Let \( A \) be a square matrix of order 2 such that $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, then \( A \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \)
Given: $\begin{bmatrix}2 & 1 \\ 3 & 2\end{bmatrix} A \begin{bmatrix}-3 & 2 \\ 5 & -3\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}.$
Let $P = \begin{bmatrix}2 & 1 \\ 3 & 2\end{bmatrix},\ Q = \begin{bmatrix}-3 & 2 \\ 5 & -3\end{bmatrix}.$
Equation becomes $P A Q = I_2 \Rightarrow A = P^{-1} I_2 Q^{-1} = P^{-1} Q^{-1}.$
Compute $P^{-1}$: $\det P = (2)(2)-(3)(1)=4-3=1$
$P^{-1} = \frac{1}{1} \begin{bmatrix}2 & -1 \\ -3 & 2\end{bmatrix} = \begin{bmatrix}2 & -1 \\ -3 & 2\end{bmatrix}.$
Compute $Q^{-1}$: $\det Q = (-3)(-3)-(5)(2)=9-10=-1$
$Q^{-1} = \frac{1}{-1} \begin{bmatrix}-3 & -2 \\ -5 & -3\end{bmatrix} = \begin{bmatrix}3 & 2 \\ 5 & 3\end{bmatrix}.$
Therefore $A = P^{-1} Q^{-1} = \begin{bmatrix}2 & -1 \\ -3 & 2\end{bmatrix} \begin{bmatrix}3 & 2 \\ 5 & 3\end{bmatrix}.$
Multiply the matrices:
First row: $(2*3 + (-1)*5, 2*2 + (-1)*3) = (6-5, 4-3) = (1,1)$
Second row: $(-3*3 + 2*5, -3*2 + 2*3) = (-9+10, -6+6) = (1,0)$
Thus $A = \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix}.$