b2+c2

=a

1	0	0
0	1	0
0	0	1

${\displaystyle ⟹\left[\begin{array}{ccc}{a}^2+b^2+c^2& ab+bc+ca& ab+bc+ca\\ ab+bc+ca& a^2+b^2+c^2& ab+bc+ca\\ ab+bc+ca& ab+bc+ca& a^2+b^2+c^2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$⟹a2+b2+c2=1andab+bc+ca=0${\displaystyle ⟹a^2+b^2+c^2=1\mathrm{a}\mathrm{n}\mathrm{d}ab+bc+ca=0}$…(ii)${\displaystyle \dots (ii)}$ Given: a,b,c${\displaystyle a,b,c}$ are positive and abc=1${\displaystyle abc=1}$We know, a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ca)${\displaystyle a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}$⟹a3+b3+c3=(a+b+c)(1-0)+3${\displaystyle ⟹a^3+b^3+c^3=(a+b+c)(1-0)+3}$⟹a3+b3+c3=(a+b+c)+3${\displaystyle ⟹a^3+b^3+c^3=(a+b+c)+3}$∴a3+b3+c3=(a+b+c)+3${\displaystyle \therefore a^3+b^3+c^3=(a+b+c)+3}$…(iii)${\displaystyle \dots (iii)}$ Now, (a+b+c)2=a2+b2+c2+2(ab+bc+ca)=1+0=1${\displaystyle (a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)=1+0=1}$…(iv)${\displaystyle \dots (iv)}$ From equation (iii),a3+b3+c3=(a+b+c)+3=1+3=4${\displaystyle a^3+b^3+c^3=(a+b+c)+3=1+3=4}$ Answer

4

${\displaystyle \boxed{4}}$Source: IIT JEE 2003

Question #19If the minimum and the maximum values of the function f:[𝜋

4,𝜋

2]→R${\displaystyle f:[\frac{𝜋}{4},\frac{𝜋}{2}]\to \mathbb{R}}$, defined by f(t)=a

-sin2t	-1-sin2t	1
-cos2t	-1-cos2t	1
12	10	-2

${\displaystyle f(t)=\left|\begin{array}{ccc}-{\sin }^{2}t& -1-{\sin }^{2}t& 1\\ -{\cos }^{2}t& -1-{\cos }^{2}t& 1\\ 12& 10& -2\end{array}\right|}$ are m${\displaystyle m}$ and M${\displaystyle M}$, respectively, then ||m+M||${\displaystyle |m+M|}$ is __ Solution Applying, R1→R1+R3${\displaystyle R_1\to R_1+R_3}$ and R2→R2+R3${\displaystyle R_2\to R_2+R_3}$, we getf(t)${\displaystyle f(t)}$=2a

6-sin2t	4-sin2t	0
6-cos2t	4-cos2t	0
6	5	-1

${\displaystyle =2\left|\begin{array}{ccc}6-{\sin }^{2}t& 4-{\sin }^{2}t& 0\\ 6-{\cos }^{2}t& 4-{\cos }^{2}t& 0\\ 6& 5& -1\end{array}\right|}$ ${\displaystyle }$=2×(-1)×((6-sin2t)(4-cos2t)-(4-sin2t)(6-cos2t))${\displaystyle =2\times (-1)\times \left(\right(6-{\sin }^{2}t\left)\right(4-{\cos }^{2}t)-(4-{\sin }^{2}t\left)\right(6-{\cos }^{2}t\left)\right)}$=-2(-2cos2t+2sin2t)${\displaystyle =-2(-2{\cos }^{2}t+2{\sin }^{2}t)}$=4cos2t${\displaystyle =4\cos 2t}$ As, t∈[𝜋

4,𝜋

2]${\displaystyle t\in [\frac{𝜋}{4},\frac{𝜋}{2}]}$⟹2t∈[𝜋

2,𝜋]${\displaystyle ⟹2t\in [\frac{𝜋}{2},𝜋]}$.∴f(t)|min=f(𝜋

2)=-4=m${\displaystyle \therefore f(t){|}_{\min }=f\left(\frac{𝜋}{2}\right)=-4=m}$ and, f(t)|max=f(𝜋

4)=0=M${\displaystyle f(t){|}_{\max }=f\left(\frac{𝜋}{4}\right)=0=M}$ ∴||m+M||=||-4+0||=4${\displaystyle \therefore |m+M|=|-4+0|=4}$ Answer

4

${\displaystyle \boxed{4}}$Source: JEE Main 2020

Question #20Let z=-1+3i

2${\displaystyle z=\frac{-1+\sqrt{3}i}{2}}$, where i=-1${\displaystyle i=\sqrt{-1}}$, and r,s∈{1,2,3}${\displaystyle r,s\in \{1,2,3\}}$. Let P=a

(-z)r	z2s
z2s	zr

${\displaystyle P=\left[\begin{array}{cc}(-z{)}^r& z^{2s}\\ z^{2s}& z^r\end{array}\right]}$ and I${\displaystyle I}$ be the identity matrix of order 2. Then, the total number of ordered pairs (r,s)${\displaystyle (r,s)}$ for which P2=-I${\displaystyle P^2=-I}$ is __ SolutionGiven: z=-1+3i

2${\displaystyle z=\frac{-1+\sqrt{3}i}{2}}$. Assume: w=-1+3i

2${\displaystyle w=\frac{-1+\sqrt{3}i}{2}}$∵P=a

(-w)r	w2s
w2s	wr

${\displaystyle \because P=\left[\begin{array}{cc}(-w{)}^r& w^{2s}\\ w^{2s}& w^r\end{array}\right]}$.⟹P2=a

(-w)r	w2s
w2s	wr

a

(-w)r	w2s
w2s	wr

=a

w2r+w4s	wr+2s((-1)r+1)
wr+2s((-1)r+1)	w4s+w2r

${\displaystyle ⟹P^2=\left[\begin{array}{cc}(-w{)}^r& w^{2s}\\ w^{2s}& w^r\end{array}\right]\left[\begin{array}{cc}(-w{)}^r& w^{2s}\\ w^{2s}& w^r\end{array}\right]=\left[\begin{array}{cc}{w}^{2r}+w^{4s}& w^{r+2s}((-1{)}^r+1)\\ w^{r+2s}((-1{)}^r+1)& w^{4s}+w^{2r}\end{array}\right]}$. Given, P2=-I${\displaystyle P^2=-I}$⟹a

w2r+w4s	wr+2s((-1)r+1)
wr+2s((-1)r+1)	w4s+w2r

=-a

1	0
0	1

${\displaystyle ⟹\left[\begin{array}{cc}{w}^{2r}+w^{4s}& w^{r+2s}((-1{)}^r+1)\\ w^{r+2s}((-1{)}^r+1)& w^{4s}+w^{2r}\end{array}\right]=-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]}$.⟹a

w2r+w4s	wr+2s((-1)r+1)
wr+2s((-1)r+1)	w4s+w2r

=a

-1	0
0	-1

${\displaystyle ⟹\left[\begin{array}{cc}{w}^{2r}+w^{4s}& w^{r+2s}((-1{)}^r+1)\\ w^{r+2s}((-1{)}^r+1)& w^{4s}+w^{2r}\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]}$. ∴w2r+w4s=-1${\displaystyle \therefore w^{2r}+w^{4s}=-1}$ and wr+2s((-1)r+1)=0${\displaystyle w^{r+2s}((-1{)}^r+1)=0}$Since, r∈{1,2,3}${\displaystyle r\in \{1,2,3\}}$ and (-1)r+1=0${\displaystyle (-1{)}^r+1=0}$⟹r∈{1,3}${\displaystyle ⟹r\in \{1,3\}}$ Also, ${\displaystyle }$w2r+w4s=-1${\displaystyle w^{2r}+w^{4s}=-1}$If r=1${\displaystyle r=1}$, then w2+w4s=-1${\displaystyle w^2+w^{4s}=-1}$, which is only possible when s=1${\displaystyle s=1}$As, w2+w4=-1${\displaystyle w^2+w^4=-1}$. ∴r=1,s=1${\displaystyle \therefore r=1,s=1}$ Again, if r=3${\displaystyle r=3}$, then w6+w4s=-1${\displaystyle w^6+w^{4s}=-1}$⟹1+w4s=-1${\displaystyle ⟹1+w^{4s}=-1}$.⟹w4s=-2${\displaystyle ⟹w^{4s}=-2}$ (Rejecting. Why?)Since w3=1${\displaystyle w^3=1}$, w4s∈{1,w,w2}${\displaystyle w^{4s}\in \{1,w,w^2\}}$ or simply, |w4s|=1≠2${\displaystyle \left|w^{4s}\right|=1\ne 2}$. So impossible.∴r≠3${\displaystyle \therefore r\ne 3}$.⟹(r,s)=(1,1)${\displaystyle ⟹(r,s)=(1,1)}$ is the only solution. Hence, total number of ordered pairs is 1. Answer

1

${\displaystyle \boxed{1}}$Source: JEE Advanced 2016

Question #21Ifa

x-4	2x	2x
2x	x-4	2x
2x	2x	x-4

=(A+Bx)(x-A)2${\displaystyle \left|\begin{array}{ccc}x-4& 2x& 2x\\ 2x& x-4& 2x\\ 2x& 2x& x-4\end{array}\right|=(A+Bx)(x-A{)}^2}$, and gcd(||A||,||B||)=1${\displaystyle \gcd (|A|,|B|)=1}$, then (||A||+B)${\displaystyle (|A|+B)}$ is __ Solutiona

x-4	2x	2x
2x	x-4	2x
2x	2x	x-4

=(A+Bx)(x-A)2${\displaystyle \left|\begin{array}{ccc}x-4& 2x& 2x\\ 2x& x-4& 2x\\ 2x& 2x& x-4\end{array}\right|=(A+Bx)(x-A{)}^2}$ Applying, R1→R1+R2+R3${\displaystyle R_1\to R_1+R_2+R_3}$=(5x-4)a

1	1	1
2x	x-4	2x
2x	2x	x-4

${\displaystyle =(5x-4)\left|\begin{array}{ccc}1& 1& 1\\ 2x& x-4& 2x\\ 2x& 2x& x-4\end{array}\right|}$ Applying, C2→C2-C1${\displaystyle C_2\to C_2-C_1}$, C3→C3-C1${\displaystyle C_3\to C_3-C_1}$=(5x-4)a

1	0	0
2x	-x-4	0
2x	0	-x-4

${\displaystyle =(5x-4)\left|\begin{array}{ccc}1& 0& 0\\ 2x& -x-4& 0\\ 2x& 0& -x-4\end{array}\right|}$=(5x-4)(x+4)2${\displaystyle =(5x-4)(x+4{)}^2}$ On comparing with (A+Bx)(x-A)2${\displaystyle (A+Bx)(x-A{)}^2}$,A=-4${\displaystyle A=-4}$, B=5${\displaystyle B=5}$ which satisfies gcd(||A||,||B||)=1${\displaystyle \gcd (|A|,|B|)=1}$ ∴(||A||+B)=(||-4||+5)=9${\displaystyle \therefore (|A|+B)=(|-4|+5)=9}$ Answer

9

${\displaystyle \boxed{9}}$Source: JEE Main 2018

Question #22The sum of the solutions of the equation a

1+sin2x	sin2x	sin2x
cos2x	1+cos2x	cos2x
4sin2x	4sin2x	1+4sin2x

=0${\displaystyle \left|\begin{array}{ccc}1+{\sin }^{2}x& {\sin }^{2}x& {\sin }^{2}x\\ {\cos }^{2}x& 1+{\cos }^{2}x& {\cos }^{2}x\\ 4\sin 2x& 4\sin 2x& 1+4\sin 2x\end{array}\right|=0}$ is m𝜋

n${\displaystyle \frac{m𝜋}{n}}$; where x∈[0,2𝜋]${\displaystyle x\in [0,2𝜋]}$, gcd(m,n)=1${\displaystyle \gcd (m,n)=1}$. The value of (m+n)${\displaystyle (m+n)}$ is ____ Solution Applying, R1→R1+R2${\displaystyle R_1\to R_1+R_2}$⟹a

1+sin2x+cos2x	sin2x+1+cos2x	sin2x+cos2x
cos2x	1+cos2x	cos2x
4sin2x	4sin2x	1+4sin2x

=0${\displaystyle ⟹\left|\begin{array}{ccc}1+{\sin }^{2}x+{\cos }^{2}x& {\sin }^{2}x+1+{\cos }^{2}x& {\sin }^{2}x+{\cos }^{2}x\\ {\cos }^{2}x& 1+{\cos }^{2}x& {\cos }^{2}x\\ 4\sin 2x& 4\sin 2x& 1+4\sin 2x\end{array}\right|=0}$.⟹a

2	2	1
cos2x	1+cos2x	cos2x
4sin2x	4sin2x	1+4sin2x

=0${\displaystyle ⟹\left|\begin{array}{ccc}2& 2& 1\\ {\cos }^{2}x& 1+{\cos }^{2}x& {\cos }^{2}x\\ 4\sin 2x& 4\sin 2x& 1+4\sin 2x\end{array}\right|=0}$. Applying, C1→C1-C2${\displaystyle C_1\to C_1-C_2}$⟹a

2-2	2	1
cos2x-1-cos2x	1+cos2x	cos2x
4sin2x-4sin2x	4sin2x	1+4sin2x

=0${\displaystyle ⟹\left|\begin{array}{ccc}2-2& 2& 1\\ {\cos }^{2}x-1-{\cos }^{2}x& 1+{\cos }^{2}x& {\cos }^{2}x\\ 4\sin 2x-4\sin 2x& 4\sin 2x& 1+4\sin 2x\end{array}\right|=0}$.⟹a

0	2	1
-1	1+cos2x	cos2x
0	4sin2x	1+4sin2x

=0${\displaystyle ⟹\left|\begin{array}{ccc}0& 2& 1\\ -1& 1+{\cos }^{2}x& {\cos }^{2}x\\ 0& 4\sin 2x& 1+4\sin 2x\end{array}\right|=0}$.⟹-(-1)a

2	1
4sin2x	1+4sin2x

=0${\displaystyle ⟹-(-1)\left|\begin{array}{cc}2& 1\\ 4\sin 2x& 1+4\sin 2x\end{array}\right|=0}$.⟹1(2+8sin2x-4sin2x)=0⟹2+4sin2x=0${\displaystyle ⟹1(2+8\sin 2x-4\sin 2x)=0⟹2+4\sin 2x=0}$.⟹sin2x=-1

2${\displaystyle ⟹\sin 2x=-\frac{1}{2}}$ Since x∈[0,2𝜋]${\displaystyle x\in [0,2𝜋]}$, 2x∈[0,4𝜋]${\displaystyle 2x\in [0,4𝜋]}$Thus, 2x∈{7𝜋

6,11𝜋

6,19𝜋

6,23𝜋

6}${\displaystyle 2x\in \{\frac{7𝜋}{6},\frac{11𝜋}{6},\frac{19𝜋}{6},\frac{23𝜋}{6}\}}$Hence, x∈{7𝜋

12,11𝜋

12,19𝜋

12,23𝜋

12}${\displaystyle x\in \{\frac{7𝜋}{12},\frac{11𝜋}{12},\frac{19𝜋}{12},\frac{23𝜋}{12}\}}$ The sum is 7+11+19+23

12=60𝜋

12=5𝜋${\displaystyle \frac{7+11+19+23}{12}=\frac{60𝜋}{12}=5𝜋}$Therefore, m=5${\displaystyle m=5}$, n=1${\displaystyle n=1}$. Hence, (m+n)=5+1=6${\displaystyle (m+n)=5+1=6}$ Answer

6

${\displaystyle \boxed{6}}$Source: JEE Main 2019

Question #23Let a${\displaystyle a}$ and b${\displaystyle b}$ be the roots of the equation x2+x+1=0${\displaystyle x^2+x+1=0}$. Then for y≠0${\displaystyle y\ne 0}$ in R${\displaystyle \mathbb{R}}$, a

y+1	a	b
a	y+b	1
b	1	y+a

${\displaystyle \left|\begin{array}{ccc}y+1& a& b\\ a& y+b& 1\\ b& 1& y+a\end{array}\right|}$ is equal to (A) y(y2-1)${\displaystyle y(y^2-1)}$(B) y(y2-3)${\displaystyle y(y^2-3)}$(C) y3${\displaystyle y^3}$(D) y3-1${\displaystyle y^3-1}$ SolutionGiven: a${\displaystyle a}$ and b${\displaystyle b}$ be the roots of the equation x2+x+1=0${\displaystyle x^2+x+1=0}$Then a+b=-1${\displaystyle a+b=-1}$ and ab=1${\displaystyle ab=1}$Also, (x2+x+1)=0${\displaystyle (x^2+x+1)=0}$⟹x∈{-1+3i

2,-1-3i

2}≡{a,b}${\displaystyle ⟹x\in \{\frac{-1+\sqrt{3}i}{2},\frac{-1-\sqrt{3}i}{2}\}\equiv \{a,b\}}$Also, a

y+1	a	b
a	y+b	1
b	1	y+a

${\displaystyle \left|\begin{array}{ccc}y+1& a& b\\ a& y+b& 1\\ b& 1& y+a\end{array}\right|}$ On applying, R1→R1+R2+R3${\displaystyle R_1\to R_1+R_2+R_3}$, =a

y+1+a+b	y+1+a+b	y+1+a+b
a	y+b	1
b	1	y+a

${\displaystyle =\left|\begin{array}{ccc}y+1+a+b& y+1+a+b& y+1+a+b\\ a& y+b& 1\\ b& 1& y+a\end{array}\right|}$=a

y	y	y
a	y+b	1
b	1	y+a

${\displaystyle =\left|\begin{array}{ccc}y& y& y\\ a& y+b& 1\\ b& 1& y+a\end{array}\right|}$; (∵a+b=-1)${\displaystyle (\because a+b=-1)}$ On applying, C2→C2-C1${\displaystyle C_2\to C_2-C_1}$ and C3→C3-C1${\displaystyle C_3\to C_3-C_1}$, we get=a

y	0	0
a	y+b-a	1-a
b	1-b	y+a-b

${\displaystyle =\left|\begin{array}{ccc}y& 0& 0\\ a& y+b-a& 1-a\\ b& 1-b& y+a-b\end{array}\right|}$=a

y	0	0
a	y+b-a	1-a
b	1-b	y+a-b

${\displaystyle =\left|\begin{array}{ccc}\boxed{y}& 0& 0\\ a& y+b-a& 1-a\\ b& 1-b& y+a-b\end{array}\right|}$ (expanding along row 1)=y((y+b-a)(y+a-b)-(1-a)(1-b))${\displaystyle =y((y+b-a)(y+a-b)-(1-a)(1-b))}$=y(y2-(b-a)2-(1-a-b+ab))${\displaystyle =y(y^2-(b-a{)}^2-(1-a-b+ab))}$.=y(y2-b2-a2+2ab-1+(a+b)-ab)${\displaystyle =y(y^2-b^2-a^2+2ab-1+(a+b)-ab)}$.=y(y2-(a+b)2+2ab+2ab-1+(a+b)-ab)${\displaystyle =y(y^2-(a+b{)}^2+2ab+2ab-1+(a+b)-ab)}$.=y(y2-1+3-1-1)${\displaystyle =y(y^2-1+3-1-1)}$. Reason: given a+b=-1${\displaystyle a+b=-1}$ and ab=1${\displaystyle ab=1}$.=y3${\displaystyle =y^3}$ Answer (C)JEE Main 2019

Question #24Let w${\displaystyle w}$ be a complex number such that 2w+1=z${\displaystyle 2w+1=z}$, where z=-3${\displaystyle z=\sqrt{-3}}$. If a

1	1	1
1	-w2-1	w2
1	w2	w7

=3k${\displaystyle \left|\begin{array}{ccc}1& 1& 1\\ 1& -w^2-1& w^2\\ 1& w^2& w^7\end{array}\right|=3k}$, then |k2|${\displaystyle \left|k^2\right|}$ SolutionGiven: 2w+1=z${\displaystyle 2w+1=z}$, where z=-3${\displaystyle z=\sqrt{-3}}$.⟹2w+1=-3${\displaystyle ⟹2w+1=\sqrt{-3}}$.⟹w=-1+3i

2${\displaystyle ⟹w=\frac{-1+\sqrt{3}i}{2}}$. Since, w${\displaystyle w}$ is a cube root of unit.∴w2=-1-3i

2${\displaystyle \therefore w^2=\frac{-1-\sqrt{3}i}{2}}$ and w3n=1${\displaystyle w^{3n}=1}$. Now, a

1	1	1
1	-w2-1	w2
1	w2	w7

=3k${\displaystyle \left|\begin{array}{ccc}1& 1& 1\\ 1& -w^2-1& w^2\\ 1& w^2& w^7\end{array}\right|=3k}$.⟹a

1	1	1
1	w	w2
1	w2	w

=3k${\displaystyle ⟹\left|\begin{array}{ccc}1& 1& 1\\ 1& w& w^2\\ 1& w^2& w\end{array}\right|=3k}$ (∵1+w+w2=0${\displaystyle (\because 1+w+w^2=0}$ and w7=(w3)2.w=w)${\displaystyle w^7=(w^3{)}^2.w=w)}$. On applying, R1→R1+R2+R3${\displaystyle R_1\to R_1+R_2+R_3}$, we get,a

3	1+w+w2	1+w+w2
1	-w2-1	w2
1	w2	w7

=3k${\displaystyle \left|\begin{array}{ccc}3& 1+w+w^2& 1+w+w^2\\ 1& -w^2-1& w^2\\ 1& w^2& w^7\end{array}\right|=3k}$.⟹a

3	0	0
1	w	w2
1	w2	w

=3k${\displaystyle ⟹\left|\begin{array}{ccc}3& 0& 0\\ 1& w& w^2\\ 1& w^2& w\end{array}\right|=3k}$.⟹3(w2-w4)=3k${\displaystyle ⟹3(w^2-w^4)=3k}$.⟹(w2-w)=k${\displaystyle ⟹(w^2-w)=k}$.⟹k=(-1-3i

2)-(-1+3i

2)${\displaystyle ⟹k=\left(\frac{-1-\sqrt{3}i}{2}\right)-\left(\frac{-1+\sqrt{3}i}{2}\right)}$.⟹k=-3i=-z${\displaystyle ⟹k=-\sqrt{3}i=-z}$. ∴|k2|=|(-3i)2|=3${\displaystyle \therefore \left|k^2\right|=\left|\right(-\sqrt{3}i{)}^2|=3}$.Answer

3

${\displaystyle \boxed{3}}$Source: JEE Main 2017

Question #25If A,B≠0${\displaystyle A,B\ne 0}$ and f(n)=An+Bn${\displaystyle f(n)=A^n+B^n}$ and a

3	1+f(1)	1+f(2)
1+f(1)	1+f(2)	1+f(3)
1+f(2)	1+f(3)	1+f(4)

=K(1-A)2(1-B)2(A-B)2${\displaystyle \left|\begin{array}{ccc}3& 1+f(1)& 1+f(2)\\ 1+f(1)& 1+f(2)& 1+f(3)\\ 1+f(2)& 1+f(3)& 1+f(4)\end{array}\right|=K(1-A{)}^2(1-B{)}^2(A-B{)}^2}$, then K${\displaystyle K}$ is equal to __ SolutionLet Δ=a

3	1+f(1)	1+f(2)
1+f(1)	1+f(2)	1+f(3)
1+f(2)	1+f(3)	1+f(4)

${\displaystyle \Delta =\left|\begin{array}{ccc}3& 1+f(1)& 1+f(2)\\ 1+f(1)& 1+f(2)& 1+f(3)\\ 1+f(2)& 1+f(3)& 1+f(4)\end{array}\right|}$.⟹Δ=a

1+1+1	1+A+B	1+A2+B2
1+A+B	1+A2+B2	1+A3+B3
1+A2+B2	1+A3+B3	1+A4+B4

${\displaystyle ⟹\Delta =\left|\begin{array}{ccc}1+1+1& 1+A+B& 1+A^2+B^2\\ 1+A+B& 1+A^2+B^2& 1+A^3+B^3\\ 1+A^2+B^2& 1+A^3+B^3& 1+A^4+B^4\end{array}\right|}$.⟹Δ=a

1.1+1.1+1.1	1.1+1.A+1.B	1.1+A.A+B.B
1.1+1.A+1.B	1.1+A.A+B.B	1.1+A.A2+B.B2
1.1+A.A+B.B	1.1+A.A2+B.B2	1.1+A2.A2+B2.B2

${\displaystyle ⟹\Delta =\left|\begin{array}{ccc}1.1+1.1+1.1& 1.1+1.A+1.B& 1.1+A.A+B.B\\ 1.1+1.A+1.B& 1.1+A.A+B.B& 1.1+A.A^2+B.B^2\\ 1.1+A.A+B.B& 1.1+A.A^2+B.B^2& 1.1+A^2.A^2+B^2.B^2\end{array}\right|}$.⟹Δ=a

1	1	1
1	A	B
1	A2	B2

a

1	1	1
1	A	B
1	A2	B2

=a

1	1	1
1	A	B
1	A2	B2

2${\displaystyle ⟹\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& A& B\\ 1& A^2& B^2\end{array}\right|\left|\begin{array}{ccc}1& 1& 1\\ 1& A& B\\ 1& A^2& B^2\end{array}\right|={\left|\begin{array}{ccc}1& 1& 1\\ 1& A& B\\ 1& A^2& B^2\end{array}\right|}^2}$. Using the

Vandermonde

determinant formula, we get:Δ=(B-A)2(1-A)2(1-B)2=(1-A)2(1-B)2(A-B)2${\displaystyle \Delta =(B-A{)}^2(1-A{)}^2(1-B{)}^2=(1-A{)}^2(1-B{)}^2(A-B{)}^2}$.But, K(1-A)2(1-B)2(A-B)2${\displaystyle K(1-A{)}^2(1-B{)}^2(A-B{)}^2}$ is given. Hence, K(1-A)2(1-B)2(A-B)2=1.(1-A)2(1-B)2(A-B)2${\displaystyle K(1-A{)}^2(1-B{)}^2(A-B{)}^2=1.(1-A{)}^2(1-B{)}^2(A-B{)}^2}$.∴K=1${\displaystyle \therefore K=1}$. Answer

1

${\displaystyle \boxed{1}}$Source: JEE Main 2014; DeepSeek

Question #26Let d∈R${\displaystyle d\in \mathbb{R}}$, and A=a

-2	4+d	(sint)-2
1	sint+2	d
5	(2sint)-d	(-sint)+2+2d

${\displaystyle A=\left[\begin{array}{ccc}-2& 4+d& (\sin t)-2\\ 1& \sin t+2& d\\ 5& (2\sin t)-d& (-\sin t)+2+2d\end{array}\right]}$, t∈[0,2𝜋]${\displaystyle t\in [0,2𝜋]}$. If the minimum value of det(A)${\displaystyle \det (A)}$ is 8${\displaystyle 8}$, then the absolute value of sum of all possible values for d${\displaystyle d}$ is __ SolutionA=a

-2	4+d	(sint)-2
1	sint+2	d
5	(2sint)-d	(-sint)+2+2d

${\displaystyle A=\left[\begin{array}{ccc}-2& 4+d& (\sin t)-2\\ 1& \sin t+2& d\\ 5& (2\sin t)-d& (-\sin t)+2+2d\end{array}\right]}$.∴det(A)=a

-2	4+d	(sint)-2
1	sint+2	d
5	(2sint)-d	(-sint)+2+2d

${\displaystyle \therefore \det (A)=\left|\begin{array}{ccc}-2& 4+d& (\sin t)-2\\ 1& \sin t+2& d\\ 5& (2\sin t)-d& (-\sin t)+2+2d\end{array}\right|}$. Applying, R3→R3-2R2+R1${\displaystyle R_3\to R_3-2R_2+R_1}$.=a

-2	4+d	(sint)-2
1	sint+2	d
1	0	0

${\displaystyle =\left|\begin{array}{ccc}-2& 4+d& (\sin t)-2\\ 1& \sin t+2& d\\ \boxed{1}& 0& 0\end{array}\right|}$.=1((4+d)d-sin2t+4)${\displaystyle =1((4+d)d-{\sin }^{2}t+4)}$=d2+4d+4-sin2t${\displaystyle =d^2+4d+4-{\sin }^{2}t}$.=(d+2)2-sin2t${\displaystyle =(d+2{)}^2-{\sin }^{2}t}$. Now det(A)${\displaystyle \det (A)}$ will be minimum if sin2t${\displaystyle {\sin }^{2}t}$ is maximum, that is, if sin2t${\displaystyle {\sin }^{2}t}$ takes value 1.∵(det(A))min=8${\displaystyle \because (\det (A){)}_{\min }=8}$ (given)So, (d+2)2-1=8${\displaystyle (d+2{)}^2-1=8}$⟹(d+2)2=9${\displaystyle ⟹(d+2{)}^2=9}$.⟹d+2=±3${\displaystyle ⟹d+2=\pm 3}$.⟹d∈{-5,1}${\displaystyle ⟹d\in \{-5,1\}}$ So, the absolute value of sum of all possible values for d${\displaystyle d}$ is |-5+1|=4${\displaystyle |-5+1|=4}$ Answer

4

${\displaystyle \boxed{4}}$Source: JEE Main 2019

Question #27 Let A=a

2	b	1
b	b2+1	b
1	b	2

${\displaystyle A=\left[\begin{array}{ccc}2& b& 1\\ b& b^2+1& b\\ 1& b& 2\end{array}\right]}$, where b>0${\displaystyle b>0}$. Then, (min(det(A)

b))×3${\displaystyle \left(\min \right(\frac{\det (A)}{b}\left)\right)\times \sqrt{3}}$ is __ SolutionGiven: A=a

2	b	1
b	b2+1	b
1	b	2

${\displaystyle A=\left[\begin{array}{ccc}2& b& 1\\ b& b^2+1& b\\ 1& b& 2\end{array}\right]}$, where b>0${\displaystyle b>0}$So, det(A)=a

2	b	1
b	b2+1	b
1	b	2

${\displaystyle \det (A)=\left|\begin{array}{ccc}2& b& 1\\ b& b^2+1& b\\ 1& b& 2\end{array}\right|}$=2(2(b2+1)-b2)-b(2b-b)+1(b2-b2-1)${\displaystyle =2\left(2\right(b^2+1)-b^2)-b(2b-b)+1(b^2-b^2-1)}$.=2(2b2+2-b2)-b2-1${\displaystyle =2(2b^2+2-b_{}^2)-b^2-1}$.=2b2+4-b2-1${\displaystyle =2b^2+4-b^2-1}$.=b2+3${\displaystyle =b^2+3}$. Now, det(A)

b=b2+3

b=b+3

b${\displaystyle \frac{\det (A)}{b}=\frac{b^2+3}{b}=b+\frac{3}{b}}$ Now, by AM≥GM${\displaystyle \mathrm{A}\mathrm{M}\ge \mathrm{G}\mathrm{M}}$, we get, ⟹(b+3

b)

2≥(b+3

b)1/2${\displaystyle ⟹\frac{(b+\frac{3}{b})}{2}\ge (b+\frac{3}{b}{)}^{1/2}}$; (∵b>0)${\displaystyle (\because b>0)}$.⟹b+3

b≥23${\displaystyle ⟹b+\frac{3}{b}\ge 2\sqrt{3}}$. So, min(det(A)

b)=23${\displaystyle \min \left(\frac{\det (A)}{b}\right)=2\sqrt{3}}$∴(min(det(A)

b))×3=23×3=6${\displaystyle \therefore \left(\min \right(\frac{\det (A)}{b}\left)\right)\times \sqrt{3}=2\sqrt{3}\times \sqrt{3}=6}$ Answer

6

${\displaystyle \boxed{6}}$JEE Main 2019

Question #28Let P=[aij]${\displaystyle P=[a_{ij}]}$ be a 3×3${\displaystyle 3\times 3}$ matrix and let Q=[bij]${\displaystyle Q=[b_{ij}]}$, where bij=2i+jaij${\displaystyle b_{ij}=2^{i+j}{a}_{ij}}$ for 1≤i${\displaystyle 1\le i}$, j≤3${\displaystyle j\le 3}$. If the det(P)=2${\displaystyle \det (P)=2}$, then log2(det(Q))${\displaystyle {\log }_2(\det (Q))}$ is __ SolutionGiven: P=[aij]${\displaystyle P=[a_{ij}]}$ be a 3×3${\displaystyle 3\times 3}$ matrixP=[aij]3×3=a

a11	a12	a13
a21	a22	a23
a31	a32	a33

${\displaystyle P=[a_{ij}{]}_{3\times 3}=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]}$; Q=[bij]3×3=a

b11	b12	b13
b21	b22	b23
b31	b32	b33

${\displaystyle Q=[b_{ij}{]}_{3\times 3}=\left[\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right]}$, where bij=2i+jaij${\displaystyle b_{ij}=2^{i+j}{a}_{ij}}$ for 1≤i${\displaystyle 1\le i}$, j≤3${\displaystyle j\le 3}$∴det(Q)=a

4a11	8a12	16a13
8a21	16a22	32a23
16a31	32a32	64a33

${\displaystyle \therefore \det (Q)=\left|\begin{array}{ccc}4a_{11}& 8a_{12}& 16a_{13}\\ 8a_{21}& 16a_{22}& 32a_{23}\\ 16a_{31}& 32a_{32}& 64a_{33}\end{array}\right|}$=4×8×16×a

a11	a12	a13
2a21	2a22	2a23
4a31	4a32	4a33

${\displaystyle =4\times 8\times 16\times \left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ 2a_{21}& 2a_{22}& 2a_{23}\\ 4a_{31}& 4a_{32}& 4a_{33}\end{array}\right|}$.=29×2×4a

a11	a12	a13
a21	a22	a23
a31	a32	a33

${\displaystyle =2^9\times 2\times 4\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|}$.=212.det(P)${\displaystyle =2^{12}.\det (P)}$.=212.21${\displaystyle =2^{12}{.2}^1}$ (given det(P)=2${\displaystyle \det (P)=2}$)=213${\displaystyle =2^{13}}$ ∴log2(det(Q))=log2(213)=13×log22=13${\displaystyle \therefore {\log }_2(\det (Q))={\log }_2\left(2^{13}\right)=13\times {\log }_{2}2=13}$ Answer

13

${\displaystyle \boxed{13}}$Source: AIEEE 2012

Question #29If A=a

a	2
2	a

${\displaystyle A=\left[\begin{array}{cc}a& 2\\ 2& a\end{array}\right]}$ and det(A3)=125${\displaystyle \det \left(A^3\right)=125}$, then the absolute value of product of all possible values for a${\displaystyle a}$ is __ SolutionWe know, det(An)=(det(A))n${\displaystyle \det \left(A^n\right)=(\det (A){)}^n}$Since, det(A3)=125${\displaystyle \det \left(A^3\right)=125}$⟹(det(A))3=53${\displaystyle ⟹(\det (A){)}^3=5^3}$.⟹det(A)=5${\displaystyle ⟹\det (A)=5}$; (rejecting the other two complex roots) Now, det(A)=5${\displaystyle \det (A)=5}$⟹a

a	2
2	a

=5${\displaystyle ⟹\left|\begin{array}{cc}a& 2\\ 2& a\end{array}\right|=5}$.⟹a2-4=5${\displaystyle ⟹a^2-4=5}$.⟹a=±3${\displaystyle ⟹a=\pm 3}$. So, the absolute value of product of all possible values for a${\displaystyle a}$ is ||-3×3||=9${\displaystyle |-3\times 3|=9}$ Answer

9

${\displaystyle \boxed{9}}$Source: AIEEE 2004

Question #30 The number of distinct real roots of a

sinx	cosx	cosx
cosx	sinx	cosx
cosx	cosx	sinx

=0${\displaystyle \left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ \cos x& \sin x& \cos x\\ \cos x& \cos x& \sin x\end{array}\right|=0}$, in the interval x∈[-𝜋

4,𝜋

4]${\displaystyle x\in [-\frac{𝜋}{4},\frac{𝜋}{4}]}$ is __ SolutionGiven: a

sinx	cosx	cosx
cosx	sinx	cosx
cosx	cosx	sinx

=0${\displaystyle \left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ \cos x& \sin x& \cos x\\ \cos x& \cos x& \sin x\end{array}\right|=0}$ Applying, C1→C1+C2+C3${\displaystyle C_1\to C_1+C_2+C_3}$⟹a

sinx+2cosx	cosx	cosx
sinx+2cosx	sinx	cosx
sinx+2cosx	cosx	sinx

=0${\displaystyle ⟹\left|\begin{array}{ccc}\sin x+2\cos x& \cos x& \cos x\\ \sin x+2\cos x& \sin x& \cos x\\ \sin x+2\cos x& \cos x& \sin x\end{array}\right|=0}$.⟹(sinx+2cosx)a

1	cosx	cosx
1	sinx	cosx
1	cosx	sinx

=0${\displaystyle ⟹(\sin x+2\cos x)\left|\begin{array}{ccc}1& \cos x& \cos x\\ 1& \sin x& \cos x\\ 1& \cos x& \sin x\end{array}\right|=0}$. Applying, R2→R2-R1${\displaystyle R_2\to R_2-R_1}$, R3→R3-R1${\displaystyle R_3\to R_3-R_1}$⟹(sinx+2cosx)a

1	cosx	cosx
0	sinx-cosx	0
0	0	sinx-cosx

=0${\displaystyle ⟹(\sin x+2\cos x)\left|\begin{array}{ccc}1& \cos x& \cos x\\ 0& \sin x-\cos x& 0\\ 0& 0& \sin x-\cos x\end{array}\right|=0}$.⟹(2cosx+sinx)(sinx-cosx)2=0${\displaystyle ⟹(2\cos x+\sin x)(\sin x-\cos x{)}^2=0}$.⟹2cosx+sinx=0or (sinx-cosx)2=0${\displaystyle ⟹2\cos x+\sin x=0\mathrm{o}\mathrm{r}(\sin x-\cos x{)}^2=0}$.⟹2cosx=-sinxor sinx=cosx${\displaystyle ⟹2\cos x=-\sin x\mathrm{o}\mathrm{r}\sin x=\cos x}$. Now, cotx=-1

2${\displaystyle \cot x=-\frac{1}{2}}$ gives no solution because x∈[-𝜋

4,𝜋

4]${\displaystyle x\in [-\frac{𝜋}{4},\frac{𝜋}{4}]}$and sinx=cosx⟹tanx=1⟹x=𝜋

4${\displaystyle \sin x=\cos x⟹\tan x=1⟹x=\frac{𝜋}{4}}$ So, The number of distinct real roots is 1${\displaystyle 1}$Answer

1

${\displaystyle \boxed{1}}$Source: AIEEE 2001

Question #31If f(x)=a

1	x	x+1
2x	x(x-1)	(x+1)x
3x(x-1)	x(x-1)(x-2)	(x+1)x(x-1)

${\displaystyle f(x)=\left|\begin{array}{ccc}1& x& x+1\\ 2x& x(x-1)& (x+1)x\\ 3x(x-1)& x(x-1)(x-2)& (x+1)x(x-1)\end{array}\right|}$, then f(100)${\displaystyle f(100)}$ is __ SolutionApplying C3→C3-(C1+C2)${\displaystyle C_3\to C_3-(C_1+C_2)}$f(x)=a

1	x	0
2x	x(x-1)	0
3x(x-1)	x(x-1)(x-2)	0

${\displaystyle f(x)=\left|\begin{array}{ccc}1& x& 0\\ 2x& x(x-1)& 0\\ 3x(x-1)& x(x-1)(x-2)& 0\end{array}\right|}$⟹f(x)=0${\displaystyle ⟹f(x)=0}$ ∴f(100)=0${\displaystyle \therefore f(100)=0}$ Answer

0

${\displaystyle \boxed{0}}$Source: AIEEE 1999

Question #32Which of the following is (are) not the square of a 3×3${\displaystyle 3\times 3}$ matrix with real entries?

(A) a 1 0 0 0 1 0 0 0 1 ${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$

(B) a 1 0 0 0 1 0 0 0 -1 ${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right]}$

(C) a 1 0 0 0 -1 0 0 0 -1 ${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right]}$

(D) a -1 0 0 0 -1 0 0 0 -1 ${\displaystyle \left[\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right]}$

SolutionTo determine which matrices are not the square of a 3×3${\displaystyle 3\times 3}$ matrix with real entries, we examine each option. Since all given matrices are diagonal, the square root matrix (if it exists with real entries) will also be diagonal with real entries. For any real square matrix A${\displaystyle A}$: det(A2)=(detA)2≥0${\displaystyle \det \left(A^2\right)=(\det A{)}^2\ge 0}$ Since, square of a real number is always non-negative If A2=B${\displaystyle A^2=B}$, then det(B)=det(A2)≥0${\displaystyle \det (B)=\det \left(A^2\right)\ge 0}$ So, if det(B)<0${\displaystyle \det (B)<0}$, B${\displaystyle B}$ cannot be the square of a real matrix. Option (A)a

1	0	0
0	1	0
0	0	1

=A2${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=A^2}$. Reason: I2=I${\displaystyle I^2=I}$ Option (B)a

1	0	0
0	1	0
0	0	-1

=A2${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right]=A^2}$⟹a

1	0	0
0	1	0
0	0	-1

=||A||2⟹||A||2=-1${\displaystyle ⟹\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right|=|A{|}^2⟹|A{|}^2=-1}$ Option (C)a

1	0	0
0	-1	0
0	0	-1

=||A||2⟹||A||2=1${\displaystyle \left|\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right|=|A{|}^2⟹|A{|}^2=1}$ Here, A=a

-1	0	0
0	0	-1
0	1	0

${\displaystyle A=\left[\begin{array}{ccc}-1& 0& 0\\ 0& 0& -1\\ 0& 1& 0\end{array}\right]}$⟹A2=a

1	0	0
0	-1	0
0	0	-1

${\displaystyle ⟹A^2=\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right]}$; Option (D)a

-1	0	0
0	-1	0
0	0	-1

=A2${\displaystyle \left[\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right]=A^2}$⟹a

-1	0	0
0	-1	0
0	0	-1

=||A||2⟹||A||2=-1${\displaystyle ⟹\left|\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right|=|A{|}^2⟹|A{|}^2=-1}$ Answer (B, D)Source: JEE Advanced 2017; Question 41 Paper 1https://jeeadv.ac.in/past_qps/2017_1.pdf

Question #33For positive numbers x${\displaystyle x}$, y${\displaystyle y}$ and z${\displaystyle z}$, the numerical value of the determinanta

1	logxy	logxz
logyx	1	logyz
logzx	logzy	1

${\displaystyle \left|\begin{array}{ccc}1& {\log }_{x}y& {\log }_{x}z\\ {\log }_{y}x& 1& {\log }_{y}z\\ {\log }_{z}x& {\log }_{z}y& 1\end{array}\right|}$ is __ SolutionLet Δ=a

1	logxy	logxz
logyx	1	logyz
logzx	logzy	1

${\displaystyle \Delta =\left|\begin{array}{ccc}1& {\log }_{x}y& {\log }_{x}z\\ {\log }_{y}x& 1& {\log }_{y}z\\ {\log }_{z}x& {\log }_{z}y& 1\end{array}\right|}$=a

1	lny lnx	lnz lnx
lnx lny	1	lnz lny
lnx lnz	lny lnz	1

${\displaystyle =\left|\begin{array}{ccc}1& \frac{\ln y}{\ln x}& \frac{\ln z}{\ln x}\\ \frac{\ln x}{\ln y}& 1& \frac{\ln z}{\ln y}\\ \frac{\ln x}{\ln z}& \frac{\ln y}{\ln z}& 1\end{array}\right|}$ On dividing and multiplying R1${\displaystyle R_1}$, R2${\displaystyle R_2}$, R3${\displaystyle R_3}$ by lnx${\displaystyle \ln x}$, lny${\displaystyle \ln y}$, lnz${\displaystyle \ln z}$, respectively, =1

lnxlnylnza

lnx	lny	lnz
lnx	lny	lnz
lnx	lny	lnz

=0${\displaystyle =\frac{1}{\ln x\ln y\ln z}\left|\begin{array}{ccc}\ln x& \ln y& \ln z\\ \ln x& \ln y& \ln z\\ \ln x& \ln y& \ln z\end{array}\right|=0}$ Answer

0

${\displaystyle \boxed{0}}$Source: JEE 1993

Question #34Let P be matrix of order 3×3${\displaystyle 3\times 3}$ such that all the entries in P are from the set {-1,0,1}${\displaystyle \{-1,0,1\}}$. Then, the maximum possible value of the determinant of P is ___ Solution Trying to compute the maximum possible value Each row of the matrix can have at most three entries of 1${\displaystyle 1}$. The length (magnitude) of such a row is 12+12+12=3${\displaystyle \sqrt{1^2+1^2+1^2}=\sqrt{3}}$ For a 3×3${\displaystyle 3\times 3}$ matrix, the maximum possible determinant is the product of the lengths of the three rows: Max possible determinant=3×3×3=33≈5.19${\displaystyle =\sqrt{3}\times \sqrt{3}\times \sqrt{3}=3\sqrt{3}\approx 5.19}$ Next, narrow down the integer options Since the entries are integers {−1,0,1}${\displaystyle \{-1,0,1\}}$ the determinant must be an integer. The largest integer less than 5.19${\displaystyle 5.19}$ could be 5 or4${\displaystyle 5\mathrm{o}\mathrm{r}4}$.det(P)≤5${\displaystyle \det (P)\le 5}$. Now, Case 1: Matrix has at least one zero. If any row has a zero, its length ≤2${\displaystyle \le 2}$. Then:det(P)≤3×3×2=32≈4.24${\displaystyle \det (P)\le \sqrt{3}\times \sqrt{3}\times \sqrt{2}=3\sqrt{2}\approx 4.24}$.So, det(P)≤4${\displaystyle \det (P)\le 4}$. Case 2: Matrix has all entries ±1${\displaystyle \pm 1}$.For a 3×3${\displaystyle 3\times 3}$ matrix with entires ±1${\displaystyle \pm 1}$, the maximum determinant is 4 (well-known result; can be verified by considering that dot products of distinct rows must be odd, so orthogonality is impossible).Therefore, 5 is impossible (in all cases) ∴det(P)≤4${\displaystyle \therefore \det (P)\le 4}$Now, Check if 4 is possibleWe just need to see if we can actually get 4. It is a known fact that the matrix with all 1s on the diagonal and (-1)${\displaystyle (-1)}$s elsewhere (or vice versa) gives 4.a

1	1	1
1	-1	1
1	1	-1

=4${\displaystyle \left|\begin{array}{ccc}1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right|=4}$ Answer

4

${\displaystyle \boxed{4}}$Source: JEE Advanced 2018; Question 8 Paper 2; Page 28https://jeeadv.ac.in/past_qps/2018_2.pdf

Question #35The value of the determinant a

1	a	a2-bc
1	b	b2-ac
1	c	c2-ab

${\displaystyle \left|\begin{array}{ccc}1& a& a^2-bc\\ 1& b& b^2-ac\\ 1& c& c^2-ab\end{array}\right|}$is __Solutiona

1	a	a2-bc
1	b	b2-ac
1	c	c2-ab

${\displaystyle \left|\begin{array}{ccc}1& a& a^2-bc\\ 1& b& b^2-ac\\ 1& c& c^2-ab\end{array}\right|}$=a

1	a	a2
1	b	b2
1	c	c2

-a

1	a	bc
1	b	ac
1	c	ab

${\displaystyle =\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|-\left|\begin{array}{ccc}1& a& bc\\ 1& b& ac\\ 1& c& ab\end{array}\right|}$.Now, a

1	a	bc
1	b	ac
1	c	ab

=1

abca

a	a2	abc
b	b2	abc
c	c2	abc

${\displaystyle \left|\begin{array}{ccc}1& a& bc\\ 1& b& ac\\ 1& c& ab\end{array}\right|=\frac{1}{abc}\left|\begin{array}{ccc}a& a^2& abc\\ b& b^2& abc\\ c& c^2& abc\end{array}\right|}$ Applying, R2→aR1${\displaystyle R_2\to a{R}_1}$, R2→bR2${\displaystyle R_2\to b{R}_2}$, R3→cR3${\displaystyle R_3\to c{R}_3}$=1

abc×abc×a

a	a2	1
b	b2	1
c	c2	1

=a

1	a	a2
1	b	b2
1	c	c2

${\displaystyle =\frac{1}{abc}\times abc\times \left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|=\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|}$ ∴a

1	a	a2-bc
1	b	b2-ac
1	c	c2-ab

=0${\displaystyle \therefore \left|\begin{array}{ccc}1& a& a^2-bc\\ 1& b& b^2-ac\\ 1& c& c^2-ab\end{array}\right|=0}$ Answer

0

${\displaystyle \boxed{0}}$Source: JEE 1988

Question #36Let M${\displaystyle M}$ and N${\displaystyle N}$ be two 3×3${\displaystyle 3\times 3}$ matrices such that MN=NM${\displaystyle MN=NM}$. Further, if M≠N2${\displaystyle M\ne N^2}$ and M2=N4${\displaystyle M^2=N^4}$, then (A) det(M2+MN2)${\displaystyle \det (M^2+M{N}^2)}$ is 0${\displaystyle 0}$(B) there is a 3×3${\displaystyle 3\times 3}$ non-zero matrix U${\displaystyle U}$ such that (M2+MN2)U${\displaystyle (M^2+M{N}^2)U}$ is zero matrix(C) det(M2+MN2)≥1${\displaystyle \det (M^2+M{N}^2)\ge 1}$(D) for a 3×3${\displaystyle 3\times 3}$ matrix U${\displaystyle U}$, if (M2+MN2)U${\displaystyle (M^2+M{N}^2)U}$ equals the zero matrix, then U${\displaystyle U}$ is the zero matrix. SolutionWe know, if A${\displaystyle A}$ and B${\displaystyle B}$ are two non zero matrices and AB=BA${\displaystyle AB=BA}$, then (A-B)(A+B)=A2-B2${\displaystyle (A-B)(A+B)=A^2-B^2}$ Also, det(AB)=det(A)×det(B)${\displaystyle \det (AB)=\det (A)\times \det (B)}$ Given, M≠N2${\displaystyle M\ne N^2}$ and M2=N4${\displaystyle M^2=N^4}$ Since, M2-N4=O${\displaystyle M^2-N^4=O}$ ⟹(M-N2)(M+N2)=O${\displaystyle ⟹(M-N^2)(M+N^2)=O}$ If det(M+N2)≠0${\displaystyle \det (M+N_{}^2)\ne 0}$, then (M+N2)${\displaystyle (M+N^2)}$ is invertible.⟹(M-N2)(M+N2)(M+N2)-1=O×(M+N2)-1${\displaystyle ⟹(M-N^2)(M+N^2)(M+N^2{)}^{-1}=O\times (M+N^2{)}^{-1}}$⟹(M-N2)I=O${\displaystyle ⟹(M-N^2)I=O}$⟹(M-N2)=O${\displaystyle ⟹(M-N^2)=O}$Hence, det(M2+MN2)=0${\displaystyle \det (M^2+M{N}^2)=0}$ Also, det(M2+MN2)=(detM)(detM+N2)=det(M)×0=0${\displaystyle \det (M^2+M{N}^2)=(\det M)(\det M+N^2)=\det (M)\times 0=0}$As, det(M2+MN2)=0${\displaystyle \det (M^2+M{N}^2)=0}$ Option (A) is correct Thus, there exists a non zero matrix U${\displaystyle U}$ such that (M2+MN2)U=O${\displaystyle (M^2+M{N}^2)U=O}$Option (B) is correct Answer (A, B)Source: JEE Advanced 2014; Question 41 Paper 1

Question #37Given that x=-9${\displaystyle x=-9}$ is a root of a

x	3	7
2	x	2
7	6	x

=0${\displaystyle \left|\begin{array}{ccc}x& 3& 7\\ 2& x& 2\\ 7& 6& x\end{array}\right|=0}$, the product of the other two roots is __SolutionApplying R1→R1+R2+R3${\displaystyle R_1\to R_1+R_2+R_3}$⟹a

x+9	x+9	x+9
2	x	2
7	6	x

=0${\displaystyle ⟹\left|\begin{array}{ccc}x+9& x+9& x+9\\ 2& x& 2\\ 7& 6& x\end{array}\right|=0}$⟹(x+9)a

1	1	1
2	x	2
7	6	x

=0${\displaystyle ⟹(x+9)\left|\begin{array}{ccc}1& 1& 1\\ 2& x& 2\\ 7& 6& x\end{array}\right|=0}$ Applying C2→C2-C1${\displaystyle C_2\to C_2-C_1}$ and C3→C3-C1${\displaystyle C_3\to C_3-C_1}$⟹(x+9)a

1	0	0
2	x-2	0
7	-1	x-7

=0${\displaystyle ⟹(x+9)\left|\begin{array}{ccc}1& 0& 0\\ 2& x-2& 0\\ 7& -1& x-7\end{array}\right|=0}$⟹(x+9)(x-2)(x-7)=0${\displaystyle ⟹(x+9)(x-2)(x-7)=0}$⟹x∈{-9,2,7}${\displaystyle ⟹x\in \{-9,2,7\}}$ So, the other two roots are {2,7}${\displaystyle \{2,7\}}$.the product of the other two roots=2×7=14${\displaystyle =2\times 7=14}$ Answer

14

${\displaystyle \boxed{14}}$Source: JEE 1983

Question #38 ***Let M=a

sin4t	-1-sin2t
1+cos2t	cos4t

=AI+BM-1${\displaystyle M=\left[\begin{array}{cc}{\sin }^{4}t& -1-{\sin }^{2}t\\ 1+{\cos }^{2}t& {\cos }^{4}t\end{array}\right]=AI+B{M}^{-1}}$,where A=A(t)${\displaystyle A=A(t)}$ and B=B(t)${\displaystyle B=B(t)}$ are real numbers, and I${\displaystyle I}$ is the 2×2${\displaystyle 2\times 2}$ identity matrix. If A*${\displaystyle A^{*}}$ is the minimum of the set {A(t):t∈[0,2𝜋)}${\displaystyle \{A(t):t\in [0,2𝜋)\}}$ and B*${\displaystyle B^{*}}$ is the minimum of the set {B(t):t∈[0,2𝜋)}${\displaystyle \{B(t):t\in [0,2𝜋)\}}$, then value of (-16×(A*+B*))${\displaystyle (-16\times (A^{*}+B^{*}\left)\right)}$ is __ SolutionM=a

sin4t	-1-sin2t
1+cos2t	cos4t

=AI+BM-1${\displaystyle M=\left[\begin{array}{cc}{\sin }^{4}t& -1-{\sin }^{2}t\\ 1+{\cos }^{2}t& {\cos }^{4}t\end{array}\right]=AI+B{M}^{-1}}$. M=AI+BM-1${\displaystyle M=AI+B{M}^{-1}}$.⟹MM=AIM+BM-1M${\displaystyle ⟹MM=AIM+B{M}^{-1}M}$.⟹M2=AM+BI${\displaystyle ⟹M^2=AM+BI}$ (∵M-1M=I)${\displaystyle (\because M^{-1}M=I)}$.⟹M2=AM+BI${\displaystyle ⟹M^2=AM+BI}$…(i)${\displaystyle \dots (i)}$. Using Cayley-Hamilton Theorem,For any 2×2${\displaystyle 2\times 2}$ matrix, M2=trace(M).M-(detM).I${\displaystyle M^2=\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{c}\mathrm{e}(M).M-(\det M).I}$Comparing with equation (i), A=trace(M)${\displaystyle A=\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{c}\mathrm{e}(M)}$…(ii)${\displaystyle \dots (ii)}$, B=-(detM)${\displaystyle B=-(\det M)}$…(iii)${\displaystyle \dots (iii)}$. Now, A=trace(M)${\displaystyle A=\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{c}\mathrm{e}(M)}$⟹A=tracea

sin4t	-1-sin2t
1+cos2t	cos4t

${\displaystyle ⟹A=\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{c}\mathrm{e}\left(\begin{array}{cc}{\sin }^4\mathrm{t}& \mathrm{-}1\mathrm{-}{\sin }^2\mathrm{t}\\ 1\mathrm{+}{\cos }^2\mathrm{t}& {\cos }^4\mathrm{t}\end{array}\right)}$⟹A=sin4t+cos4t${\displaystyle ⟹A={\sin }^{4}t+{\cos }^{4}t}$. Now, sin4t+cos4t${\displaystyle {\sin }^{4}t+{\cos }^{4}t}$=(sin2t)2+(cos2t)2${\displaystyle =({\sin }^{2}t{)}^2+({\cos }^{2}t{)}^2}$=(sin2t+cos2t)-2sin2tcos2t${\displaystyle =({\sin }^{2}t+{\cos }^{2}t)-2{\sin }^{2}t{\cos }^{2}t}$=1-2sin2tcos2t${\displaystyle =1-2{\sin }^{2}t{\cos }^{2}t}$=1-1

2sin2(2t)${\displaystyle =1-\frac{1}{2}{\sin }^2(2t)}$.So, Amin=min(1-1

2sin2(2t))=1-1

2(1)=1

2${\displaystyle A_{\min }=\min (1-\frac{1}{2}{\sin }^2(2t))=1-\frac{1}{2}(1)=\frac{1}{2}}$. Reason: max(sin2(2t))=1${\displaystyle \max ({\sin }^2(2t))=1}$ Now, det(M)${\displaystyle \det (M)}$=sin4t+cos4t+1+sin2t+cos2t+sin2tcos2t${\displaystyle ={\sin }^{4}t+{\cos }^{4}t+1+{\sin }^{2}t+{\cos }^{2}t+{\sin }^{2}t{\cos }^{2}t}$=sin4tcos4t+sin2tcos2t+2${\displaystyle ={\sin }^{4}t{\cos }^{4}t+{\sin }^{2}t{\cos }^{2}t+2}$ Using (iii), B=-(detM)${\displaystyle B=-(\det M)}$⟹B=-(sin4tcos4t+sin2tcos2t+2)${\displaystyle ⟹B=-({\sin }^{4}t{\cos }^{4}t+{\sin }^{2}t{\cos }^{2}t+2)}$ Let x=sin2tcos2t${\displaystyle x={\sin }^{2}t{\cos }^{2}t}$. So, B=-(x2-x+2)${\displaystyle B=-(x^2-x+2)}$To compute the min value, we need to make (x2-x+2)${\displaystyle (x^2-x+2)}$ as max as possible.Max value of x=sin2tcos2t${\displaystyle x={\sin }^{2}t{\cos }^{2}t}$? We know sint.cost=1

2sin(2t)${\displaystyle \sin t.\cos t=\frac{1}{2}\sin (2t)}$So, x=(1

2sin(2t))2=1

4sin2(2t)${\displaystyle x=(\frac{1}{2}\sin (2t){)}^2=\frac{1}{4}{\sin }^2(2t)}$The max value of sin2(2t)=1${\displaystyle {\sin }^2(2t)=1}$. So, the maximum of x=1

4${\displaystyle x=\frac{1}{4}}$.Now, put x=1

4${\displaystyle x=\frac{1}{4}}$ in -(x2-x+2)${\displaystyle -(x^2-x+2)}$B=-((1

4)2-(1

4)+2)⟹B=-37

16${\displaystyle B=-\left(\right(\frac{1}{4}{)}^2-\left(\frac{1}{4}\right)+2)⟹B=-\frac{37}{16}}$ So, -16×(A*+B*)=-16×(1

2-37

16)=-16×(-29

16)=29${\displaystyle -16\times (A^{*}+B^{*})=-16\times (\frac{1}{2}-\frac{37}{16})=-16\times (-\frac{29}{16})=29}$ Answer

29

${\displaystyle \boxed{29}}$Source: JEE Advanced 2019

Question #39If B=a

5	2C	1
0	2	1
C	3	-1

${\displaystyle B=\left[\begin{array}{ccc}5& 2C& 1\\ 0& 2& 1\\ C& 3& -1\end{array}\right]}$ is the inverse of a 3×3${\displaystyle 3\times 3}$ matrix A${\displaystyle A}$, then the sum of all possible values of C${\displaystyle C}$ for which det(A)+1=0${\displaystyle \det (A)+1=0}$ is __ Solution;Given matrix B${\displaystyle B}$ is the inverse of a 3×3${\displaystyle 3\times 3}$ matrix A${\displaystyle A}$, where B=a

5	2C	1
0	2	1
C	3	-1

${\displaystyle B=\left[\begin{array}{ccc}5& 2C& 1\\ 0& 2& 1\\ C& 3& -1\end{array}\right]}$We know that, det(A)×det(B)=1${\displaystyle \det (A)\times \det (B)=1}$. Reason det(A-1)=1

det(A)${\displaystyle \det \left(A^{-1}\right)=\frac{1}{\det (A)}}$.⟹det(A)=1

det(B)${\displaystyle ⟹\det (A)=\frac{1}{\det (B)}}$. Since, det(A)+1=0${\displaystyle \det (A)+1=0}$ (given)⟹1

det(B)+1=0${\displaystyle ⟹\frac{1}{\det (B)}+1=0}$.⟹det(B)=-1${\displaystyle ⟹\det (B)=-1}$.⟹5(-2-3)-2C(0-C)+1(0-2C)=-1${\displaystyle ⟹5(-2-3)-2C(0-C)+1(0-2C)=-1}$⟹-25+2C2-2C=-1${\displaystyle ⟹-25+2C^2-2C=-1}$.⟹C2-C-12=0${\displaystyle ⟹C^2-C-12=0}$.⟹(C-4)(C+3)=0${\displaystyle ⟹(C-4)(C+3)=0}$.⟹C∈{-3,4}${\displaystyle ⟹C\in \{-3,4\}}$. So, the sum of all possible values of C=-3+4=1${\displaystyle C=-3+4=1}$ Answer

1

${\displaystyle \boxed{1}}$Source: JEE Main 2019

Question #40If A=a

2	-3
-4	1

${\displaystyle A=\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]}$, then abs(det(adj(3A2+12A)))${\displaystyle \mathrm{a}\mathrm{b}\mathrm{s}\left(\det \right(\mathrm{a}\mathrm{d}\mathrm{j}(3A^2+12A)\left)\right)}$ is equal to __ SolutionA=a

2	-3
-4	1

${\displaystyle A=\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]}$.Now compute, 3A2=3a

2	-3
-4	1

a

2	-3
-4	1

=3a

16	-9
-4	1

=a

48	-27
-12	3

${\displaystyle 3A^2=3\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]=3\left[\begin{array}{cc}16& -9\\ -4& 1\end{array}\right]=\left[\begin{array}{cc}48& -27\\ -12& 3\end{array}\right]}$Now, 12A=12a

2	-3
-4	1

=a

24	-36
-48	12

${\displaystyle 12A=12\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]=\left[\begin{array}{cc}24& -36\\ -48& 12\end{array}\right]}$ 3A2+12A${\displaystyle 3A^2+12A}$=a

48	-27
-12	3

+a

24	-36
-48	12

=a

72	-63
-84	51

${\displaystyle =\left[\begin{array}{cc}48& -27\\ -12& 3\end{array}\right]+\left[\begin{array}{cc}24& -36\\ -48& 12\end{array}\right]=\left[\begin{array}{cc}72& -63\\ -84& 51\end{array}\right]}$. Now, adj(3A2+12A)=adj(a

72	-63
-84	51

)=a

51	63
84	72

${\displaystyle \mathrm{a}\mathrm{d}\mathrm{j}(3A^2+12A)=\mathrm{a}\mathrm{d}\mathrm{j}\left(\left[\begin{array}{cc}72& -63\\ -84& 51\end{array}\right]\right)=\left[\begin{array}{cc}51& 63\\ 84& 72\end{array}\right]}$.Reason. adj(a

a	b
c	d

)=a

d	-b
-c	a

${\displaystyle \mathrm{a}\mathrm{d}\mathrm{j}\left(\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\right)=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]}$. Next, det(adj(3A2+12A))=51×72-63×84=-1620${\displaystyle \det \left(\mathrm{a}\mathrm{d}\mathrm{j}\right(3A^2+12A\left)\right)=51\times 72-63\times 84=-1620}$∴abs(det(adj(3A2+12A)))=abs(-1620)=1620${\displaystyle \therefore \mathrm{a}\mathrm{b}\mathrm{s}\left(\det \right(\mathrm{a}\mathrm{d}\mathrm{j}(3A^2+12A)\left)\right)=\mathrm{a}\mathrm{b}\mathrm{s}(-1620)=1620}$. Answer

1620

${\displaystyle \boxed{1620}}$Source: JEE Main 2017