Question

Find the inverse of matrix with rows 19,45,43 and 71,81,61 and 32,41,44

Answer 1

$\begin{bmatrix} -1063/18826 & 217/18826 & 369/9413 \\ 586/9413 & 270/9413 & -947/9413 \\ -319/18826 & -661/18826 & 828/9413 \end{bmatrix}$

Answer 2

Let the given matrix be $A = \begin{bmatrix} 19 & 45 & 43 \\ 71 & 81 & 61 \\ 32 & 41 & 44 \end{bmatrix}$.

We find the determinant of $A$: $\det(A) = 19(81 \times 44 - 61 \times 41) - 45(71 \times 44 - 61 \times 32) + 43(71 \times 41 - 81 \times 32)$ $\det(A) = 19(3564 - 2501) - 45(3124 - 1952) + 43(2911 - 2592)$ $\det(A) = 19(1063) - 45(1172) + 43(319)$ $\det(A) = 20197 - 52740 + 13717 = 33914 - 52740 = -18826$. Since $\det(A) \neq 0$, the inverse exists.

We find the cofactor matrix $C$: $C = \begin{bmatrix} 1063 & -1172 & 319 \\ -217 & -540 & 661 \\ -738 & 1894 & -1656 \end{bmatrix}$

We find the adjoint matrix $\text{adj}(A) = C^T$: $\text{adj}(A) = \begin{bmatrix} 1063 & -217 & -738 \\ -1172 & -540 & 1894 \\ 319 & 661 & -1656 \end{bmatrix}$

The inverse matrix is $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$: $A^{-1} = \frac{1}{-18826} \begin{bmatrix} 1063 & -217 & -738 \\ -1172 & -540 & 1894 \\ 319 & 661 & -1656 \end{bmatrix}$ $A^{-1} = \begin{bmatrix} -1063/18826 & 217/18826 & 738/18826 \\ 1172/18826 & 540/18826 & -1894/18826 \\ -319/18826 & -661/18826 & 1656/18826 \end{bmatrix}$

Simplifying the fractions: $A^{-1} = \begin{bmatrix} -1063/18826 & 217/18826 & 369/9413 \\ 586/9413 & 270/9413 & -947/9413 \\ -319/18826 & -661/18826 & 828/9413 \end{bmatrix}$