Using matrix method, solve the following system of equations:

OR

Using elementary transformations, find the inverse of the matrix

Let

2u +3v + 10w =4

4u – 6v + 5w = 1

6u + 9v – 20w = 2

Writing equation as AX = B

=

A = X = B =

= (2) – (-3) + (10)

= 2 (75) – 3 (-110) + 10 (72)

= 150 + 330 + 720

= 1200 0

its inverse exists and the system of equations is consistent and has a unique solution.

AX = B

X=A^-1B

A^-1 = adj (A)

Adj (A) = =

A =

M11 = 120 – 45 = 75

M12 = (-80 – 30) = -110

M13 = (36 + 36) = 72

M21 = (-60 – 90) = -150

M22 =(-40 – 60) = -100

M23 = (18 – 18) = 0

M31 = (15 + 60) = 75

M32 = (10 – 40) = - 30

M33 = (-12 – 12) = - 24

(Taken from above mentioned calculations)

A11 = (-1)¹⁺¹ M11 = (-1)² . 75 = 75

A12 = 110

A13 = 72

A21 = 150

A22 = -100

A23 = 0

A31 = 75

A32 =30

A33 =-24

Adj. A =

X = A^-1 B

Putting Values,

Hence, putting back the values,

X = 2, y = 3 & z = 5.

OR

Using elementary transformation,

A = IA