<s>[INST] System of equations based question in matrix form If $\begin{bmatrix}1^2&2^2&3^2\\2^2&3^2&4^2\\3^2&4^2&5^2\\4^2&5^2&6^2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\4\\10\\\lambda\end{bmatrix}$,then $\lambda=$
$(A)17\hspace{1cm}(B)18\hspace{1cm}(C)19\hspace{1cm}(D)20$

I tried solving this question.
$1^2x+2^2y+3^2z=1$
$2^2x+3^2y+4^2z=4$
$3^2x+4^2y+5^2z=10$
I solved this system to get $x=\frac{25}{8},y=\frac{-5}{2},z=\frac{7}{8}$
$4^2x+5^2y+6^2z=\lambda$
I substituted $x,y,z$ in the above equation to get $\lambda=19$.
But this method is long and tedious.My book gives one method but that i cannot understand.Please help me understand this.Book's method is not self explanatory.
Book's solution:
$(n+3)^2=n^2-3(n+1)^2+3(n+2)^2$
$\lambda=1-3\times4+3\times 10=19$

OR if someone knows any better and elegant way to solve this question,please tell me,like using eigen values etc.Thanks.
 [/INST] You can solve this by row reducing the augmented matrix into echelon form. The augmented matrix is given by:
$$ \left[
    \begin{array}{ccc|c}
      1^2&2^2&3^2&1\\
      2^2&3^2&4^2&4\\
      3^2&4^2&5^2&10\\
      4^2&5^2&6^2&\lambda\\
    \end{array}
\right] $$
Row reducing leads to:
$$ \left[
    \begin{array}{ccc|c}
      1^2&2^2&3^2&1\\
      0&-7&-20&0\\
      0&0&\frac{8}{7}&1\\
      0&0&0&\lambda-19\\
    \end{array}
\right] $$
This system is only consistent if $\lambda - 19 = 0$ , therefore $\lambda = 19$.
</s>