I am preparing for my Linear Algebra paper which is coming up in 4 weeks so I am working through questions in each paper.
I am puzzled with one question I have come across and perhaps someone can help me see the light.
Express the following matrix equation
$$x_1
\begin{bmatrix}
2\\
5\\
\end{bmatrix}+x_2
\begin{bmatrix}
3\\-4
\end{bmatrix}
+x_3
\begin{bmatrix}
-2\\2
\end{bmatrix}
=\begin{bmatrix}
5\\6
\end{bmatrix}$$
into a system of linear equations and show $x_1 = 2, x_2 = 3, x_3 = 4$ is a solution to the matrix equation and to the resulting system of linear equations.
So having an understanding of matrix multiplication and linear combinations I figure I'll go ahead and find the reduced row echelon form of the matrix by applying a series of elementary row operations.
\begin{bmatrix}2&3&-2&5\\5&-4&2&6\end{bmatrix}
\begin{bmatrix}1&3/2&-1&5/2\\5&-4&2&6\end{bmatrix}
\begin{bmatrix}1&3/2&-1&5/2\\0&-23/2&7&-13/2\end{bmatrix}
\begin{bmatrix}1&3/2&-1&5/2\\0&1&-14/2&13/23\end{bmatrix}
\begin{bmatrix}1&0&-2/23&38/23\\0&1&-14/23&13/23\end{bmatrix}
I cannot see how I could get $x_1 = 2, x_2 = 3, x_3 = 4$ as solutions to this matrix.
Please could someone advise if I am overlooking something here?
Thanks
Best Answer
Hint : Your final equations are
$$x_1-\frac{2}{23}x_3=\frac{38}{23}$$
and
$$x_2-\frac{14}{23}x_3=\frac{13}{23}$$
Set $x_3=4$ and determine $x_1$ and $x_2$. You get one of infinite many solutions this way.
Choosing another real value for $x_3$, you get another solution this way.