Q. 265.0( 2 Votes )

Using matrices, s

Answer :

Given: equations are x + y + z = 6, x + 2z = 7, 3x + y + z = 12


Given: equation can be written in matrix form as


AX = B X = A-1B … (1)


Here


Now |A| = 1 (0 – 2) – 1 (1 – 6) + 1 (1 – 0)


= -2 + 5 + 1


= 4 ≠ 0


A-1 exists.


Now for adj A:


A11 = -2


A11 = -(-5) = 5


A11 = 1


A11 = -(1 – 1) = 0


A11 = 1 – 3 = -2


A11 = -(1 – 3) = 2


A11 = 2 – 0 = 2


A11 = -(2 – 1) = -1


A11 = 0 – 1 = -1




From (1), X = A-1B





x = 3, y = 1 and z = 2


OR


Given:


A = IA



Applying R1 1/3 R1, we get



Applying R2 R2 – 2R1, we get



Applying R2 1/3 R2, we get



Applying R3 R3 – 4R2, we get



Applying R3 9R3, we get



Applying R2 R2 – 2/9 R3 and R1 1/3 R3, we get




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