If A=⎡⎢⎣122212221⎤⎥⎦ , then show that A2−4A−5I=0, and hence find A2
OR
If A=⎡⎢⎣20−1510013⎤⎥⎦, then find A−1 using elementary row operations.
If A=⎡⎢⎣122212221⎤⎥⎦ , then show that A2−4A−5I=0, and hence find A2
OR
If A=⎡⎢⎣20−1510013⎤⎥⎦, then find A−1 using elementary row operations.
A=⎡⎢⎣122212221⎤⎥⎦
A2=⎡⎢⎣122212221⎤⎥⎦ ⎡⎢⎣122212221⎤⎥⎦ =⎡⎢⎣1×1+2×2+2×21×2+2×1+2×21×2+2×2+2×12×1+1×2+2×22×2+1×1+2×22×2+2×2+2×12×1+2×2+1×22×2+2×1+1×22×2+2×2+1×1⎤⎥⎦ =⎡⎢⎣1+4+42+2+42+4+22+2+44+1+44+2+22+4+24+2+24+4+1⎤⎥⎦ =⎡⎢⎣988898889⎤⎥⎦ Consider A2−4A−5I ⎡⎢⎣988898889⎤⎥⎦−4 ⎡⎢⎣122212221⎤⎥⎦−5⎡⎢⎣100010001⎤⎥⎦ ⎡⎢⎣988898889⎤⎥⎦−⎡⎢⎣488848884⎤⎥⎦−⎡⎢⎣500050005⎤⎥⎦ =⎡⎢⎣9−98−88−88−89−98−88−88−89−9⎤⎥⎦ =⎡⎢⎣000000000⎤⎥⎦ NowA2−4A−5I=0A2−4A=5IA2A1−4A A−1=5I A−1(Postmultiply by A−1)A−4I=5 A−1 ⎡⎢⎣122212221⎤⎥⎦−⎡⎢⎣400040004⎤⎥⎦=5A−1⎡⎢⎣−3222−3222−3⎤⎥⎦=5A−1A−1=⎡⎢
⎢
⎢
⎢⎣−35252525−35252525−35⎤⎥
⎥
⎥
⎥⎦ OR |A|=∣∣
∣∣20−1510013∣∣
∣∣=⎡⎢⎣100010001⎤⎥⎦ Applying R1→(12)R1A−1⎡⎢
⎢⎣10−12510013⎤⎥
⎥⎦ =⎡⎢
⎢⎣1200010001⎤⎥
⎥⎦ Applying R2→R2+(−5)R1A−1 ⎡⎢
⎢⎣1012510013⎤⎥
⎥⎦= ⎡⎢
⎢⎣1200010001⎤⎥
⎥⎦ Applying R2→R2+(−5) ⎡⎢
⎢
⎢⎣10−120152013⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣−1200−−5210001⎤⎥
⎥
⎥⎦ Applying R3→R3+(−1)R2⎡⎢
⎢
⎢⎣10−120152012⎤⎥
⎥
⎥⎦= ⎡⎢
⎢
⎢⎣1200−−521052−11⎤⎥
⎥
⎥⎦Applying R3→(2)R3 ⎡⎢
⎢
⎢⎣10−12−−5210011⎤⎥
⎥
⎥⎦= ⎡⎢
⎢
⎢⎣−1200−52105−22⎤⎥
⎥
⎥⎦ Applying R1→R1+(12)R3R2→R2+(−52)R3A−1 ⎡⎢⎣100010001⎤⎥⎦=A−1⎡⎢⎣3−11−156−55−22⎤⎥⎦A−1=A−1⎡⎢⎣3−11−156−55−22⎤⎥⎦
A=⎡⎢⎣122212221⎤⎥⎦
A2=⎡⎢⎣122212221⎤⎥⎦ ⎡⎢⎣122212221⎤⎥⎦ =⎡⎢⎣1×1+2×2+2×21×2+2×1+2×21×2+2×2+2×12×1+1×2+2×22×2+1×1+2×22×2+2×2+2×12×1+2×2+1×22×2+2×1+1×22×2+2×2+1×1⎤⎥⎦ =⎡⎢⎣1+4+42+2+42+4+22+2+44+1+44+2+22+4+24+2+24+4+1⎤⎥⎦ =⎡⎢⎣988898889⎤⎥⎦ Consider A2−4A−5I ⎡⎢⎣988898889⎤⎥⎦−4 ⎡⎢⎣122212221⎤⎥⎦−5⎡⎢⎣100010001⎤⎥⎦ ⎡⎢⎣988898889⎤⎥⎦−⎡⎢⎣488848884⎤⎥⎦−⎡⎢⎣500050005⎤⎥⎦ =⎡⎢⎣9−98−88−88−89−98−88−88−89−9⎤⎥⎦ =⎡⎢⎣000000000⎤⎥⎦ NowA2−4A−5I=0A2−4A=5IA2A1−4A A−1=5I A−1(Postmultiply by A−1)A−4I=5 A−1 ⎡⎢⎣122212221⎤⎥⎦−⎡⎢⎣400040004⎤⎥⎦=5A−1⎡⎢⎣−3222−3222−3⎤⎥⎦=5A−1A−1=⎡⎢ ⎢ ⎢ ⎢⎣−35252525−35252525−35⎤⎥ ⎥ ⎥ ⎥⎦ OR |A|=∣∣ ∣∣20−1510013∣∣ ∣∣=⎡⎢⎣100010001⎤⎥⎦ Applying R1→(12)R1A−1⎡⎢ ⎢⎣10−12510013⎤⎥ ⎥⎦ =⎡⎢ ⎢⎣1200010001⎤⎥ ⎥⎦ Applying R2→R2+(−5)R1A−1 ⎡⎢ ⎢⎣1012510013⎤⎥ ⎥⎦= ⎡⎢ ⎢⎣1200010001⎤⎥ ⎥⎦ Applying R2→R2+(−5) ⎡⎢ ⎢ ⎢⎣10−120152013⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢⎣−1200−−5210001⎤⎥ ⎥ ⎥⎦ Applying R3→R3+(−1)R2⎡⎢ ⎢ ⎢⎣10−120152012⎤⎥ ⎥ ⎥⎦= ⎡⎢ ⎢ ⎢⎣1200−−521052−11⎤⎥ ⎥ ⎥⎦Applying R3→(2)R3 ⎡⎢ ⎢ ⎢⎣10−12−−5210011⎤⎥ ⎥ ⎥⎦= ⎡⎢ ⎢ ⎢⎣−1200−52105−22⎤⎥ ⎥ ⎥⎦ Applying R1→R1+(12)R3R2→R2+(−52)R3A−1 ⎡⎢⎣100010001⎤⎥⎦=A−1⎡⎢⎣3−11−156−55−22⎤⎥⎦A−1=A−1⎡⎢⎣3−11−156−55−22⎤⎥⎦
