Certainly! To prove that the matrices [tex]\(\left[\begin{array}{cc}-3 & -2 \\ 5 & 3\end{array}\right]\)[/tex] and [tex]\(\left[\begin{array}{cc}3 & 2 \\ -5 & -3\end{array}\right]\)[/tex] are inverses of each other, we need to show that their product equals the identity matrix.
The identity matrix for [tex]\(2 \times 2\)[/tex] matrices is:
[tex]\[
I = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]
\][/tex]
Let's denote the matrices as:
[tex]\[
A = \left[\begin{array}{cc}-3 & -2 \\ 5 & 3\end{array}\right]
\][/tex]
[tex]\[
B = \left[\begin{array}{cc}3 & 2 \\ -5 & -3\end{array}\right]
\][/tex]
To prove [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are inverses, we must show that [tex]\(AB = I\)[/tex].
Calculating the product [tex]\(AB\)[/tex]:
[tex]\[
AB = \left[\begin{array}{cc}-3 & -2 \\ 5 & 3\end{array}\right] \left[\begin{array}{cc}3 & 2 \\ -5 & -3\end{array}\right]
\][/tex]
To find the element in the first row and first column of the product [tex]\(AB\)[/tex]:
[tex]\[
(-3 \times 3) + (-2 \times -5) = -9 + 10 = 1
\][/tex]
To find the element in the first row and second column of the product [tex]\(AB\)[/tex]:
[tex]\[
(-3 \times 2) + (-2 \times -3) = -6 + 6 = 0
\][/tex]
To find the element in the second row and first column of the product [tex]\(AB\)[/tex]:
[tex]\[
(5 \times 3) + (3 \times -5) = 15 - 15 = 0
\][/tex]
To find the element in the second row and second column of the product [tex]\(AB\)[/tex]:
[tex]\[
(5 \times 2) + (3 \times -3) = 10 - 9 = 1
\][/tex]
Putting all these elements together, we get:
[tex]\[
AB = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]
\][/tex]
Since [tex]\(AB = I\)[/tex], we have shown that matrix [tex]\(A\)[/tex] is the inverse of matrix [tex]\(B\)[/tex], and equivalently, matrix [tex]\(B\)[/tex] is the inverse of matrix [tex]\(A\)[/tex]. Thus, we have proven that the matrices [tex]\(\left[\begin{array}{cc}-3 & -2 \\ 5 & 3\end{array}\right]\)[/tex] and [tex]\(\left[\begin{array}{cc}3 & 2 \\ -5 & -3\end{array}\right]\)[/tex] are indeed inverses of each other.
