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