If AX=B,where A=begin{bmatrix} 1 & 3 & 3 1 & | vedclass

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(A) Given the matrix equation $AX=B$:$\begin{bmatrix} 1 & 3 & 3 \ 1 & 4 & 4 \ 1 & 3 & 4 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \b

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) Given the matrix equation $AX=B$:$\begin{bmatrix} 1 & 3 & 3 \ 1 & 4 & 4 \ 1 & 3 & 4 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 12 \ 15 \ 13 \end{bmatrix}$Applying row operations $R_{2} ightarrow R_{2}-R_{1}$ and $R_{3} ightarrow R_{3}-R_{1}$:$\begin{bmatrix} 1 & 3 & 3 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 12 \ 3 \ 1 \end{bmatrix}$This gives the system of equations:$x + 3y + 3z = 12$$y + z = 3$$z = 1$From $z = 1$,substitute into $y + z = 3$ to get $y + 1 = 3$,so $y = 2$.Substitute $y = 2$ and $z = 1$ into $x + 3y + 3z = 12$:$x + 3(2) + 3(1) = 12$$x + 6 + 3 = 12$$x + 9 = 12 \Rightarrow x = 3$Finally,calculate $x^{2} + y^{2} + z^{2} = 3^{2} + 2^{2} + 1^{2} = 9 + 4 + 1 = 14$.

If $AX=B$,where $A=\begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 4 \\ 1 & 3 & 4 \end{bmatrix}$,$X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ and $B=\begin{bmatrix} 12 \\ 15 \\ 13 \end{bmatrix}$,then $x^{2}+y^{2}+z^{2}=$

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