Match List – I with List – II.
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LIST I |
LIST II |
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A. If $A^{-1}=\left[\begin{array}{ll}3 & 4 \\ 5 & 6\end{array}\right]$, then A is |
I. $\left[\begin{array}{ll}5 & 2 \\ 7 & 3\end{array}\right]$ |
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B. If $A=\left[\begin{array}{cc}2 & -1 \\ 3 & 1\end{array}\right]$, then $A A^T$ is |
II. $\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$ |
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C. If $A^{-1}=\left[\begin{array}{ll}5 & 7 \\ 2 & 3\end{array}\right]$, then $\left(A^T\right)^{-1}$ is |
III. $\left[\begin{array}{cc}-3 & 2 \\ \frac{5}{2} & -\frac{3}{2}\end{array}\right]$ |
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D. If $A=\left[a_{i j}\right]_{2 \times 2}$ where $a_{i j}=(i-j)^2$, then A is |
IV. $\left[\begin{array}{cc}5 & 5 \\ 5 & 10\end{array}\right]$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-III, B-IV, C-I, D-II
(A) $A^{-1}=\left[\begin{array}{ll}3 & 4 \\ 5 & 6\end{array}\right]$
$⇒A=(A^{-1})^{-1}=\left[\begin{array}{ll}-3 & 2 \\ \frac{5}{2} & -\frac{3}{2}\end{array}\right]$
(B) $A=\left[\begin{array}{cc}2 & -1 \\ 3 & 1\end{array}\right]$
$⇒A^T=\left[\begin{array}{cc}2 & 3 \\ -1 & 1\end{array}\right]$
$⇒AA^T=\left[\begin{array}{cc}2 & -1 \\ 3 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ -1 & 1\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
Hence, there are called orthogonal matrix.
(C) $A^{-1}=\left[\begin{array}{ll}5 & 7 \\ 2 & 3\end{array}\right]$
$⇒(A^T)^{-1}=(A^{-1})^T=\left[\begin{array}{ll}5 & 2 \\ 7 & 3\end{array}\right]$
(D) $A=\left[a_{i j}\right]_{2 \times 2},a_{i j}=(i-j)^2$
$⇒A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$