1. Determinant of a 3x3 matrix:
\[|A| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})\]2. Adjoint matrix:
\[adj(A) = \begin{bmatrix} C_{11} & C_{21} & C_{31} \ C_{12} & C_{22} & C_{32} \ C_{13} & C_{23} & C_{33} \end{bmatrix}^T\]Where \(C_{ij}\) is the cofactor of element \(a_{ij}\).
3. Inverse matrix:
\[A^{-1} = \frac{1}{|A|} adj(A)\]We can verify our result by multiplying A with its inverse (if it exists):
Since the determinant is zero, the matrix is not invertible, and we cannot perform this verification.
The inverse of a square matrix A, denoted as A^(-1), is a matrix that, when multiplied with A, results in the identity matrix. For a 3x3 matrix A, if A^(-1) exists:
\[A \cdot A^{-1} = A^{-1} \cdot A = I_3\]Where I_3 is the 3x3 identity matrix.
For a 3x3 matrix A, its inverse is given by:
\[A^{-1} = \frac{1}{|A|} \cdot adj(A)\]Where:|A| is the determinant of A adj(A) is the adjoint (adjugate) of A
The determinant of a 3x3 matrix A = [a_ij] is calculated as:
\[|A| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})\]The adjoint of a 3x3 matrix A is the transpose of its cofactor matrix:
\[adj(A) = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix}\]Where C_ij is the cofactor of element a_ij.
Let's find the inverse of matrix A:
\[A = \begin{bmatrix} 4 & 7 & 2 \\ 2 & 6 & 3 \\ 1 & 5 & 8 \end{bmatrix}\]1. Calculate the determinant:
\[|A| = 4(6 \cdot 8 - 3 \cdot 5) - 7(2 \cdot 8 - 3 \cdot 1) + 2(2 \cdot 5 - 6 \cdot 1) = 4(48 - 15) - 7(16 - 3) + 2(10 - 6) = 132 - 91 + 8 = 49\]2. Calculate the cofactor matrix:
\[C = \begin{bmatrix} 33 & -3 & -9 \\ -31 & 30 & -5 \\ 13 & -18 & 10 \end{bmatrix}\]3. Transpose to get the adjoint:
\[adj(A) = \begin{bmatrix} 33 & -31 & 13 \\ -3 & 30 & -18 \\ -9 & -5 & 10 \end{bmatrix}\]4. Divide by the determinant to get the inverse:
\[A^{-1} = \frac{1}{49} \begin{bmatrix} 33 & -31 & 13 \\ -3 & 30 & -18 \\ -9 & -5 & 10 \end{bmatrix} = \begin{bmatrix} \frac{33}{49} & -\frac{31}{49} & \frac{13}{49} \\ -\frac{3}{49} & \frac{30}{49} & -\frac{18}{49} \\ -\frac{9}{49} & -\frac{5}{49} & \frac{10}{49} \end{bmatrix}\]We can create a free, personalized calculator just for you!
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