To determine which of the matrices [tex]\( B \)[/tex] satisfies the equation [tex]\( AB = I \)[/tex] for a given matrix [tex]\( A \)[/tex], we need to check each candidate matrix [tex]\( B \)[/tex] and verify that the product [tex]\( AB \)[/tex] yields the identity matrix.
Given:
[tex]\[ A = \begin{bmatrix} -1 & 4 \\ -3 & 8 \end{bmatrix} \][/tex]
The candidate matrices are:
[tex]\[ B_1 = \begin{bmatrix} \frac{1}{4} & 1 \\ -\frac{3}{4} & -2 \end{bmatrix} \][/tex]
[tex]\[ B_2 = \begin{bmatrix} -\frac{1}{4} & -1 \\ \frac{3}{4} & 2 \end{bmatrix} \][/tex]
[tex]\[ B_3 = \begin{bmatrix} -2 & 1 \\ -\frac{3}{4} & \frac{1}{4} \end{bmatrix} \][/tex]
We need to calculate the matrix products [tex]\( AB_1 \)[/tex], [tex]\( AB_2 \)[/tex], and [tex]\( AB_3 \)[/tex], and see which one equals the [tex]\( 2 \times 2 \)[/tex] identity matrix [tex]\( I \)[/tex]:
[tex]\[ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \][/tex]
### Checking [tex]\( AB_1 \)[/tex]:
[tex]\[ AB_1 = \begin{bmatrix} -1 & 4 \\ -3 & 8 \end{bmatrix} \begin{bmatrix} \frac{1}{4} & 1 \\ -\frac{3}{4} & -2 \end{bmatrix} \][/tex]
Calculating the elements:
[tex]\[
\begin{align*}
(1,1) & : -1 \cdot \frac{1}{4} + 4 \cdot -\frac{3}{4} = -\frac{1}{4} - 3 = -\frac{1}{4} - \frac{12}{4} = -\frac{13}{4} \\
(1,2) & : -1 \cdot 1 + 4 \cdot -2 = -1 - 8 = -9 \\
(2,1) & : -3 \cdot \frac{1}{4} + 8 \cdot -\frac{3}{4} = -\frac{3}{4} - 6 = -\frac{3}{4} - \frac{24}{4} = -\frac{27}{4} \\
(2,2) & : -3 \cdot 1 + 8 \cdot -2 = -3 - 16 = -19 \\
\end{align*}
\][/tex]
Since [tex]\( AB_1 \neq I \)[/tex], [tex]\( B_1 \)[/tex] is not the correct matrix.
### Checking [tex]\( AB_2 \)[/tex]:
[tex]\[ AB_2 = \begin{bmatrix} -1 & 4 \\ -3 & 8 \end{bmatrix} \begin{bmatrix} -\frac{1}{4} & -1 \\ \frac{3}{4} & 2 \end{bmatrix} \][/tex]
Calculating the elements:
[tex]\[
\begin{align*}
(1,1) & : -1 \cdot -\frac{1}{4} + 4 \cdot \frac{3}{4} = \frac{1}{4} + 3 = \frac{1}{4} + \frac{12}{4} = \frac{13}{4} \\
(1,2) & : -1 \cdot -1 + 4 \cdot 2 = 1 + 8 = 9 \\
(2,1) & : -3 \cdot -\frac{1}{4} + 8 \cdot \frac{3}{4} = \frac{3}{4} + 6 = \frac{3}{4} + \frac{24}{4} = \frac{27}{4} \\
(2,2) & : -3 \cdot -1 + 8 \cdot 2 = 3 + 16 = 19 \\
\end{align*}
\][/tex]
Since [tex]\( AB_2 \neq I \)[/tex], [tex]\( B_2 \)[/tex] is not the correct matrix.
### Checking [tex]\( AB_3 \)[/tex]:
[tex]\[ AB_3 = \begin{bmatrix} -1 & 4 \\ -3 & 8 \end{bmatrix} \begin{bmatrix} -2 & 1 \\ -\frac{3}{4} & \frac{1}{4} \end{bmatrix} \][/tex]
Calculating the elements:
[tex]\[
\begin{align*}
(1,1) & : -1 \cdot -2 + 4 \cdot -\frac{3}{4} = 2 - 3 = -1 \\
(1,2) & : -1 \cdot 1 + 4 \cdot \frac{1}{4} = -1 + 1 = 0 \\
(2,1) & : -3 \cdot -2 + 8 \cdot -\frac{3}{4} = 6 - 6 = 0 \\
(2,2) & : -3 \cdot 1 + 8 \cdot \frac{1}{4} = -3 + 2 = -1 \\
\end{align*}
\][/tex]
Thus:
[tex]\[
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I
\][/tex]
So [tex]\( AB_3 = I \)[/tex], which means that the correct matrix [tex]\( B \)[/tex] is:
[tex]\[ \begin{bmatrix} -2 & 1 \\ -\frac{3}{4} & \frac{1}{4} \end{bmatrix} \][/tex]
Therefore, the correct value of [tex]\( B \)[/tex] is:
[tex]\[ \boxed{\begin{bmatrix} -2 & 1 \\ -\frac{3}{4} & \frac{1}{4} \end{bmatrix}} \][/tex]
