# Can't solve 3 variables - Systems of Linear Equations

I've was asked to solve this (as homework):

$$2x + y + z = 3$$ $$4x + 2z = 10$$ $$2x + 2y = -2$$

I need to solve it with matrices and I have NO IDEA how to do so.

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What have you learned so far? –  EuYu Nov 27 '12 at 22:05
Hints: Can you write Aw = b? A is a 3x3 matrix (can you write out the equations using x, y, z)? What is b? How have you learned how to find a solution given this matrix form? Of course, maybe you have just learned some row-reduced echelon form and can do it that way and a half-dozen other ways too. –  Amzoti Nov 27 '12 at 22:07
It's most unusual to assign homework on a topic not studied. Have you no notes on how the teacher said to do this? Have you no textbook to refer to? –  Gerry Myerson Nov 27 '12 at 22:14
@GerryMyerson I find that to be generally untrue, depending on the discipline and instructor. ;) –  Arkamis Nov 27 '12 at 22:39

$$2x + y + z = 3 \tag{1}$$ $$4x + 2z = 10 \tag{2}$$ $$2x + 2y = -2 \tag{3}$$

A clever way to attack this problem is to note that we can divide through by $2$ for equation (2) to get $2x + z = 5$. Note that $2x+z$ is a part of equation (1), so we can rewrite (1) as:

$$5 + y = 3 \implies y = -2$$

Plug $y = -2$ into equation (3) to get: $$2x + 2(-2) = -2 \implies 2x = 2 \implies x = 1$$

Plug our known values for $x$ and $y$ into either equation (1) or (2) to find the value for $z$. I'll use equation (2).

$$4(1) + 2z = 10 \implies 2z = 6 \implies z = 3$$

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Perfect answer, very clear to understand . How couldn't I think about it? thanks you very very much! –  Billie Nov 27 '12 at 22:51
No problem, glad to help. Feel free to click the up arrow to the left of an answer if you find it helpful :) –  Joe Nov 27 '12 at 23:01
I'd love to. But unfortunately I'm not allowed to do so duo to my low points sum.. –  Billie Nov 27 '12 at 23:16
But but but --- it says, "I need to solve it with matrices." This is a very good way to solve the system, but it doesn't answer the question as asked. –  Gerry Myerson Nov 27 '12 at 23:31
I thought it is the only way. but it doesn't - his way better. @GerryMyerson –  Billie Nov 27 '12 at 23:55

Your system of equations is equivalent to: $$\left( \begin{array}{ccc} 2 & 1 & 1 \\ 4 & 0 & 2 \\ 2 & 2 & 0 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 3 \\ 10 \\ -2 \end{array} \right).$$ You want to add/subtract multiples of the three rows (see Gaussian elimination) until you get to $$\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} a \\ b \\ c \end{array} \right).$$

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Angel - Thanks you! very helpful. –  Billie Nov 27 '12 at 22:54

You have:

$$A=\pmatrix{2&1&1\\4&0&2\\2&2&0}, b = \pmatrix{3\\10\\-2}$$

Do you know how to use the method of Gaussian elimination? If not, there are plenty of YouTube videos that demonstrate it. If you row reduce you get:

$$\pmatrix{1\\-2\\3}$$

$\left( \begin{array}{ccc} 2 & 1 & 1 \\ 4 & 0 & 2 \\ 2 & 2 & 0 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 3 \\ 10 \\ -2 \end{array} \right)$
Form the augmented matrix and then use the following row operations $R_2= 1/2R_2-R_1$ $R_3= R_3-R_1$, which gives the matrix as $\left( \begin{array}{ccc} 2 & 1 & 1 & 3 \\ 0 & -1 & 0 & 2 \\ 0 & 1 & -1 & -5 \end{array} \right)$
Then use the row operation $R_3 = R_3+R_2$ which gives the new matrix as $\left( \begin{array}{ccc} 2 & 1 & 1 & 3 \\ 0 & -1 & 0 & 2 \\ 0 & 0 & -1 & -3 \end{array} \right)$