Solve for xy and z in the matrix equation

  3x-3     2        3                 2    u    3 

    3        9      y-3       =       3    9    8  

   3z      -2        3                  9   -2   3

Matrix calculations are different from regular calculations of scalar numbers hence, a set of matrix rules are followed to solve these questions.

Answer: The values of u, x, y, z in the given matrix equation are 2, 5 / 3, 11 and 3 respectively.

Let's look into the stepwise solution.

Explanation:

Given matrix:

  3x-3      2        3                  2    u    3 

    3         9      y-3       =        3    9    8  

   3z       -2        3                  9   -2    3

When two matrices are equal, every element situated at the same positions have equal values in both the matrix.

Therefore,

3x - 3 = 2

⇒ x = 5 / 3

y - 3 = 8

⇒ y = 11

3z = 9

⇒ z = 3

u = 2

Thus, the values of u, x, y, z are 2, 5 / 3, 11 and 3 respectively.

$\begingroup$

I have two matrices $A$ and $B$ and I'm trying to figure out what $x$, $y$, and $z$ are.

$$\begin{bmatrix}x+2y&x\\-x+y&2x-y\end{bmatrix} = \begin{bmatrix}10&2x-3y\\-4&10\end{bmatrix}$$

What I have so far is:

$x + 2y = 10$

$-x+y= -4$

$x=2x-3y$

$2x-y = 10$

I don't know how to proceed on from here as each equation has two unknowns. I'm really bad at algebra so any help would be much appreciated!

I also do not have any idea why the question asks to find $z$ as there is no $z$ variable to work with.

asked Jan 21, 2014 at 23:47

$\endgroup$

2

$\begingroup$

Adding the first and second equations yields

$$ 3y = 6 \Longrightarrow y = 2$$

Not put $y=2$ in the second equation and you will get $$-x + 2 = -4 \Longrightarrow x = 6$$

Now you have to check, if for those values of $x$ and $y$ the 3rd and 4th equation is fulfilled as well,

answered Jan 21, 2014 at 23:54

user127.0.0.1user127.0.0.1

6,9976 gold badges28 silver badges44 bronze badges

$\endgroup$

$\begingroup$

From the third equation, you get $x = 3y$, from the first you get $y=2$, and so you get $x = 6$.

answered Jan 21, 2014 at 23:54

nbubisnbubis

32k7 gold badges77 silver badges136 bronze badges

$\endgroup$

Hi there! This page is only going to make sense when you know a little about Systems of Linear Equations and Matrices, so please go and learn about those if you don't know them already!

The Example

One of the last examples on Systems of Linear Equations was this one:

Example: Solve

• x + y + z = 6
• 2y + 5z = −4
• 2x + 5y − z = 27

We then went on to solve it using "elimination" ... but we can solve it using Matrices!

Using Matrices makes life easier because we can use a computer program (such as the Matrix Calculator) to do all the "number crunching".

But first we need to write the question in Matrix form.

In Matrix Form?

OK. A Matrix is an array of numbers, right?

A Matrix

Well, think about the equations:

They could be turned into a table of numbers like this:

We could even separate the numbers before and after the "=" into:

Now it looks like we have 2 Matrices.

In fact we have a third one, which is [x y z]:

Why does [x y z] go there? Because when we Multiply Matrices the left side becomes:

Which is the original left side of our equations above (you might like to check that).

The Matrix Solution

We can write this:

like this:

AX = B

where

• A is the 3x3 matrix of x, y and z coefficients
• X is x, y and z, and
• B is 6, −4 and 27

Then (as shown on the Inverse of a Matrix page) the solution is this:

X = A-1B

What does that mean?

It means that we can find the values of x, y and z (the X matrix) by multiplying the inverse of the A matrix by the B matrix.

So let's go ahead and do that.

First, we need to find the inverse of the A matrix (assuming it exists!)

Using the Matrix Calculator we get this:

(I left the 1/determinant outside the matrix to make the numbers simpler)

Then multiply A-1 by B (we can use the Matrix Calculator again):

And we are done! The solution is:

x = 5,
y = 3,
z = −2

Just like on the Systems of Linear Equations page.

Quite neat and elegant, and the human does the thinking while the computer does the calculating.

Just For Fun ... Do It Again!

For fun (and to help you learn), let us do this all again, but put matrix "X" first.

I want to show you this way, because many people think the solution above is so neat it must be the only way.

So we will solve it like this:

XA = B

And because of the way that matrices are multiplied we need to set up the matrices differently now. The rows and columns have to be switched over ("transposed"):

And XA = B looks like this:

The Matrix Solution

Then (also shown on the Inverse of a Matrix page) the solution is this:

X = BA-1

This is what we get for A-1:

In fact it is just like the Inverse we got before, but Transposed (rows and columns swapped over).

Next we multiply B by A-1:

And the solution is the same:

x = 5, y = 3 and z = −2

It didn't look as neat as the previous solution, but it does show us that there is more than one way to set up and solve matrix equations. Just be careful about the rows and columns!

How do you find the value of XY and Z in a matrix?