Solve the following system of equations by matrix method:
x + y + z = 3
2x – y + z = – 1
2x + y – 3z = – 9
The given system can be written in matrix form as:
or A X = B
A =
, X =
and B = ![]()
Now, |A| = 1![]()
= (3 – 1) – 1(– 6 – 2) + 1(2 + 2)
= 2 + 8 + 4
= 14
So, the above system has a unique solution, given by
X = A – 1B
Cofactors of A are:
C11 = (– 1)1 + 1 3 – 1 = 2
C21 = (– 1)2 + 1 – 3 – 1 = 4
C31 = (– 1)3 + 1 1 + 1 = 2
C12 = (– 1)1 + 2 – 6 – 2 = 8
C22 = (– 1)2 + 1 – 3 – 2 = – 5
C32 = (– 1)3 + 1 1 – 2 = 1
C13 = (– 1)1 + 2 2 + 2 = 4
C23 = (– 1)2 + 1 1 – 2 = 1
C33 = (– 1)3 + 1 – 1 – 2 = – 3
adj A = 
= 
Now, X = A – 1B = 
X = 
X = ![]()
Hence, X =
,Y =
and Z = ![]()
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