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Sagot :
To find the inverse of a 2x2 matrix, we begin with a given matrix:
[tex]\[ A = \begin{pmatrix} 14 & 1 \\ 16 & 1 \\ \end{pmatrix} \][/tex]
The formula for finding the inverse of a 2x2 matrix
[tex]\[ A = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \][/tex]
is given by:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix} \][/tex]
Here, we identify [tex]\(a = 14\)[/tex], [tex]\(b = 1\)[/tex], [tex]\(c = 16\)[/tex], and [tex]\(d = 1\)[/tex]. We need to calculate the determinant [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = ad - bc = (14 \times 1) - (16 \times 1) = 14 - 16 = -2 \][/tex]
Since the determinant [tex]\(\Delta \neq 0\)[/tex], the matrix has an inverse.
Now, substituting [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(d\)[/tex] along with [tex]\(\Delta\)[/tex] into the inverse formula, we get:
[tex]\[ A^{-1} = \frac{1}{-2} \begin{pmatrix} 1 & -1 \\ -16 & 14 \\ \end{pmatrix} = \begin{pmatrix} \frac{1}{-2} & \frac{-1}{-2} \\ \frac{-16}{-2} & \frac{14}{-2} \\ \end{pmatrix} = \begin{pmatrix} -0.5 & 0.5 \\ 8 & -7 \\ \end{pmatrix} \][/tex]
Thus, the inverse of the given matrix
[tex]\[ A = \begin{pmatrix} 14 & 1 \\ 16 & 1 \\ \end{pmatrix} \][/tex]
is
[tex]\[ A^{-1} = \begin{pmatrix} -0.5 & 0.5 \\ 8 & -7 \\ \end{pmatrix} \][/tex]
This result is already rounded to the nearest hundredth. Therefore, the inverse of the matrix is:
[tex]\[ \begin{pmatrix} -0.5 & 0.5 \\ 8 & -7 \\ \end{pmatrix} \][/tex]
[tex]\[ A = \begin{pmatrix} 14 & 1 \\ 16 & 1 \\ \end{pmatrix} \][/tex]
The formula for finding the inverse of a 2x2 matrix
[tex]\[ A = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \][/tex]
is given by:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix} \][/tex]
Here, we identify [tex]\(a = 14\)[/tex], [tex]\(b = 1\)[/tex], [tex]\(c = 16\)[/tex], and [tex]\(d = 1\)[/tex]. We need to calculate the determinant [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = ad - bc = (14 \times 1) - (16 \times 1) = 14 - 16 = -2 \][/tex]
Since the determinant [tex]\(\Delta \neq 0\)[/tex], the matrix has an inverse.
Now, substituting [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(d\)[/tex] along with [tex]\(\Delta\)[/tex] into the inverse formula, we get:
[tex]\[ A^{-1} = \frac{1}{-2} \begin{pmatrix} 1 & -1 \\ -16 & 14 \\ \end{pmatrix} = \begin{pmatrix} \frac{1}{-2} & \frac{-1}{-2} \\ \frac{-16}{-2} & \frac{14}{-2} \\ \end{pmatrix} = \begin{pmatrix} -0.5 & 0.5 \\ 8 & -7 \\ \end{pmatrix} \][/tex]
Thus, the inverse of the given matrix
[tex]\[ A = \begin{pmatrix} 14 & 1 \\ 16 & 1 \\ \end{pmatrix} \][/tex]
is
[tex]\[ A^{-1} = \begin{pmatrix} -0.5 & 0.5 \\ 8 & -7 \\ \end{pmatrix} \][/tex]
This result is already rounded to the nearest hundredth. Therefore, the inverse of the matrix is:
[tex]\[ \begin{pmatrix} -0.5 & 0.5 \\ 8 & -7 \\ \end{pmatrix} \][/tex]
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