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

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 $$a$$?

• There are definitely more values of $a$. Have you seen determinants yet? Commented Feb 24, 2022 at 20:52
• 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?) Commented Feb 24, 2022 at 20:53
• @TheoBendit Yes, I have Commented Feb 24, 2022 at 20:55
• @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! Commented Feb 24, 2022 at 21:14
• @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). Commented Feb 24, 2022 at 21:21

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*}
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 $$a\ne 0, 2, 4$$.