To find the inverse of a given 2x2 matrix, we use the following formula:
For a matrix [tex]\( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex], the inverse is given by:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
This formula can be used if and only if the determinant [tex]\( ad - bc \neq 0 \)[/tex].
Let's apply this formula to the given matrix:
[tex]\[ A = \begin{pmatrix} 15 & 6 \\ 12 & 4 \end{pmatrix} \][/tex]
Firstly, we need to find the determinant of matrix [tex]\( A \)[/tex]:
[tex]\[ \text{det}(A) = (15 \times 4) - (6 \times 12) \][/tex]
[tex]\[ \text{det}(A) = 60 - 72 \][/tex]
[tex]\[ \text{det}(A) = -12 \][/tex]
Since the determinant is non-zero ([tex]\(-12 \neq 0\)[/tex]), the inverse exists. Now we use the inverse formula:
[tex]\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
Substituting the values [tex]\( a = 15 \)[/tex], [tex]\( b = 6 \)[/tex], [tex]\( c = 12 \)[/tex], and [tex]\( d = 4 \)[/tex]:
[tex]\[ A^{-1} = \frac{1}{-12} \begin{pmatrix} 4 & -6 \\ -12 & 15 \end{pmatrix} \][/tex]
Next, we scalar multiply each element in the matrix by [tex]\(\frac{1}{-12}\)[/tex]:
[tex]\[ A^{-1} = \begin{pmatrix} 4 \times \frac{1}{-12} & -6 \times \frac{1}{-12} \\ -12 \times \frac{1}{-12} & 15 \times \frac{1}{-12} \end{pmatrix} \][/tex]
[tex]\[ A^{-1} = \begin{pmatrix} \frac{4}{-12} & \frac{-6}{-12} \\ \frac{-12}{-12} & \frac{15}{-12} \end{pmatrix} \][/tex]
[tex]\[ A^{-1} = \begin{pmatrix} -\frac{1}{3} & \frac{1}{2} \\ 1 & -\frac{5}{4} \end{pmatrix} \][/tex]
Thus, the inverse of the matrix is:
[tex]\[ A^{-1} = \begin{pmatrix} -0.333333333333333 & 0.4999999999999995 \\ 0.9999999999999991 & -1.249999999999999 \end{pmatrix} \][/tex]
