Problem solving three equations using matrices.

Question

Hi I'm not able to solve this system of equations, I use elementary row operations but it's pretty complicated, will you guys help me?

$$-x-5y-5z=2$$ $$4x-5y+4z=19$$ $$x+5y-z=-20$$

Answer 1

The matrix corresponding to the given equation is this: $$ \left(\begin{array}{rrr|r} -1 &-5 &-5 & 2 \\ 4 &-5 & 4 &19 \\ 1 & 5 &-1 &-20 \end{array}\right) $$

If possible we want to have $1$ in the upper left corner. So we multiply the first row by $-1$ $$\left(\begin{array}{rrr|r} -1 &-5 &-5 & 2 \\ 4 &-5 & 4 &19 \\ 1 & 5 &-1 &-20 \end{array}\right)\sim \left(\begin{array}{rrr|r} 1 & 5 & 5 & -2 \\ 4 &-5 & 4 & 19 \\ 1 & 5 &-1 &-20 \end{array}\right)$$

Now the remaining elements in the first column should be zeroes. So we add to the second row $(-4)$-multiple of the first row. We add to the third row $(-1)$-multiple of the first row. (In the other words, we subtract appropriate multiple of the first row.)
We get $$ \left(\begin{array}{rrr|r} 1 & 5 & 5 & -2 \\ 4 &-5 & 4 & 19 \\ 1 & 5 &-1 &-20 \end{array}\right)\sim \left(\begin{array}{rrr|r} 1 & 5 & 5 & -2 \\ 0 &-25&-16& 27 \\ 1 & 5 &-1 &-20 \end{array}\right)\sim \left(\begin{array}{rrr|r} 1 & 5 & 5 & -2 \\ 0 &-25&-16& 27 \\ 0 & 0 &-6 &-18 \end{array}\right)$$

Since the third row is very simple, we can get it to the form where the pivot is equal to $1$. $$ \left(\begin{array}{rrr|r} 1 & 5 & 5 & -2 \\ 0 &-25&-16& 27 \\ 0 & 0 &-6 &-18 \end{array}\right)\sim \left(\begin{array}{rrr|r} 1 & 5 & 5 & -2 \\ 0 &-25&-16& 27 \\ 0 & 0 & 1 & 3 \end{array}\right)$$

Can you now use the third row to obtain zeroes in the third column in the first two rows? After that you are almost done. Can you continue from there?

If I am not mistaken, the solution should be $x=-2$, $y=-3$, $z=3$.

Answer 2

It's pretty simple.

Add the first and third equations to get $$-6z=-18$$ or $z=3$.

Then adding the second and the third,

$$5x+3z=-1,$$

or $x=-2$.

Finally, $y=-3$.

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In matrix notation, you can sum the third row to the second, and the first to the third.

$$ \left(\begin{array}{rrr|r} -1 &-5 &-5 & 2 \\ 5 &0 & 3 &-1\\ 0 & 0 &-6 &-18\end{array}\right) $$ You obtain a triangular system which is better solved directly.