To find the inverse of a [tex]\(2 \times 2\)[/tex] matrix, we utilize the following matrix inversion formula. For a matrix [tex]\(A\)[/tex] represented as:
[tex]\[
A = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\][/tex]
its inverse [tex]\(A^{-1}\)[/tex] can be calculated by:
[tex]\[
A^{-1} = \frac{1}{ad - bc} \begin{pmatrix}
d & -b \\
-c & a
\end{pmatrix}
\][/tex]
provided that [tex]\(ad - bc \neq 0\)[/tex] (which ensures that the matrix is invertible).
Given the matrix
[tex]\[
A = \begin{pmatrix}
5 & 6 \\
7 & 8
\end{pmatrix}
\][/tex]
we identify:
[tex]\(a = 5\)[/tex], [tex]\(b = 6\)[/tex], [tex]\(c = 7\)[/tex], and [tex]\(d = 8\)[/tex].
First, we compute the determinant of [tex]\(A\)[/tex], denoted by [tex]\( \text{det}(A) \)[/tex]:
[tex]\[
\text{det}(A) = (5)(8) - (6)(7) = 40 - 42 = -2
\][/tex]
Since the determinant is non-zero, the inverse exists and can be calculated as follows:
[tex]\[
A^{-1} = \frac{1}{-2} \begin{pmatrix}
8 & -6 \\
-7 & 5
\end{pmatrix}
\][/tex]
Next, distribute [tex]\(\frac{1}{-2}\)[/tex] across all elements of the matrix:
[tex]\[
A^{-1} = \begin{pmatrix}
8 \cdot \frac{1}{-2} & -6 \cdot \frac{1}{-2} \\
-7 \cdot \frac{1}{-2} & 5 \cdot \frac{1}{-2}
\end{pmatrix} = \begin{pmatrix}
-4 & 3 \\
3.5 & -2.5
\end{pmatrix}
\][/tex]
Finally, rounding the elements to the nearest hundredth, we get:
[tex]\[
A^{-1} \approx \begin{pmatrix}
-4.00 & 3.00 \\
3.50 & -2.50
\end{pmatrix}
\][/tex]
So, the inverse of the matrix
[tex]\[
\begin{pmatrix}
5 & 6 \\
7 & 8
\end{pmatrix}
\][/tex]
is
[tex]\[
\begin{pmatrix}
-4 & 3 \\
3.5 & -2.5
\end{pmatrix}
\][/tex]
rounded to the nearest hundredth.
