To solve the matrix equation [tex]\(5(X - C) = D\)[/tex] for the unknown matrix [tex]\(X\)[/tex], we can follow these detailed steps:
1. Rearrange the equation to solve for [tex]\(X - C\)[/tex]:
[tex]\[
X - C = \frac{D}{5}
\][/tex]
This simplification comes from dividing both sides of the equation by 5.
2. Calculate [tex]\(\frac{D}{5}\)[/tex]:
[tex]\[
\frac{D}{5} = \frac{1}{5} \begin{bmatrix} 10 & 20 \\ 10 & 20 \\ 30 & 0 \end{bmatrix} = \begin{bmatrix} \frac{10}{5} & \frac{20}{5} \\ \frac{10}{5} & \frac{20}{5} \\ \frac{30}{5} & \frac{0}{5} \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 2 & 4 \\ 6 & 0 \end{bmatrix}
\][/tex]
3. Add matrix [tex]\(C\)[/tex] to [tex]\(\frac{D}{5}\)[/tex] to isolate [tex]\(X\)[/tex]:
[tex]\[
X = C + \frac{D}{5}
\][/tex]
To perform this addition, we add each corresponding element in the matrices [tex]\(C\)[/tex] and [tex]\(\frac{D}{5}\)[/tex]:
[tex]\[
C = \begin{bmatrix} 8 & 6 \\ 8 & 0 \\ 0 & 1 \end{bmatrix}
\][/tex]
[tex]\[
\frac{D}{5} = \begin{bmatrix} 2 & 4 \\ 2 & 4 \\ 6 & 0 \end{bmatrix}
\][/tex]
4. Perform matrix addition:
[tex]\[
X = \begin{bmatrix} 8 & 6 \\ 8 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 2 & 4 \\ 2 & 4 \\ 6 & 0 \end{bmatrix} = \begin{bmatrix} 8+2 & 6+4 \\ 8+2 & 0+4 \\ 0+6 & 1+0 \end{bmatrix} = \begin{bmatrix} 10 & 10 \\ 10 & 4 \\ 6 & 1 \end{bmatrix}
\][/tex]
So, the matrix [tex]\(X\)[/tex] is:
[tex]\[
X = \begin{bmatrix} 10 & 10 \\ 10 & 4 \\ 6 & 1 \end{bmatrix}
\][/tex]
