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$A= \begin{bmatrix} 1 &k &1 \\ 1 &k+1 &k+2 \\ 1& k+2 & 2k+4 \end{bmatrix}$

When trying to find the determinant I get:

$\det(A)= 1((2k^2 +6k +4)-(k^2 +4k+4))-k((2k+4)-(k+2))-1((k+2)-(k+1))$

$=(k^2 +2k)-(k^2 +2k)-1$

$=-1$

What do I do from here?
I am used to equalling my equation to $0$ and finding the values which $k$ can't be but in this situation I can not.
Any help would be greatly appreciated.

2 Answers 2

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Your idea was to set the equality $\det(A) = 0$ in order to decide which values of $k$ make the equation true, and hence cause $A$ to be non-invertible. This is the correct idea.

According to your work (which seems correct), this yields the equation $-1 = 0$. There are no values of $k$ for which $-1 = 0$. Thus, there is no value of $k$ for which $A$ is non-invertible. So, your conclusion should be that $A$ is invertible for every value of $k$.

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    this has been of great help. thank you so much – 2017-02-28
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\begin{align}\begin{vmatrix} 1 & k & 1 \\ 1 & k+1 & k+2 \\ 1 & k+2 & 2k+4\end{vmatrix} &= \begin{vmatrix} 1 & k & 1 \\ 0 & 1 & k+1 \\ 0 & 2 & 2k+3\end{vmatrix} \\ &= \begin{vmatrix} 1 & k & 1 \\ 0 & 1 & k+1 \\ 0 & 0 & 1\end{vmatrix} \\ &= 1 \end{align}

The determinant is $1$ regardless of the value of $k$. Hence it is always intertible.

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    thank you for your help! cant believe how dumb i can be sometimes lol but thanks again – 2017-02-28