Solve Matrix Equations
-Let $A = \left[ \begin{array} { r r } 5 & - 2 \ - 1 & 2 \end{array} \right]$ and $B = \left[ \begin{array} { r r } - 1 & - 5 \ 5 & 5 \end{array} \right] ; \quad X + A = B$
A)
$X = \left[ \begin{array} { r r } - 6 & - 3 \ 6 & 3 \end{array} \right]$
B)

$X = \left[ \begin{array} { r r } - 3 & - 6 \ 3 & 6 \end{array} \right]$
C)
$X = \left[ \begin{array} { r r } 6 & 3 \ - 6 & - 3 \end{array} \right]$
D)
$X = \left[ \begin{array} { r r } 3 & - 6 \ - 3 & 6 \end{array} \right]$

Question

Solve Matrix Equations
-Let $A = \left[ \begin{array} { r r } 5 & - 2 \ - 1 & 2 \end{array} \right]$ and $B = \left[ \begin{array} { r r } - 1 & - 5 \ 5 & 5 \end{array} \right] ; \quad X + A = B$
A)
$X = \left[ \begin{array} { r r } - 6 & - 3 \ 6 & 3 \end{array} \right]$
B)

$X = \left[ \begin{array} { r r } - 3 & - 6 \ 3 & 6 \end{array} \right]$
C)
$X = \left[ \begin{array} { r r } 6 & 3 \ - 6 & - 3 \end{array} \right]$
D)
$X = \left[ \begin{array} { r r } 3 & - 6 \ - 3 & 6 \end{array} \right]$

Quizplus · Accepted Answer

To find $X$, subtract $A$ from both sides of the equation $X + A = B$, resulting in $X = B - A$. Calculating $B - A$ gives $\left[ \begin{array} { r r } - 1 & - 5 \ 5 & 5 \end{array} \right] - \left[ \begin{array} { r r } 5 & - 2 \ - 1 & 2 \end{array} \right] = \left[ \begin{array} { r r } - 6 & - 3 \ 6 & 3 \end{array} \right]$.

Solve Matrix Equations -Let A=[5−2−12]A = \left[ \begin{array} { r r } 5 & - 2 \\ - 1 & 2 \end{array} \right]A=[5−1​−22​]

Solve Matrix Equations
-Let $A = \left[ \begin{array} { r r } 5 & - 2 \\ - 1 & 2 \end{array} \right]$