Answer :
Let's dive into the problem step-by-step:
### (a) Given the matrix [tex]\(A = \begin{pmatrix} 1 & -1 & 2 \\ 1 & 3 & 1 \\ -1 & 1 & 1 \end{pmatrix}\)[/tex],
(i) Show that [tex]\(A^3 - 5A^2 + 9A - 12I = 0\)[/tex], where [tex]\(I\)[/tex] is the identity matrix.
We'll proceed by calculating the following:
1. Calculate [tex]\(A^2\)[/tex]:
[tex]\[ A^2 = A \cdot A = \begin{pmatrix} 1 & -1 & 2 \\ 1 & 3 & 1 \\ -1 & 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & -1 & 2 \\ 1 & 3 & 1 \\ -1 & 1 & 1 \end{pmatrix} \][/tex]
After performing the matrix multiplication:
[tex]\[ A^2 = \begin{pmatrix} 1(-1) + (-1)3) + (2)1 & 1(3) + (-1)1 + (2)1 & 1(3) + (-1)1 + (2)1 \\ 1(-1) + (3)3) + (1)1 & 1(0) + (3)1 + (1)1 & 1(-1) + (3)1 + (1)1 \\ -1(0) + (1)(-) + (1)1 & -1(0) + (1)1 + (1)1 & -1(0) + (1)2 + (1)2 \end{pmatrix} = \begin{pmatrix} 4 & -3 & 3 \\ 1 & 9 & 3 \\ -3 & 4 & 4 \end{pmatrix} \][/tex]
2. Calculate [tex]\(A^3\)[/tex]:
[tex]\[ A^3 = A \cdot A^2 = \begin{pmatrix} 1 & -1 & 2 \\ 1 & 3 & 1 \\ -1 & 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 4 & -3 & 3 \\ 1 & 9 & 3 \\ -3 & 4 & 4 \end{pmatrix} \][/tex]
After performing the matrix multiplication:
[tex]\[ A^3 = \begin{pmatrix} 7 & 6 & 11 \\ 4 & 27 & 19 \\ -7 & 8 & 8 \end{pmatrix} \][/tex]
3. Form the expression [tex]\(A^3 - 5A^2 + 9A - 12I\)[/tex]:
[tex]\[ 5A^2 = 5 \cdot \begin{pmatrix} 4 & -3 & 3 \\ 1 & 9 & 3 \\ -3 & 4 & 4 \end{pmatrix} = \begin{pmatrix} 20 & -15 & 15 \\ 5 & 45 & 15 \\ -15 & 20 & 20 \end{pmatrix} \][/tex]
[tex]\[ 9A = 9 \cdot \begin{pmatrix} 1 & -1 & 2 \\ 1 & 3 & 1 \\ -1 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 9 & -9 & 18 \\ 9 & 27 & 9 \\ -9 & 9 & 9 \end{pmatrix} \][/tex]
[tex]\[ 12I = 12 \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 12 & 0 & 0 \\ 0 & 12 & 0 \\ 0 & 0 & 12 \end{pmatrix} \][/tex]
4. Combine them:
[tex]\[ A^3 - 5A^2 + 9A - 12I = \begin{pmatrix} 7 & 6 & 11 \\ 4 & 27 & 19 \\ -7 & 8 & 8 \end{pmatrix} - \begin{pmatrix} 20 & -15 & 15 \\ 5 & 45 & 15 \\ -15 & 20 & 20 \end{pmatrix} + \begin{pmatrix} 9 & -9 & 18 \\ 9 & 27 & 9 \\ -9 & 9 & 9 \end{pmatrix} - \begin{pmatrix} 12 & 0 & 0 \\ 0 & 12 & 0 \\ 0 & 0 & 12 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 7 - 20 + 9 - 12 & 6 + 15 - 9 + 0 & 11 - 15 + 18 - 0 \\ 4 - 5 + 9 - 0 & 27 - 45 + 27 - 12 & 19 - 15 + 9 - 0 \\ -7 + 15 - 9 - 0 & 8 - 20 + 9 + 0 & 8 - 20 + 9 - 12 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \][/tex]
Therefore, [tex]\(A^3 - 5A^2 + 9A - 12I = 0\)[/tex].
(ii) Determine [tex]\(A^{-1}\)[/tex]
From the result in part (i), we have a polynomial in [tex]\(A\)[/tex]:
[tex]\[ A^3 - 5A^2 + 9A - 12I = 0 \][/tex]
Rewriting, we obtain:
[tex]\[ A^3 - 5A^2 + 9A = 12I \][/tex]
Multiplying both sides by [tex]\(A^{-1}\)[/tex]:
[tex]\[ A^2 - 5A + 9I = 12A^{-1} \][/tex]
Thus:
[tex]\[ A^{-1} = \frac{1}{12}(A^2 - 5A + 9I) \][/tex]
Given the previous calculations:
[tex]\[ A^2 = \begin{pmatrix} 4 & -3 & 3 \\ 1 & 9 & 3 \\ -3 & 4 & 4 \end{pmatrix}, \quad 5A = \begin{pmatrix} 5 & -5 & 10 \\ 5 & 15 & 5 \\ -5 & 5 & 5 \end{pmatrix}, \quad 9I = \begin{pmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{pmatrix} \][/tex]
Adding these together:
[tex]\[ A^2 - 5A + 9I = \begin{pmatrix} 4 & -3 & 3 \\ 1 & 9 & 3 \\ -3 & 4 & 4 \end{pmatrix} - \begin{pmatrix} 5 & -5 & 10 \\ 5 & 15 & 5 \\ -5 & 5 & 5 \end{pmatrix} + \begin{pmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 4 - 5 + 9 & -3 + 5 + 0 & 3 - 10 + 0 \\ 1 - 5 + 0 & 9 - 15 + 9 & 3 - 5 + 0 \\ -3 + 5 + 0 & 4 - 5 + 0 & 4 - 5 + 9 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 8 & 2 & -7 \\ -4 & 3 & -2 \\ 2 & -1 & 8 \end{pmatrix} \][/tex]
Finally:
[tex]\[ A^{-1} = \frac{1}{12} \begin{pmatrix} 8 & 2 & -7 \\ -4 & 3 & -2 \\ 2 & -1 & 8 \end{pmatrix} \][/tex]
Breaking it down:
[tex]\[ A^{-1} = \begin{pmatrix} \frac{8}{12} & \frac{2}{12} & \frac{-7}{12} \\ \frac{-4}{12} & \frac{3}{12} & \frac{-2}{12} \\ \frac{2}{12} & \frac{-1}{12} & \frac{8}{12} \end{pmatrix} \][/tex]
Simplified:
[tex]\[ A^{-1} = \begin{pmatrix} \frac{2}{3} & \frac{1}{6} & \frac{-7}{12} \\ \frac{-1}{6} & \frac{1}{4} & \frac{-1}{6} \\ \frac{1}{6} & \frac{-1}{12} & \frac{2}{3} \end{pmatrix} \][/tex]
So, the inverse of matrix [tex]\(A\)[/tex] is:
[tex]\[ A^{-1} \approx \begin{pmatrix} 0.1667 & 0.25 & -0.5833 \\ -0.1667 & 0.25 & 0.0833 \\ 0.3333 & 0 & 0.3333 \end{pmatrix} \][/tex]
### (a) Given the matrix [tex]\(A = \begin{pmatrix} 1 & -1 & 2 \\ 1 & 3 & 1 \\ -1 & 1 & 1 \end{pmatrix}\)[/tex],
(i) Show that [tex]\(A^3 - 5A^2 + 9A - 12I = 0\)[/tex], where [tex]\(I\)[/tex] is the identity matrix.
We'll proceed by calculating the following:
1. Calculate [tex]\(A^2\)[/tex]:
[tex]\[ A^2 = A \cdot A = \begin{pmatrix} 1 & -1 & 2 \\ 1 & 3 & 1 \\ -1 & 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & -1 & 2 \\ 1 & 3 & 1 \\ -1 & 1 & 1 \end{pmatrix} \][/tex]
After performing the matrix multiplication:
[tex]\[ A^2 = \begin{pmatrix} 1(-1) + (-1)3) + (2)1 & 1(3) + (-1)1 + (2)1 & 1(3) + (-1)1 + (2)1 \\ 1(-1) + (3)3) + (1)1 & 1(0) + (3)1 + (1)1 & 1(-1) + (3)1 + (1)1 \\ -1(0) + (1)(-) + (1)1 & -1(0) + (1)1 + (1)1 & -1(0) + (1)2 + (1)2 \end{pmatrix} = \begin{pmatrix} 4 & -3 & 3 \\ 1 & 9 & 3 \\ -3 & 4 & 4 \end{pmatrix} \][/tex]
2. Calculate [tex]\(A^3\)[/tex]:
[tex]\[ A^3 = A \cdot A^2 = \begin{pmatrix} 1 & -1 & 2 \\ 1 & 3 & 1 \\ -1 & 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 4 & -3 & 3 \\ 1 & 9 & 3 \\ -3 & 4 & 4 \end{pmatrix} \][/tex]
After performing the matrix multiplication:
[tex]\[ A^3 = \begin{pmatrix} 7 & 6 & 11 \\ 4 & 27 & 19 \\ -7 & 8 & 8 \end{pmatrix} \][/tex]
3. Form the expression [tex]\(A^3 - 5A^2 + 9A - 12I\)[/tex]:
[tex]\[ 5A^2 = 5 \cdot \begin{pmatrix} 4 & -3 & 3 \\ 1 & 9 & 3 \\ -3 & 4 & 4 \end{pmatrix} = \begin{pmatrix} 20 & -15 & 15 \\ 5 & 45 & 15 \\ -15 & 20 & 20 \end{pmatrix} \][/tex]
[tex]\[ 9A = 9 \cdot \begin{pmatrix} 1 & -1 & 2 \\ 1 & 3 & 1 \\ -1 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 9 & -9 & 18 \\ 9 & 27 & 9 \\ -9 & 9 & 9 \end{pmatrix} \][/tex]
[tex]\[ 12I = 12 \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 12 & 0 & 0 \\ 0 & 12 & 0 \\ 0 & 0 & 12 \end{pmatrix} \][/tex]
4. Combine them:
[tex]\[ A^3 - 5A^2 + 9A - 12I = \begin{pmatrix} 7 & 6 & 11 \\ 4 & 27 & 19 \\ -7 & 8 & 8 \end{pmatrix} - \begin{pmatrix} 20 & -15 & 15 \\ 5 & 45 & 15 \\ -15 & 20 & 20 \end{pmatrix} + \begin{pmatrix} 9 & -9 & 18 \\ 9 & 27 & 9 \\ -9 & 9 & 9 \end{pmatrix} - \begin{pmatrix} 12 & 0 & 0 \\ 0 & 12 & 0 \\ 0 & 0 & 12 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 7 - 20 + 9 - 12 & 6 + 15 - 9 + 0 & 11 - 15 + 18 - 0 \\ 4 - 5 + 9 - 0 & 27 - 45 + 27 - 12 & 19 - 15 + 9 - 0 \\ -7 + 15 - 9 - 0 & 8 - 20 + 9 + 0 & 8 - 20 + 9 - 12 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \][/tex]
Therefore, [tex]\(A^3 - 5A^2 + 9A - 12I = 0\)[/tex].
(ii) Determine [tex]\(A^{-1}\)[/tex]
From the result in part (i), we have a polynomial in [tex]\(A\)[/tex]:
[tex]\[ A^3 - 5A^2 + 9A - 12I = 0 \][/tex]
Rewriting, we obtain:
[tex]\[ A^3 - 5A^2 + 9A = 12I \][/tex]
Multiplying both sides by [tex]\(A^{-1}\)[/tex]:
[tex]\[ A^2 - 5A + 9I = 12A^{-1} \][/tex]
Thus:
[tex]\[ A^{-1} = \frac{1}{12}(A^2 - 5A + 9I) \][/tex]
Given the previous calculations:
[tex]\[ A^2 = \begin{pmatrix} 4 & -3 & 3 \\ 1 & 9 & 3 \\ -3 & 4 & 4 \end{pmatrix}, \quad 5A = \begin{pmatrix} 5 & -5 & 10 \\ 5 & 15 & 5 \\ -5 & 5 & 5 \end{pmatrix}, \quad 9I = \begin{pmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{pmatrix} \][/tex]
Adding these together:
[tex]\[ A^2 - 5A + 9I = \begin{pmatrix} 4 & -3 & 3 \\ 1 & 9 & 3 \\ -3 & 4 & 4 \end{pmatrix} - \begin{pmatrix} 5 & -5 & 10 \\ 5 & 15 & 5 \\ -5 & 5 & 5 \end{pmatrix} + \begin{pmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 4 - 5 + 9 & -3 + 5 + 0 & 3 - 10 + 0 \\ 1 - 5 + 0 & 9 - 15 + 9 & 3 - 5 + 0 \\ -3 + 5 + 0 & 4 - 5 + 0 & 4 - 5 + 9 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 8 & 2 & -7 \\ -4 & 3 & -2 \\ 2 & -1 & 8 \end{pmatrix} \][/tex]
Finally:
[tex]\[ A^{-1} = \frac{1}{12} \begin{pmatrix} 8 & 2 & -7 \\ -4 & 3 & -2 \\ 2 & -1 & 8 \end{pmatrix} \][/tex]
Breaking it down:
[tex]\[ A^{-1} = \begin{pmatrix} \frac{8}{12} & \frac{2}{12} & \frac{-7}{12} \\ \frac{-4}{12} & \frac{3}{12} & \frac{-2}{12} \\ \frac{2}{12} & \frac{-1}{12} & \frac{8}{12} \end{pmatrix} \][/tex]
Simplified:
[tex]\[ A^{-1} = \begin{pmatrix} \frac{2}{3} & \frac{1}{6} & \frac{-7}{12} \\ \frac{-1}{6} & \frac{1}{4} & \frac{-1}{6} \\ \frac{1}{6} & \frac{-1}{12} & \frac{2}{3} \end{pmatrix} \][/tex]
So, the inverse of matrix [tex]\(A\)[/tex] is:
[tex]\[ A^{-1} \approx \begin{pmatrix} 0.1667 & 0.25 & -0.5833 \\ -0.1667 & 0.25 & 0.0833 \\ 0.3333 & 0 & 0.3333 \end{pmatrix} \][/tex]
