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If $\mathrm{A}=\left(\begin{array}{ccc}1 & 5 & 3 \\ 2 & 4 & 0 \\ 3 & -1 & -5\end{array}\right), \mathrm{B}=\left(\begin{array}{c}-1 \\ -2 \\ 4\end{array}\right)$ and $[\mathrm{x} \mathrm{y} \mathrm{z}] \mathrm{A}^{\mathrm{T}}=\mathrm{B}^{\mathrm{T}}$, then
$x+y+z=$
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$x+y+z=$
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2701 Upvotes
Verified Answer
The correct answer is:
$6$
$\begin{aligned} & \Rightarrow\left[\begin{array}{ll}x y z\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 3 \\ 5 & 4 & -1 \\ 3 & 0 & -5\end{array}\right]=\left[\begin{array}{lll}-1 & -24\end{array}\right] \\ & \Rightarrow \quad[x+5 y+3 z \quad 2 x+4 y \quad 3 x-y-5 z]=\left[\begin{array}{lll}-1 & -2 & 4\end{array}\right]\end{aligned}$
$\therefore \quad x+5 y+3 z=-1$ ...(i)
$2 x+4 y=-2, \Rightarrow x+2 y=-1$ ...(ii)
$3 x-y-5 z=4$ ...(iii)
Solving eqs. (i), (ii) $\&$ (iii), we get
$x=6, y=-7 / 2, z=7 / 2$
$\therefore x+y+z=6-\frac{7}{2}+\frac{7}{2}=6$
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