Using elementary row operations (transformations), find the inverse of ⎛⎜⎝012123310⎞⎟⎠ OR If A = ⎡⎢⎣067−6087−80⎤⎥⎦, B = ⎡⎢⎣011102120⎤⎥⎦, C= ⎡⎢⎣2−23⎤⎥⎦, then calculate AC, BC and (A+B) C.
Also verify that (A+B)C = AC+BC.
Using elementary row operations (transformations), find the inverse of ⎛⎜⎝012123310⎞⎟⎠ OR If A = ⎡⎢⎣067−6087−80⎤⎥⎦, B = ⎡⎢⎣011102120⎤⎥⎦, C= ⎡⎢⎣2−23⎤⎥⎦, then calculate AC, BC and (A+B) C.
Also verify that (A+B)C = AC+BC.
Let A =⎡⎢⎣012123310⎤⎥⎦
By using elementary row transformations, A = IA
⇒⎡⎢⎣012123310⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A By R1↔R2
⇒⎡⎢⎣123012310⎤⎥⎦=⎡⎢⎣010100001⎤⎥⎦A By R3→R3−3R1
⇒⎡⎢⎣1230120−5−9⎤⎥⎦=⎡⎢⎣0101000−31⎤⎥⎦A By R3→R3+5R2
⇒⎡⎢⎣123012001⎤⎥⎦=⎡⎢⎣0101005−31⎤⎥⎦A By R1→R1−2R2
⇒⎡⎢⎣10−1012001⎤⎥⎦=⎡⎢⎣−2101005−31⎤⎥⎦A By R1→R1−2R3
⇒⎡⎢⎣10−1010001⎤⎥⎦=⎡⎢⎣−210−96−25−31⎤⎥⎦A By R1→R1+R3
⇒⎡⎢⎣100010001⎤⎥⎦=⎡⎢⎣3−21−96−25−31⎤⎥⎦A∵I=A−1A∴A−1=⎡⎢⎣3−21−96−25−31⎤⎥⎦ OR We have AC = ⎡⎢⎣067−6087−80⎤⎥⎦=⎡⎢⎣2−23⎤⎥⎦=⎡⎢⎣0−12+21−120+2414+16+0⎤⎥⎦=⎡⎢⎣91230⎤⎥⎦,
And, BC= ⎡⎢⎣011102120⎤⎥⎦=⎡⎢⎣2−23⎤⎥⎦=⎡⎢⎣0−2+32+0+62+−4+0⎤⎥⎦=⎡⎢⎣18−2⎤⎥⎦.
Also A+B=⎡⎢⎣067−6087−80⎤⎥⎦+⎡⎢⎣011102120⎤⎥⎦=⎡⎢⎣078−50108−60⎤⎥⎦
⇒(A+B)C=⎡⎢⎣078−50108−60⎤⎥⎦⎡⎢⎣2−23⎤⎥⎦=⎡⎢⎣0−14+24−10+0+3016+120⎤⎥⎦=⎡⎢⎣102028⎤⎥⎦...(i)And,AC+BC=⎡⎢⎣91230⎤⎥⎦+⎡⎢⎣18−2⎤⎥⎦=⎡⎢⎣102028⎤⎥⎦...(ii)
By (i) & (ii), it is clear that (A+B)C= AC+BC.
Let A =⎡⎢⎣012123310⎤⎥⎦
By using elementary row transformations, A = IA
⇒⎡⎢⎣012123310⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A By R1↔R2
⇒⎡⎢⎣123012310⎤⎥⎦=⎡⎢⎣010100001⎤⎥⎦A By R3→R3−3R1
⇒⎡⎢⎣1230120−5−9⎤⎥⎦=⎡⎢⎣0101000−31⎤⎥⎦A By R3→R3+5R2
⇒⎡⎢⎣123012001⎤⎥⎦=⎡⎢⎣0101005−31⎤⎥⎦A By R1→R1−2R2
⇒⎡⎢⎣10−1012001⎤⎥⎦=⎡⎢⎣−2101005−31⎤⎥⎦A By R1→R1−2R3
⇒⎡⎢⎣10−1010001⎤⎥⎦=⎡⎢⎣−210−96−25−31⎤⎥⎦A By R1→R1+R3
⇒⎡⎢⎣100010001⎤⎥⎦=⎡⎢⎣3−21−96−25−31⎤⎥⎦A∵I=A−1A∴A−1=⎡⎢⎣3−21−96−25−31⎤⎥⎦ OR We have AC = ⎡⎢⎣067−6087−80⎤⎥⎦=⎡⎢⎣2−23⎤⎥⎦=⎡⎢⎣0−12+21−120+2414+16+0⎤⎥⎦=⎡⎢⎣91230⎤⎥⎦,
And, BC= ⎡⎢⎣011102120⎤⎥⎦=⎡⎢⎣2−23⎤⎥⎦=⎡⎢⎣0−2+32+0+62+−4+0⎤⎥⎦=⎡⎢⎣18−2⎤⎥⎦.
Also A+B=⎡⎢⎣067−6087−80⎤⎥⎦+⎡⎢⎣011102120⎤⎥⎦=⎡⎢⎣078−50108−60⎤⎥⎦
⇒(A+B)C=⎡⎢⎣078−50108−60⎤⎥⎦⎡⎢⎣2−23⎤⎥⎦=⎡⎢⎣0−14+24−10+0+3016+120⎤⎥⎦=⎡⎢⎣102028⎤⎥⎦...(i)And,AC+BC=⎡⎢⎣91230⎤⎥⎦+⎡⎢⎣18−2⎤⎥⎦=⎡⎢⎣102028⎤⎥⎦...(ii)
By (i) & (ii), it is clear that (A+B)C= AC+BC.
