Solving Using Matrices and Cramer’s Rule Part 2, 3 Variables with 3 Equations
Solving Using Matrices and Cramer’s Rule Part 2, 3 Variables with 3 Equations

This method can be applied only when the coefficient matrix is a square matrix and non-singular. Consider the matrix equation AX = B ,

Pre-multiplying both sides of (1) by A−1, we get

A−1(AX) = A−1B

(A−1A) X = A−1B

IX = A−1B

X = A−1B

Question 1 :

Solve the following system of linear equations by matrix inversion method:

(i) 2x + 5y = −2, x + 2y = −3

Solution :

X = A-1 B

|A| = 4 – 5 = -1

Hence the value of x and y are -11 and 4 respectively.

Question 2 :

(ii) 2x − y = 8, 3x + 2y = −2

Solution :

X = A-1 B

|A| = 4 + 3 = 7

x = 14/7 = 2

y = -28/7 = -4

Hence the values of x and y are 2 and -4 respectively.

Question 3 :

(iii) 2x + 3y − z = 9, x + y + z = 9, 3x − y − z = −1

Solution :

|A| = 2(-1 + 1) – 3(-1 – 3) – 1(-1 -3)

= 2(0) – 3(-4) – 1(-4)

= 12 + 4

|A| = 16

x = 32/16 = 2

y = 48/16 = 3

z = 64/16 = 4

Hence the values of x, y and z are 2, 3 and 4 respectively.

Question 4 :

(iv) x + y + z − 2 = 0, 6x − 4y + 5z − 31 = 0, 5x + 2y + 2z =13

Solution :

|A| = 1(-8 – 10) – 1(12 – 25) + 1(12 + 20)

= 1(-18) – 1(-13) + 1(32)

= -18 + 13 + 32

= 27

x = 81/27 = 3

y = -54/27 = -2

z = 27/27 = 1

Hence the values of x, y and z are 3, -2 and 1 respectively.

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