To find the inverse of a [tex]\( 2 \times 2 \)[/tex] matrix, you can use a standard formula, given the matrix is:
[tex]\[ \mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \][/tex]
The inverse, [tex]\(\mathbf{A}^{-1}\)[/tex], is given by:
[tex]\[ \mathbf{A}^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
Given the matrix:
[tex]\[ \mathbf{A} = \begin{pmatrix} 20 & 2 \\ -40 & -3 \end{pmatrix} \][/tex]
Let's identify the elements as:
[tex]\[ a = 20, \quad b = 2, \quad c = -40, \quad d = -3 \][/tex]
First, compute the determinant [tex]\( ad - bc \)[/tex]:
[tex]\[
ad - bc = (20 \times -3) - (2 \times -40) = -60 + 80 = 20
\][/tex]
Next, substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], [tex]\( c \)[/tex], and [tex]\( d \)[/tex] into the formula for the inverse:
[tex]\[
\mathbf{A}^{-1} = \frac{1}{20} \begin{pmatrix} -3 & -2 \\ 40 & 20 \end{pmatrix} = \begin{pmatrix} -\frac{3}{20} & -\frac{2}{20} \\ 2 & 1 \end{pmatrix}
\][/tex]
Simplify the elements:
[tex]\[
\mathbf{A}^{-1} = \begin{pmatrix} -0.15 & -0.1 \\ 2 & 1 \end{pmatrix}
\][/tex]
Thus, the inverse of the matrix:
[tex]\[ \begin{pmatrix} 20 & 2 \\ -40 & -3 \end{pmatrix} \][/tex]
rounded to the nearest hundredth, is:
[tex]\[ \begin{pmatrix} -0.15 & -0.1 \\ 2 & 1 \end{pmatrix} \][/tex]
So, the completed solution is:
[tex]\[ \mathbf{A}^{-1} = \begin{pmatrix} -0.15 & -0.1 \\ 2 & 1 \end{pmatrix} \][/tex]
