Answer to Question #204705 in Linear Algebra for umme habiba sumi

Question

If A = ( 1,2,3) (5,6,7) (0,1,4) and B = (1,0,3) ( 5,6,1) (2,1,4) then find i) AB ii) Inverse of A & B

Answer 1

i)

$$AB=\begin{pmatrix} 1 & 2 & 3 \\ 5 & 6 & 7 \\ 0 & 1 & 4 \\ \end{pmatrix}\begin{pmatrix} 1 & 0 & 3 \\ 5 & 6 & 1 \\ 2 & 1 & 4 \\ \end{pmatrix}$$

$$=\begin{pmatrix} 1(1)+2(5)+3(2) & 1(0)+2(6)+3(1) & 1(3)+2(1)+3(4) \\ 5(1)+6(5)+7(2) & 5(0)+6(6)+7(1) & 5(3)+6(1)+7(4) \\ 0(1)+1(5)+4(2) & 0(0)+1(6)+4(1) & 0(3)+1(1)+4(4) \\ \end{pmatrix}$$

$$=\begin{pmatrix} 17 & 15 & 17 \\ 49 & 43 & 49 \\ 13 & 10 & 17 \\ \end{pmatrix}$$

$$AB=\begin{pmatrix} 17 & 15 & 17 \\ 49 & 43 & 49 \\ 13 & 10 & 17 \\ \end{pmatrix}$$

i)

Augment the matrix $AB$ with the identity matrix:

$$\begin{pmatrix} 17 & 15 & 17 & & 1 & 0 & 0 \\ 49 & 43 & 49 & & 0 & 1 & 0 \\ 13 & 10 & 17 & & 0 & 0 & 1 \\ \end{pmatrix}$$

$R_1=R_1/17$

$$\begin{pmatrix} 1 & 15/17 & 1 & & 1/17 & 0 & 0 \\ 49 & 43 & 49 & & 0 & 1 & 0 \\ 13 & 10 & 17 & & 0 & 0 & 1 \\ \end{pmatrix}$$

$R_2=R_2-49R_1$

$$\begin{pmatrix} 1 & 15/17 & 1 & & 1/17 & 0 & 0 \\ 0 & -4/17 & 0 & & -49/17 & 1 & 0 \\ 13 & 10 & 17 & & 0 & 0 & 1 \\ \end{pmatrix}$$

$R_3=R_3-13R_1$

$$\begin{pmatrix} 1 & 15/17 & 1 & & 1/17 & 0 & 0 \\ 0 & -4/17 & 0 & & -49/17 & 1 & 0 \\ 0 & -25/17 & 4 & & -13/17 & 0 & 1 \\ \end{pmatrix}$$

$R_2=-(17/4)R_2$

$$\begin{pmatrix} 1 & 15/17 & 1 & & 1/17 & 0 & 0 \\ 0 & 1 & 0 & & 49/4 & -17/4 & 0 \\ 0 & -25/17 & 4 & & -13/17 & 0 & 1 \\ \end{pmatrix}$$

$R_1=R_1-(15/17)R_2$

$$\begin{pmatrix} 1 & 0 & 1 & & -43/4 & 15/4 & 0 \\ 0 & 1 & 0 & & 49/4 & -17/4 & 0 \\ 0 & -25/17 & 4 & & -13/17 & 0 & 1 \\ \end{pmatrix}$$

$R_3=R_3+(25/17)R_2$

$$\begin{pmatrix} 1 & 0 & 1 & & -43/4 & 15/4 & 0 \\ 0 & 1 & 0 & & 49/4 & -17/4 & 0 \\ 0 & 0 & 4 & & 69/4 & -25/4 & 1 \\ \end{pmatrix}$$

$R_3=R_3/4$

$$\begin{pmatrix} 1 & 0 & 1 & & -43/4 & 15/4 & 0 \\ 0 & 1 & 0 & & 49/4 & -17/4 & 0 \\ 0 & 0 & 1 & & 69/16 & -25/16 & 1/4 \\ \end{pmatrix}$$

$R_1=R_1-R_3$

$$\begin{pmatrix} 1 & 0 & 0 & & -241/16 & 85/16 & -1/4 \\ 0 & 1 & 0 & & 49/4 & -17/4 & 0 \\ 0 & 0 & 1 & & 69/16 & -25/16 & 1/4 \\ \end{pmatrix}$$

On the left is the identity matrix. On the right is the inverse matrix.

$$(AB)^{-1}=\begin{pmatrix} -241/16 & 85/16 & -1/4 \\ 49/4 & -17/4 & 0 \\ 69/16 & -25/16 & 1/4 \\ \end{pmatrix}$$

Augment the matrix $A$ with the identity matrix:

$$\begin{pmatrix} 1 & 2 & 3 & & 1 & 0 & 0 \\ 5 & 6 & 7 & & 0 & 1 & 0 \\ 0 & 1 & 4 & & 0 & 0 & 1 \\ \end{pmatrix}$$

$R_2=R_2-5R_1$

$$\begin{pmatrix} 1 & 2 & 3 & & 1 & 0 & 0 \\ 0 & -4 & -8 & & -5 & 1 & 0 \\ 0 & 1 & 4 & & 0 & 0 & 1 \\ \end{pmatrix}$$

$R_2=-R_2/4$

$$\begin{pmatrix} 1 & 2 & 3 & & 1 & 0 & 0 \\ 0 & 1 & 2 & & 5/4 & -1/4 & 0 \\ 0 & 1 & 4 & & 0 & 0 & 1 \\ \end{pmatrix}$$

$R_1=R_1-2R_2$

$$\begin{pmatrix} 1 & 0 & -1 & & -3/2 & 1/2 & 0 \\ 0 & 1 & 2 & & 5/4 & -1/4 & 0 \\ 0 & 1 & 4 & & 0 & 0 & 1 \\ \end{pmatrix}$$

$R_3=R_3-R_2$

$$\begin{pmatrix} 1 & 0 & -1 & & -3/2 & 1/2 & 0 \\ 0 & 1 & 2 & & 5/4 & -1/4 & 0 \\ 0 & 0 & 2 & & -5/4 & 1/4 & 1 \\ \end{pmatrix}$$

$R_3=R_3/2$

$$\begin{pmatrix} 1 & 0 & -1 & & -3/2 & 1/2 & 0 \\ 0 & 1 & 2 & & 5/4 & -1/4 & 0 \\ 0 & 0 & 1 & & -5/8 & 1/8 & 1/2 \\ \end{pmatrix}$$

$R_1=R_1+R_3$

$$\begin{pmatrix} 1 & 0 & 0 & & -17/8 & 5/8 & 1/2 \\ 0 & 1 & 2 & & 5/4 & -1/4 & 0 \\ 0 & 0 & 1 & & -5/8 & 1/8 & 1/2 \\ \end{pmatrix}$$

$R_2=R_2-2R_3$

$$\begin{pmatrix} 1 & 0 & 0 & & -17/8 & 5/8 & 1/2 \\ 0 & 1 & 0 & & 5/2 & -1/2 & -1 \\ 0 & 0 & 1 & & -5/8 & 1/8 & 1/2 \\ \end{pmatrix}$$

On the left is the identity matrix. On the right is the inverse matrix.

$$(A)^{-1}=\begin{pmatrix} -17/8 & 5/8 & 1/2 \\ 5/2 & -1/2 & -1 \\ -5/8 & 1/8 & 1/2 \\ \end{pmatrix}$$

Augment the matrix $b$ with the identity matrix:

$$\begin{pmatrix} 1 & 0 & 3 & & 1 & 0 & 0 \\ 5 & 6 & 1 & & 0 & 1 & 0 \\ 2 & 1 & 4 & & 0 & 0 & 1 \\ \end{pmatrix}$$

$R_2=R_2-5R_1$

$$\begin{pmatrix} 1 & 0 & 3 & & 1 & 0 & 0 \\ 0 & 6 & -14 & & -5 & 1 & 0 \\ 2 & 1 & 4 & & 0 & 0 & 1 \\ \end{pmatrix}$$

$R_3=R_3-2R_1$

$$\begin{pmatrix} 1 & 0 & 3 & & 1 & 0 & 0 \\ 0 & 6 & -14 & & -5 & 1 & 0 \\ 0 & 1 & -2 & & -2 & 0 & 1 \\ \end{pmatrix}$$

$R_2=R_2/6$

$$\begin{pmatrix} 1 & 0 & 3 & & 1 & 0 & 0 \\ 0 & 1 & -7/3 & & -5/6 & 1/6 & 0 \\ 0 & 1 & -2 & & -2 & 0 & 1 \\ \end{pmatrix}$$

$R_3=R_3-R_2$

$$\begin{pmatrix} 1 & 0 & 3 & & 1 & 0 & 0 \\ 0 & 1 & -7/3 & & -5/6 & 1/6 & 0 \\ 0 & 0 & 1/3 & & -7/6 & -1/6 & 1 \\ \end{pmatrix}$$

$R_3=3R_3$

$$\begin{pmatrix} 1 & 0 & 3 & & 1 & 0 & 0 \\ 0 & 1 & -7/3 & & -5/6 & 1/6 & 0 \\ 0 & 0 & 1 & & -7/2 & -1/2& 3 \\ \end{pmatrix}$$

$R_1=R_1-3R_3$

$$\begin{pmatrix} 1 & 0 & 0 & & 23/2 & 3/2 & -9 \\ 0 & 1 & -7/3 & & -5/6 & 1/6 & 0 \\ 0 & 0 & 1 & & -7/2 & -1/2& 3 \\ \end{pmatrix}$$

$R_2=R_2+(7/3)R_3$

$$\begin{pmatrix} 1 & 0 & 0 & & 23/2 & 3/2 & -9 \\ 0 & 1 & 0 & & -9 & -1 & 7 \\ 0 & 0 & 1 & & -7/2 & -1/2& 3 \\ \end{pmatrix}$$

On the left is the identity matrix. On the right is the inverse matrix.

$$(B)^{-1}=\begin{pmatrix} 23/2 & 3/2 & -9 \\ -9 & -1 & 7 \\ -7/2 & -1/2 & 3 \\ \end{pmatrix}$$