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To find for which values of a does the following matrix have an inverse, should I row reduce it or use the determinant? I am not sure how to use the determinant here.

\begin{bmatrix} 2&-1&0\\ -1&1&1\\ 0&a&2 \end{bmatrix}

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    $\begingroup$ Do both. See if you get the same answer. $\endgroup$ – bubba Feb 29 '16 at 13:58
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Compute the determinant ($=2-2 a$) and set it $\neq 0$. [A square matrix that is not invertible is called singular or degenerate. A square matrix is singular if and only if its determinant is $0$].

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  • $\begingroup$ I computed the determinant and got the same result as you. However I also need an explanation if I use the determinant method. How can I explain? $\endgroup$ – NeoXx Feb 29 '16 at 14:10
  • $\begingroup$ You can explain using exactly what is written in the parentheses: A square matrix is not invertible if and only if its determinant is 0. You should have proven this result in class, or it should be written somewhere in the textbook. $\endgroup$ – Omnomnomnom Feb 29 '16 at 14:16
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    $\begingroup$ If you have not talked about this fact in class, however, I would guess that you should be using the row-reduction method. $\endgroup$ – Omnomnomnom Feb 29 '16 at 14:17
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Either approach works. Since another answer explained how to use the determinant, I'll use row reduction.

Proceed with as many steps as possible that avoid division by any expression involving $a$: $$ \pmatrix{ 2&-1&0\\ -1&1&1\\ 0&a&2\\ } \to\\ \pmatrix{ 1&-1&-1\\ 2&-1&0\\ 0&a&2\\ } \to \\ \pmatrix{ 1&-1&-1\\ 0&1&2\\ 0&a&2\\ } \to \\ \pmatrix{ 1&-1&-1\\ 0&1&2\\ 0&0&2-2a\\ } $$ Now, in order to get that last pivot to be a $1$, we would divide the last row by $2 - 2a$. So, whenever $2 - 2a \neq 0$, we can conclude that the matrix is invertible.

On the other hand, if $2 - 2a = 0$, then we've row-reduced the matrix and produced a row of zeros. So, in this case, the matrix is not invertible.

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There is a third way: by simple inspection. The matrix is not invertible if and only if its three columns (let's call them $u,v$ and $w$) are linearly dependent. Clearly, neither of them is a scalar multiple of another. So, if some nontrivial linear combination of them is zero, the coefficients must all be nonzero and we may assume that $u=\lambda v+\mu w$ for some scalars $\lambda$ and $\mu$. By comparing the first, second and third entries of both sides one by one, we see that $\lambda=2,\ \mu=-1$ and $2a-2=0$. Hence the matrix is non-invertible if and only if $a=1$, i.e. it has an inverse iff $a\ne1$.

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Another method :

Note that the first two rows are linearly independent, so the matrix is not invertible if the third row is a linear combination of the first two. This means that there exist two numbers $(x,y) \ne (0,0)$ such that: $$ x(2,-1,0)+y(-1,1,1)=(0,a,2) \iff (2x-y,-x+y,y)=(0,a,2) $$ so we have:

$y=2$,

$2x-y=2x-2=0 \rightarrow x=1$

$y-x=a \rightarrow a=1$

So, if $a \ne 1$ the third row is linearly independent and the matrix is invertible.

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  • $\begingroup$ Just the same method as user1551 ... two minute after! ( +1 for rapidity) . $\endgroup$ – Emilio Novati Feb 29 '16 at 15:01
  • $\begingroup$ Hmm, but there's a small gap in your argument. Consider, e.g. the matrix $\operatorname{diag}(0,0,1)$. It is not invertible, but the third row is not a linear combination of the first two. Your argument works in the OP's case because the first two rows are linearly independent, but this is not mentioned in your answer. $\endgroup$ – user1551 Feb 29 '16 at 15:10
  • $\begingroup$ Good point! Thanks, I add to my answer. $\endgroup$ – Emilio Novati Feb 29 '16 at 15:16

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