Show that the equation $$\pmatrix{3&-7&0\\2&2&5\\1&3&4}\pmatrix{x\\y\\z} = \pmatrix{3\\2\\1}$$ does not have a unique solution, and give a geometrical interpretation.
Now, the first part of the question is fairly straightforward. I've shown that the determinant of the 3×3 matrix is 0; and thus the system of equation does not have a unique solution.
After that, I've rewritten the matrices into a set of linear equations.
$\begin{cases}3x-7y+0z=3\\2x+2y+5z=2\\
x+3y+4z=1\end{cases}$
Now this is where I'm slightly stumped, I've attempted this part of the question but have no idea if I'm correct.
Through eliminating $x$ from pairs of equations, I've gotten $-4y-3z =0$ from 2 pairs and $-16y-22z=0$ from the other. What does this mean exactly?
Best Answer
The three planes pass through the same line $r$ of equation $$ r:\begin{cases} 3 x-7 y-3=0\\ 4 y+3 z=0\\ \end{cases} $$