Question:
If \(A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}\), then find \(A^{-1}\).
If \(A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}\), then find \(A^{-1}\).
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When calculating cofactors, be meticulous with the sign pattern: \(\begin{pmatrix} + & - & +
- & + & -
+ & - & + \end{pmatrix}\). A single sign error will make the entire inverse incorrect. Double-check your determinant calculation first, as if it's zero, you can stop immediately.
- & + & -
+ & - & + \end{pmatrix}\). A single sign error will make the entire inverse incorrect. Double-check your determinant calculation first, as if it's zero, you can stop immediately.
Updated On: Sep 6, 2025
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Solution and Explanation
Step 1: Understanding the Concept:
To find the inverse of a square matrix \(A\), we use the formula \(A^{-1} = \frac{1}{\det(A)} \text{adj}(A)\), where \(\det(A)\) is the determinant of \(A\) and \(\text{adj}(A)\) is the adjugate (or adjoint) of \(A\). The inverse exists only if the determinant is non-zero.
Step 2: Key Formula or Approach:
The process involves three main steps: 1. Calculate the determinant of \(A\), \(\det(A)\). 2. Find the matrix of cofactors, then transpose it to get the adjugate matrix, \(\text{adj}(A)\). 3. Substitute these values into the inverse formula.
Step 3: Detailed Explanation:
1. Calculate the Determinant of A: \[ \det(A) = 1(4 \cdot 4 - 3 \cdot 3) - 3(1 \cdot 4 - 1 \cdot 3) + 3(1 \cdot 3 - 1 \cdot 4) \] \[ = 1(16 - 9) - 3(4 - 3) + 3(3 - 4) \] \[ = 7 - 3 - 3 = 1 \] Since \(\det(A) = 1 \neq 0\), the inverse exists.
2. Find the Adjugate of A:
The adjugate is the transpose of the cofactor matrix. Cofactors \(C_{ij}\) are:
\[ C_{11} = +(16-9) = 7, \quad C_{12} = -(4-3) = -1, \quad C_{13} = +(3-4) = -1 \] \[ C_{21} = -(3\cdot4 - 3\cdot3) = -3, \quad C_{22} = +(1\cdot4 - 1\cdot3) = 1, \quad C_{23} = -(1\cdot3 - 1\cdot3) = 0 \] \[ C_{31} = +(3\cdot3 - 4\cdot3) = -3, \quad C_{32} = -(1\cdot3 - 1\cdot3) = 0, \quad C_{33} = +(1\cdot4 - 1\cdot3) = 1 \] Cofactor matrix: \[ C = \begin{bmatrix} 7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix} \] Adjugate matrix: \[ \text{adj}(A) = C^T = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \] 3. Calculate the Inverse: \[ A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = 1 \cdot \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \] Step 4: Final Answer:
The inverse of the matrix \(A\) is \[ A^{-1} = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}. \]
To find the inverse of a square matrix \(A\), we use the formula \(A^{-1} = \frac{1}{\det(A)} \text{adj}(A)\), where \(\det(A)\) is the determinant of \(A\) and \(\text{adj}(A)\) is the adjugate (or adjoint) of \(A\). The inverse exists only if the determinant is non-zero.
Step 2: Key Formula or Approach:
The process involves three main steps: 1. Calculate the determinant of \(A\), \(\det(A)\). 2. Find the matrix of cofactors, then transpose it to get the adjugate matrix, \(\text{adj}(A)\). 3. Substitute these values into the inverse formula.
Step 3: Detailed Explanation:
1. Calculate the Determinant of A: \[ \det(A) = 1(4 \cdot 4 - 3 \cdot 3) - 3(1 \cdot 4 - 1 \cdot 3) + 3(1 \cdot 3 - 1 \cdot 4) \] \[ = 1(16 - 9) - 3(4 - 3) + 3(3 - 4) \] \[ = 7 - 3 - 3 = 1 \] Since \(\det(A) = 1 \neq 0\), the inverse exists.
2. Find the Adjugate of A:
The adjugate is the transpose of the cofactor matrix. Cofactors \(C_{ij}\) are:
\[ C_{11} = +(16-9) = 7, \quad C_{12} = -(4-3) = -1, \quad C_{13} = +(3-4) = -1 \] \[ C_{21} = -(3\cdot4 - 3\cdot3) = -3, \quad C_{22} = +(1\cdot4 - 1\cdot3) = 1, \quad C_{23} = -(1\cdot3 - 1\cdot3) = 0 \] \[ C_{31} = +(3\cdot3 - 4\cdot3) = -3, \quad C_{32} = -(1\cdot3 - 1\cdot3) = 0, \quad C_{33} = +(1\cdot4 - 1\cdot3) = 1 \] Cofactor matrix: \[ C = \begin{bmatrix} 7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix} \] Adjugate matrix: \[ \text{adj}(A) = C^T = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \] 3. Calculate the Inverse: \[ A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = 1 \cdot \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \] Step 4: Final Answer:
The inverse of the matrix \(A\) is \[ A^{-1} = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}. \]
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