Matrix help!

[Einsteinwannabe]

Hi all, I'm trying to solve the following problem to find y:


The answer is meant to be y=(1,2,0), which I'm not getting. I know you can find the solution via Gaussian elimination but the lecturer was talking as if it should be a quick and easy thing to solve. I thought I could get the answers by solving it line by line (i.e. 5y=5, and then 6y=9 etc.), but that doesn't get the right answer...

I feel like I'm missing something very obvious, so any help would be greatly appreciated, thanks!

 

[Harry_the_cat]

Y is a 3x1 matrix. Call it (a, b, c) written vertically. Then multiply it out. Because of the position of the 0s, you will solve for a, then b, then c.

 

[HallsofIvy]

That is a "lower triangular matrix". As Harry the Cat suggests, write the vector y as \(\displaystyle \begin{pmatrix}x \\ y \\ z \end{pmatrix}\) and do the multiplication:
\(\displaystyle \begin{pmatrix}5 & 0 & 0 \\ 3 & 3 & 0 \\ -1 & 1 & 3 \end{pmatrix}\begin{pmatrix}x \\ y \\ z \end{pmatrix}= \begin{pmatrix}5x \\ 3x+ 3y \\ -x+ y+ 3z\end{pmatrix}= \begin{pmatrix} 5 \\ 9 \\ 1 \end{pmatrix}\).

So that is the same as 5x= 5, 3x+ 3y= 9, and -x+ y+ 3z= 1.

It is easy to solve 5x= 5 for x. Put that value of x into 3x+ 3y= 9 and you get an equation for y, then put those values of x and y into -x+ y+ 3z= 1 to get an equation for z.