Question:
If \( A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \), then verify that \( A \cdot \text{adj}(A) = |A| I \) and find \( A^{-1} \).
If \( A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \), then verify that \( A \cdot \text{adj}(A) = |A| I \) and find \( A^{-1} \).
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The relationship \( A \cdot \text{adj}(A) = |A| I \) is a cornerstone of matrix algebra and directly leads to the formula for the inverse. Verifying it serves as a good check on your determinant and adjugate calculations before finding the inverse.
Updated On: Sep 3, 2025
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Solution and Explanation
Step 1: Understanding the Concept:
This problem involves two parts. First, we need to verify a fundamental property of matrices which states that the product of a matrix and its adjugate is equal to the determinant of the matrix times the identity matrix. Second, we need to find the inverse of the matrix A using the formula involving the determinant and the adjugate.
Step 2: Key Formula or Approach:
1. Calculate the determinant of A, denoted as \( |A| \).
2. Find the matrix of cofactors, then find the adjugate of A, \( \text{adj}(A) \), which is the transpose of the cofactor matrix.
3. Calculate the product \( A \cdot \text{adj}(A) \).
4. Calculate the product \( |A| I \), where I is the 3x3 identity matrix.
5. Compare the results from steps 3 and 4 to verify the property.
6. Use the formula \( A^{-1} = \frac{1}{|A|} \text{adj}(A) \) to find the inverse.
Step 3: Detailed Explanation or Calculation:
Given \( A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \).
1. Calculate \( |A| \):
\[ |A| = 1(16 - 9) - 3(4 - 3) + 3(3 - 4) = 1(7) - 3(1) + 3(-1) = 7 - 3 - 3 = 1 \]
2. Find \( \text{adj}(A) \):
The cofactors are:
\( C_{11} = 7, C_{12} = -1, C_{13} = -1 \)
\( C_{21} = -3, C_{22} = 1, C_{23} = 0 \)
\( C_{31} = -3, C_{32} = 0, C_{33} = 1 \)
The cofactor matrix is \( C = \begin{bmatrix} 7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix} \).
The adjugate matrix is \( \text{adj}(A) = C^T = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \).
3. Calculate \( A \cdot \text{adj}(A) \):
\[ A \cdot \text{adj}(A) = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \]
\[ = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I \]
4. Calculate \( |A| I \):
\[ |A| I = (1) \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I \]
5. Verification:
From steps 3 and 4, we see that \( A \cdot \text{adj}(A) = |A| I \) is verified.
6. Find \( A^{-1} \):
\[ A^{-1} = \frac{1}{|A|} \text{adj}(A) = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \]
Step 4: Final Answer:
The property is verified. The inverse of A is:
\[ A^{-1} = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \]
This problem involves two parts. First, we need to verify a fundamental property of matrices which states that the product of a matrix and its adjugate is equal to the determinant of the matrix times the identity matrix. Second, we need to find the inverse of the matrix A using the formula involving the determinant and the adjugate.
Step 2: Key Formula or Approach:
1. Calculate the determinant of A, denoted as \( |A| \).
2. Find the matrix of cofactors, then find the adjugate of A, \( \text{adj}(A) \), which is the transpose of the cofactor matrix.
3. Calculate the product \( A \cdot \text{adj}(A) \).
4. Calculate the product \( |A| I \), where I is the 3x3 identity matrix.
5. Compare the results from steps 3 and 4 to verify the property.
6. Use the formula \( A^{-1} = \frac{1}{|A|} \text{adj}(A) \) to find the inverse.
Step 3: Detailed Explanation or Calculation:
Given \( A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \).
1. Calculate \( |A| \):
\[ |A| = 1(16 - 9) - 3(4 - 3) + 3(3 - 4) = 1(7) - 3(1) + 3(-1) = 7 - 3 - 3 = 1 \]
2. Find \( \text{adj}(A) \):
The cofactors are:
\( C_{11} = 7, C_{12} = -1, C_{13} = -1 \)
\( C_{21} = -3, C_{22} = 1, C_{23} = 0 \)
\( C_{31} = -3, C_{32} = 0, C_{33} = 1 \)
The cofactor matrix is \( C = \begin{bmatrix} 7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix} \).
The adjugate matrix is \( \text{adj}(A) = C^T = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \).
3. Calculate \( A \cdot \text{adj}(A) \):
\[ A \cdot \text{adj}(A) = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \]
\[ = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I \]
4. Calculate \( |A| I \):
\[ |A| I = (1) \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I \]
5. Verification:
From steps 3 and 4, we see that \( A \cdot \text{adj}(A) = |A| I \) is verified.
6. Find \( A^{-1} \):
\[ A^{-1} = \frac{1}{|A|} \text{adj}(A) = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \]
Step 4: Final Answer:
The property is verified. The inverse of A is:
\[ A^{-1} = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \]
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