Q. 244.7( 3 Votes )

# If<im

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, Rate this question :

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