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Let \(\alpha\) be a root of the equation \(x^2+x+1=0\)
and the matrix \(A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha^4\end{array}\right]\), then the matrix \(A^{31}\) is equal to:
as \(\alpha\) is root of \(x^2+x+1\)
\(\therefore \alpha=\frac{-1 \pm \sqrt{3}}{2}\)
\(\therefore \alpha=w\) or \(w^2\)
\(\therefore A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha^4\end{array}\right]\)
put \(\alpha=w\)
\(A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & w & w^2 \\ 1 & w^2 & w^4\end{array}\right]\)
We know that \(w^3=1\)
So, \(w^4=w^3 \cdot w=w\)
\(
A^2=\frac{1}{(\sqrt{3})^2}\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & w & w^2 \\
1 & w^2 & w
\end{array}\right]\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & w & w^2 \\
1 & w^2 & w
\end{array}\right]
\)
\(
A^2=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array}\right]
\)
\(
A^4=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array}\right]
\)
\(
\begin{aligned}
& =\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
& =I \\
& A^{31}=A^{28} A^3 \\
& =\left(A^4\right)^7 \cdot A^3
\end{aligned}
\)
\(
\begin{aligned}
& =I \cdot A^3 \\
& =A^3
\end{aligned}
\)
If \(A=\left[\begin{array}{cc}\cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta\end{array}\right],\left(\theta=\frac{\pi}{24}\right)\) and \(A^5=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\), where \(i=\sqrt{-1}\), then which one of the following is not true?
\(
\begin{aligned}
& \text { (d) } \because A=\left[\begin{array}{cc}
\cos \theta & i \sin \theta \\
i \sin \theta & \cos \theta
\end{array}\right] \\
& \therefore A^n=\left[\begin{array}{cc}
\cos n \theta & i \sin n \theta \\
i \sin n \theta & \cos n \theta
\end{array}\right], n \in \mathbf{N} \\
& \because A^5=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \\
& \therefore A^5=\left[\begin{array}{cc}
\cos 5 \theta & i \sin 5 \theta \\
i \sin 5 \theta & \cos 5 \theta
\end{array}\right]=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \\
& \therefore a=\cos 5 \theta, b=i \sin 5 \theta=c, d=\cos 5 \theta \\
& \therefore a^2-b^2=\cos ^2 5 \theta+\sin ^2 5 \theta=1 \\
& a^2-c^2=\cos ^2 5 \theta+\sin ^2 5 \theta=1 \\
& a^2-d^2=\cos ^2 5 \theta-\cos ^2 5 \theta=1 \\
& a^2+b^2=\cos ^2 5 \theta-\sin ^2 5 \theta=\cos 10 \theta=\cos \frac{10 \pi}{24}
\end{aligned}
\)
\(
\text { and } \begin{aligned}
0 & <\cos \frac{5 \pi}{12}<1 \Rightarrow 0 \leq a^2+b^2 \leq 1 \\
\therefore a^2-b^2 & =\frac{1}{2} \text { is wrong. }
\end{aligned}
\)
Let \(A=\left[\begin{array}{cr}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right],(\alpha \in R)\) such that \(A^{32}=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\). Then a value of \(\alpha\) is :
\(
\begin{aligned}
& A=\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right] \\
& A^2=\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right] \\
& =\left[\begin{array}{cc}
\cos 2 \alpha & -\sin 2 \alpha \\
\sin 2 \alpha & \cos 2 \alpha
\end{array}\right] \\
& A^3=\left[\begin{array}{cc}
\cos 2 \alpha & -\sin 2 \alpha \\
\sin 2 \alpha & \cos 2 \alpha
\end{array}\right]\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right] \\
& =\left[\begin{array}{ll}
\cos 3 \alpha & -\sin 3 \alpha \\
\sin 3 \alpha & \cos 3 \alpha
\end{array}\right]
\end{aligned}
\)
Similarly
\(
A^{32}=\left[\begin{array}{cc}
\cos 32 \alpha & -\sin 32 \alpha \\
\sin 32 \alpha & \cos 32 \alpha
\end{array}\right]=\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]
\)
\(\Rightarrow \cos 32 \alpha=0\) and \(\sin 32 \alpha=1\)
\(
\Rightarrow 32 \alpha=(4 n+1) \frac{\pi}{2}, n \in I
\)
\(
\alpha=(4 n+1) \frac{\pi}{64}, n \in I
\)
\(
\alpha=\frac{\pi}{64} \text { for } n=0
\)
Let \(P=\left[\begin{array}{lll}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1\end{array}\right]\) and \(\mathrm{Q}=\left[\mathrm{q}_{\mathrm{ij}}\right]\) be two \(3 \times 3\) matrices such that \(\mathrm{Q}-\) \(\mathrm{P}^5=\mathrm{I}_3\). Then \(\frac{\mathrm{q}_{21}+\mathrm{q}_{31}}{\mathrm{q}_{32}}\) is equal to :
\(
\begin{aligned}
& P=\left[\begin{array}{lll}
1 & 0 & 0 \\
3 & 1 & 0 \\
9 & 3 & 1
\end{array}\right] \\
& \begin{aligned}
P^2 & =\left[\begin{array}{ccc}
1 & 0 & 0 \\
3+3 & 1 & 0 \\
9+9+9 & 3+3 & 1
\end{array}\right] \\
P^3 & =\left[\begin{array}{ccc}
1 & 0 & 0 \\
3+3+3 & 1 & 0 \\
6.9 & 3+3+3 & 1
\end{array}\right]
\end{aligned} \\
& P^n=\left[\begin{array}{ccc}
1 & 0 & 0 \\
3 n & 1 & 0 \\
\frac{n(n+1)}{2} 3^2 & 3 n & 1
\end{array}\right] \\
& P^5=\left[\begin{array}{ccc}
1 & 0 & 0 \\
5.3 & 1 & 0 \\
15.9 & 5.3 & 1
\end{array}\right] \\
& Q=P^5+I_3 \\
& Q=\left[\begin{array}{ccc}
2 & 0 & 0 \\
15 & 2 & 0 \\
135 & 15 & 2
\end{array}\right] \\
& \frac{q_{21}+q_{31}}{q_{32}}=\frac{15+135}{15}=10 \\
&
\end{aligned}
\)
Let \(A=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1\end{array}\right]\) and \(B=A^{20}\). Then the sum of the elements of the first column of \(B\) is?
\(
\text { Here } A=\left[\begin{array}{lll}
1 & 0 & 0 \\
1 & 1 & 0 \\
1 & 1 & 1
\end{array}\right]
\)
\(
\therefore \quad A^2=A \cdot A=\left[\begin{array}{lll}
1 & 0 & 0 \\
1 & 1 & 0 \\
1 & 1 & 1
\end{array}\right] \times\left[\begin{array}{lll}
1 & 0 & 0 \\
1 & 1 & 0 \\
1 & 1 & 1
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right]
\)
\(
\text { and, } A^4=A^3 \cdot A=\left[\begin{array}{lll}
1 & 0 & 0 \\
3 & 1 & 0 \\
6 & 3 & 1
\end{array}\right] \times\left[\begin{array}{lll}
1 & 0 & 0 \\
1 & 1 & 0 \\
1 & 1 & 1
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
4 & 1 & 0 \\
10 & 4 & 1
\end{array}\right]
\)
On observing the pattern, we come to a conclusion that,
\(
A=\left[\begin{array}{ccc}
1 & 0 & 0 \\
n & 1 & 0 \\
\frac{n(n+1)}{2} & n & 1
\end{array}\right]
\)
\(
\therefore A^{20}=\left[\begin{array}{ccc}
1 & 0 & 0 \\
20 & 1 & 0 \\
210 & 20 & 1
\end{array}\right]
\)
Therefore, sum of first column of \(A^{20}=[1+20+210]=231\)
How many \(3 \times 3\) matrices \(M\) with entries from \(\{0,1,2\}\) are there, for which the sum of the diagonal entries of \(\mathrm{M}^{\mathrm{T}} \mathrm{M}\) is 5?
(b) \(\quad\) Let \(\mathrm{M}=\left[\begin{array}{lll}\mathrm{a}_1 & \mathrm{a}_2 & \mathrm{a}_3 \\ \mathrm{a}_4 & \mathrm{a}_5 & \mathrm{a}_6 \\ \mathrm{a}_7 & \mathrm{a}_8 & \mathrm{a}_9\end{array}\right]\) where \(\mathrm{a}_{\mathrm{i}} \in\{0,1,2\}\)
Then \(M^T M=\left[\begin{array}{lll}a_1 & a_4 & a_7 \\ a_2 & a_5 & a_8 \\ a_3 & a_6 & a_9\end{array}\right]\left[\begin{array}{lll}a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9\end{array}\right]\)
Sum of the dlagonal entrles \(\ln M^{\top} M=5\)
\(
\begin{aligned}
\Rightarrow & \left(a_1^2+a_4^2+a_7^2\right)+\left(a_2^2+a_5^2+a_8^2\right) \\
& +\left(a_3^2+a_6^2+a_9^2\right)=5
\end{aligned}
\)
It is possible when
Case I: \(5 \mathrm{a}_{\mathrm{i}}\) ‘s are 1 and \(4 \mathrm{a}_{\mathrm{i}}\) ‘s are zero Which can be done in
\(
{ }^9 \mathrm{C}_4 \text { ways }=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}=126
\)
Case II: \(1 \mathrm{~a}_{\mathrm{i}}\) is 1 and \(1 \mathrm{~a}_{\mathrm{i}}\) is 2 and rest \(7 \mathrm{a}_{\mathrm{i}}{ }\) ‘s are zero
It can be done in \({ }^9 \mathrm{C}_1 \times{ }^8 \mathrm{C}_1=9 \times 8=72\) ways
\(\therefore\) Total no. of ways \(=126+72=198\).
If \(A=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\), then which one of the following statements is not correct?
Given
\(
\begin{aligned}
& A=\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right] \\
& A^2=\left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right] \Rightarrow A^2=-I \\
& A^3=\left[\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array}\right] \\
& A^4=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=I \\
& A^2+I=A^3-A \\
& -I+I=A^3-A \\
& A^3 \neq A
\end{aligned}
\)
If \(A=\left[\begin{array}{ccc}1 & 2 & x \\ 3 & -1 & 2\end{array}\right]\) and \(B=\left[\begin{array}{l}y \\ x \\ 1\end{array}\right]\) be such that \(\mathrm{AB}=\left[\begin{array}{l}6 \\ 8\end{array}\right]\), then
Let \(A=\left[\begin{array}{ccc}1 & 2 & x \\ 3 & -1 & 2\end{array}\right]\) and \(B=\left[\begin{array}{l}y \\ x \\ 1\end{array}\right]\)
\(
\begin{aligned}
& \mathrm{AB}=\left[\begin{array}{ccc}
1 & 2 & x \\
3 & -1 & 2
\end{array}\right]\left[\begin{array}{l}
y \\
x \\
1
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{l}
6 \\
8
\end{array}\right]=\left[\begin{array}{l}
y+2 x+x \\
3 y-x+2
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{l}
6 \\
8
\end{array}\right]=\left[\begin{array}{l}
y+3 x \\
3 y-x+2
\end{array}\right] \\
& \Rightarrow y+3 x=6 \text { and } 3 y-x=6
\end{aligned}
\)
On solving, we get
\(
\begin{aligned}
& x=\frac{6}{5} \text { and } y=\frac{12}{5} \\
& \Rightarrow y=2 x
\end{aligned}
\)
If \(p, q, r\) are 3 real numbers satisfying the matrix equation, \([p q r]\left[\begin{array}{lll}3 & 4 & 1 \\ 3 & 2 & 3 \\ 2 & 0 & 2\end{array}\right]=\left[\begin{array}{lll}3 & 0 & 1\end{array}\right]\) then \(2 p+q-r\) equals :
(a) Given
\(
\left[\begin{array}{lll}
p & q & r
\end{array}\right]\left[\begin{array}{lll}
3 & 4 & 1 \\
3 & 2 & 3 \\
2 & 0 & 2
\end{array}\right]=\left[\begin{array}{lll}
3 & 0 & 1
\end{array}\right]
\)
\(
\Rightarrow\left[\begin{array}{lll}
3 p+3 q+2 r & 4 p+2 q & p+3 q+2 r
\end{array}\right]=\left[\begin{array}{lll}
3 & 0 & 1
\end{array}\right]
\)
\(
\begin{aligned}
\Rightarrow 3 p+3 q+2 r & =3 \dots(i) \\
4 p+2 q & =0 \Rightarrow q=-2 p \dots(ii) \\
p+3 q+2 r & =1 \dots(iii)
\end{aligned}
\)
On solving (i), (ii) and (iii), we get
\(
\begin{aligned}
& p=1, q=-2, r=3 \\
& \therefore 2 p+q-r=2(1)+(-2)-(3)=-3 .
\end{aligned}
\)
The matrix \(A^2+4 A-5 I\), where \(I\) is identity matrix and \(A=\left[\begin{array}{cc}1 & 2 \\ 4 & -3\end{array}\right]\), equals
\(
\text { (a) } \mathrm{A}^2+4 \mathrm{~A}-5 \mathrm{I}=\mathrm{A} \times \mathrm{A}+4 \mathrm{~A}-5 \mathrm{I}
\)
\(
=\left[\begin{array}{cc}
1 & 2 \\
4 & -3
\end{array}\right] \times\left[\begin{array}{cc}
1 & 2 \\
4 & -3
\end{array}\right]+4\left[\begin{array}{cc}
1 & 2 \\
4 & -3
\end{array}\right]-5\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]
\)
\(
=\left[\begin{array}{cc}
9 & -4 \\
-8 & 17
\end{array}\right]+\left[\begin{array}{cc}
4 & 8 \\
16 & -12
\end{array}\right]-\left[\begin{array}{cc}
5 & 0 \\
0 & 5
\end{array}\right]
\)
\(
=\left[\begin{array}{cc}
9+4-5 & -4+8-0 \\
-8+16-0 & 17-12-5
\end{array}\right]=\left[\begin{array}{ll}
8 & 4 \\
8 & 0
\end{array}\right]
\)
\(
=4\left[\begin{array}{ll}
2 & 1 \\
2 & 0
\end{array}\right]
\)
If \(A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 2 & 1 & 0 \\ -3 & 2 & 1\end{array}\right]\) and \(B=\left[\begin{array}{ccc}1 & 0 & 0 \\ -2 & 1 & 0 \\ 7 & -2 & 1\end{array}\right]\) then \(A B\) equals
\(
\begin{aligned}
& A=\left[\begin{array}{ccc}
1 & 0 & 0 \\
2 & 1 & 0 \\
-3 & 2 & 1
\end{array}\right], B=\left[\begin{array}{ccc}
1 & 0 & 0 \\
-2 & 1 & 0 \\
7 & -2 & 1
\end{array}\right] \\
& A B=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=I
\end{aligned}
\)
If \(\omega \neq 1\) is the complex cube root of unity and matrix \(H=\left[\begin{array}{ll}\omega & 0 \\ 0 & \omega\end{array}\right]\), then \(\mathrm{H}^{70}\) is equal to
\(
H^2=\left[\begin{array}{ll}
\omega & 0 \\
0 & \omega
\end{array}\right]\left[\begin{array}{ll}
\omega & 0 \\
0 & \omega
\end{array}\right]=\left[\begin{array}{ll}
\omega^2 & 0 \\
0 & \omega^2
\end{array}\right]
\)
If \(H^k=\left[\begin{array}{cc}\omega^k & 0 \\ 0 & \omega\end{array}\right]\) then \(\mathrm{H}^{k+1}=\left[\begin{array}{cc}\omega^{k+1} & 0 \\ 0 & \omega^{k+1}\end{array}\right]\)
So by principle of mathematical induction,
\(
H^{70}=\left[\begin{array}{cc}
\omega^{70} & 0 \\
0 & \omega^{70}
\end{array}\right]=\left[\begin{array}{cc}
\omega^{69} \omega & 0 \\
0 & \omega^{69} \omega
\end{array}\right]=\left[\begin{array}{cc}
\omega & 0 \\
0 & \omega
\end{array}\right]=H
\)
Note: As we know that, if \(\omega\) is a cube root of unity
The number of \(3 \times 3\) non-singular matrices, with four entries as 1 and all other entries as 0, is
Find the number of matrices formed based on given conditions
Form the matrix,
\(
\left[\begin{array}{lll}
1 & * & * \\
* & 1 & * \\
* & * & 1
\end{array}\right]
\)
We have four 1 ‘s. In the above matrix, three are filled. So, remaining 1 can be placed in any of the \(6^*\) positions and 0 elsewhere.
Therefore, we get 6 non-singular matrices
Now consider
\(
\left[\begin{array}{lll}
* & * & 1 \\
* & 1 & * \\
1 & * & *
\end{array}\right]
\)
Similarly, we get another set of 6 non-singular matrices.
So, we can easily form multiple matrices based on given conditions.
So, required cases are more than 7 , non-singular \(3 \times 3\) matrices.
Let \(A=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right)\) and \(B=\left(\begin{array}{ll}a & 0 \\ 0 & b\end{array}\right), a, b \in N\). Then
\(
\begin{aligned}
& A=\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right] \quad B=\left[\begin{array}{ll}
a & 0 \\
0 & b
\end{array}\right] \\
& A B=\left[\begin{array}{cc}
a & 2 b \\
3 a & 4 b
\end{array}\right] \\
& B A=\left[\begin{array}{ll}
a & 0 \\
0 & b
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]=\left[\begin{array}{cc}
a & 2 a \\
3 b & 4 b
\end{array}\right]
\end{aligned}
\)
Hence, \(A B=B A\) only when \(a=b\)
\(\therefore\) There can be infinitely many \(B\) ‘s for which \(A B=B A\)
If \(A\) and \(B\) are square matrices of size \(n \times n\) such that \(A^2-B^2=(A-B)(A+B)\), then which of the following will be always true?
\(
\begin{aligned}
& A^2-B^2=(A-B)(A+B) \\
& A^2-B^2=A^2+A B-B A-B^2 \\
& \Rightarrow A B=B A
\end{aligned}
\)
If \(A=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]\) and \(I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\), then which one of the following holds for all \(n \geq 1\), by the principle of mathematical induction
We observe that \(A^2=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right], A^3=\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right]\) and we can prove by induction that \(A^n=\left[\begin{array}{ll}1 & 0 \\ n & 1\end{array}\right]\)
Now
\(
\begin{aligned}
n A-(n-1) I & =\left[\begin{array}{ll}
n & 0 \\
n & n
\end{array}\right]-\left[\begin{array}{cc}
n-1 & 0 \\
0 & n-1
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & 0 \\
n & 1
\end{array}\right]=A^n
\end{aligned}
\)
\(
\therefore n A-(n-1) I=A^n
\)
If \(A=\left[\begin{array}{ll}a & b \\ b & a\end{array}\right]\) and \(A^2=\left[\begin{array}{ll}\alpha & \beta \\ \beta & \alpha\end{array}\right]\), then
\(
\begin{aligned}
A^2 & =\left[\begin{array}{ll}
\alpha & \beta \\
\beta & \alpha
\end{array}\right]=\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right] \\
& =\left[\begin{array}{cc}
a^2+b^2 & 2 a b \\
2 a b & a^2+b^2
\end{array}\right] \\
\alpha & =a^2+b^2 ; \beta=2 a b
\end{aligned}
\)
For two \(3 \times 3\) matrices \(A\) and \(B\), let \(A+B=2 B^T\) and \(3 A+2 B=I_3\), where \(B^T\) is the transpose of \(B\) and \(I_3\) is \(3 \times 3\) identity matrix. Then:
\(
\begin{aligned}
& \mathrm{A}^{\mathrm{T}}+\mathrm{B}^{\mathrm{T}}=2 \mathrm{~B} \\
& \because \quad\left[(\mathrm{A}+\mathrm{B})^{\mathrm{T}}=\left(2 \mathrm{~B}^{\mathrm{T}}\right)^{\mathrm{T}}\right] \\
\Rightarrow & \mathrm{B}=\frac{\mathrm{A}^{\mathrm{T}}+\mathrm{B}^{\mathrm{T}}}{2}=\mathrm{A}+\left(\frac{\mathrm{B}^{\mathrm{T}}+\mathrm{A}^{\mathrm{T}}}{2}\right)=2 \mathrm{~B}^{\mathrm{T}} \\
\Rightarrow & \quad 2 \mathrm{~A}+\mathrm{A}^{\mathrm{T}}=3 \mathrm{~B}^{\mathrm{T}} \\
\Rightarrow & \mathrm{A}=\frac{3 \mathrm{~B}^{\mathrm{T}}-\mathrm{A}^{\mathrm{T}}}{2} \\
& \text { Also, } 3 \mathrm{~A}+2 \mathrm{~B}=\mathrm{I}_3 dots(i) \\
\Rightarrow & 3\left(\frac{3 \mathrm{~B}^{\mathrm{T}}-\mathrm{A}^{\mathrm{T}}}{2}\right)+2\left(\frac{\mathrm{A}^{\mathrm{T}}+\mathrm{B}^{\mathrm{T}}}{2}\right)=\mathrm{I}_3
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad 11 \mathrm{~B}^{\mathrm{T}}-\mathrm{A}^{\mathrm{T}}=2 I_3 \dots(ii) \\
& \text { Add (i) and (ii) } \\
& \quad 35 \mathrm{~B}=7 \mathrm{I}_3 \\
& \Rightarrow \quad \mathrm{B}=\frac{\mathrm{I}_3}{5} \\
& \Rightarrow \quad 11 \frac{\mathrm{I}_3}{5}-\mathrm{A}=2 I_3 \\
& \Rightarrow \quad 11 \frac{\mathrm{I}_3}{5}-2 \mathrm{I}_3=\mathrm{A} \\
& \Rightarrow \quad \mathrm{A}=\frac{\mathrm{I}_3}{5} \\
& \because \quad 5 \mathrm{~A}=5 \mathrm{~B}=\mathrm{I}_3 \\
& \Rightarrow \quad 10 \mathrm{~A}+5 \mathrm{~B}=3 \mathrm{I}_3
\end{aligned}
\)
If \(P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) and \(Q=\operatorname{PAP}^T\), then \(\mathrm{P}^{\mathrm{T}} \mathrm{Q}^{2015} \mathrm{P}\) is :
\(
P=\left[\begin{array}{cc}
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
-\frac{1}{2} & \frac{\sqrt{3}}{2}
\end{array}\right] P^T=\left[\begin{array}{cc}
\frac{\sqrt{3}}{2} & \frac{-1}{2} \\
\frac{1}{2} & \frac{\sqrt{3}}{2}
\end{array}\right]
\)
\(
\begin{aligned}
& P P^T=P^T P=\mathrm{I} \\
& \mathrm{Q}^{2015}=\left(P A P^T\right)\left(P A P^T\right)—-(2015 \text { terms })
\end{aligned}
\)
\(
\begin{aligned}
& =P A^{2015} P^T \\
& P^T \mathrm{Q}^{2015} P=A^{2015}
\end{aligned}
\)
\(
\begin{aligned}
& A^2=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right] \\
& A^3=\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right] \\
& A^{2015}=\left[\begin{array}{cc}
1 & 2015 \\
0 & 1
\end{array}\right]
\end{aligned}
\)
\(
\text { If } A=\left[\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
a & 2 & b
\end{array}\right]
\)
is a matrix satisfying the equation
\(\mathrm{AA}^{\mathrm{T}}=9 \mathrm{I}\), where \(\mathrm{I}\) is \(3 \times 3\) identity matrix, then the ordered pair \((a, b)\) is equal to:
\(
\begin{aligned}
& {\left[\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
\mathrm{a} & 2 & \mathrm{~b}
\end{array}\right]\left[\begin{array}{ccc}
1 & 2 & \mathrm{a} \\
2 & 1 & 2 \\
2 & -2 & \mathrm{~b}
\end{array}\right]=\left[\begin{array}{ccc}
9 & 0 & 0 \\
0 & 9 & 0 \\
0 & 0 & 9
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{ccc}
1+4+4 & 2+2-4 & \mathrm{a}+4+2 \mathrm{~b} \\
2+2-4 & 4+1+4 & 2 \mathrm{a}+2-2 \mathrm{~b} \\
\mathrm{a}+4+2 \mathrm{~b} & 2 \mathrm{a}+2-2 \mathrm{~b} & \mathrm{a}^2+4+\mathrm{b}^2
\end{array}\right]=\left[\begin{array}{ccc}
9 & 0 & 0 \\
0 & 9 & 0 \\
0 & 0 & 9
\end{array}\right]
\end{aligned}
\)
\(
\begin{aligned}
\Rightarrow \mathrm{a}+4+2 \mathrm{~b}= & 0 \Rightarrow \mathrm{a}+2 \mathrm{~b}=-4 \dots(i) \\
2 \mathrm{a}+2-2 \mathrm{~b}=0 & \Rightarrow 2 \mathrm{a}-2 \mathrm{~b}=-2 \\
& \Rightarrow \mathrm{a}-\mathrm{b}=-1 \dots(ii)
\end{aligned}
\)
On solving (i) and (ii) we get
\(
-1+b+2 b=-4 \dots(iii)
\)
\(
\begin{aligned}
-1+3 b & =-4 \\
3 b & =-3 \\
b & =-1
\end{aligned}
\)
\(
\begin{aligned}
& \text { and } a=-2 \\
& (a, b)=(-2,-1)
\end{aligned}
\)
Let \(A\) and \(B\) be any two \(3 \times 3\) matrices. If \(A\) is symmetric and \(B\) is skewsymmetric, then the matrix \(A B-B A\) is:
Let \(A\) be symmetric matrix and \(B\) be skew symmetric matrix.
\(
\therefore \mathrm{A}^{\mathrm{T}}=\mathrm{A} \text { and } \mathrm{B}^{\mathrm{T}}=-\mathrm{B}
\)
Consider
\(
\begin{aligned}
& (A B-B A)^T=(A B)^T-(B A)^T \\
& =B^T A^T-A^T B^T \\
& =(-B)(A)-(A)(-B) \\
& =-B A+A B=A B-B A
\end{aligned}
\)
This shows \(A B-B A\) is symmetric matrix.
If \(A=\left(\begin{array}{c}\alpha-1 \\ 0 \\ 0\end{array}\right), B=\left(\begin{array}{c}\alpha+1 \\ 0 \\ 0\end{array}\right)\) be two matrices, then \(A B^T\) is a non-zero matrix for \(|\alpha|\) not equal to
\(
\text { Let } A=\left(\begin{array}{c}
\alpha-1 \\
0 \\
0
\end{array}\right), B=\left(\begin{array}{c}
\alpha+1 \\
0 \\
0
\end{array}\right)
\)
be two matrices.
\(
\begin{aligned}
& A B^T=\left(\begin{array}{c}
\alpha-1 \\
0 \\
0
\end{array}\right)\left(\begin{array}{lll}
\alpha+1 & 0 & 0
\end{array}\right) \\
& =\left(\begin{array}{ccc}
\alpha^2-1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right) \\
&
\end{aligned}
\)
Thus, \(A B^T\) is non-zero matrix for \(|\alpha| \neq 1\)
Let \(A\) and \(B\) be two symmetric matrices of order 3.
Statement-1: \(A(B A)\) and \((A B) A\) are symmetric matrices.
Statement-2: \(A B\) is symmetric matrix if matrix multiplication of \(A\) with \(B\) is commutative.
\(
\begin{aligned}
& \therefore \mathrm{A}^{\prime}=\mathrm{A} \\
& \mathrm{B}^{\prime}=\mathrm{B}
\end{aligned}
\)
Now \((\mathrm{A}(\mathrm{BA}))^{\prime}=(\mathrm{BA})^{\prime} \mathrm{A}^{\prime}\)
\(
=\left(\mathrm{A}^{\prime} \mathrm{B}^{\prime}\right) \mathrm{A}^{\prime}=(\mathrm{AB}) \mathrm{A}=\mathrm{A}(\mathrm{BA})
\)
Similarly \(((A B) A)^{\prime}=(A B) A\)
So, \(A(B A)\) and \((A B) A\) are symmetric matrices.
\(
\text { Again }(A B)^{\prime}=B^{\prime} A^{\prime}=B A
\)
Now if \(B A=A B\), then \(A B\) is symmetric matrix.
Let \(\omega \neq 1\) be a cube root of unity and \(\mathrm{S}\) be the set of all non-singular matrices of the form \(\left[\begin{array}{ccc}1 & a & b \\ \omega & 1 & c \\ \omega^2 & \omega & 1\end{array}\right]\)
where each of \(a, b\) and \(c\) is either \(\omega\) or \(\omega^2\). Then the number of distinct matrices in the set \(\mathrm{S}\) is
(a) For the given matrix to be non-singular \(\left|\begin{array}{ccc}1 & a & b \\ \omega & 1 & c \\ \omega^2 & \omega & 1\end{array}\right| \neq 0\)
\(
\begin{aligned}
& \Rightarrow 1(1-c \omega)-a\left(\omega-c \omega^2\right)+b\left(\omega^2-\omega^2\right) \neq 0 \\
& \Rightarrow 1-c \omega-a \omega+a c \omega^2 \neq 0 \\
& \Rightarrow(1-c \omega)(1-a \omega) \neq 0 \\
& \Rightarrow a \neq \frac{1}{\omega} c \neq \frac{1}{\omega} \\
& \Rightarrow a \neq \omega^2, c \neq \omega^2
\end{aligned}
\)
As \(a, b\) and \(c\) are complex cube roots of unity
\(\therefore \quad a\) and \(c\) can take only one value i.e. \(\omega\) while \(b\) can take 2 values i.e. \(\omega\) and \(\omega^2\).
\(\therefore \quad\) Total number of distinct matrices in the set \(\mathrm{S}\) \(=1 \times 1 \times 2=2\)
If \(A\) and \(B\) are square matrices of equal degree, then which one is correct among the following?
(a) Since \(A\) and \(B\) are square matrices of the same degree, therefore matrices \(A\) and \(B\) can be added or subtracted or multiplied. By algebra of matrices the only correct option is \(A+B=B+A\).
Let \(M\) be a \(3 \times 3\) matrix satisfying \(M\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{l}-1 \\ 2 \\ 3\end{array}\right], M\left[\begin{array}{l}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ -1\end{array}\right]\), and \(M\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 12\end{array}\right]\). Then the sum of the diagonal entries of \(M\) is
Let \(\mathbf{M}=\left[\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right]\)
then \(\left[\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right]\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right]\)
\(
\Rightarrow \mathrm{b}_1=-1, \mathrm{~b}_2=2, \mathrm{~b}_3=3
\)
\(\left[\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & c_3 & c_3\end{array}\right]\left[\begin{array}{l}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ -1\end{array}\right]\)
\(\Rightarrow \mathrm{a}_1-\mathrm{b}_1=1, \mathrm{a}_2-\mathrm{b}_2=1, \mathrm{a}_3-\mathrm{b}_3=-1\)
\(\Rightarrow \mathrm{a}_1=0, \mathrm{a}_2=3, \mathrm{a}_3=2\)
and \(\left[\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{c}0 \\ 0 \\ 12\end{array}\right]\)
\(\Rightarrow a_3+b_3+c_3=12 \Rightarrow c_3=7\)
\(\therefore\) Sum of diagonal elements \(=\mathrm{a}_1+\mathrm{b}_2+\mathrm{c}_3=0+2+7=9\)
Let \(A=\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right], x \in \mathbf{R}\) and \(A^4=\left[a_{i j}\right]\). If \(a_{11}=109\), then \(a_{22}\) is equal to _____.
\(
\begin{aligned}
& A^2=\left[\begin{array}{ll}
x & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{ll}
x & 1 \\
1 & 0
\end{array}\right]=\left[\begin{array}{cc}
x^2+1 & x \\
x & 1
\end{array}\right] \\
& A^4=\left[\begin{array}{cc}
x^2+1 & x \\
x & 1
\end{array}\right]\left[\begin{array}{cc}
x^2+1 & x \\
x & 1
\end{array}\right] \\
= & {\left[\begin{array}{cc}
\left(x^2+1\right)^2+x^2 & x\left(x^2+1\right)+x \\
x\left(x^2+1\right)+x & x^2+1
\end{array}\right] }
\end{aligned}
\)
Given that \(\left(x^2+1\right)^2+x^2=109\)
\(
\begin{aligned}
& x^4+3 x^2-108=0 \\
& \Rightarrow\left(x^2+12\right)\left(x^2-9\right)=0 \\
& \therefore x^2=9 \\
& a_{22}=x^2+1=9+1=10 .
\end{aligned}
\)
Let \(a, b, c \in \mathbf{R}\) be all non-zero and satisfy \(a^3+b^3+c^3=2\). If the matrix \(A=\left(\begin{array}{ccc}a & b & c \\ b & c & a \\ c & a & b\end{array}\right)\) satisfies \(A^T A=I\), then a value of \(a b c\) can be :
\(
\begin{aligned}
& \mathrm{A}^{\mathrm{T}} \mathrm{A}=\mathrm{I} \\
& \Rightarrow \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2=1 \\
& \text { and } \mathrm{ab}+\mathrm{bc}+\mathrm{ca}=0 \\
& \text { Now, }(\mathrm{a}+\mathrm{b}+\mathrm{c})^2=1 \\
& \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}= \pm 1 \\
& \text { So, } \mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3-3 \mathrm{abc} \\
& =(\mathrm{a}+\mathrm{b}+\mathrm{c})\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca}\right) \\
& = \pm 1(1-0)= \pm 1 \\
& \Rightarrow 3 \mathrm{abc}=2 \pm 1=3,1 \\
& \Rightarrow \mathrm{abc}=1, \frac{1}{3}
\end{aligned}
\)
If \(A=\left(\begin{array}{ll}2 & 2 \\ 9 & 4\end{array}\right)\) and \(I=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\), then \(10 A^{-1}\) is equal to:
(c) Characteristics equation of matrix ‘ \(A\) ‘ is \(|A-\lambda I|=0\)
\(
\left|\begin{array}{cc}
2-\lambda & 2 \\
9 & 4-\lambda
\end{array}\right|=0
\)
\(
\Rightarrow \lambda^2-6 \lambda-10=0
\)
\(
\therefore \quad A^2-6 A-10 I=0
\)
\(
\Rightarrow \quad A^{-1}\left(A^2\right)-6 A A^{-1}-10 I A^{-1}=0
\)
\(
\Rightarrow \quad 10 A^{-1}=A-6 I
\)
If \(\mathrm{A}\) is a symmetric matrix and \(\mathrm{B}\) is a skew-symmetrix matrix such that \(\mathrm{A}+\mathrm{B}=\left[\begin{array}{cc}2 & 3 \\ 5 & -1\end{array}\right]\), then \(\mathrm{AB}\) is equal to :
\(
\text { Given } A=A^{\top} \text { and } B=-B^{\top}
\)
\(
\because A+B=\left[\begin{array}{cc}
2 & 3 \\
5 & -1
\end{array}\right] \dots(i)
\)
\(
(A+B)^{\top}=\left[\begin{array}{cc}
2 & 3 \\
5 & -1
\end{array}\right]^{\top}=\left[\begin{array}{cc}
2 & 5 \\
3 & -1
\end{array}\right]
\)
\(
A^{\top}+B^{\top}=A-B=\left[\begin{array}{cc}
2 & 5 \\
3 & -1
\end{array}\right] \dots(ii)
\)
Solving (i) and (ii) we get
\(
\begin{aligned}
& A=\left[\begin{array}{cc}
2 & 4 \\
4 & -1
\end{array}\right] \text { and } B=\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right] \\
& \therefore A B=\left[\begin{array}{cc}
4 & -2 \\
-1 & -4
\end{array}\right]
\end{aligned}
\)
The total number of matrices \(\mathrm{A}=\left(\begin{array}{ccc}0 & 2 y & 1 \\ 2 x & y & -1 \\ 2 x & -y & 1\end{array}\right)\), \((x, y \in \mathrm{R}, x \neq y)\) for which \(\mathrm{A}^{\mathrm{T}} \mathrm{A}=3 \mathrm{I}_3\) is:
\(
\text { (d) Given, } \mathrm{A}^{\mathrm{T}} \mathrm{A}=3 \mathrm{I}
\)
\(
\left[\begin{array}{ccc}
0 & 2 x & 2 x \\
2 y & y & -y \\
1 & -1 & 1
\end{array}\right]\left[\begin{array}{ccc}
0 & 2 y & 1 \\
2 x & y & -1 \\
2 x & -y & 1
\end{array}\right]=3I
\)
\(
\Rightarrow\left[\begin{array}{ccc}
8 x^2 & 0 & 0 \\
0 & 6 y^2 & 0 \\
0 & 0 & 3
\end{array}\right]=\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{array}\right]
\)
\(\Rightarrow 8 x^2=3\) and \(6 y^2=3 \Rightarrow x= \pm \sqrt{\frac{3}{8}}\) and \(y= \pm \sqrt{\frac{1}{2}}\) Number of combinations of \((x, y)=2 \times 2=4\)
Let \(A=\left(\begin{array}{ccc}0 & 2 \mathrm{q} & r \\ p & q & -r \\ p & -q & r\end{array}\right)\). If \(\mathrm{AA}^{\mathrm{T}}=\mathrm{I}_3\), then \(|\mathrm{p}|\) is :
\(
A \times A^T=\left[\begin{array}{ccc}
0 & 2 q & r \\
p & q & -r \\
p & -q & r
\end{array}\right] \times\left[\begin{array}{ccc}
0 & p & p \\
2 q & q & -q \\
r & -r & r
\end{array}\right]
\)
\(
=\left[\begin{array}{ccc}
4 q^2+r^2 & 2 q^2-r^2 & -2 q^2+r^2 \\
2 q^2-r^2 & p^2+q^2+r^2 & p^2-q^2-r^2 \\
-2 q^2+r^2 & p^2-q^2-r^2 & p^2+q^2+r^2
\end{array}\right]
\)
\(
\text { Given, } A A^T=I
\)
\(
\begin{aligned}
& 4 q^2+r^2=p^2+q^2+r^2=1 \\
& p^2-3 q^2=0 \text { and } r^2=1-4 q^2
\end{aligned}
\)
\(
2 q^2-r^2=0; r^2=2 q^2
\)
\(
p^2=\frac{1}{2}, q^2=\frac{1}{6} \text { and } r^2=\frac{1}{3}
\)
\(
|p|=\frac{1}{\sqrt{2}}
\)
If \(P\) is a \(3 \times 3\) matrix such that \(P^T=2 P+I\), where \(P^T\) is the transpose of \(P\) and \(I\) is the \(3 \times 3\) identity matrix, then there exists a column matrix \(X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right] \neq\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]\) such that
\(
\begin{aligned}
& \text { (d) } P^T=2 P+I \\
& \Rightarrow P=2 P^T+I \Rightarrow P=2(2 P+I)+I \\
& \Rightarrow P=4 P+3 I \Rightarrow P+I=0 \\
& \Rightarrow P X+X=0 \Rightarrow P X=-X
\end{aligned}
\)
If \(P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]\) and \(A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) and \(Q=P A P^T\) and \(X=P^T Q^{2005} P\) then \(X\) is equal to
(a) Given : \(P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\),
\(
\begin{gathered}
Q=P A P^T \text { and } X=P^T Q^{2005} P \\
\text { Now } Q=P A P^T \Rightarrow Q^2=\left(P A P^T\right)\left(P A P^T\right) \\
=P A\left(P^T P\right) A P^T=P A(I A) P^T=P A^2 P^T
\end{gathered}
\)
Proceeding in the same way, \(Q^{2005}=P A^{2005} P^T\) Now, \(A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] \Rightarrow A^2=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]\)
Proceeding in the same way, \(A^{2005}=\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]\)
Now, \(X=P^{\mathrm{T}} Q^{2005} P=P^T\left(P A^{2005} P^T\right) P\)
\(
\begin{aligned}
& =\left(P^T P\right) A^{2005}\left(P^T P\right)=I A^{2005} I \\
& =A^{2005}=\left[\begin{array}{cc}
1 & 2005 \\
0 & 1
\end{array}\right]
\end{aligned}
\)
The number of all \(3 \times 3\) matrices \(A\), with entries from the set \(\{-1,0,1\}\) such that the sum of the diagonal elements of \(A A^T\) is 3,____.
\(
A=\left[a_{i j}\right]_{3 \times 3}
\)
It is given that sum of diagonal elements of \(\mathrm{AA}^{\mathrm{T}}\) is 3
\(\operatorname{trace}\left(\mathrm{AA}^{\mathrm{T}}\right)=\Sigma \mathrm{a}_{\mathrm{ij}}^2=3\)
\(
a_{11}^2+a_{12}^2+a_{13}^2+a_{21}^2+\ldots . .+a_{33}^2=3
\)
So out of 9 elements \(\left(a_{i j}\right)^{\prime} s, 3\) elements must be equal to 1 or -1 and rest elements must be 0.
Possible cases are
\(
\begin{array}{cccr}
0,0,0,0,0,0,1,1,1 & \rightarrow & 6\left(0^{\prime} s\right) \text { and } 3\left(1^{\prime} s\right) \\
0,0,0,0,0,0,-1,-1,-1 & \rightarrow & 6\left(0^{\prime} s\right) \text { and } 3\left(-1^{\prime} s\right) \\
0,0,0,0,0,0,1,1,-1 & \rightarrow & 6\left(0^{\prime} s\right) \text { and } 2\left(1^{\prime} s\right) \text { and } 1\left(-1^{\prime} s\right) \\
0,0,0,0,0,0,-1,1,-1 & \rightarrow & 6\left(0^{\prime} s\right) \text { and } 2\left(-1^{\prime} s\right) \text { and } 1\left(1^{\prime} s\right)
\end{array}
\)
\(
\begin{aligned}
&\Rightarrow 9_{C_6} \times(1+1+3+3)\\
&=9_{C_6} \times 8\\
&=84 \times 8\\
&=672
\end{aligned}
\)
Hence, number of such matrices 672.
Let \(A=\left[a_{i j}\right]\) and \(B=\left[b_{i j}\right]\) be two \(3 \times 3\) real matrices such that \(b_{i j}=(3)^{(i+j-2)} a_{j i}\), where \(\mathrm{i}, \mathrm{j}=1,2,3\). If the determinant of \(\mathrm{B}\) is 81 , then the determinant of \(\mathrm{A}\) is :
\(
b_{i j}=(3)^{(i+j-2)} a_{i j}
\)
\(
\mathbf{B}=\left[\begin{array}{ccc}
\mathrm{a}_{11} & 3 \mathrm{a}_{12} & 3^2 \mathrm{a}_{13} \\
3 \mathrm{a}_{21} & 3 \mathrm{a}_{22} & 3 \mathrm{a}_{23} \\
3^2 \mathrm{a}_{31} & 3^2 \mathrm{a}_{32} & 3^2 \mathrm{a}_{33}
\end{array}\right]
\)
\(
\Rightarrow|\mathrm{B}|=3 \times 3^2\left|\begin{array}{ccc}
\mathrm{a}_{11} & \mathrm{a}_{12} & \mathrm{a}_{13} \\
3 \mathrm{a}_{21} & 3 \mathrm{a}_{22} & 3 \mathrm{a}_{23} \\
3^2 \mathrm{a}_{31} & 3^2 \mathrm{a}_{32} & 3^2 \mathrm{a}_{33}
\end{array}\right|
\)
\(
\begin{aligned}
& =3^6|\mathrm{~A}| \\
& \Rightarrow|\mathrm{A}|=\frac{81}{27 \times 27}=\frac{1}{9}
\end{aligned}
\)
If the matrices \(A=\left[\begin{array}{ccc}1 & 1 & 2 \\ 1 & 3 & 4 \\ 1 & -1 & 3\end{array}\right], B=\operatorname{adj} A\) and \(\mathrm{C}=3 \mathrm{~A}\), then \(\frac{|\operatorname{adj} \mathrm{B}|}{|\mathrm{C}|}\) is equal to :
\(
A=\left[\begin{array}{ccc}
1 & 1 & 2 \\
1 & 3 & 4 \\
1 & -1 & 3
\end{array}\right]
\)
\(
\Rightarrow \quad|\mathrm{A}|=6
\)
\(
\frac{|\operatorname{adj} \mathrm{B}|}{|\mathrm{c}|}=\frac{|\operatorname{adj}(\operatorname{adj} \mathrm{A})|}{|9 \mathrm{~A}|}=\frac{|\mathrm{A}|^4}{3^3|\mathrm{~A}|}=\frac{|\mathrm{A}|^3}{3^3}
\)
\(
=\frac{(6)^3}{(3)^3}=8
\)
Let \(A\) be a \(2 \times 2\) real matrix with entries from \(\{0,1\}\) and \(|A| \neq 0\). Consider the following two statements:
(P) If \(A \neq I_2\), then \(|A|=-1\)
(Q) If \(|A|=1\), then \(\operatorname{tr}(A)=2\),
where \(\mathrm{I}_2\) denotes \(2 \times 2\) identity matrix and \(\operatorname{tr}(\mathrm{A})\) denotes the sum of the diagonal entries of \(\mathrm{A}\). Then:
\(|\mathrm{A}| \neq 0\)
For \((\mathrm{P}): \mathrm{A} \neq \mathrm{I}_2\)
So, \(A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\) or \(\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]\) or \(\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]\) or \(\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) or \(\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]\)
\(|\mathrm{A}|\) can be -1 or 1
So \((\mathrm{P})\) is false.
For (Q); \(|\mathrm{A}|=1\)
\(
\begin{aligned}
& A=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \text { or }\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] \text { or }\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right] \\
& \Rightarrow \operatorname{tr}(A)=2 \\
& \Rightarrow Q \text { is true }
\end{aligned}
\)
Let \(A=\left\{X=(x, y, z)^T: P X=0\right.\) and \(\left.\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2=1\right\}\) where \(\mathrm{P}=\left[\begin{array}{ccc}1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1\end{array}\right]\) then the set A :
Given \(P=\left[\begin{array}{ccc}1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & 1\end{array}\right]\), Here \(|P|=0\) & also given \(\mathrm{PX}=0\)
\(
\Rightarrow\left[\begin{array}{ccc}
1 & 2 & 1 \\
-2 & 3 & -4 \\
1 & 9 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=0
\)
\(
\left.\Rightarrow \begin{array}{c}
x+2 y+z=0 \\
-2 x+3 y-4 z=0 \\
x+9 y-z=0
\end{array}\right\} \because D=0 \text {, so system have infinite many solutions,}
\)
By solving these equation
we get \(x=\frac{-11 \lambda}{2} ; y=\lambda ; z=\frac{7 \lambda}{2}\)
Also given, \(\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2=1\)
\(
\begin{aligned}
& \Rightarrow\left(\frac{-11 \lambda}{2}\right)^2+(\lambda)^2+\left(\frac{7 \lambda}{2}\right)^2=1 \\
& \Rightarrow \lambda= \pm \frac{1}{\sqrt{\frac{121}{4}+1+\frac{49}{4}}}
\end{aligned}
\)
so, there are 2 values of \(\lambda\).
\(\therefore\) so, there are 2 solution set of \((\mathrm{x}, \mathrm{y}, \mathrm{z})\).
Let \(A\) be a \(3 \times 3\) matrix such that adj \(A=\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 0 & 2 \\ 1 & -2 & -1\end{array}\right]\) and \(B=\operatorname{adj}(\operatorname{adj} A)\). If \(|\mathrm{A}|=\lambda\) and \(\left|\left(\mathrm{B}^{-1}\right)^{\mathrm{T}}\right|=\mu\), then the ordered pair, \((|\lambda|, \mu)\) is equal to :
\(
C=\operatorname{adj} A=\left|\begin{array}{ccc}
+2 & -1 & 1 \\
-1 & 0 & 2 \\
1 & -2 & -1
\end{array}\right|
\)
\(
\mid \mathrm{C}\mid=\mid {adj} \mathrm{A} \mid=+2(0+4)+1 .(1-2)+1 .(2-0)=+8-1+2
\)
\(
\left.|{adj} \mathrm{A}|=| \mathrm{A}\right|^2=9=9
\)
\(
\begin{aligned}
& \lambda=|\mathrm{A}|= \pm 3 \\
& |\lambda|=3
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{B}=\operatorname{adj} C \\
& |\mathrm{~B}|=|\operatorname{adj} \mathrm{C}|=|\mathrm{C}|^2=81
\end{aligned}
\)
\(
\left|\left(\mathrm{B}^{-1}\right)^{\mathrm{T}}\right|=|\mathrm{B}|^{-1}=\frac{1}{81}
\)
\(
(|\lambda|, \mu)=\left(3, \frac{1}{81}\right)
\)
Suppose the vectors \(\mathrm{x}_1, \mathrm{x}_2\) and \(\mathrm{x}_3\) are the solutions of the system of linear equations, \(\mathrm{Ax}=\mathrm{b}\) when the vector \(\mathrm{b}\) on the right side is equal to \(b_1, b_2\) and \(b_3\) respectively. If \(\mathrm{x}_1=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right], \mathrm{x}_2=\left[\begin{array}{l}0 \\ 2 \\ 1\end{array}\right], \mathrm{x}_3=\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right], \mathrm{b}_1=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]\)
\(b_2=\left[\begin{array}{l}0 \\ 2 \\ 0\end{array}\right]\) and \(b_3=\left[\begin{array}{l}0 \\ 0 \\ 2\end{array}\right]\), then the determinant of \(\mathrm{A}\) is equal to :
\(
\begin{aligned}
& \mathrm{Ax}_1=\mathrm{b}_1 \\
& \mathrm{Ax}_2=\mathrm{b}_2 \\
& \mathrm{Ax}_3=\mathrm{b}_3 \\
& \Rightarrow|\mathrm{A}|\left|\begin{array}{lll}
1 & 0 & 0 \\
1 & 2 & 0 \\
1 & 1 & 1
\end{array}\right|=\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{array}\right| \\
& \Rightarrow|\mathrm{A}|=\frac{4}{2}=2
\end{aligned}
\)
Let \(\theta=\frac{\pi}{5}\) and \(A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\).
If \(\mathrm{B}=\mathrm{A}+\mathrm{A}^4\), then \(\operatorname{det}(\mathrm{B})\) :
\(
\begin{aligned}
& A=\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right] \\
& A^2=\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right] \\
& A^2=\left[\begin{array}{cc}
\cos 2 \theta & \sin 2 \theta \\
-\sin 2 \theta & \cos 2 \theta
\end{array}\right]
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{B}=\mathrm{A}+\mathrm{A}^4 \\
& =\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]+\left[\begin{array}{cc}
\cos 4 \theta & \sin 4 \theta \\
-\sin 4 \theta & \cos 4 \theta
\end{array}\right] \\
& \mathrm{B}=\left[\begin{array}{cc}
(\cos \theta+\cos 4 \theta) & (\sin \theta+\sin 4 \theta) \\
-(\sin \theta+\sin 4 \theta) & (\cos \theta+\cos 4 \theta)
\end{array}\right] \\
& |\mathrm{B}|=(\cos \theta+\cos 4 \theta)^2+(\sin \theta+\sin 4 \theta)^2
\end{aligned}
\)
\(|B|=2+2 \cos 3 \theta\), when \(\theta=\frac{\pi}{5}\)
\(
\begin{aligned}
& |B|=2+2 \cos \frac{3 \pi}{5}=2(1-\sin 18) \\
& |B|=2\left(1-\frac{\sqrt{5}-1}{4}\right)=2\left(\frac{5-\sqrt{5}}{4}\right)=\frac{5-\sqrt{5}}{2}
\end{aligned}
\)
If \(\Delta=\left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \\ 2 x-3 & 3 x-4 & 4 x-5 \\ 3 x-5 & 5 x-8 & 10 x-17\end{array}\right|=A x^3+B x^2+C x+D\), then \(B+C\) is equal to :
\(
\Delta=\left|\begin{array}{ccc}
x-2 & 2 x-3 & 3 x-4 \\
2 x-3 & 3 x-4 & 4 x-5 \\
3 x-5 & 5 x-8 & 10 x-17
\end{array}\right|=A x^3+B x^2+\mathrm{Cx}+\mathrm{D}
\)
\(
\begin{aligned}
& R_2 \rightarrow R_2-R_1 \quad \quad \quad R_3 \rightarrow R_3-R_2 \\
& \Delta=\left|\begin{array}{ccc}
x-2 & 2 x-3 & 3 x-4 \\
x-1 & x-1 & x-1 \\
x-2 & 2(x-2) & 6(x-2)
\end{array}\right| \\
& =(x-1)(x-2)\left|\begin{array}{ccc}
x-2 & 2 x-3 & 3 x-4 \\
1 & 1 & 1 \\
1 & 2 & 6
\end{array}\right| \\
& =-3(x-1)^2(x-2)=-3 x^3+12 x^2-15 x+6 \\
& \therefore B+C=12-15=-3
\end{aligned}
\)
Let \(\mathrm{a}-2 \mathrm{~b}+\mathrm{c}=1\).
If \(f(x)=\left|\begin{array}{lll}x+a & x+2 & x+1 \\ x+b & x+3 & x+2 \\ x+c & x+4 & x+3\end{array}\right|\), then :
\(
\begin{aligned}
& \mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_3-2 \mathrm{R}_2 \\
& f(\mathrm{x})=\left|\begin{array}{ccc}
\mathrm{a}+\mathrm{c}-2 \mathrm{~b} & 0 & 0 \\
\mathrm{x}+\mathrm{b} & \mathrm{x}+3 & \mathrm{x}+2 \\
\mathrm{x}+\mathrm{c} & \mathrm{x}+4 & \mathrm{x}+3
\end{array}\right| \\
& =(\mathrm{a}+\mathrm{c}-2 \mathrm{~b})\left((\mathrm{x}+3)^2-(\mathrm{x}+2)(\mathrm{x}+4)\right) \\
& =\mathrm{x}^2+6 \mathrm{x}+9-\mathrm{x}^2-6 \mathrm{x}-8=1 \\
& \Rightarrow f(\mathrm{x})=1 \Rightarrow f(50)=1
\end{aligned}
\)
If \(\Delta_1=\left|\begin{array}{ccc}\mathrm{x} & \sin \theta & \cos \theta \\ -\sin \theta & -\mathrm{x} & 1 \\ \cos \theta & 1 & \mathrm{x}\end{array}\right|\) and
\(\Delta_2=\left|\begin{array}{ccc}x & \sin 2 \theta & \cos 2 \theta \\ -\sin 2 \theta & -x & 1 \\ \cos 2 \theta & 1 & x\end{array}\right|, x \neq 0\); then for all \(\theta \in\left(0, \frac{\pi}{2}\right)\) :
\(
\text { (d) } \Delta_1=\left|\begin{array}{ccc}
x & \sin \theta & \cos \theta \\
-\sin \theta & -x & 1 \\
\cos \theta & 1 & x
\end{array}\right|
\)
\(
=\left(x-x^2-1\right)-\sin \mathrm{\theta}(-x \sin \mathrm{\theta}-\cos \mathrm{\theta})+\cos \theta(-\sin \theta+x \cos \theta)
\)
\(
\begin{aligned}
& =-x^3-x+x \sin ^2 \mathrm{\theta}+\sin \mathrm{\theta} \cos \mathrm{\theta}-\cos \mathrm{\theta} \sin \mathrm{\theta}+x \cos ^2 \mathrm{\theta} \\
& =-x^3-x+x=-x^3
\end{aligned}
\)
\(
\text { Similarly, } \Delta_2=-x^3Â
\)
Then, \(\Delta_1+\Delta_2=-2 x^3\)
Let \(A=\left[\begin{array}{ccc}2 & b & 1 \\ b & b^2+1 & b \\ 1 & b & 2\end{array}\right]\) where \(b>0\). Then the minimum value of \(\frac{\operatorname{det}(\mathrm{A})}{\mathrm{b}}\) is :
\(
\begin{aligned}
& A=\left[\begin{array}{ccc}
2 & b & 1 \\
b & b^2+1 & b \\
1 & b & 2
\end{array}\right] \text { where } b>0 \\
& \Rightarrow|A|=2\left[2\left(b^2+1\right)-b^2\right]-b[2 b-b]+1\left[b^2-b^2-1\right] \\
& \Rightarrow|A|=2 b^2+4-b^2-1 \\
& \Rightarrow|A|=b^2+3
\end{aligned}
\)
We have to calculate, \(\min \left(\frac{\operatorname{det}(A)}{b}\right)\)
\(
\Rightarrow \frac{b^2+3}{b}=b+\frac{3}{b}
\)
Let \(f(b)=b+\frac{3}{b}\)
\(\Rightarrow f^{\prime}(b)=1-\frac{3}{b^2}\)
\(
\Rightarrow f^{\prime \prime}(b)=\frac{6}{b^3}
\)
For maxima and minima, \(f^{\prime}(b)=0\)
\(
\begin{aligned}
& \therefore \mathrm{b}=\sqrt{3} \quad(\because \mathrm{b}>0) \\
& \text { and } \mathrm{f}^{\prime \prime}(\sqrt{3})>0
\end{aligned}
\)
Therefore, \(f(b)\) is minimum at \(b=\sqrt{3}\)
\(\therefore\) Minimum value of \(f(b)=\frac{\operatorname{det}(A)}{b}\) is
\(
\begin{aligned}
& f(b)=f(\sqrt{3})=\sqrt{3}+\frac{3}{\sqrt{3}} . \\
& \Rightarrow f(\sqrt{3})=2 \sqrt{3}
\end{aligned}
\)
Minimum value of \(\frac{\operatorname{det}(A)}{b}\) is \(2 \sqrt{3}\)
If \(\left|\begin{array}{lll}x-4 & 2 x & 2 x \\ 2 x & x-4 & 2 x \\ 2 x & 2 x & x-4\end{array}\right|=(A+B x)(x-A)^2\), then the ordered pair \((A, B)\) is equal to :
\(
\text { (b) Here, }\left|\begin{array}{ccc}
x-4 & 2 x & 2 x \\
2 x & x-4 & 2 x \\
2 x & 2 x & x-4
\end{array}\right|=(A+B x)(x-A)^2
\)
Put
\(
\begin{aligned}
x & =0 \Rightarrow\left|\begin{array}{ccc}
-4 & 0 & 0 \\
0 & -4 & 0 \\
0 & 0 & -4
\end{array}\right|=A^3 \Rightarrow A^3=(-4)^3 \\
& \Rightarrow A=-4 \\
& \Rightarrow\left|\begin{array}{ccc}
x-4 & 2 x & 2 x \\
2 x & x-4 & 2 x \\
2 x & 2 x & x-4
\end{array}\right|=(B x-4)(x+4)^2
\end{aligned}
\)
Now take \(\mathrm{x}\) common from both the sides
\(
\therefore\left|\begin{array}{ccc}
1-\frac{4}{x} & 2 x & 2 x \\
2 x & 1-\frac{4}{x} & 2 x \\
2 x & 2 x & 1-\frac{4}{x}
\end{array}\right|=\left(B-\frac{4}{x}\right)\left(1+\frac{4}{x}\right)^2
\)
\(
\text { Now take } x \rightarrow \infty \text {, then } \frac{1}{x} \rightarrow 0
\)
\(
\Rightarrow\left|\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right|=\mathrm{B} \Rightarrow \mathrm{B}=5
\)
\(\therefore \quad\) ordered pair \((\mathrm{A}, \mathrm{B})\) is \((-4,5)\)
If \(S=\left\{x \in[0,2 \pi]:\left|\begin{array}{ccc}0 & \cos x & -\sin x \\ \sin x & 0 & \cos x \\ \cos x & \sin x & 0\end{array}\right|=0\right\}\), then
\(\sum_{x \in S} \tan \left(\frac{\pi}{3}+x\right)\) is equal to
(c) Since the given determinant is equal to zero.
\(
\begin{array}{ll}
\Rightarrow & 0(0-\cos x \sin x)-\cos x\left(0-\cos ^2 x\right)-\sin x \\
& \left(\sin ^2 x-0\right)=0 \\
\Rightarrow & \cos ^3 x-\sin ^3 x=0 \\
\Rightarrow & \tan ^3=1 \Rightarrow \tan x=1
\end{array}
\)
\(
\therefore \quad \sum_{\mathrm{x} \in \mathrm{s}} \tan \left(\frac{\pi}{3}+\mathrm{x}\right)=\sum_{x \in s} \frac{\tan \pi / 3+\tan x}{1-\tan \pi / 3 \cdot \tan x}
\)
\(
=\sum_{x \in s} \frac{\sqrt{3}+1}{1-\sqrt{3}}
\)
\(
\begin{gathered}
\sum_{x \in s} \frac{\sqrt{3}+1}{1-\sqrt{3}} \times \frac{1+\sqrt{3}}{1+\sqrt{3}} \Rightarrow \sum_{x \in s} \frac{1+3+2 \sqrt{3}}{-2} \\
=-2-\sqrt{3}
\end{gathered}
\)
If \(A=\left[\begin{array}{ll}-4 & -1 \\ 3 & 1\end{array}\right]\), then the determinant of the matrix \(\left(A^{2016}-2 A^{2015}-A^{2014}\right)\) is :
\(
\begin{aligned}
\text { (d) } & A=\left[\begin{array}{cc}
-4 & -1 \\
3 & 1
\end{array}\right] \Rightarrow A^2=\left[\begin{array}{cc}
-4 & -1 \\
3 & 1
\end{array}\right]\left[\begin{array}{cc}
-4 & -1 \\
3 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
13 & 3 \\
-9 & -2
\end{array}\right] \text { and }|A|=1 . \\
& \text { Now, } A^{2016}-2 A^{2015}-A^{2014}=A^{2014}\left(A^2-2 A-I\right) \\
\Rightarrow & \left|A^{2016}-2 A^{2015}-A^{2014}\right|=\left|A^{2014}\right|\left|A^2-2 A-I\right| \\
& =|A|^{2014}\left|\begin{array}{cc}
20 & 5 \\
-15 & -5
\end{array}\right|=-25
\end{aligned}
\)
if \(\left|\begin{array}{ccc}x^2+x & x+1 & x-2 \\ 2 x^2+3 x-1 & 3 x & 3 x-3 \\ x^2+2 x+3 & 2 x-1 & 2 x-1\end{array}\right|=a x-12\), then ‘ \(a\) ‘ is equal to
(a) Let \(\left|\begin{array}{ccc}x^2+x & x+1 & x-2 \\ 2 x^2+3 x-1 & 3 x & 3 x-3 \\ x^2+2 x+3 & 2 x-1 & 2 x-1\end{array}\right|=a x-12\)
Put \(x=-1\), we get
\(
\begin{aligned}
& \left|\begin{array}{ccc}
0 & 0 & -3 \\
-2 & -3 & 0 \\
2 & -3 & -3
\end{array}\right|=-a-12 \\
& \Rightarrow-3(6+6)=-a-12 \Rightarrow-36+12=a \\
& \Rightarrow a=24
\end{aligned}
\)
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