Let A = \left(\right. 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \left.\right). If u1 and u2 are column matrices such that \left(Au\right)_{1} = \left(\right. 1 \\ 0 \\ 0 \left.\right) \textrm{ }\textrm{ } and \textrm{ }\textrm{ } \left(Au\right)_{2} = \left(\right. 0 \\ 1 \\ 0 \left.\right) then u1 + u2 is equal to :

Engineering

Mathematics

Inverse of a Matrix

Question

Let $A = (\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{matrix})$ . If u₁ and u₂ are column matrices such that ${Au}_{1} = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) and {Au}_{2} = (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix})$ then u₁ + u₂ is equal to :

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

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

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

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

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$u_{1} = [\begin{matrix} x \\ y \\ z \end{matrix}]$

$[\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{array}{l} 1 \\ 0 \\ 0 \end{array}]$

$u_{1} = [\begin{array}{l} 1 \\ - 2 \\ 1 \end{array}]; u_{2} = [\begin{array}{l} 1 \\ 0 \\ 0 \end{array}]$

$u_{1} + u_{2} = [\begin{array}{l} 1 \\ - 1 \\ - 1 \end{array}]$