To solve for the matrix [tex]\( C \)[/tex] in the equation [tex]\( 3C + A = B \)[/tex], follow these steps:
1. Start by isolating [tex]\( 3C \)[/tex]:
Given the original equation:
[tex]\[
3C + A = B
\][/tex]
Subtract [tex]\( A \)[/tex] from both sides to isolate [tex]\( 3C \)[/tex]:
[tex]\[
3C = B - A
\][/tex]
2. Compute [tex]\( B - A \)[/tex]:
First, recall the matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[
A = \begin{pmatrix} 1 & -2 \\ 0 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} -3 & 4 \\ 2 & 1 \end{pmatrix}
\][/tex]
Next, perform the matrix subtraction [tex]\( B - A \)[/tex]:
[tex]\[
B - A = \begin{pmatrix} -3 & 4 \\ 2 & 1 \end{pmatrix} - \begin{pmatrix} 1 & -2 \\ 0 & 3 \end{pmatrix}
\][/tex]
Subtract the corresponding elements to get:
[tex]\[
B - A = \begin{pmatrix} -3 - 1 & 4 - (-2) \\ 2 - 0 & 1 - 3 \end{pmatrix} = \begin{pmatrix} -4 & 6 \\ 2 & -2 \end{pmatrix}
\][/tex]
3. Solve for [tex]\( C \)[/tex]:
The equation now simplifies to:
[tex]\[
3C = \begin{pmatrix} -4 & 6 \\ 2 & -2 \end{pmatrix}
\][/tex]
To find [tex]\( C \)[/tex], divide each element of the resulting matrix by 3:
[tex]\[
C = \frac{1}{3} \begin{pmatrix} -4 & 6 \\ 2 & -2 \end{pmatrix} = \begin{pmatrix} -\frac{4}{3} & 2 \\ \frac{2}{3} & -\frac{2}{3} \end{pmatrix}
\][/tex]
4. Final Answer:
Therefore, the matrix [tex]\( C \)[/tex] is:
[tex]\[
C = \begin{pmatrix} -\frac{4}{3} & 2 \\ \frac{2}{3} & -\frac{2}{3} \end{pmatrix}
\][/tex]
Or equivalently:
[tex]\[
C = \begin{pmatrix} -1.33333333 & 2 \\ 0.66666667 & -0.66666667 \end{pmatrix}
\][/tex]
Thus, we have solved for matrix [tex]\( C \)[/tex] in the equation [tex]\( 3C + A = B \)[/tex].
