Question

If , find A −1 . Using A −1 solve the system of equations

Answer 1

The given matrix is,

A=[ 2 −3 5 3 2 −4 1 1 −2 ]

The determinant of A is,

| A |=2( −4+4 )+3( −6+4 )+5( 3−2 ) =0−6+5 =−1

Since | A |≠0, thus, A −1 exists and,

A −1 = adjA | A |

The co-factors of each elements of the elements of the matrix are,

A 11 = ( −1 ) 1+1 [ −4+4 ] =0

A 12 = ( −1 ) 1+2 [ −6+4 ] =−( −2 ) =2

A 13 = ( −1 ) 1+3 [ 3−2 ] =1

A 21 = ( −1 ) 2+1 [ 6−5 ] =−1

A 22 = ( −1 ) 2+2 [ −4−5 ] =−9

A 23 = ( −1 ) 2+3 [ 2+3 ] =−5

A 31 = ( −1 ) 3+1 [ 12−10 ] =2

A 32 = ( −1 ) 3+2 [ −8−15 ] =−( −23 ) =23

A 33 = ( −1 ) 3+3 [ 4+9 ] =13

So, the value of adjA is,

adjA=[ A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 ] =[ 0 −1 2 2 −9 23 1 −5 13 ]

Since | A |=−1, thus,

A −1 =−1[ 0 −1 2 2 −9 23 1 −5 13 ] =[ 0 1 −2 −2 9 −23 −1 5 −13 ]

Since X= A −1 B, thus,

[ x y z ]=[ 0 1 −2 −2 9 −23 −1 5 −13 ][ 11 −5 −3 ] =[ 0−5+6 −22−45+69 −11−25+39 ] =[ 1 2 3 ]

Thus,

x=1, y=2 and z=3.