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I am supposed to find for what values the following matrix is invertible I have the following matrix:

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

So after calculating the determinant I end up with determinant = 2. But where do I go from here? Normally when calculating the determinant I have an expression like $(x-2)(x-3)$ and I can easily see that that matrix is invertible when $x$ is not 2, 3 or 0. But what do I do with the matrix above when the determinant is 2? How do I find the value for which it is (or isn't) invertible?

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    If the determinant is 2 (i don't believe it) your matrix is always invertible independent of $a$ – 2017-02-09
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    @the_architect, I didn't believe it either at first but it's correct. Easiest to calculate with cofactors along the bottom row. All terms with $a$ cancel. – 2017-02-09
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    I got 2, and I checked with a determinant calculator and it is indeed 2. But my answer sheet says it is invertible for all numbers except when a = -2? – 2017-02-09
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    @OfeliavanAnalhard then your answer sheet must be wrong – 2017-02-09
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    The answer sheet is indeed wrong. Try substituting $a = -2$ into the matrix and then calculate the determinant. You'll still get $2$. – 2017-02-09
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    @adjan I doubt that, then it also asks me to find the inverse for all the a:s where an inverse exists. Im very confused... – 2017-02-09
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    @OfeliavanAnalhard There is no doubt. There exists an inverse for any $a$. – 2017-02-09
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    Why doubt it? It's easy to verify by hand and with a determinant calculator. If $a = -2$ then the matrix is still invertible. These things (answer sheets) do have mistakes sometimes. – 2017-02-09

2 Answers 2

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If you've found that the determinant is $2$, then this means that the determinant will always be $2$ regardless of the value of $a$. Therefore all possible values of $a$ give you an invertible matrix.

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The inverse is $$\begin{bmatrix}a-1&-a&\dfrac{a+1}2\\2-a&a-1&-\dfrac a2\\1-a&a&-\dfrac a2\end{bmatrix}.$$

It is indeed defined for all $a$.