To determine whether matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are inverses of each other, we need to compute their product and verify if the result is the identity matrix.
Given:
[tex]\[
A = \begin{bmatrix}
2 & -6 \\
1 & -1
\end{bmatrix}
\][/tex]
[tex]\[
B = \begin{bmatrix}
-3 & 2 \\
2 & -1
\end{bmatrix}
\][/tex]
First, let's calculate the product [tex]\( A \cdot B \)[/tex]:
[tex]\[
A \cdot B = \begin{bmatrix}
2 & -6 \\
1 & -1
\end{bmatrix} \cdot \begin{bmatrix}
-3 & 2 \\
2 & -1
\end{bmatrix}
\][/tex]
We'll multiply the matrices step by step:
The element at the first row, first column of the product:
[tex]\[
(2 \times -3) + (-6 \times 2) = -6 + (-12) = -18
\][/tex]
The element at the first row, second column of the product:
[tex]\[
(2 \times 2) + (-6 \times -1) = 4 + 6 = 10
\][/tex]
The element at the second row, first column of the product:
[tex]\[
(1 \times -3) + (-1 \times 2) = -3 + (-2) = -5
\][/tex]
The element at the second row, second column of the product:
[tex]\[
(1 \times 2) + (-1 \times -1) = 2 + 1 = 3
\][/tex]
Therefore, the product of [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is:
[tex]\[
A \cdot B = \begin{bmatrix}
-18 & 10 \\
-5 & 3
\end{bmatrix}
\][/tex]
The identity matrix for 2x2 matrices is:
[tex]\[
I = \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
\][/tex]
Since:
[tex]\[
A \cdot B \ne I = \begin{bmatrix}
-18 & 10 \\
-5 & 3
\end{bmatrix} \ne \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
\][/tex]
Thus, the product [tex]\( A \cdot B \)[/tex] is not the identity matrix.
Therefore, matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are not inverse matrices.
The correct answer is:
False
