If matrix A =
, find A–1. Use it to solve the system of equations.
2x – 3y + 5z = 11
3x + 2y – 4z = 0
x + y – 2z = – 3.
OR
Using elementary row transformations, find the inverse of the matrix A =
.
A = 
∴ A = 2( – 4 + 4) + 3( – 6 + 4) + 5(3 – 2) = 0 – 6 + 5 = – 1
0
Now, ![]()
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…… (i)
Now the given system of equations can be written in the form of AX = B, where
A =
, X =
, B = ![]()
The solution of the system of equations is given by X = ![]()
X = ![]()
[Using (i)]


Hence, x = 1, y = 2, z = 3.
OR
A = 
|A| = 1( – 25 + 28) – 2( – 10 + 14) + 3( – 8 + 100)
= 3 – 2(4) + 3(2)
= 9 – 8
= 1
0
exists.
A.
= I

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= 
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