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

a) Find $\det(A)$.

b) Find all values of $k$ (if any) such that $A^{-1}$ does not exist

I'm having some trouble figuring out how to reduce the matrix into lower triangular form.

Can anyone help me with the steps to get this into lower triangular form?

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  • $\begingroup$ You don't need to row reduce to complete this problem. If you have found the determinant, then simply factor the polynomial. As for finding the determinant, Laplace Expansion along the first or last column would be a good first step. $\endgroup$ – JMoravitz Oct 19 '16 at 17:14
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As mentioned in the comments, you don't need to row reduce to find the determinant! If you don't know how, you should definitely read more about this.

However, in this case, it can easily be done.

Replace the first column $C_1$ by $C'_1 = C_1 - C_4$ and the second column $C_2$ by $C'_2 = C_2 - C_C$. You then get $$ A'= \begin{pmatrix} k-1 & -1 & 1 & 1 \\ 0 & 0 & k & 1 \\ 0 & -k-3 & k+2 & 0 \\ 0 & 0 & 1 & 0 \\ \end{pmatrix} $$ Now permute columns $C'_2, C'_3$ and $C'_4$ to obtain $$ A''= \begin{pmatrix} k-1 & 1&-1 & 1 \\ 0 & 1&0 & k \\ 0 & 0&-k-3 & k+2 \\ 0 & 0&0 & 1 \\ \end{pmatrix} $$ Now, $\det(A'') = -(k-1)(k+3)$. However, we swapped columns, so you still have to check if this equals $\det(A)$ or $-\det(A)$, which depends on the number of swaps!

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