Answer :

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.