Let A=1 0 02 1 03 2 1. If u1 and u1 are column matrices such that Au1=100 and Au2=010 then u1+u2 is equal to

A=1 0 02 1 03 2 1,Au1=100 and Au2=010 Au1+Au2=100+010=1+00+10+0Au1+u2=110A−1Au1+u2=A−1110⇒u1+u2=100210321−1×110⇒ A−1=1 0 0−2 1 01 −2 1 u1+u2=1002⋅10321−1×110=100−2101−21×110u1+u2=1+0+0−2+1+01−2+0=1−1−1

Let A=1 0 02 1 03 2 1. If u1 and u1 are column matrices such that Au1=100 and Au2=010 then u1+u2 is equal to

A=1 0 02 1 03 2 1,Au1=100 and Au2=010 Au1+Au2=100+010=1+00+10+0Au1+u2=110A−1Au1+u2=A−1110⇒u1+u2=100210321−1×110⇒ A−1=1 0 0−2 1 01 −2 1 u1+u2=1002⋅10321−1×110=100−2101−21×110u1+u2=1+0+0−2+1+01−2+0=1−1−1

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$Let A = (\begin{array}{lll} 1 0 0 \\ 2 1 0 \\ 3 2 1 \end{array}) . If u_{1} and u_{1} are column matrices such that A u_{1} = (\begin{array}{l} 1 \\ 0 \\ 0 \end{array}) and A u_{2} = (\begin{array}{l} 0 \\ 1 \\ 0 \end{array}) then u_{1} + u_{2} is equal to$

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$(\begin{matrix} - 1 \\ 1 \\ 0 \end{matrix})$

$(\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix})$

$(\begin{array}{r} 1 \\ - 1 \\ - 1 \end{array})$

answer is D.

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Detailed Solution

$A = [\begin{array}{lll} 1 0 0 \\ 2 1 0 \\ 3 2 1 \end{array}], A u_{1} = [\begin{array}{l} 1 \\ 0 \\ 0 \end{array}] and A u_{2} = [\begin{array}{l} 0 \\ 1 \\ 0 \end{array}] \begin{array}{l} A u_{1} + A u_{2} = [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}] + [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}] = [\begin{matrix} 1 + 0 \\ 0 + 1 \\ 0 + 0 \end{matrix}] \\ A (u_{1} + u_{2}) = [\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}] \end{array}$
$\begin{array}{l} A^{- 1} A (u_{1} + u_{2}) = A^{- 1} [\begin{array}{l} 1 \\ 1 \\ 0 \end{array}] \\ \Rightarrow u_{1} + u_{2} = {[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array}]}^{- 1} \times [\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}] \end{array}$
$\Rightarrow A^{- 1} = [\begin{array}{rrr} 1 0 0 \\ - 2 1 0 \\ 1 - 2 1 \end{array}] \begin{aligned} u_{1} + u_{2} = {[\begin{array}{lll} 1 & 0 & 0 \\ 2 & \cdot 1 & 0 \\ 3 & 2 & 1 \end{array}]}^{- 1} \times [\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}] \\ = [\begin{array}{rrr} 1 & 0 & 0 \\ - 2 & 1 & 0 \\ 1 & - 2 & 1 \end{array}] \times [\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}] \\ u_{1} + u_{2} = [\begin{array}{r} 1 + 0 + 0 \\ - 2 + 1 + 0 \\ 1 - 2 + 0 \end{array}] = [\begin{array}{r} 1 \\ - 1 \\ - 1 \end{array}] \end{aligned}$