Inverse Matrix 3 by 3
Inverse Matrix 3×3 Determinant
Inverse of a 3×3 Matrix
The inverse of a matrix \(A\) is another matrix \(A^{-1}\) such that when multiplied, the result is the identity matrix \(I\), i.e.,
\[
A \times A^{-1} = A^{-1} \times A = I
\]
A matrix has an inverse if and only if its determinant is non-zero. The inverse of a 3×3 matrix is calculated using the following formula:
Given the matrix
\[
A = \begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{pmatrix}
\]
the inverse is:
\[
A^{-1} = \frac{1}{\text{det}(A)} \times \begin{pmatrix}
\text{det}(A_{11}) & -\text{det}(A_{12}) & \text{det}(A_{13}) \\
-\text{det}(A_{21}) & \text{det}(A_{22}) & -\text{det}(A_{23}) \\
\text{det}(A_{31}) & -\text{det}(A_{32}) & \text{det}(A_{33})
\end{pmatrix}
\]
Where \(A_{ij}\) represents the \(2 \times 2\) submatrix formed by removing the \(i\)-th row and \(j\)-th column of matrix \(A\).
Step-by-Step Calculation:
1. Find the determinant of \(A\) (as described earlier).
2. Calculate the cofactors: For each element of the matrix, calculate the cofactor by finding the determinant of the corresponding \(2 \times 2\) submatrix and applying the appropriate sign.
3. Form the adjugate matrix: The adjugate matrix is the transpose of the cofactor matrix.
4. Multiply by \(1/\text{det}(A)\): The inverse matrix is the adjugate matrix multiplied by the reciprocal of the determinant of \(A\).
Example 1:
Given the matrix
\[
A = \begin{pmatrix}
2 & 1 & 3 \\
0 & 4 & 5 \\
1 & 1 & 6
\end{pmatrix}
\]
Step 1: Find the determinant of \(A\)
\[
\text{det}(A) = 2(4 \times 6 – 5 \times 1) – 1(0 \times 6 – 5 \times 1) + 3(0 \times 1 – 4 \times 1)
\]
\[
\text{det}(A) = 2(24 – 5) – 1(0 – 5) + 3(0 – 4)
\]
\[
\text{det}(A) = 2(19) – 1(-5) + 3(-4)
\]
\[
\text{det}(A) = 38 + 5 – 12 = 31
\]
So, the determinant of \(A\) is \(31\).
Step 2: Find the cofactors:
– \(A_{11} = \begin{vmatrix} 4 & 5 \\ 1 & 6 \end{vmatrix} = (4 \times 6) – (5 \times 1) = 24 – 5 = 19\)
– \(A_{12} = \begin{vmatrix} 0 & 5 \\ 1 & 6 \end{vmatrix} = (0 \times 6) – (5 \times 1) = -5\)
– \(A_{13} = \begin{vmatrix} 0 & 4 \\ 1 & 1 \end{vmatrix} = (0 \times 1) – (4 \times 1) = -4\)
– \(A_{21} = \begin{vmatrix} 1 & 3 \\ 1 & 6 \end{vmatrix} = (1 \times 6) – (3 \times 1) = 6 – 3 = 3\)
– \(A_{22} = \begin{vmatrix} 2 & 3 \\ 1 & 6 \end{vmatrix} = (2 \times 6) – (3 \times 1) = 12 – 3 = 9\)
– \(A_{23} = \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = (2 \times 1) – (1 \times 1) = 2 – 1 = 1\)
– \(A_{31} = \begin{vmatrix} 1 & 3 \\ 4 & 5 \end{vmatrix} = (1 \times 5) – (3 \times 4) = 5 – 12 = -7\)
– \(A_{32} = \begin{vmatrix} 2 & 3 \\ 0 & 5 \end{vmatrix} = (2 \times 5) – (3 \times 0) = 10\)
– \(A_{33} = \begin{vmatrix} 2 & 1 \\ 0 & 4 \end{vmatrix} = (2 \times 4) – (1 \times 0) = 8\)
Step 3: Form the cofactor matrix:
\[
\text{Cofactor}(A) = \begin{pmatrix}
19 & -5 & -4 \\
-3 & 9 & 1 \\
-7 & 10 & 8
\end{pmatrix}
\]
Step 4: Transpose the cofactor matrix to get the adjugate matrix:
\[
\text{Adj}(A) = \begin{pmatrix}
19 & -3 & -7 \\
-5 & 9 & 10 \\
-4 & 1 & 8
\end{pmatrix}
\]
Step 5: Multiply by \(1/\text{det}(A)\):
\[
A^{-1} = \frac{1}{31} \times \begin{pmatrix}
19 & -3 & -7 \\
-5 & 9 & 10 \\
-4 & 1 & 8
\end{pmatrix}
\]
\[
A^{-1} = \begin{pmatrix}
\frac{19}{31} & \frac{-3}{31} & \frac{-7}{31} \\
\frac{-5}{31} & \frac{9}{31} & \frac{10}{31} \\
\frac{-4}{31} & \frac{1}{31} & \frac{8}{31}
\end{pmatrix}
\]
Thus, the inverse of matrix \(A\) is:
\[
A^{-1} = \begin{pmatrix}
\frac{19}{31} & \frac{-3}{31} & \frac{-7}{31} \\
\frac{-5}{31} & \frac{9}{31} & \frac{10}{31} \\
\frac{-4}{31} & \frac{1}{31} & \frac{8}{31}
\end{pmatrix}
\]
The inverse of a 3×3 matrix is computed by:
- Finding the determinant.
- Calculating the cofactors.
- Forming the cofactor matrix.
- Taking the transpose to get the adjugate.
- Multiplying by the reciprocal of the determinant.
If the determinant is zero, the matrix does not have an inverse, and it is called singular.
