If $A X=B$, where $A=\left[\begin{array}{lll}1 & 3 & 3 \ 1 & 4 & 4 \ 1 & 3 & 4\end{array} ight], X=\left[\begin{array}{l}x \ y \ z\end{array} ight]$ and $B=\left[\begin{array}{l}12 \ 15 \ 13\end{array} ight]$, then $x^{2}+y^{2}+z^{2}=$

Question

If $A X=B$, where $A=\left[\begin{array}{lll}1 & 3 & 3 \ 1 & 4 & 4 \ 1 & 3 & 4\end{array}ight], X=\left[\begin{array}{l}x \ y \ z\end{array}ight]$ and $B=\left[\begin{array}{l}12 \ 15 \ 13\end{array}ight]$, then $x^{2}+y^{2}+z^{2}=$

Aicharya · Accepted Answer

Given $\mathrm{AX}=\mathrm{B}$ $\left[\begin{array}{lll} 1 & 3 & 3 \ 1 & 4 & 4 \ 1 & 3 & 4 \end{array}ight]\left[\begin{array}{l} x \ y \ z \end{array}ight]=\left[\begin{array}{c} 12 \ 15 \ 13 \end{array}ight]$ $\mathrm{R}_{2} ightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$ and $\mathrm{R}_{3} ightarrow \mathrm{R}_{3}-\mathrm{R}_{1}$ $\left[\begin{array}{lll}1 & 3 & 3 \ 0 & 1 & 1 \ 0 & 0 & 1\end{array}ight]\left[\begin{array}{l}x \ y \ z\end{array}ight]=\left[\begin{array}{l}12 \ 3 \ 1\end{array}ight] \Rightarrow\left[\begin{array}{l}x+3 y+3 z \ y+z \ z\end{array}ight]=\left[\begin{array}{l}12 \ 3 \ 1\end{array}ight]$ $\begin{aligned} 	herefore & x+3 y+3 z=12 \ & y+z=3 \ & z=1 \end{aligned}$ Thus $z=1, y=2, x=3$ $	herefore x^{2}+y^{2}+z^{2}=9+4+1=14$

If $A X=B$, where $A=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 4 \\ 1 & 3 & 4\end{array}\right],…

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