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# Solve the following system of equations by matrix method:2y – z = 1x – y + z = 22x – y = 0

The given system can be written in matrix form as: AX = B

Now, |A| = 0 = 0 + 4 – 1

= 3

So, the above system has a unique solution, given by

X = A – 1B

Cofactors of A are:

C11 = (– 1)1 + 1 1 – 0 = 1

C21 = (– 1)2 + 11 – 2 = 1

C31 = (– 1)3 + 10 + 1 = 1

C12 = (– 1)1 + 2 – 2 – 0 = 2

C22 = (– 1)2 + 1 – 1 – 0 = – 1

C32 = (– 1)3 + 1 0 – 2 = 2

C13 = (– 1)1 + 2 4 – 0 = 4

C23 = (– 1)2 + 1 2 – 0 = – 2

C33 = (– 1)3 + 1 – 1 + 2 = 1

adj A = = A – 1 = Now, X = A – 1B = X = X = X = Hence, X = 1,Y = 2 and Z = 3

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