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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?

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

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