(B) : Given that, \(\mathrm{f}=\left(\begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 4 & 5 & 1 & 6 \end{array}\right)\) We know that, \(\mathrm{f}^{-1}\) by using inverse permutation, \(\begin{array}{ll} \mathrm{f}^{-1}(1)=1 \rightarrow 3 \rightarrow 4 \rightarrow 5 \rightarrow 1 \\ \mathrm{f}^{-1}(1)=5 & \begin{array}{l} \text { because value } 1 \text { again } \\ \text { repeated opposite to } 5 . \end{array} \\ \mathrm{f}^{-1}(2)=2 \rightarrow 2 & \text { because } 2 \text { is opposite to } 2 . \\ \mathrm{f}^{-1}(2)=2 & \begin{array}{l} \text { because value } 3 \text { again } \\ \mathrm{f}^{-1}(3)=3 \rightarrow 4 \rightarrow 5 \rightarrow 1 \rightarrow 3 \end{array} \\ \mathrm{f}^{-1}(3)=1 & \begin{array}{l} \text { repeated opposite of } 1 . \end{array} \\ \mathrm{f}^{-1}(4)=4 \rightarrow 5 \rightarrow 1 \rightarrow 3 \rightarrow 4 \end{array}\) \(\begin{array}{ll} \mathrm{f}^{-1}(4)=3 & \begin{array}{l} \text { because value } 4 \text { again } \\ \text { repeated opposite to } 3 . \end{array} \\ \mathrm{f}^{-1}(5)=5 \rightarrow 1 \rightarrow 3 \rightarrow 4 \rightarrow 5 \\ \mathrm{f}^{-1}(5)=4 & \begin{array}{l} 3 \rightarrow 4 \rightarrow 5 \\ \text { because value } 5 \text { again } \end{array} \end{array}\) \(=6 \rightarrow 6\) \(\mathrm{f}^{-1}(6)=6\) because value of 6 is opposite to 6 . Hence, \(\mathrm{f}^{-1}=\left(\begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 2 & 1 & 3 & 4 & 6 \end{array}\right) .\)
[JCECE-2017]
Matrix and Determinant
78845
If \(A=\left[\begin{array}{cc}1 & \tan x \\ -\tan x & 1\end{array}\right]\), then the value of \(\left|\mathbf{A}^{\mathrm{T}} \mathbf{A}^{-1}\right|\) is
1 \(\cos 4 \mathrm{x}\)
2 \(\sec ^{2} x\)
3 \(-\cos 4 x\)
4 1
Explanation:
(B) : Given, \(A=\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right] \Rightarrow|A|=1+\tan ^{2} x=\sec ^{2} x\) \(A^{T}=\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right]\) \(\operatorname{adj}(A)=\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right]\) And, \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}\) \(=\frac{1}{\sec ^{2} \mathrm{x}}\left[\begin{array}{cc} 1 & -\tan \mathrm{x} \\ \tan \mathrm{x} & 1 \end{array}\right]\) Now, \(\left|A^{T} A^{-1}\right| =\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \times \frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right]\) \(=\frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1-\tan ^{2} x & -\tan x-\tan x \\ \tan x+\tan x & 1-\tan ^{2} x \end{array}\right]\) \(=\frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1-\tan ^{2} x & -2 \tan x \\ 2 \tan x & 1-\tan ^{2} x \end{array}\right]\) \(=\frac{\left(1-\tan ^{2} x\right)^{2}+4 \tan ^{2} x}{\sec ^{2} x}\) \(=\frac{\left(1+\tan ^{2} x\right)^{2}}{\sec ^{2} x}=\frac{\left(\sec ^{2} x\right)^{2}}{\sec ^{2} x}=\sec ^{2} x\)
CG PET-2011
Matrix and Determinant
78846
If \(A=\left[\begin{array}{cc}4 & 11 \\ 2 & 6\end{array}\right]\), then \(A^{-1}\) is equal to
(B) : Given that, \(\mathrm{f}=\left(\begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 4 & 5 & 1 & 6 \end{array}\right)\) We know that, \(\mathrm{f}^{-1}\) by using inverse permutation, \(\begin{array}{ll} \mathrm{f}^{-1}(1)=1 \rightarrow 3 \rightarrow 4 \rightarrow 5 \rightarrow 1 \\ \mathrm{f}^{-1}(1)=5 & \begin{array}{l} \text { because value } 1 \text { again } \\ \text { repeated opposite to } 5 . \end{array} \\ \mathrm{f}^{-1}(2)=2 \rightarrow 2 & \text { because } 2 \text { is opposite to } 2 . \\ \mathrm{f}^{-1}(2)=2 & \begin{array}{l} \text { because value } 3 \text { again } \\ \mathrm{f}^{-1}(3)=3 \rightarrow 4 \rightarrow 5 \rightarrow 1 \rightarrow 3 \end{array} \\ \mathrm{f}^{-1}(3)=1 & \begin{array}{l} \text { repeated opposite of } 1 . \end{array} \\ \mathrm{f}^{-1}(4)=4 \rightarrow 5 \rightarrow 1 \rightarrow 3 \rightarrow 4 \end{array}\) \(\begin{array}{ll} \mathrm{f}^{-1}(4)=3 & \begin{array}{l} \text { because value } 4 \text { again } \\ \text { repeated opposite to } 3 . \end{array} \\ \mathrm{f}^{-1}(5)=5 \rightarrow 1 \rightarrow 3 \rightarrow 4 \rightarrow 5 \\ \mathrm{f}^{-1}(5)=4 & \begin{array}{l} 3 \rightarrow 4 \rightarrow 5 \\ \text { because value } 5 \text { again } \end{array} \end{array}\) \(=6 \rightarrow 6\) \(\mathrm{f}^{-1}(6)=6\) because value of 6 is opposite to 6 . Hence, \(\mathrm{f}^{-1}=\left(\begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 2 & 1 & 3 & 4 & 6 \end{array}\right) .\)
[JCECE-2017]
Matrix and Determinant
78845
If \(A=\left[\begin{array}{cc}1 & \tan x \\ -\tan x & 1\end{array}\right]\), then the value of \(\left|\mathbf{A}^{\mathrm{T}} \mathbf{A}^{-1}\right|\) is
1 \(\cos 4 \mathrm{x}\)
2 \(\sec ^{2} x\)
3 \(-\cos 4 x\)
4 1
Explanation:
(B) : Given, \(A=\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right] \Rightarrow|A|=1+\tan ^{2} x=\sec ^{2} x\) \(A^{T}=\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right]\) \(\operatorname{adj}(A)=\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right]\) And, \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}\) \(=\frac{1}{\sec ^{2} \mathrm{x}}\left[\begin{array}{cc} 1 & -\tan \mathrm{x} \\ \tan \mathrm{x} & 1 \end{array}\right]\) Now, \(\left|A^{T} A^{-1}\right| =\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \times \frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right]\) \(=\frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1-\tan ^{2} x & -\tan x-\tan x \\ \tan x+\tan x & 1-\tan ^{2} x \end{array}\right]\) \(=\frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1-\tan ^{2} x & -2 \tan x \\ 2 \tan x & 1-\tan ^{2} x \end{array}\right]\) \(=\frac{\left(1-\tan ^{2} x\right)^{2}+4 \tan ^{2} x}{\sec ^{2} x}\) \(=\frac{\left(1+\tan ^{2} x\right)^{2}}{\sec ^{2} x}=\frac{\left(\sec ^{2} x\right)^{2}}{\sec ^{2} x}=\sec ^{2} x\)
CG PET-2011
Matrix and Determinant
78846
If \(A=\left[\begin{array}{cc}4 & 11 \\ 2 & 6\end{array}\right]\), then \(A^{-1}\) is equal to
(B) : Given that, \(\mathrm{f}=\left(\begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 4 & 5 & 1 & 6 \end{array}\right)\) We know that, \(\mathrm{f}^{-1}\) by using inverse permutation, \(\begin{array}{ll} \mathrm{f}^{-1}(1)=1 \rightarrow 3 \rightarrow 4 \rightarrow 5 \rightarrow 1 \\ \mathrm{f}^{-1}(1)=5 & \begin{array}{l} \text { because value } 1 \text { again } \\ \text { repeated opposite to } 5 . \end{array} \\ \mathrm{f}^{-1}(2)=2 \rightarrow 2 & \text { because } 2 \text { is opposite to } 2 . \\ \mathrm{f}^{-1}(2)=2 & \begin{array}{l} \text { because value } 3 \text { again } \\ \mathrm{f}^{-1}(3)=3 \rightarrow 4 \rightarrow 5 \rightarrow 1 \rightarrow 3 \end{array} \\ \mathrm{f}^{-1}(3)=1 & \begin{array}{l} \text { repeated opposite of } 1 . \end{array} \\ \mathrm{f}^{-1}(4)=4 \rightarrow 5 \rightarrow 1 \rightarrow 3 \rightarrow 4 \end{array}\) \(\begin{array}{ll} \mathrm{f}^{-1}(4)=3 & \begin{array}{l} \text { because value } 4 \text { again } \\ \text { repeated opposite to } 3 . \end{array} \\ \mathrm{f}^{-1}(5)=5 \rightarrow 1 \rightarrow 3 \rightarrow 4 \rightarrow 5 \\ \mathrm{f}^{-1}(5)=4 & \begin{array}{l} 3 \rightarrow 4 \rightarrow 5 \\ \text { because value } 5 \text { again } \end{array} \end{array}\) \(=6 \rightarrow 6\) \(\mathrm{f}^{-1}(6)=6\) because value of 6 is opposite to 6 . Hence, \(\mathrm{f}^{-1}=\left(\begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 2 & 1 & 3 & 4 & 6 \end{array}\right) .\)
[JCECE-2017]
Matrix and Determinant
78845
If \(A=\left[\begin{array}{cc}1 & \tan x \\ -\tan x & 1\end{array}\right]\), then the value of \(\left|\mathbf{A}^{\mathrm{T}} \mathbf{A}^{-1}\right|\) is
1 \(\cos 4 \mathrm{x}\)
2 \(\sec ^{2} x\)
3 \(-\cos 4 x\)
4 1
Explanation:
(B) : Given, \(A=\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right] \Rightarrow|A|=1+\tan ^{2} x=\sec ^{2} x\) \(A^{T}=\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right]\) \(\operatorname{adj}(A)=\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right]\) And, \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}\) \(=\frac{1}{\sec ^{2} \mathrm{x}}\left[\begin{array}{cc} 1 & -\tan \mathrm{x} \\ \tan \mathrm{x} & 1 \end{array}\right]\) Now, \(\left|A^{T} A^{-1}\right| =\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \times \frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right]\) \(=\frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1-\tan ^{2} x & -\tan x-\tan x \\ \tan x+\tan x & 1-\tan ^{2} x \end{array}\right]\) \(=\frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1-\tan ^{2} x & -2 \tan x \\ 2 \tan x & 1-\tan ^{2} x \end{array}\right]\) \(=\frac{\left(1-\tan ^{2} x\right)^{2}+4 \tan ^{2} x}{\sec ^{2} x}\) \(=\frac{\left(1+\tan ^{2} x\right)^{2}}{\sec ^{2} x}=\frac{\left(\sec ^{2} x\right)^{2}}{\sec ^{2} x}=\sec ^{2} x\)
CG PET-2011
Matrix and Determinant
78846
If \(A=\left[\begin{array}{cc}4 & 11 \\ 2 & 6\end{array}\right]\), then \(A^{-1}\) is equal to
(B) : Given that, \(\mathrm{f}=\left(\begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 4 & 5 & 1 & 6 \end{array}\right)\) We know that, \(\mathrm{f}^{-1}\) by using inverse permutation, \(\begin{array}{ll} \mathrm{f}^{-1}(1)=1 \rightarrow 3 \rightarrow 4 \rightarrow 5 \rightarrow 1 \\ \mathrm{f}^{-1}(1)=5 & \begin{array}{l} \text { because value } 1 \text { again } \\ \text { repeated opposite to } 5 . \end{array} \\ \mathrm{f}^{-1}(2)=2 \rightarrow 2 & \text { because } 2 \text { is opposite to } 2 . \\ \mathrm{f}^{-1}(2)=2 & \begin{array}{l} \text { because value } 3 \text { again } \\ \mathrm{f}^{-1}(3)=3 \rightarrow 4 \rightarrow 5 \rightarrow 1 \rightarrow 3 \end{array} \\ \mathrm{f}^{-1}(3)=1 & \begin{array}{l} \text { repeated opposite of } 1 . \end{array} \\ \mathrm{f}^{-1}(4)=4 \rightarrow 5 \rightarrow 1 \rightarrow 3 \rightarrow 4 \end{array}\) \(\begin{array}{ll} \mathrm{f}^{-1}(4)=3 & \begin{array}{l} \text { because value } 4 \text { again } \\ \text { repeated opposite to } 3 . \end{array} \\ \mathrm{f}^{-1}(5)=5 \rightarrow 1 \rightarrow 3 \rightarrow 4 \rightarrow 5 \\ \mathrm{f}^{-1}(5)=4 & \begin{array}{l} 3 \rightarrow 4 \rightarrow 5 \\ \text { because value } 5 \text { again } \end{array} \end{array}\) \(=6 \rightarrow 6\) \(\mathrm{f}^{-1}(6)=6\) because value of 6 is opposite to 6 . Hence, \(\mathrm{f}^{-1}=\left(\begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 2 & 1 & 3 & 4 & 6 \end{array}\right) .\)
[JCECE-2017]
Matrix and Determinant
78845
If \(A=\left[\begin{array}{cc}1 & \tan x \\ -\tan x & 1\end{array}\right]\), then the value of \(\left|\mathbf{A}^{\mathrm{T}} \mathbf{A}^{-1}\right|\) is
1 \(\cos 4 \mathrm{x}\)
2 \(\sec ^{2} x\)
3 \(-\cos 4 x\)
4 1
Explanation:
(B) : Given, \(A=\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right] \Rightarrow|A|=1+\tan ^{2} x=\sec ^{2} x\) \(A^{T}=\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right]\) \(\operatorname{adj}(A)=\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right]\) And, \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}\) \(=\frac{1}{\sec ^{2} \mathrm{x}}\left[\begin{array}{cc} 1 & -\tan \mathrm{x} \\ \tan \mathrm{x} & 1 \end{array}\right]\) Now, \(\left|A^{T} A^{-1}\right| =\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \times \frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right]\) \(=\frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1-\tan ^{2} x & -\tan x-\tan x \\ \tan x+\tan x & 1-\tan ^{2} x \end{array}\right]\) \(=\frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1-\tan ^{2} x & -2 \tan x \\ 2 \tan x & 1-\tan ^{2} x \end{array}\right]\) \(=\frac{\left(1-\tan ^{2} x\right)^{2}+4 \tan ^{2} x}{\sec ^{2} x}\) \(=\frac{\left(1+\tan ^{2} x\right)^{2}}{\sec ^{2} x}=\frac{\left(\sec ^{2} x\right)^{2}}{\sec ^{2} x}=\sec ^{2} x\)
CG PET-2011
Matrix and Determinant
78846
If \(A=\left[\begin{array}{cc}4 & 11 \\ 2 & 6\end{array}\right]\), then \(A^{-1}\) is equal to
