Question:
Find the inverse of the matrix \( A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} \).
Find the inverse of the matrix \( A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} \).
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To verify your answer, you can multiply the original matrix A by its calculated inverse \( A^{-1} \). The result should be the identity matrix I. \( A \cdot A^{-1} = I \). This is a great way to catch calculation errors in an exam.
Updated On: Sep 3, 2025
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Solution and Explanation
Step 1: Understanding the Concept:
To find the inverse of a 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:
1. Calculate the determinant of A, \( \det(A) \).
2. Find the matrix of cofactors of A.
3. Find the adjugate of A by transposing the cofactor matrix, \( \text{adj}(A) = C^T \).
4. Calculate the inverse using the formula \( A^{-1} = \frac{1}{\det(A)} \text{adj}(A) \).
Step 3: Detailed Explanation or Calculation:
1. Calculate the Determinant:
\[ \det(A) = \begin{vmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{vmatrix} \] Expanding along the first row:
\[ \det(A) = 2 \begin{vmatrix} 1 & 0 \\ 1 & 3 \end{vmatrix} - 0 \begin{vmatrix} 5 & 0 \\ 0 & 3 \end{vmatrix} + (-1) \begin{vmatrix} 5 & 1 \\ 0 & 1 \end{vmatrix} \] \[ \det(A) = 2( (1)(3) - (0)(1) ) - 0 + (-1)( (5)(1) - (1)(0) ) \] \[ \det(A) = 2(3) - 1(5) = 6 - 5 = 1 \] Since \( \det(A) = 1 \neq 0 \), the inverse exists.
2. Find the Cofactor Matrix:
The cofactors \( C_{ij} = (-1)^{i+j} M_{ij} \), where \( M_{ij} \) is the minor of the element \( a_{ij} \).
\( C_{11} = + \begin{vmatrix} 1 & 0 \\ 1 & 3 \end{vmatrix} = 3 \)
\( C_{12} = - \begin{vmatrix} 5 & 0 \\ 0 & 3 \end{vmatrix} = -15 \)
\( C_{13} = + \begin{vmatrix} 5 & 1 \\ 0 & 1 \end{vmatrix} = 5 \)
\( C_{21} = - \begin{vmatrix} 0 & -1 \\ 1 & 3 \end{vmatrix} = -(0 - (-1)) = -1 \)
\( C_{22} = + \begin{vmatrix} 2 & -1 \\ 0 & 3 \end{vmatrix} = 6 \)
\( C_{23} = - \begin{vmatrix} 2 & 0 \\ 0 & 1 \end{vmatrix} = -2 \)
\( C_{31} = + \begin{vmatrix} 0 & -1 \\ 1 & 0 \end{vmatrix} = (0 - (-1)) = 1 \)
\( C_{32} = - \begin{vmatrix} 2 & -1 \\ 5 & 0 \end{vmatrix} = -(0 - (-5)) = -5 \)
\( C_{33} = + \begin{vmatrix} 2 & 0 \\ 5 & 1 \end{vmatrix} = 2 \)
The cofactor matrix is \( C = \begin{bmatrix} 3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2 \end{bmatrix} \).
3. Find the Adjugate Matrix:
\[ \text{adj}(A) = C^T = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \] 4. Calculate the Inverse:
\[ A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{1} \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \] Step 4: Final Answer:
The inverse of the matrix A is:
\[ A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \]
To find the inverse of a 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:
1. Calculate the determinant of A, \( \det(A) \).
2. Find the matrix of cofactors of A.
3. Find the adjugate of A by transposing the cofactor matrix, \( \text{adj}(A) = C^T \).
4. Calculate the inverse using the formula \( A^{-1} = \frac{1}{\det(A)} \text{adj}(A) \).
Step 3: Detailed Explanation or Calculation:
1. Calculate the Determinant:
\[ \det(A) = \begin{vmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{vmatrix} \] Expanding along the first row:
\[ \det(A) = 2 \begin{vmatrix} 1 & 0 \\ 1 & 3 \end{vmatrix} - 0 \begin{vmatrix} 5 & 0 \\ 0 & 3 \end{vmatrix} + (-1) \begin{vmatrix} 5 & 1 \\ 0 & 1 \end{vmatrix} \] \[ \det(A) = 2( (1)(3) - (0)(1) ) - 0 + (-1)( (5)(1) - (1)(0) ) \] \[ \det(A) = 2(3) - 1(5) = 6 - 5 = 1 \] Since \( \det(A) = 1 \neq 0 \), the inverse exists.
2. Find the Cofactor Matrix:
The cofactors \( C_{ij} = (-1)^{i+j} M_{ij} \), where \( M_{ij} \) is the minor of the element \( a_{ij} \).
\( C_{11} = + \begin{vmatrix} 1 & 0 \\ 1 & 3 \end{vmatrix} = 3 \)
\( C_{12} = - \begin{vmatrix} 5 & 0 \\ 0 & 3 \end{vmatrix} = -15 \)
\( C_{13} = + \begin{vmatrix} 5 & 1 \\ 0 & 1 \end{vmatrix} = 5 \)
\( C_{21} = - \begin{vmatrix} 0 & -1 \\ 1 & 3 \end{vmatrix} = -(0 - (-1)) = -1 \)
\( C_{22} = + \begin{vmatrix} 2 & -1 \\ 0 & 3 \end{vmatrix} = 6 \)
\( C_{23} = - \begin{vmatrix} 2 & 0 \\ 0 & 1 \end{vmatrix} = -2 \)
\( C_{31} = + \begin{vmatrix} 0 & -1 \\ 1 & 0 \end{vmatrix} = (0 - (-1)) = 1 \)
\( C_{32} = - \begin{vmatrix} 2 & -1 \\ 5 & 0 \end{vmatrix} = -(0 - (-5)) = -5 \)
\( C_{33} = + \begin{vmatrix} 2 & 0 \\ 5 & 1 \end{vmatrix} = 2 \)
The cofactor matrix is \( C = \begin{bmatrix} 3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2 \end{bmatrix} \).
3. Find the Adjugate Matrix:
\[ \text{adj}(A) = C^T = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \] 4. Calculate the Inverse:
\[ A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{1} \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \] Step 4: Final Answer:
The inverse of the matrix A is:
\[ A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \]
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