Let [tex][tex]$A=\left(\begin{array}{cc}1 & -2 \\ 0 & 3\end{array}\right)$[/tex][/tex] and [tex][tex]$B=\left(\begin{array}{cc}-3 & 4 \\ 2 & 1\end{array}\right)$[/tex][/tex].

Solve the matrix [tex][tex]$C$[/tex][/tex] in the equation [tex][tex]$3C + A = B$[/tex][/tex].



Answer :

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].