Consider the matrices A = begin{bmatrix} 2 &a | vedclass

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(A) Given $A = \begin{bmatrix} 2 & -2 \ 4 & -2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 9 \ 1 & 3 \end{bmatrix}$.
First, calcula

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(A) Given $A = \begin{bmatrix} 2 & -2 \ 4 & -2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 9 \ 1 & 3 \end{bmatrix}$.
First, calculate the determinant of $A$: $|A| = (2)(-2) - (-2)(4) = -4 + 8 = 4$.
Since $|A| 
eq 0$, $A^{-1}$ exists. $A^{-1} = \frac{1}{4} \begin{bmatrix} -2 & 2 \ -4 & 2 \end{bmatrix} = \begin{bmatrix} -0.5 & 0.5 \ -1 & 0.5 \end{bmatrix}$.
Given $PA = B \implies P = BA^{-1} = \begin{bmatrix} 3 & 9 \ 1 & 3 \end{bmatrix} \begin{bmatrix} -0.5 & 0.5 \ -1 & 0.5 \end{bmatrix} = \begin{bmatrix} -1.5-9 & 1.5+4.5 \ -0.5-3 & 0.5+1.5 \end{bmatrix} = \begin{bmatrix} -10.5 & 6 \ -3.5 & 2 \end{bmatrix}$.
Given $AQ = B \implies Q = A^{-1}B = \begin{bmatrix} -0.5 & 0.5 \ -1 & 0.5 \end{bmatrix} \begin{bmatrix} 3 & 9 \ 1 & 3 \end{bmatrix} = \begin{bmatrix} -1.5+0.5 & -4.5+1.5 \ -3+0.5 & -9+1.5 \end{bmatrix} = \begin{bmatrix} -1 & -3 \ -2.5 & -7.5 \end{bmatrix}$.
Now, $P+Q = \begin{bmatrix} -10.5-1 & 6-3 \ -3.5-2.5 & 2-7.5 \end{bmatrix} = \begin{bmatrix} -11.5 & 3 \ -6 & -5.5 \end{bmatrix}$.
Then $2(P+Q) = \begin{bmatrix} -23 & 6 \ -12 & -11 \end{bmatrix}$.
The sum of the diagonal elements is $-23 + (-11) = -34$. The absolute value is $|-34| = 34$.

Consider the matrices $A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix}$. If matrices $P$ and $Q$ are such that $PA = B$ and $AQ = B$, then the absolute value of the sum of the diagonal elements of $2(P+Q)$ is . . . . . . .

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