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Question

If A =235324112, find A1. Use it to solve the system of equation 2x-3y+5z = 11, 3x+2y-4z = -5, x+y-2z =-3. 

OR 

Using elementary row transformations, find the inverse of the matrix

A =123257245


Solution

Here A =235324112 

|A|=A=235324112 = 0+3(-2) +5(1)= -1 0 

A1 exists.

Consider Cij be the cofactor of corresponding element aij of matrix A.

C11=0,C12=2,C13=1,C21=1,C22=9,C23=5,C31=2,C32=2,C33=13,adj.A=01229231513A1=adj.A|A|=01229231513

Now 2x-3y+5z = 11, 3x+2y-4z= -5, x+y - 2z= -3

Let A = 235324112,X=xyzand B = 1153

As AX = B     X=A1B=012292315131153xyz=123

By equality of matrices , we get : x = 1, y = 2, z = 3.

OR Given that A =123257245

Using elementary row operation we have : A = IA i.e., 123257245=100010001A

By R2R22R1,R3R3+2R1,123011001=100210201ABy R1R12R2,R2R2+R3,101010001=520411201AByR1R1R3,101010001=321411201A 

Using I = A1 A, we get A1=321411201

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