If A=⎡⎢⎣abcbcacab⎤⎥⎦,abc=ATA=1, then find the value of a3+b3+c3.
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Solution
A=⎡⎢⎣abcbcacab⎤⎥⎦,AT=⎡⎢⎣abcbcacab⎤⎥⎦ given AAT=I∴∣∣AAT∣∣=|I|=1 or ∣∣
∣∣abcbcacab∣∣
∣∣2=1or∣∣
∣∣abcbcacab∣∣
∣∣=±1 or 3abc−a3−b3−c3=±1 or 3∓1=a3+b3+c3 ∴a3+b3+c3=2 or 4