To find the determinant of [tex]\(A^2 + 5A^{-1} - 14I\)[/tex] where [tex]\( A \)[/tex] is given by
[tex]\[ A = \begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}, \][/tex]
and [tex]\(I\)[/tex] is the [tex]\(2 \times 2\)[/tex] identity matrix, follow these steps:
1. Calculate [tex]\(A^2\)[/tex]:
[tex]\[ A^2 = A \cdot A = \begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix} \cdot \begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix} \][/tex]
Perform the matrix multiplication:
[tex]\[ A^2 = \begin{pmatrix} 3(3) + 5(1) & 3(5) + 5(2) \\ 1(3) + 2(1) & 1(5) + 2(2) \end{pmatrix} = \begin{pmatrix} 9 + 5 & 15 + 10 \\ 3 + 2 & 5 + 4 \end{pmatrix} = \begin{pmatrix} 14 & 25 \\ 5 & 9 \end{pmatrix} \][/tex]
2. Calculate [tex]\(A^{-1}\)[/tex]:
For a [tex]\(2 \times 2\)[/tex] 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]
For [tex]\( A = \begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix} \)[/tex],
[tex]\[ \text{det}(A) = 3 \cdot 2 - 5 \cdot 1 = 6 - 5 = 1 \][/tex]
Thus,
[tex]\[ A^{-1} = \frac{1}{1} \begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix} \][/tex]
3. Calculate [tex]\(5A^{-1}\)[/tex]:
[tex]\[ 5A^{-1} = 5 \begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 10 & -25 \\ -5 & 15 \end{pmatrix} \][/tex]
4. Calculate [tex]\(14I\)[/tex]:
[tex]\[ 14I = 14 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 14 & 0 \\ 0 & 14 \end{pmatrix} \][/tex]
5. Combine the matrices:
First, add [tex]\(A^2\)[/tex] and [tex]\(5A^{-1}\)[/tex]:
[tex]\[ A^2 + 5A^{-1} = \begin{pmatrix} 14 & 25 \\ 5 & 9 \end{pmatrix} + \begin{pmatrix} 10 & -25 \\ -5 & 15 \end{pmatrix} = \begin{pmatrix} 14 + 10 & 25 - 25 \\ 5 - 5 & 9 + 15 \end{pmatrix} = \begin{pmatrix} 24 & 0 \\ 0 & 24 \end{pmatrix} \][/tex]
Next, subtract [tex]\(14I\)[/tex]:
[tex]\[ A^2 + 5A^{-1} - 14I = \begin{pmatrix} 24 & 0 \\ 0 & 24 \end{pmatrix} - \begin{pmatrix} 14 & 0 \\ 0 & 14 \end{pmatrix} = \begin{pmatrix} 10 & 0 \\ 0 & 10 \end{pmatrix} \][/tex]
6. Calculate the determinant:
The determinant of a diagonal matrix [tex]\(\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}\)[/tex] is simply the product of its diagonal entries:
[tex]\[ \det\left(\begin{pmatrix} 10 & 0 \\ 0 & 10 \end{pmatrix}\right) = 10 \cdot 10 = 100 \][/tex]
Thus, the determinant of [tex]\(A^2 + 5A^{-1} - 14I\)[/tex] is [tex]\(100\)[/tex].
