To determine which value of [tex]\( t \)[/tex] makes the two matrices inverses of each other, we need to verify if the product of the two matrices results in the identity matrix. We will consider the given matrices:
[tex]\[
A = \begin{pmatrix}
-4 & 6 \\
3 & -4
\end{pmatrix}
\][/tex]
and
[tex]\[
B = \begin{pmatrix}
2 & 3 \\
1.5 & 4
\end{pmatrix}
\][/tex]
The problem specifies testing specific values for [tex]\( t \)[/tex]. However, let's first check the product of [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
Perform matrix multiplication:
[tex]\[
A \cdot B = \begin{pmatrix}
-4 & 6 \\
3 & -4
\end{pmatrix} \cdot \begin{pmatrix}
2 & 3 \\
1.5 & 4
\end{pmatrix}
\][/tex]
Calculate element-wise:
1. First row, first column:
[tex]\[
(-4 \cdot 2) + (6 \cdot 1.5) = -8 + 9 = 1
\][/tex]
2. First row, second column:
[tex]\[
(-4 \cdot 3) + (6 \cdot 4) = -12 + 24 = 12
\][/tex]
3. Second row, first column:
[tex]\[
(3 \cdot 2) + (-4 \cdot 1.5) = 6 - 6 = 0
\][/tex]
4. Second row, second column:
[tex]\[
(3 \cdot 3) + (-4 \cdot 4) = 9 - 16 = -7
\][/tex]
This forms the resultant matrix:
[tex]\[
A \cdot B = \begin{pmatrix}
1 & 12 \\
0 & -7
\end{pmatrix}
\][/tex]
For [tex]\( A \)[/tex] and [tex]\( B \)[/tex] to be inverses of each other, this product should be the identity matrix:
[tex]\[
I = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\][/tex]
Upon examining the resultant matrix, it is evident that:
[tex]\[
\begin{pmatrix}
1 & 12 \\
0 & -7
\end{pmatrix}
\][/tex]
is not the identity matrix.
Thus, matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are not inverses of each other under any normal circumstances. Hence, none of the provided options ([tex]\(-3, -2, 2, 3\)[/tex]) can make these matrices inverses of each other.
So, the final answer is:
[tex]\[
None \text{ of the values make the matrices inverses of each other.}
\][/tex]
