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Answer :
To solve for the matrix [tex]\( X \)[/tex] in the equation [tex]\(-\frac{1}{3} A - \frac{1}{3} X = B\)[/tex], let's break down the steps in a straightforward manner.
1. Understand the equation:
We are given:
[tex]\[
-\frac{1}{3} A - \frac{1}{3} X = B
\][/tex]
We need to solve for [tex]\( X \)[/tex].
2. Isolate [tex]\(-\frac{1}{3} X\)[/tex]:
Rearrange the equation to get:
[tex]\[
-\frac{1}{3} X = B + \frac{1}{3} A
\][/tex]
This step involves moving [tex]\(-\frac{1}{3} A\)[/tex] to the other side of the equation by adding it to [tex]\( B \)[/tex].
3. Calculate [tex]\(\frac{1}{3} A\)[/tex]:
Given:
[tex]\[
A = \begin{bmatrix} 9 & -6 \\ -9 & -6 \end{bmatrix}
\][/tex]
Multiply each element in [tex]\( A \)[/tex] by [tex]\(-\frac{1}{3}\)[/tex]:
[tex]\[
-\frac{1}{3} A = \begin{bmatrix} \frac{-1}{3} \times 9 & \frac{-1}{3} \times (-6) \\ \frac{-1}{3} \times (-9) & \frac{-1}{3} \times (-6) \end{bmatrix} = \begin{bmatrix} -3 & 2 \\ 3 & 2 \end{bmatrix}
\][/tex]
4. Substitute into the equation:
Now we have:
[tex]\[
-\frac{1}{3} X = B + \begin{bmatrix} -3 & 2 \\ 3 & 2 \end{bmatrix}
\][/tex]
Given:
[tex]\[
B = \begin{bmatrix} -9 & 8 \\ 6 & 6 \end{bmatrix}
\][/tex]
Add matrices [tex]\( B \)[/tex] and [tex]\(\begin{bmatrix} -3 & 2 \\ 3 & 2 \end{bmatrix}\)[/tex]:
[tex]\[
-\frac{1}{3} X = \begin{bmatrix} -9 + (-3) & 8 + 2 \\ 6 + 3 & 6 + 2 \end{bmatrix} = \begin{bmatrix} -12 & 10 \\ 9 & 8 \end{bmatrix}
\][/tex]
5. Solve for [tex]\( X \)[/tex]:
Multiply both sides by [tex]\(-3\)[/tex] to isolate [tex]\( X \)[/tex]:
[tex]\[
X = -3 \times \begin{bmatrix} -12 & 10 \\ 9 & 8 \end{bmatrix} = \begin{bmatrix} 18 & -18 \\ -9 & -12 \end{bmatrix}
\][/tex]
Thus, the matrix [tex]\( X \)[/tex] is:
[tex]\[
X = \begin{bmatrix} 18 & -18 \\ -9 & -12 \end{bmatrix}
\][/tex]
1. Understand the equation:
We are given:
[tex]\[
-\frac{1}{3} A - \frac{1}{3} X = B
\][/tex]
We need to solve for [tex]\( X \)[/tex].
2. Isolate [tex]\(-\frac{1}{3} X\)[/tex]:
Rearrange the equation to get:
[tex]\[
-\frac{1}{3} X = B + \frac{1}{3} A
\][/tex]
This step involves moving [tex]\(-\frac{1}{3} A\)[/tex] to the other side of the equation by adding it to [tex]\( B \)[/tex].
3. Calculate [tex]\(\frac{1}{3} A\)[/tex]:
Given:
[tex]\[
A = \begin{bmatrix} 9 & -6 \\ -9 & -6 \end{bmatrix}
\][/tex]
Multiply each element in [tex]\( A \)[/tex] by [tex]\(-\frac{1}{3}\)[/tex]:
[tex]\[
-\frac{1}{3} A = \begin{bmatrix} \frac{-1}{3} \times 9 & \frac{-1}{3} \times (-6) \\ \frac{-1}{3} \times (-9) & \frac{-1}{3} \times (-6) \end{bmatrix} = \begin{bmatrix} -3 & 2 \\ 3 & 2 \end{bmatrix}
\][/tex]
4. Substitute into the equation:
Now we have:
[tex]\[
-\frac{1}{3} X = B + \begin{bmatrix} -3 & 2 \\ 3 & 2 \end{bmatrix}
\][/tex]
Given:
[tex]\[
B = \begin{bmatrix} -9 & 8 \\ 6 & 6 \end{bmatrix}
\][/tex]
Add matrices [tex]\( B \)[/tex] and [tex]\(\begin{bmatrix} -3 & 2 \\ 3 & 2 \end{bmatrix}\)[/tex]:
[tex]\[
-\frac{1}{3} X = \begin{bmatrix} -9 + (-3) & 8 + 2 \\ 6 + 3 & 6 + 2 \end{bmatrix} = \begin{bmatrix} -12 & 10 \\ 9 & 8 \end{bmatrix}
\][/tex]
5. Solve for [tex]\( X \)[/tex]:
Multiply both sides by [tex]\(-3\)[/tex] to isolate [tex]\( X \)[/tex]:
[tex]\[
X = -3 \times \begin{bmatrix} -12 & 10 \\ 9 & 8 \end{bmatrix} = \begin{bmatrix} 18 & -18 \\ -9 & -12 \end{bmatrix}
\][/tex]
Thus, the matrix [tex]\( X \)[/tex] is:
[tex]\[
X = \begin{bmatrix} 18 & -18 \\ -9 & -12 \end{bmatrix}
\][/tex]
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