If A=begin{bmatrix} 1 & 5 & 3 2 & 4 & 0 3 & | vedclass

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(C) Given $[x \ y \ z] A^{T} = B^{T}$. Taking transpose on both sides,we get $([x \ y \ z] A^{T})^{T} = (B^{T})^{T}$.This implies $A [x \ y \ z]^{T} =

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$6$

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(C) Given $[x \ y \ z] A^{T} = B^{T}$. Taking transpose on both sides,we get $([x \ y \ z] A^{T})^{T} = (B^{T})^{T}$.This implies $A [x \ y \ z]^{T} = B$.Substituting the matrices:$\begin{bmatrix} 1 & 5 & 3 \ 2 & 4 & 0 \ 3 & -1 & -5 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} -1 \ -2 \ 4 \end{bmatrix}$.This gives the system of linear equations:$x + 5y + 3z = -1$ ...$(i)$$2x + 4y = -2 \implies x + 2y = -1 \implies x = -1 - 2y$ ...(ii)$3x - y - 5z = 4$ ...(iii)Substituting $x = -1 - 2y$ into $(i)$:$(-1 - 2y) + 5y + 3z = -1 \implies 3y + 3z = 0 \implies y = -z$.Substituting $x = -1 - 2y$ and $y = -z$ into (iii):$3(-1 - 2(-z)) - (-z) - 5z = 4$$3(-1 + 2z) + z - 5z = 4$$-3 + 6z - 4z = 4 \implies 2z = 7 \implies z = 3.5$.Then $y = -3.5$ and $x = -1 - 2(-3.5) = -1 + 7 = 6$.Thus,$x + y + z = 6 - 3.5 + 3.5 = 6$.

If $A=\begin{bmatrix} 1 & 5 & 3 \\ 2 & 4 & 0 \\ 3 & -1 & -5 \end{bmatrix}$,$B=\begin{bmatrix} -1 \\ -2 \\ 4 \end{bmatrix}$ and $[x \ y \ z] A^{T}=B^{T}$,then $x+y+z=$

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