Question

# If A =⎡⎢⎣2−3532−411−2⎤⎥⎦, find A−1. 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 =⎡⎢⎣123257−2−4−5⎤⎥⎦

Solution

## Here A =⎡⎢⎣2−3532−411−2⎤⎥⎦  ∴|A|=A=⎡⎢⎣2−3532−411−2⎤⎥⎦ = 0+3(-2) +5(1)= -1 ≠0  ∴A−1 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=⎡⎢⎣0−122−9231−513⎤⎥⎦⇒A−1=adj.A|A|=⎡⎢⎣01−2−29−23−15−13⎤⎥⎦ Now 2x-3y+5z = 11, 3x+2y-4z= -5, x+y - 2z= -3 Let A = ⎡⎢⎣2−3532−411−2⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦and B = ⎡⎢⎣11−5−3⎤⎥⎦ As AX = B     ⇒X=A−1B=⎡⎢⎣01−2−29−23−15−13⎤⎥⎦⎡⎢⎣11−5−3⎤⎥⎦⇒⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣123⎤⎥⎦ By equality of matrices , we get : x = 1, y = 2, z = 3. OR Given that A =⎡⎢⎣123257−2−4−5⎤⎥⎦ Using elementary row operation we have : A = IA i.e., ⎡⎢⎣123257−2−4−5⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A By R2→R2−2R1,R3→R3+2R1,⎡⎢⎣123011001⎤⎥⎦=⎡⎢⎣100−210201⎤⎥⎦ABy R1→R1−2R2,R2→R2+R3,⎡⎢⎣101010001⎤⎥⎦=⎡⎢⎣5−20−41−1201⎤⎥⎦AByR1⇒R1−R3,⎡⎢⎣101010001⎤⎥⎦=⎡⎢⎣3−2−1−41−1201⎤⎥⎦A  Using I = A−1 A, we get A−1=⎡⎢⎣3−2−1−41−1201⎤⎥⎦

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