Question

Find the inverse of matrix A given below using the formula A ^-1 =C^T/│A│

A=$\begin{vmatrix} -6 & -4 & -4 \\ -1 & -4 & 6 \\\ -6 & 5 & -1 \end{vmatrix}$

Answer 1

$A=\begin{vmatrix} -6 & -4 & -4 \\ -1 & -4 & 6 \\ -6 & 5 & -1 \end{vmatrix}$

Main determinant

$|A|=-6*((-4)*(-1) - 5*6) - (-1)*((-4)*(-1) - 5*(-4)) + -6*((-4)*6 - (-4)*(-4)) = 420$

The determinant is non-zero, therefore, the matrix is non-degenerate and it is possible to find the inverse matrix A^-1 for it.

The inverse matrix will look like this:

$A^{-1}=\frac{1}{420}\begin{pmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \\ \end{pmatrix}$

where A_ij are algebraic additions.

The transposed matrix.

$A_{11}=((-4)*(-1) - 6*5) = -26\\ A_{12}= -((-4)*(-1) - (-4)*5) = -24\\ A_{13}= ((-4)*6 - (-4)*(-4)) = -40\\ A_{21}= -((-1)*(-1) - 6*(-6)) = -37\\ A_{22}= ((-6)*(-1) - (-4)*(-6)) = -18\\ A_{32}= -((-6)*6 - (-4)*(-1)) = 40\\ A_{31}= ((-1)*5 - (-4)*(-6)) = -29\\ A_{32}= -((-6)*5 - (-4)*(-6)) = 54\\ A_{33}= ((-6)*(-4) - (-4)*(-1)) = 20\\$

$A^{-1}=\frac{1}{420}\begin{pmatrix} -26 & -24 & -40 \\ -37 & -18 & 40 \\ -29 & 54 & 20 \\ \end{pmatrix}$

$A^{-1}=\begin{pmatrix} \frac{-13}{210} & \frac{-2}{35} & \frac{-2}{21} \\ \frac{-37}{420} & \frac{-3}{70} & \frac{2}{21} \\ \frac{-29}{420} & \frac{9}{70} & \frac{1}{21} \\ \end{pmatrix}$