Answer to Question #150834 in Linear Algebra for shanto

Question

if A= [ 1 -3 2 and B= [ 2 1 5
4 1 -1 -1 -2 -2
-3 2 5 ] 3 1 2 ]

then find (AB)^-1

Answer 1

Given matrices are

$A = \begin{bmatrix} 1 & -3 & 2 \\ 4 & 1 & -1 \\ 3 & 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 & 5 \\ -1 & -2 & -2 \\ 3 & 1 & 2 \end{bmatrix}$

Now, $AB = \begin{bmatrix} 1 & -3 & 2 \\ 4 & 1 & -1 \\ 3 & 2 & 5 \end{bmatrix} \begin{bmatrix} 2 & 1 & 5 \\ -1 & -2 & -2 \\ 3 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 11 & 9 & 15 \\ 4 & 1 & 16 \\ 19 & 4 & 21 \end{bmatrix}$

Determinant of the matrix AB is 1471.

Co-factors are:

$a_{11} = (21-64) = -43$

$a_{12}=-(84-304)=220$

$a_{13}= (16-19) = -3$

$a_{21}=-(189-60)=-129$

$a_{22}=(231-285)=-54$

$a_{23}=-(44-171)=127$

$a_{31}=(144-15)=129$

$a_{32}=-(176-60)=-116$

$a_{33}=(11-36)=-25$

$adj (AB) = \begin{bmatrix} -43 & -129 & 129 \\ 220 & -54 & -116 \\ -3 & 127 & -25 \end{bmatrix}$

$adj (AB) =$

Then inverse of the matrix AB is,

$(AB)^{-1} = \frac{1}{1471}\begin{bmatrix} -43 & -129 & 129 \\ 220 & -54 & -116 \\ -3 & 127 & -25 \end{bmatrix}$