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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 $a$?

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    $\begingroup$ There are definitely more values of $a$. Have you seen determinants yet? $\endgroup$
    – Theo Bendit
    Commented Feb 24, 2022 at 20:52
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    $\begingroup$ you have different options. You can find the determinant of the matrix and see, for which values of a, it becomes zero. Or, you can go for elementary row operations and see, for which values of a, you may face a row whose elements are zero. Moreover, you can find two other values of a in a first glance. (Hint: when two rows or two columns become same?) $\endgroup$
    – khashayar
    Commented Feb 24, 2022 at 20:53
  • $\begingroup$ @TheoBendit Yes, I have $\endgroup$
    – brandon_ducks
    Commented Feb 24, 2022 at 20:55
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    $\begingroup$ @brandon_ducks A nice feature of the determinant: the determinant is a degree $n$ polynomial of the entries of an $n \times n$ matrix. From this, you can see the determinant is a degree $3$ polynomial of $a$. If you can eyeball $3$ distinct values of $a$ where the determinant is $0$, then you've found all of them! $\endgroup$
    – Theo Bendit
    Commented Feb 24, 2022 at 21:14
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    $\begingroup$ @brandon_ducks I wasn't trying to duplicate the other user's comment, just augment it with a nice observation. If you find $3$ different values, as the user suggested, then prima facie, there could be others that you're missing. But, if you know that the determinant is of degree $3$, you can say with certainty that there are no more (without having to actually compute the determinant). $\endgroup$
    – Theo Bendit
    Commented Feb 24, 2022 at 21:21

2 Answers 2

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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?

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  • $\begingroup$ Yes, thank you so much! $\endgroup$
    – brandon_ducks
    Commented Feb 25, 2022 at 0:10
  • $\begingroup$ @brandon_ducks you are welcome ! I am glad I could help. $\endgroup$
    – Átila Correia
    Commented Feb 25, 2022 at 0:10
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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 $a\ne 0, 2, 4$.

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