If \[ A = \begin{bmatrix} 1 & 0 & 2\\ 2 & 1 & 3 \\3 & 2 & 4 \end{bmatrix}, \] then evaluate \( A^2 - 5A + 6I \)=
If \[ A = \begin{bmatrix} 1 & 0 & 2\\ 2 & 1 & 3 \\3 & 2 & 4 \end{bmatrix}, \] then evaluate \( A^2 - 5A + 6I \)=
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\( \begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 12 \end{bmatrix} \)
\( \begin{bmatrix} 8 & 4 & 0 \\ 3 & 6 & 4 \\ 4 & 0 & 14 \end{bmatrix} \)
\( \begin{bmatrix} 8 & 6 & 0 \\ 3 & 8 & 4 \\ 2 & 0 & 14 \end{bmatrix} \)
\( \begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 14 \end{bmatrix} \)
The Correct Option is D
Approach Solution - 1
We are given the matrix: \[ A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix}. \]
Step 1: Compute \( A^2 \) \[ A^2 = A \times A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix}. \]
Performing matrix multiplication: \[ A^2 = \begin{bmatrix} (1 \times 1 + 0 \times 2 + 2 \times 3) & (1 \times 0 + 0 \times 1 + 2 \times 2) & (1 \times 2 + 0 \times 3 + 2 \times 4) \\ (2 \times 1 + 1 \times 2 + 3 \times 3) & (2 \times 0 + 1 \times 1 + 3 \times 2) & (2 \times 2 + 1 \times 3 + 3 \times 4) \\ (3 \times 1 + 2 \times 2 + 4 \times 3) & (3 \times 0 + 2 \times 1 + 4 \times 2) & (3 \times 2 + 2 \times 3 + 4 \times 4) \end{bmatrix} \] \[ = \begin{bmatrix} 1 + 0 + 6 & 0 + 0 + 4 & 2 + 0 + 8 \\ 2 + 2 + 9 & 0 + 1 + 6 & 4 + 3 + 12 \\ 3 + 4 + 12 & 0 + 2 + 8 & 6 + 6 + 16 \end{bmatrix} \] \[ = \begin{bmatrix} 7 & 4 & 10 \\ 13 & 7 & 19 \\ 19 & 10 & 28 \end{bmatrix}. \]
Step 2: Compute \( 5A \) \[ 5A = 5 \times \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 10 \\ 10 & 5 & 15 \\ 15 & 10 & 20 \end{bmatrix}. \]
Step 3: Compute \( 6I \) \[ 6I = 6 \times \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\0 & 0 & 6 \end{bmatrix}. \]
Step 4: Compute \( A^2 - 5A + 6I \) \[ A^2 - 5A + 6I = \begin{bmatrix} 7 & 4 & 10 \\ 13 & 7 & 19 \\19 & 10 & 28 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 10 \\ 10 & 5 & 15 \\ 15 & 10 & 20 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}. \] \[ = \begin{bmatrix} (7 - 5 + 6) & (4 - 0 + 0) & (10 - 10 + 0) \\ (13 - 10 + 0) & (7 - 5 + 6) & (19 - 15 + 0) \\ (19 - 15 + 0) & (10 - 10 + 0) & (28 - 20 + 6) \end{bmatrix} \] \[ = \begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 14 \end{bmatrix}. \]
Thus, the correct answer is: \[ \boxed{\begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 14 \end{bmatrix}} \]
Approach Solution -2
\[ A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} \]
Step 1: Compute \( A^2 = A \times A \):
\[ A^2 = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} \]
Calculate each element of \( A^2 \):
- Row 1:
\[ (1)(1) + (0)(2) + (2)(3) = 1 + 0 + 6 = 7 \] \[ (1)(0) + (0)(1) + (2)(2) = 0 + 0 + 4 = 4 \] \[ (1)(2) + (0)(3) + (2)(4) = 2 + 0 + 8 = 10 \]
- Row 2:
\[ (2)(1) + (1)(2) + (3)(3) = 2 + 2 + 9 = 13 \] \[ (2)(0) + (1)(1) + (3)(2) = 0 + 1 + 6 = 7 \] \[ (2)(2) + (1)(3) + (3)(4) = 4 + 3 + 12 = 19 \]
- Row 3:
\[ (3)(1) + (2)(2) + (4)(3) = 3 + 4 + 12 = 19 \] \[ (3)(0) + (2)(1) + (4)(2) = 0 + 2 + 8 = 10 \] \[ (3)(2) + (2)(3) + (4)(4) = 6 + 6 + 16 = 28 \]
So,
\[ A^2 = \begin{bmatrix} 7 & 4 & 10 \\ 13 & 7 & 19 \\ 19 & 10 & 28 \end{bmatrix} \]
Step 2: Compute \( 5A \):
\[ 5A = 5 \times \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 10 \\ 10 & 5 & 15 \\ 15 & 10 & 20 \end{bmatrix} \]
Step 3: Compute \( 6I \):
\[ 6I = 6 \times \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} \]
Step 4: Calculate \( A^2 - 5A + 6I \):
\[ \begin{bmatrix} 7 & 4 & 10 \\ 13 & 7 & 19 \\ 19 & 10 & 28 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 10 \\ 10 & 5 & 15 \\ 15 & 10 & 20 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = \] \[ \begin{bmatrix} 7 - 5 + 6 & 4 - 0 + 0 & 10 - 10 + 0 \\ 13 - 10 + 0 & 7 - 5 + 6 & 19 - 15 + 0 \\ 19 - 15 + 0 & 10 - 10 + 0 & 28 - 20 + 6 \end{bmatrix} = \] \[ \begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 14 \end{bmatrix} \]
Therefore,
\[ \boxed{ \begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 14 \end{bmatrix} } \]
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