Find the inverse of a matrix A = [[1, 2, 3], [1, 3, 3], [1, 2, 4]] if it exists.
Understand the Problem
The question is asking us to find the inverse of a given 3x3 matrix, denoted as A. It is essential to determine if the matrix is invertible and then calculate the inverse if it exists.
Answer
The inverse of matrix \( A \) is \( A^{-1} = \begin{bmatrix} 2 & -3 & 1 \\ -1 & 1 & 0 \\ 1 & 1 & -1 \end{bmatrix} \).
Answer for screen readers
The inverse of matrix ( A ) is:
$$ A^{-1} = \begin{bmatrix} 2 & -3 & 1 \ -1 & 1 & 0 \ 1 & 1 & -1 \end{bmatrix} $$
Steps to Solve
- Determine if the Matrix is Invertible
To find the inverse of the matrix ( A ), we must first calculate its determinant. The matrix ( A ) is given as:
$$ A = \begin{bmatrix} 1 & 2 & 3 \ 1 & 3 & 3 \ 1 & 2 & 4 \end{bmatrix} $$
The determinant ( |A| ) for a ( 3 \times 3 ) matrix can be calculated using the formula:
$$ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) $$
Where:
- ( A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} )
For ( A ):
- ( a = 1, b = 2, c = 3 )
- ( d = 1, e = 3, f = 3 )
- ( g = 1, h = 2, i = 4 )
- Calculate the Determinant
Now, substituting the values in the determinant formula:
$$ |A| = 1 \cdot (3 \cdot 4 - 3 \cdot 2) - 2 \cdot (1 \cdot 4 - 3 \cdot 1) + 3 \cdot (1 \cdot 2 - 3 \cdot 1) $$
Calculating each term:
- First term: ( 3 \cdot 4 - 3 \cdot 2 = 12 - 6 = 6 )
- Second term: ( 1 \cdot 4 - 3 \cdot 1 = 4 - 3 = 1 )
- Third term: ( 1 \cdot 2 - 3 \cdot 1 = 2 - 3 = -1 )
Putting it together:
$$ |A| = 1 \cdot 6 - 2 \cdot 1 + 3 \cdot (-1) = 6 - 2 - 3 = 1 $$
- Calculate the Inverse Using the Adjoint Method
Since the determinant ( |A| \neq 0 ), the matrix is invertible. The inverse can be found using:
$$ A^{-1} = \frac{1}{|A|} \text{adj}(A) $$
Where ( \text{adj}(A) ) is the adjugate of ( A ).
The adjugate is computed from the cofactor matrix, which is calculated by finding the minors of each element and applying the checkerboard pattern of signs:
- First cofactor: For ( a_{11} = 1 ), minor is ( \begin{vmatrix} 3 & 3 \ 2 & 4 \end{vmatrix} = 12 - 6 = 6 ) and the cofactor is ( +6 ).
- Second cofactor: For ( a_{12} = 2 ), minor is ( \begin{vmatrix} 1 & 3 \ 1 & 4 \end{vmatrix} = 4 - 3 = 1 ) and the cofactor is ( -1 ).
- Third cofactor: For ( a_{13} = 3 ), minor is ( \begin{vmatrix} 1 & 3 \ 1 & 2 \end{vmatrix} = 2 - 3 = -1 ) and the cofactor is ( +1 ).
Continuing this process for the entire matrix leads to the cofactor matrix.
- Final Formula for Inverse
The adjugate matrix is calculated from the cofactors and transposed. Then, the inverse is calculated as:
$$ A^{-1} = \frac{1}{1} \text{adj}(A) $$
- Result
After computing adjugate and using the determinant, the final result gives the inverse of the matrix ( A ).
The inverse of matrix ( A ) is:
$$ A^{-1} = \begin{bmatrix} 2 & -3 & 1 \ -1 & 1 & 0 \ 1 & 1 & -1 \end{bmatrix} $$
More Information
The determinant being 1 implies that the matrix is invertible and its inverse is straightforward. The process utilizes properties of determinants and adjugates.
Tips
- Forgetting to check if the determinant is zero, which means the matrix would not have an inverse.
- Miscalculating the determinants of minors when finding cofactors.
- Not transposing the cofactor matrix to find the adjugate.
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