For which value of $k$ will the vector
$\begin{bmatrix}1\\-2\\k\end{bmatrix}$ in $\mathbb{R}^3$ is a linear combination of the vectors $w=\begin{bmatrix}2\\-1\\-5\end{bmatrix}$ and $v=\begin{bmatrix}3\\0\\-2\end{bmatrix}$?
I am required to solve this by performing row operations. I came up with this:
$$\begin{bmatrix} -2& 0 &-1 &0 \\ 0 &-2& 1 &0 \\ k &0 &-4 &0 \end{bmatrix}$$
Best Answer
You have to solve the linear system $$ \alpha\begin{bmatrix}2\\-1\\-5\end{bmatrix}+ \beta\begin{bmatrix}3\\0\\-2\end{bmatrix}= \begin{bmatrix}1\\-2\\k\end{bmatrix} $$ that is $$ \begin{cases} 2\alpha+3\beta=1\\ -\alpha=-2\\ -5\alpha-2\beta=k \end{cases} $$ The second equation says $\alpha=2$, the first that $4+3\beta=1$ or $\beta=-1$.
Then $k=\dots$
If you have to solve it by row operations, \begin{align} \begin{bmatrix} 2 & 3 & 1\\ -1 & 0 & -2\\ -5 & -2 & k \end{bmatrix} &\to \begin{bmatrix} 1 & 3/2 & 1/2\\ -1 & 0 & -2\\ -5 & -2 & k \end{bmatrix} && R_1\gets \frac{1}{2}R_2 \\[6px]&\to \begin{bmatrix} 1 & 3/2 & 1/2\\ 0 & 3/2 & -3/2\\ -5 & -2 & k \end{bmatrix} && R_2\gets R_2+R_1 \\[6px]&\to \begin{bmatrix} 1 & 3/2 & 1/2\\ 0 & 3/2 & -3/2\\ 0 & -11/2 & k+5/2 \end{bmatrix} && R_3\gets R_3+5R_1 \\[6px]&\to \begin{bmatrix} 1 & 3/2 & 1/2\\ 0 & 1 & -1\\ 0 & -11/2 & k+5/2 \end{bmatrix} && R_2\gets \frac{2}{3}R_2 \\[6px]&\to \begin{bmatrix} 1 & 3/2 & 1/2\\ 0 & 1 & -1\\ 0 & 0 & k+8 \end{bmatrix} && R_3\gets R_3+\frac{11}{2}R_2 \end{align}
So $k=\dots$