# Determining how to make a matrix have less pivots than columns (Example)

I'm studying the following example and I can't figure out why the answer is what it is:

"Find the value of $$a$$ that will give less that 3 pivots". $$A = \begin{bmatrix} 2 & a & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \\ \end{bmatrix}$$

I got lucky and put the matrix with $$a=3$$ in Wolfram Alpha and saw that 3 is the only value that meets that condition, but I have no idea why. Could someone explain why that is and what I should be thinking in order to come to that conclusion?

• You want the determinant to be zero. – David Peterson May 17 at 5:47
• @DavidPeterson Oof, total brain fart on my end, but thank you nonetheless! Cheers! – Swoldier May 17 at 5:52

We can take a direct approach here by row reducing. \begin{align*} \left[\begin{array}{rrr} 2 & a & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{array}\right] \xrightarrow{R_2-\frac{1}{2}\cdot R_1\to R_2}\left[\begin{array}{rrr} 2 & a & 0 \\ 0 & -\frac{1}{2} \, a + 2 & 1 \\ 0 & 1 & 2 \end{array}\right] \\ \xrightarrow{R_2\leftrightarrow R_3}\left[\begin{array}{rrr} 2 & a & 0 \\ 0 & 1 & 2 \\ 0 & -\frac{1}{2} \, a + 2 & 1 \end{array}\right] \\ \xrightarrow{R_3-\left(\frac{1}{2}\,a-2\right)\cdot R_2\to R_3}\left[\begin{array}{rrr} 2 & a & 0 \\ 0 & 1 & 2 \\ 0 & 0 & a - 3 \end{array}\right] \end{align*} These reductions show that our matrix $$A$$ is row equivalent to $$\left[\begin{array}{rrr} 2 & a & 0 \\ 0 & 1 & 2 \\ 0 & 0 & a - 3 \end{array}\right]$$ What is the rank of this new matrix?