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So, the question is indeed asking for what you just read. I have the following matrix for which I have to find the values of $k$ in order to make it not invertible. I have understood that the inverse of a matrix is the result of $[A | I]$, so since I have to solve in row echelon form in order to find its inverse, I got stuck when the time came to put it from row echelon to reduced row echelon form because I have $k$ in the lowest right-hand corner.

$$A=\begin{bmatrix}1&2&-1\\2&0&2\\-1&2&k\end{bmatrix}$$

This is how the matrix looks like in row echelon form. The problem is that I don't know how to continue, even if I was to know how to get into reduced row echelon form, I wouldn't know how to find a number that would make the matrix not invertible. Am I supposed to give the value of k that would be anything but one? (because the matrix in the end has to be equal to the identity right?) Please, I would greatly appreciate your help.

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  • $\begingroup$ When you reduce, matrix is not invertible if we get a row of all zeros. $\endgroup$ – André Nicolas Nov 1 '14 at 1:25
  • $\begingroup$ I actually don't know how to get the determinant for a 3x3 matrix, and so if I get for example k-1 and have to add another number, the one that changes is only the number 1? Thank you for your prompt replies! $\endgroup$ – Irais Nov 1 '14 at 1:33
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The determinant is given by

$$\det A = \begin{vmatrix} 1 & 2 & -1 \\ \color{red}2 & \color{red}0 & \color{red}2\\ -1 & 2 & k \end {vmatrix} = -\color{red} 2\begin{vmatrix}2 & -1 \\ 2 & k \end {vmatrix} +\color{red} 0\begin{vmatrix}1 & -1 \\ -1 & k \end {vmatrix} -\color{red} 2\begin{vmatrix}1 & 2 \\ -1 & 2 \end {vmatrix} \\= -2(2k+2) - 2(2 + 2) = -4k - 12$$

Now $A$ is not invertible $\Leftrightarrow \det A = 0$. Then $k = -3$.

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  • $\begingroup$ thank you, though i don't get the process, the final answer helped me to understand better! $\endgroup$ – Irais Nov 1 '14 at 2:52
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    $\begingroup$ Colored to explain the process. @Aaron: feel free to rollback to the previous version. $\endgroup$ – Omnomnomnom Nov 1 '14 at 4:53
  • $\begingroup$ Great idea pointing out the process, @Omnomnomnom, thanks. $\endgroup$ – Aaron Maroja Nov 1 '14 at 18:31
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The determinant is equal (e.g. from the Rule of Sarrus) to $(0-4-4)-(0+4+4k)=4(-3-k)$. It is equal to 0 for $k=-3$ and then $A$ is not invertible.

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  • $\begingroup$ thank you! though I had to find out the reason behind those numbers, it helped me a lot! $\endgroup$ – Irais Nov 1 '14 at 2:53

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