Given the matrix vector system
$$\begin{bmatrix}
a & 2 & 3\\
a & a & 4\\
a & a & a
\end{bmatrix}
\cdot
\begin{bmatrix}
x\\ y\\ z
\end{bmatrix}=
\begin{bmatrix}
b_1\\b_2\\b_3
\end{bmatrix}$$
The only value I can think of, without plugging in values explicitly and performing row operations, is $a=0$. How can I show that this is either the only value for $a$ in which the system fails to have $3$ pivots, or how can I show there are more values of $a$?
Best Answer
HINT
You can also calculate the expression of the determinant related to the corresponding matrix: \begin{align*} \begin{vmatrix} a & 2 & 3\\ a & a & 4\\ a & a & a \end{vmatrix} & = \begin{vmatrix} a & 2 & 3\\ a & a & 4\\ 0 & 0 & a - 4 \end{vmatrix} = \begin{vmatrix} a & 2 & 3\\ 0 & a - 2 & 1\\ 0 & 0 & a - 4 \end{vmatrix} = a(a-2)(a-4) \end{align*}
Can you take it from here?