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Miscellaneous Exercise 4(B) · Q182

Q.If A=diag[2 −3 −5]A=\text{diag}[2\ {-3}\ {-5}], B=diag[4 −6 −3]B=\text{diag}[4\ {-6}\ {-3}] and C=diag[−3 4 1]C=\text{diag}[-3\ 4\ 1] then find 2A+B−5C2A+B-5C

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A=diag[2,−3,−5]A=\text{diag}[2,-3,-5], B=diag[4,−6,−3]B=\text{diag}[4,-6,-3], C=diag[−3,4,1]C=\text{diag}[-3,4,1].

Entry 1: 2(2)+4−5(−3)=4+4+15=232(2)+4-5(-3)=4+4+15=23.

Entry 2: 2(−3)+(−6)−5(4)=−6−6−20=−322(-3)+(-6)-5(4)=-6-6-20=-32. …

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