Q. 244.7( 3 Votes )

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Answer :


|A| = 3(-2)-1(3) +2(-4)


= -6 -3 -8


= -17


As, |A| is not equal to zero.


So, A-1 exists.


Element of a co-factor matrix is given by,


Cij = (-1)i+j(Mij)


Where Mij is minor of Aij.


Mij is equal to the determinant of the matrix obtained after removing ith row and jth column.


So, C11 = (-1)1+1


C11 = -2 + 0


C11 = -2


C12 = (-1)1+2


C12 = -1(-3 + 6)


C12 = -3


C13 = (-1)1+3


C13 = 0 – 4


C13 = -4


C21 = (-1)2+1


C21 = (-1)(-1 + 0)


C21 = 1


C22 = (-1)2+2


C22 = -3 – 4


C22 = -7


C23 = (-1)2+3


C23 = (-1)(0 – (-2))


C23 = (-1)(2)


C23 = 2


C31 = (-1)3+1


C31 = -3 – 4


C31 = -7


C32 = (-1)3+2


C32 = (-1)(-9 – 6)


C32 = (-1)(-15)


C32 = 15


C33 = (-1)3+3


C33 = 6 – 3


C33 = 3


So, co-factor matrix of A =





Now for given system of equations.



Clearly, (At)X = B


X = (At)-1 B


As, (At)-1 = (A-1)t


X = (A-1)t B






Therefore, x = 2, y = 1, and z = –4


OR


Let,


As A = IA


So,


R1R1+R3



R1 (-1) R1



R2 R2 + 5R1


R3 R3 + 3R1



Interchanging R2 and R3



R2 (-1) R2



R1 R1 + R2


R3 R3 + 2R2



R1 R1 - 9R3


R2 R2 - 152R3



Therefore,


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