If A=α011 and B=9abc and A2=B, then the value of a+ b + c is

We have A2=α 01 1 α011=α20α+11=9 ab c ⇒ we get α2=9α=±3 and a=0,c=1,b=a+1=3+1=4 or b=−3+1=−2 So a+b+c=(0+4+1)=5 or (0−2+1)=−1

If A=α011 and B=9abc and A2=B, then the value of a+ b + c is

We have A2=α 01 1 α011=α20α+11=9 ab c ⇒ we get α2=9α=±3 and a=0,c=1,b=a+1=3+1=4 or b=−3+1=−2 So a+b+c=(0+4+1)=5 or (0−2+1)=−1

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If $A = [\begin{array}{cc} α & 0 \\ 1 & 1 \end{array}| and B = [\begin{array}{cc} 9 & a \\ b & c^{} \end{array}]$ and $A^{2} = B$ , then the value of a+ b + c is

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5 or -1

1 or -1

No real values

5 or 1

answer is B.

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Detailed Solution

We have

$A^{2} = [{\begin{array}{ll} α 0 \\ 1 1 \end{array}|}^{} [\begin{array}{cc} α & 0 \\ 1 & 1 \end{array}] = [\begin{array}{cc} α^{2} & 0 \\ α + 1 & 1 \end{array} = [\begin{array}{ll} 9 a \\ b c^{} \end{array}]$

$\Rightarrow$ we get $α^{2} = 9$

$\begin{array}{ll} α = \pm 3 and a = 0, c = 1, b = a + 1 = 3 + 1 = 4 \\ or b = - 3 + 1 = - 2 \\ So a + b + c = (0 + 4 + 1) = 5 or (0 - 2 + 1) = - 1 \end{array}$