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Exercise 7.5 · Q23

Q.The matrix AA satisfying the equation [1301]A=[110−1]\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} A = \begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix} is

(1) [14−10]\begin{bmatrix} 1 & 4 \\ -1 & 0 \end{bmatrix}
(2) [1−410]\begin{bmatrix} 1 & -4 \\ 1 & 0 \end{bmatrix}
(3) [140−1]\begin{bmatrix} 1 & 4 \\ 0 & -1 \end{bmatrix}
(4) [1−411]\begin{bmatrix} 1 & -4 \\ 1 & 1 \end{bmatrix}
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Solve PA=QPA = Q by A=P−1QA = P^{-1}Q.

This tests solving a matrix equation.

Step 1. Let P=[1301]P = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}. Since det⁡P=1\det P = 1, P−1=[1−301]P^{-1} = \begin{bmatrix} 1 & -3 \\ 0 & 1 \end{bmatrix}.

Step 2. Then A=P−1Q=[1−301][110−1]A = P^{-1}Q = \begin{bmatrix} 1 & -3 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix}. …

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