If A=|4 1| |3 2|then prove that A*2=6A + 5 = 0. Created by adityashukla90093808.

If A=|4 1| |3 2|then prove that A*2=6A + 5 = 0

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adityashukla90093808 @adityashukla90093808

August 2023 1 2 Report

If A=|4 1|
|3 2|
then prove that A*2=6A + 5 = 0

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aravindramesh2013

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Answer:A=[23−34]A2=[23−34]×[23−34]=[4−96+12−6−12−9+16]=[−518−187]6A=[1218−1824]A2−6A+17I2=[−5−12+1718−18−18+187−24+17]

Answer:A=[23−34]A2=[23−34]×[23−34]=[4−96+12−6−12−9+16]=[−518−187]6A=[1218−1824]A2−6A+17I2=[−5−12+1718−18−18+187−24+17]=[0000]=0 [Proved]

Answer:A=[23−34]A2=[23−34]×[23−34]=[4−96+12−6−12−9+16]=[−518−187]6A=[1218−1824]A2−6A+17I2=[−5−12+1718−18−18+187−24+17]=[0000]=0 [Proved]A−1=171[43−32]

A=[23−34]A2=[23−34]×[23−34]=[4−96+12−6−12−9+16]=[−518−187]6A=[1218−1824]A2−6A+17I2=[−5−12+1718−18−18+187−24+17]

A=[23−34]A2=[23−34]×[23−34]=[4−96+12−6−12−9+16]=[−518−187]6A=[1218−1824]A2−6A+17I2=[−5−12+1718−18−18+187−24+17]=[0000]=0 [Proved]

A=[23−34]A2=[23−34]×[23−34]=[4−96+12−6−12−9+16]=[−518−187]6A=[1218−1824]A2−6A+17I2=[−5−12+1718−18−18+187−24+17]=[0000]=0 [Proved]A−1=171[43−32]

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