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Question

If possible, using elementary row transformations, find the inverse of the following matrix.
213531323

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Solution

For getting the inverse of the given matrix A by row elementary operations we may write the given matrix as A=IA
And, we know AA1=I
Now |A|=2×7+1×(12)+3×(1)=14123=10
So, A1 is possible.

213531323=100010001A

[R2R2+R1]
213324323=100110001A

[R3R3R2]
213324001=100110111A

[R1R1+R2]
117324001=210110111A

[R2R23R1]
1170117001=210520111A

[R1R1+R2andR31×R3]

10100117001=310520111A

[R1R1+10R3andR2R2+17R3]

100010001=7910121517111A

[R11R1andR21R2]

100010001=7910121517111A

As, I=A1A=AA1
So, the inverse of A is A1=7910121517111

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