Answer :

We are given with a matrix A = and we have to find the inverse of this matrix which will be

A - 1, so the formula to find the inverse of a matrix is,

A - 1 = Adjoint(A)

First we will find the which is,

= 2[2( - 2) - 1( - 4)] – ( - 3)[3( - 2) - 1( - 4)] + 5[3(1) - 2(1)]

= 0 – 6 + 5 = - 1

Now we have to find the adjoint matrix, and for that, we have to first find the cofactor matrix and take its transpose.

To find the cofactor matrix, we have to take determinant similar method, like this,

For finding the cofactor of the a11 element, we will take the determinant of and a sign of positive will be there when the sum of row the number and column number is even and, the sign will be negative when the sum is odd. In this case, the sum is even, so a positive sign will come.

Therefore the cofactor matrix is,

Taking transpose of the matrix, we get,

which is the Adjoint(A),

Hence A - 1 =

To solve linear equations with the help of the matrix method first write the coefficients in the form,

A X = B,

In A matrix write all the coefficients and in the X matrix write the variables involved and in the B matrix write the constants.

2x - 3y + 5z = 11

3x + 2y - 4z = - 5

x + y - 2z = - 3

The A matrix obtained from these equations is the same as given at the starting of the question,

Hence pre-multiplying the general equation by A - 1, we get,

X = A - 1 B

By multiplying these matrices, we will get the values of the variables,

As the column number of the first matrix is equal to the row number of the second matrix, they are feasible to be multiplied.

Hence x = 1, y = 2, z = 3


Given:- 3 x 3 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = InA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor In on the RHS till we obtain the result

In = BA

(iv) Write A-1 = B


We have,

A = I3A

Where I3 is 3 x 3 elementary matrix




Applying and



Hence , it is of the form

I = BA

So, as we know that

I = A-1A


A-1 = B

inverse of A

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