For which values of $a$ will this matrix system fail to have 3 pivots?

Question

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 $a$?

Answer 1

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?

Answer 2

Gauss elimination. If $a=0$, everything is clear. If $a\ne 0$, subtract row $1$ from rows $2$ and $3$, then row $2$ from row $3$. Then divide row $1$ by $a$, row $2$ by $a-2$ if $a\ne 2$ and row $3$ by $a-4$, provided $a\ne 4$. You will get $3$ pivotes precisely if $a\ne 0, 2, 4$.