# Row reduction of augmented matrix with unknowns

I've been stumped on this question for the past few days.

The question asks that the following augmented matrix be row-reduced to a 'goal' matrix:

\begin{matrix} 1 & 2 & -1 &| -3\\ 3 & 5 & k &| -4\\ 9 & (k+13) & 6 &|+9\\ \end{matrix}

Needs to be reduced to:

\begin{matrix} 1 & 2 & -1 &| -3\\ 0 & 1 & -k-3 &| -5\\ 0 & 0 & k^2-2k &|5k+11\\ \end{matrix}

I must have tried this upwards of 15 times - I can get the three zeroes just fine usually, but the unknowns (k) are rarely anywhere near the 'goal' matrix. I have tried getting the three zeroes in different orders, but that doesn't seem to help either.

The closest I've come is

\begin{matrix} 1 & 2 & -1 &| -3\\ 0 & 1 & -k+3 &| +5\\ 0 & 0 & -k^2-8k+12 &|-5k+7\\ \end{matrix}

I feel like I'm really close, but I just can't get the correct unknowns...

Any help would be greatly appreciated.

Thanks,

John

• Treat $k$ as an arbitrary constant, then you'll be able to use elementary row operations involving terms with $k$. – user124862 Mar 6 '14 at 2:33
• @Amzoti I tried what you wrote, but the became too convoluted after R3=R1/9.. – user133422 Mar 6 '14 at 8:05
• @MitchKnight That's the problem I'm having though; I can never get the constant k to be as shown in the goal matrix :( – user133422 Mar 6 '14 at 8:06

I was struggling with this one, so I used software to compute the reduced row echelon forms (which are unique to each linear system) for each and found them to be equal to each other, thus I need that the first matrix could be reduced to the second. Now, $$\begin{pmatrix} 1&2&-1&-3\\ 3&5&k&-4\\9&(k+13)&6&6\end{pmatrix} -3R_1+R_2\to R_2 \begin{pmatrix} 1&2&-1&-3\\ 0&-1&(3+k)&5\\9&(k+13)&6&9\end{pmatrix}$$ $$9R_1 -R_3 \to R_3 \begin{pmatrix} 1&2&-1&-3\\ 0&-1&(3+k)&5\\0&(5-k)&-15&-36\end{pmatrix}$$ $$-R_2 \to R_2 \begin{pmatrix} 1&2&-1&-3\\ 0&1&(-3-k)&-5\\0&(5-k)&-15&-36\end{pmatrix}$$ $$(5-k)R_2-R_3 \to R_3 \begin{pmatrix} 1&2&-1&-3\\ 0&1&(-3-k)&-5\\0&0&(k^2-2k)&(5k+11)\end{pmatrix}.$$