I have the following problem:

I have 2 3x3 matrices, on called M and the other N^-1 (this one has a few unknown numbers, k). I am asked to find the following:

(MN)^-1. Now we have NOT learned how to do the inverse of 3x3 matrices yet, so there must be some other way of solving this...but how?

M=
(3 0 8
0 2 0
1 0 4)
and N^-1=
(1 -2 3
3 k 0
2 4 k)

2. Originally Posted by HelenaStage
I have the following problem:

I have 2 3x3 matrices, on called M and the other N^-1 (this one has a few unknown numbers, k). I am asked to find the following:

(MN)^-1. Now we have NOT learned how to do the inverse of 3x3 matrices yet, so there must be some other way of solving this...but how?

M=
(3 0 8
0 2 0
1 0 4)
and N^-1=
(1 -2 3
3 k 0
2 4 k)

hi

$\displaystyle (MN)^{-1}=N^{-1}M^{-1}$

$\displaystyle N^{-1}$ is given , you will need to find the inverse of M . Since you haven't been taught on finding its inverse , you shouldn't be doing this question . Do you know how to use the calculator to get its inverse ?

hi

$\displaystyle (MN)^{-1}=N^{-1}M^{-1}$

$\displaystyle N^{-1}$ is given , you will need to find the inverse of M . Since you haven't been taught on finding its inverse , you shouldn't be doing this question . Do you know how to use the calculator to get its inverse ?
Thanks a lot for the helpful formula...
The only problem is that this question is one our teacher has given us, meaning we SHOULD be able to solve it...without a calculator. And I'm actually not supposed to use the formula you said either...(just realised the question states this)

Isn't there some way I could do that?

4. HGello, Helena!

Given: .$\displaystyle M \:=\:\begin{pmatrix}3&0&8 \\ 0&2&0 \\ 1&0&4 \end{pmatrix}\qquad N^{-1} \;=\; \begin{pmatrix}1 & \text{-}2 & 3 \\ 3&k&0 \\ 2&4&k \end{pmatrix}$

Find $\displaystyle (MN)^{-1}$

Now we have NOT learned how to do the inverse of 3x3 matrices yet, . (Really?)
so there must be some other way of solving this ... but how?
If that's true, we can still solve it.

We know: .$\displaystyle (MN)^{-1} \;=\;N^{-1}M^{-1}$ . . . and we need $\displaystyle M^{-1}.$

We can find it the old-fashioned way . . .

Let: .$\displaystyle M^{-1} \:=\:\begin{pmatrix}a&b&c\\d&e&f \\ g&h&i\end{pmatrix}$

Then: .$\displaystyle \begin{pmatrix}3&0&8\\0&2&0\\1&0&4\end{pmatrix} \begin{pmatrix}a&b&c\\d&e&f\\g&h&1\end{pmatrix} \;=\;\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatri x}$

Hence: .$\displaystyle \begin{pmatrix}3a+8g & 3b + 8h & 3c+ 8i \\ 2d & 2e & 2f \\ a+4g & b+4h & c + 4i\end{pmatrix} \;=\;\begin{pmatrix}1&0&0 \\ 0&1&0\\0&0&1 \end{pmatrix}$

And we have a system of equations: .$\displaystyle \begin{array}{ccccc} 3a+8g \:=\:1 && 3b+8h \:=\:0 && 3c + 8i \:=\:0 \\ 2d \:=\:0 && 2e \:=\:1 && 2f \:=\:0 \\ a+4g \:=\:0 && b+4h \:=\:0 && c+4i \:=\:1 \end{array}$

. . which we can solve: .$\displaystyle \begin{Bmatrix}a \:=\:1 & b\:=\:0 & c\:=\:\text{-}2 \\ \\[-4mm] d\:=\:0 & e\:=\:\frac{1}{2} & f \:=\:0 \\ \\[-4mm] g\:=\:\text{-}\frac{1}{4} & h\:=\:0 & i \:=\:\frac{3}{4} \end{Bmatrix}$

And we have: .$\displaystyle M^{-1} \;=\;\begin{pmatrix} 1 & 0 & \text{-}2 \\ \\[-4mm] 0 & \frac{1}{2} & 0 \\ \\[-3mm] \text{-}\frac{1}{4} & 0 & \frac{3}{4} \end{pmatrix}$

Got it?

5. Originally Posted by Soroban
HGello, Helena!

If that's true, we can still solve it.

We know: .$\displaystyle (MN)^{-1} \;=\;N^{-1}M^{-1}$ . . . and we need $\displaystyle M^{-1}.$

We can find it the old-fashioned way . . .

Let: .$\displaystyle M^{-1} \:=\:\begin{pmatrix}a&b&c\\d&e&f \\ g&h&i\end{pmatrix}$

Then: .$\displaystyle \begin{pmatrix}3&0&8\\0&2&0\\1&0&4\end{pmatrix} \begin{pmatrix}a&b&c\\d&e&f\\g&h&1\end{pmatrix} \;=\;\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatri x}$

Hence: .$\displaystyle \begin{pmatrix}3a+8g & 3b + 8h & 3c+ 8i \\ 2d & 2e & 2f \\ a+4g & b+4h & c + 4i\end{pmatrix} \;=\;\begin{pmatrix}1&0&0 \\ 0&1&0\\0&0&1 \end{pmatrix}$

And we have a system of equations: .$\displaystyle \begin{array}{ccccc} 3a+8g \:=\:1 && 3b+8h \:=\:0 && 3c + 8i \:=\:0 \\ 2d \:=\:0 && 2e \:=\:1 && 2f \:=\:0 \\ a+4g \:=\:0 && b+4h \:=\:0 && c+4i \:=\:1 \end{array}$

. . which we can solve: .$\displaystyle \begin{Bmatrix}a \:=\:1 & b\:=\:0 & c\:=\:\text{-}2 \\ \\[-4mm] d\:=\:0 & e\:=\:\frac{1}{2} & f \:=\:0 \\ \\[-4mm] g\:=\:\text{-}\frac{1}{4} & h\:=\:0 & i \:=\:\frac{3}{4} \end{Bmatrix}$

And we have: .$\displaystyle M^{-1} \;=\;\begin{pmatrix} 1 & 0 & \text{-}2 \\ \\[-4mm] 0 & \frac{1}{2} & 0 \\ \\[-3mm] \text{-}\frac{1}{4} & 0 & \frac{3}{4} \end{pmatrix}$

Got it?

I do! Thanks a lot. So there is actually no other way of finding (MN)^-1 other than using that formula?