Answer to Question #150948 in Linear Algebra for shanto

Question

If

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

B=[ 2 1 5
-1 -2 -2
3 1 2 ]

then find (AB)^-1

Answer 1

Solution:

1) We should multiply both matrix to each other in order to make C matrix:

$AB=A\cdot B=\begin{bmatrix} 1 & -3 &2\\ 4 & 1&-1\\-3&2&5 \end{bmatrix} \;\cdot\begin{bmatrix} 2 & 1 &5\\ -1 & -2&-2\\3&1&2 \end{bmatrix} =\begin{bmatrix} 11 & 9 &15\\ 4 & 1&16\\7&-2&-9 \end{bmatrix}$

The components of the matrix AB is calculated as:

c₁₁ = a₁₁ · b₁₁ + a₁₂ · b₂₁ + a₁₃ · b₃₁ = 1 · 2 + (-3) · (-1) + 2 · 3 = 2 + 3 + 6 = 11

c₁₂ = a₁₁ · b₁₂ + a₁₂ · b₂₂ + a₁₃ · b₃₂ = 1 · 1 + (-3) · (-2) + 2 · 1 = 1 + 6 + 2 = 9

c₁₃ = a₁₁ · b₁₃ + a₁₂ · b₂₃ + a₁₃ · b₃₃ = 1 · 5 + (-3) · (-2) + 2 · 2 = 5 + 6 + 4 = 15

c₂₁ = a₂₁ · b₁₁ + a₂₂ · b₂₁ + a₂₃ · b₃₁ = 4 · 2 + 1 · (-1) + (-1) · 3 = 8 - 1 - 3 = 4

c₂₂ = a₂₁ · b₁₂ + a₂₂ · b₂₂ + a₂₃ · b₃₂ = 4 · 1 + 1 · (-2) + (-1) · 1 = 4 - 2 - 1 = 1

c₂₃ = a₂₁ · b₁₃ + a₂₂ · b₂₃ + a₂₃ · b₃₃ = 4 · 5 + 1 · (-2) + (-1) · 2 = 20 - 2 - 2 = 16

c₃₁ = a₃₁ · b₁₁ + a₃₂ · b₂₁ + a₃₃ · b₃₁ = (-3) · 2 + 2 · (-1) + 5 · 3 = (-6) - 2 + 15 = 7

c₃₂ = a₃₁ · b₁₂ + a₃₂ · b₂₂ + a₃₃ · b₃₂ = (-3) · 1 + 2 · (-2) + 5 · 1 = (-3) - 4 + 5 = -2

c₃₃ = a₃₁ · b₁₃ + a₃₂ · b₂₃ + a₃₃ · b₃₃ = (-3) · 5 + 2 · (-2) + 5 · 2 = (-15) - 4 + 10 = -9

2) In order to find the inverse of AB, we should use the cofactor method:

I) Find determinant of AB matrix:

Use the triangle's rule to calculate the determinant of the matrix with size 3×3:

det AB = 1360

The determinant of АB is not zero, therefore the inverse matrix (AB)^-1 exists. To calculate the inverse matrix find additional minors and cofactors of matrix АB:

• Find the minor M₁₁ and the cofactor C₁₁. In the matrix, AB crosses out row 1 and column 1.

$$M_{11}=\begin{vmatrix} 1 & 16 \\ -2 & -9 \end{vmatrix}=23$$

$C_{11} = (-1)^{1+1}M_{11} = 23$

• Find the minor M₁₂ and the cofactor C₁₂. In the matrix, AB crosses out row 1 and column 2.

$$M_{12}=\begin{vmatrix} 4 & 16 \\ 7 & -9 \end{vmatrix}=-148$$

• Find the minor M₁₃ and the cofactor C₁₃. In matrix AB cross out row 1 and column 3.

$$M_{13}=\begin{vmatrix} 4 & 1 \\ 7 & 2 \end{vmatrix}=-15$$

• Find the minor M₂₁ and the cofactor C₂₁. In the matrix, AB crosses out row 2 and column 1.

$$M_{21}=\begin{vmatrix} 9 & 15 \\ -2 & -9 \end{vmatrix}=-51$$

• Find the minor M₂₂ and the cofactor C₂₂. In a matrix, AB crosses out row 2 and column 2.

$$M_{22}=\begin{vmatrix} 11 & 15 \\ 7 & -9 \end{vmatrix}=-204$$

• Find the minor M₂₃ and the cofactor C₂₃. In matrix AB cross out row 2 and column 3.

$$M_{23}=\begin{vmatrix} 11 & 9 \\ 7 & -2 \end{vmatrix}=-85$$

• Find the minor M₃₁ and the cofactor C₃₁. In matrix AB cross out row 3 and column 1.

$$M_{31}=\begin{vmatrix} 9 & 15 \\ 1 & 16 \end{vmatrix}=129$$

• Find the minor M₃₂ and the cofactor C₃₂. In matrix AB cross out row 3 and column 2.

$$M_{32}=\begin{vmatrix} 11 & 15 \\ 4 & 16 \end{vmatrix}=116$$

• Find the minor M₃₃ and the cofactor C₃₃. In matrix AB cross out row 3 and column 3.

$$M_{33}=\begin{vmatrix} 11 & 9 \\ 4 & 1 \end{vmatrix}=25$$

3) Write the matrix of cofactors:

$AB=\begin{bmatrix} 23 & -148 &-15\\ 51 & 204&85\\129&-116&-25 \end{bmatrix}$

4) Transposed matrix of cofactors:

$AB^T=\begin{bmatrix} 23 & 51 &129\\ 148 & -204&-116\\-15&85&-25 \end{bmatrix}$

5) Find the inverse matrix:

$AB^{-1}=\frac{AB^T}{det\;(AB)}=\begin{bmatrix} \frac{23}{1360} & \frac{3}{80} &\frac{129}{1360}\\ \frac{37}{340} & -\frac{3}{20}&-\frac{29}{340}\\-\frac{3}{272}&\frac{1}{16}&-\frac{5}{272} \end{bmatrix}$