Question

If A = `[(2,1,-3),(3,2,1),(1,2,-1)]`, find A^–1 and hence solve the following system of equations:

2x + y – 3z=13

3x + 2y + z = 4

x + 2y –  z = 8

Answer 1

Given A = `[(2,1,-3),(3,2,1),(1,2,-1)]`

|A| = 2(−2−2) − 1(−3−1) − 3(6−2)

= 2(−4) − 1(−4) − 3(4)

= −8 + 4 − 12

|A| = −16

Minor of Matrices A	Cofactor
M11 = (−2−2) = −4	A11 = −4
M12 = (−3−1) = −4	A12 = 4
M13 = (6−2) = 4	A13 = 4
M21 = (−1+6) = 5	A21 = −5
M22 = (−2+3) = 1	A22 = 1
M23 = (4−1) = 3	A23 = −3
M31 = (1+6) = 7	A31 = 7
M32 = (2+9) = 11	A32 = −11
M33 = (4−3) = 1	A33 = 1

`adj  A = [(-4,4,4),(-5,1,-3),(7,-11,1)]^"T"`

`adj  A = [(-4,-5,7),(4,1,-11),(4,-3,1)]`

`A^-1 = ((adj A))/|A|`

`A^-1 = 1/-16 [(-4,-5,7),(4,1,-11),(4,-3,1)]`

Given eqn can be written in matrix form as

AX = B

X = A^-1B   ...(i)

Where, A = `[(2,1,-3),(3,2,1),(1,2,-1)], X= [(x),(y),(z)] B= [(13),(4),(8)]`

from (i)

`[(x),(y),(z)]=1/-16 [(-4,-5,7),(4,1,-11),(4,-3,1)][(13),(4),(8)]`

`=1/-16 [(-52-20+56),(52+4-88),(52-12+8)]`

`= 1/-16 [(-16),(-32),(48)]`

`[(x),(y),(z)]=[(1),(2),(-3)]`

∴ x = 1, y = 2, z = −3