The inverse of a matrix is given as underneath: $$ A^{-1} = \frac{Adj\left(A\right)}{\begin{vmatrix}A\end{vmatrix} \\} $$
Where:
$$ Adj \left(A\right) = \begin{bmatrix}d & -b\\ -c & a \\\end{bmatrix} $$ For $$ A = \begin{bmatrix}a & b\\ c & d \\\end{bmatrix} $$ $$ det A = \begin{vmatrix}a & b \\c & d\end{vmatrix} \\ = ad - bc $$
For inverse of matrix, following circumstance need to be glad
we will check with the aid of the inverse matrix calculator whether the matrix is satisfying the above conditions or no longer. What approximately if you think of deredining the adjoint and inverse of 3x3 matrix. If $$ A= \begin{vmatrix}a & b & c \\d & e & f \\g & h & i\end{vmatrix} $$
Then $$ Adj A = \begin{vmatrix}M_{11} & M_{12} & M_{13} \\M_{12} & M_{22} & M_{23} \\M_{31} & M_{32} & M_{33}\end{vmatrix}^{t}$$ To determine the inverse of a 3x3 matrix we have to cope with the concept of minors and cofactors.
The minor is defined for each and each detail of the matrix. The minor of precise element is the determinant acquired after casting off the row and column containing that detail.
The cofactor of an element is determined with the aid of multiplying the minor with the exponent sum of the row and column of the unique element. $$ Cofactor\ of\ a_{ij} = (-1)^{i+j} \times \text{minor of}\ a_{ij} $$ For the given matrix \( A \) below, we calculate the minor and the cofactor of the matrix: $$ A = \begin{vmatrix}2 & 4 & 6 \\ 1 & 3 & 5 \\ 7 & 9 & 8\end{vmatrix} $$ ### Calculating Minors and Cofactors: $$ M_{11} = (-1)^{1+1} \times \begin{vmatrix}3 & 5 \\9 & 8\end{vmatrix} = 3(8) - 5(9) = 24 - 45 = -21 $$ $$ M_{12} = (-1)^{1+2} \times \begin{vmatrix}1 & 5 \\7 & 8\end{vmatrix} = -(1(8) - 5(7)) = -(8 - 35) = 27 $$ $$ M_{13} = (-1)^{1+3} \times \begin{vmatrix}1 & 3 \\7 & 9\end{vmatrix} = 1(9) - 3(7) = 9 - 21 = -12 $$ $$ M_{21} = (-1)^{2+1} \times \begin{vmatrix}4 & 6 \\9 & 8\end{vmatrix} = -(4(8) - 6(9)) = -(32 - 54) = 22 $$ $$ M_{22} = (-1)^{2+2} \times \begin{vmatrix}2 & 6 \\7 & 8\end{vmatrix} = 2(8) - 6(7) = 16 - 42 = -26 $$ $$ M_{23} = (-1)^{2+3} \times \begin{vmatrix}2 & 4 \\7 & 9\end{vmatrix} = -(2(9) - 4(7)) = -(18 - 28) = 10 $$ $$ M_{31} = (-1)^{3+1} \times \begin{vmatrix}4 & 6 \\3 & 5\end{vmatrix} = 4(5) - 6(3) = 20 - 18 = 2 $$ $$ M_{32} = (-1)^{3+2} \times \begin{vmatrix}2 & 6 \\1 & 5\end{vmatrix} = -(2(5) - 6(1)) = -(10 - 6) = -4 $$ $$ M_{33} = (-1)^{3+3} \times \begin{vmatrix}2 & 4 \\1 & 3\end{vmatrix} = 2(3) - 4(1) = 6 - 4 = 2 $$ ### Cofactor Matrix: $$ \text{Cofactor Matrix} = \begin{vmatrix}-21 & 27 & -12 \\22 & -26 & 10 \\2 & -4 & 2\end{vmatrix} $$ ### Transpose and Adjoint: The transpose of the cofactor matrix gives the adjoint of \( A \): $$ A^{t} = \begin{vmatrix}-21 & 22 & 2 \\27 & -26 & -4 \\-12 & 10 & 2\end{vmatrix} $$ The inverse matrix of \( A \) can be calculated using the formula: $$ \text{Inverse of } A = \frac{1}{\text{det}(A)} \cdot \text{Adj}(A) $$ The whole calculation of the inverse of a 3x3 matrix can be done swiftly by the inverse matrix calculator.
Example:
Calculate and solve the inverse of a 3x3 matrix by the Gauss-Jordan Elimination method: $$ \begin{bmatrix}4 & 3 & 2 \\ 1 & 2 & 5 \\ 7 & 1 & 6\end{bmatrix}\\ $$ Now find the determinant: We are going to make the matrix an identity matrix by applying the row operations. $$ \left[\begin{array}{ccc|ccc}4 & 3 & 2 & 1 & 0 & 0 \\ 1 & 2 & 5 & 0 & 1 & 0 \\ 7 & 1 & 6 & 0 & 0 & 1\end{array}\right]\\ $$ Our final inverse matrix by the Gauss-Jordan elimination method is: $$ \text{Inverse matrix} = \begin{bmatrix}0.25 & -0.0625 & -0.0625 \\ -0.375 & 0.1875 & 0.125 \\ 0.5 & -0.125 & -0.125 \end{bmatrix} $$ The inverse of matrix calculator can find the inverse of a 3x3 matrix in a matter of seconds.
A matrix having an inverse of matrix, it ought to be non- singular and rectangular in nature.
We aren't able to find the inverse of all of the matrices, best the invertible matrix inverse may be determined through the matrix inverse calculator.
