Question

A, B and C are 2x2 matrices
a) Given that AB=AC and that A is not singular, prove that B=C
b) Given that AB=AC where A=(3 6) and B=(1 5)
(1 2) (0 1)
find a matrix C whose elements are all non-zero.
First bit is fine but surely that means C is B and can't have non zero elements?

Answer 1

\[B= \left[\begin{matrix}1 & 5 \\0 & 1\end{matrix}\right] A= \left[\begin{matrix}3 & 6 \\ 1 & 2\end{matrix}\right]\]

Answer 2

"First bit is fine but surely that means C is B and can't have non zero elements?"

Can you rephrase your question? Are you talking about part a) or part b)?

Note that part a) assumes A is non-singular. But in part b) det(A)=3*2-1*6=0, so A is singular.

Answer 3

Well I was able to get why B=C. But don't worry, if A is singular then C isn't B. How then do I get C from B and A?

Answer 4

for part (b), notice that the 2nd col of A is twice the 1st column

AB
\[\left[\begin{matrix}3 & 6 \\ 1 & 2\end{matrix}\right]\left[\begin{matrix}1 & 5 \\ 0 & 1\end{matrix}\right]\]gives
\[\left[\begin{matrix}3 & 21 \\ 1 & 7\end{matrix}\right]\]

one way to interpret matrix multiplication is the 1st column of B says that the 1st column in the product matrix will be a linear combination of 1st col of A scaled by 1 and the 2nd col of A scaled by 0. i.e.
\[\left[\begin{matrix}3 \\ 1 \end{matrix}\right]\]

now, because the 2nd col of A is a scaled version (twice) of the 1st column we could say: use the 2nd col of A and subtract the 1st col of A . This has the effect of 2*1st col - 1*1st col (because 2nd col = 2*1st col) = 1st col A (which is what we want)
so one of many answers is
\[C=\left[\begin{matrix}-1 & 5 \\ 1 & 1\end{matrix}\right]\]

Answer 5

Ahh, clever. I get you. Not really technical maths, just realising that 6-3 is the same as 3-0. What other versions of C could there be? Surely thats the only one that works with both 3 and 6 and 1 and 2

Answer 6

if we call the 1st col v the 2nd col is 2v
what combinations of
av+2bv= v ?
where a and b are the entries in the 1st col of matrix C
solving: a+2b=1
a= 1-2b

there are an infinite number of choices: if b =0 a=1,
if b=1, a= -1, if b=1/2, a=0...

Answer 7

Yes, I understand that you can get infinite for 3 and 6 or 1 and 2 but not both surely.

Answer 8

If I am not clear: matrix A has column v and column 2v
we want the answer matrix C to have v as its first column

Answer 9

You mean AC to have v as it's first column

Answer 10

Give me an example of C other than the one you gave where it would give the same matrix as AB

Answer 11

yes, I mean AC

if you are asking about the 2nd column of AC, yes there are infinite combinations for it also

Answer 12

No, I'm saying that there are not an infinite number of combinations as you have 6 and 3 and 1 and 2

Answer 13

So having -3 and 4 in the first column of C would give 3 as the top of the first column in AC but not 1 in the bottom of it.

Answer 14

\[\left[\begin{matrix}21 & 5 \\ -10 &1\end{matrix}\right]\]

Answer 15

it would be -7 4 for the 1st col of C

Answer 16

Fair enough, so explain to me the first column of C.

Answer 17

Does the top have to be the negative double plus or minus 1?

Answer 18

If the bottom is negative, then plus 1

Answer 19

If the top is negative then minus 1

Answer 20

\[\left[\begin{matrix}1-2b & 7-2c\\ b & c\end{matrix}\right]\]

Answer 21

But the second column doesn't matter as that is already non zero.

Answer 22

just being complete

Answer 23

Would this be right for the first column?
\[\left(\begin{matrix}-a \\-2(a)-1\end{matrix}\right)\]

Answer 24

Or \[\left(\begin{matrix}-2(-b)+1 \\-b\end{matrix}\right)\]

Answer 25

\[\left(\begin{matrix}2b+ \\ -b\end{matrix}\right)\]

Answer 26

\[\left(\begin{matrix}2b+1 \\ -b\end{matrix}\right)\]

Answer 27

how are getting these? no.

Answer 28

If b is 10?

Answer 29

21 and -10

Answer 30

\[\left(\begin{matrix}-a \\ 2a-1\end{matrix}\right)\]

Answer 31

Or in that instance if a is 1 then -1 and 1

Answer 32

Both examples you gave.

Answer 33

for your last -a 2a-1 if a=0 you get 0 -1 and that as your 1st col gives -6 -2 not 3 1

Answer 34

If the negative is on top then it's multiplying the 3 so the bottom one has to be the negative double minus 1. If the negative is on the bottom it's multiplying the 6 so the bottom has to be the negative double plus 1.

Answer 35

Only cause a is 0

Answer 36

In every other instance it works doesn't it?

Answer 37

-1 1, -2 3, -3 5,

Answer 38

think about it some more....I am not leading you astray (though you seem to want to wander off into the woods on this)....

Answer 39

I'm not saying you are, but my explanation makes more sense to me.

Answer 40

If the 6 is negative, you need to add one more than double the number of 3s to get it to three.

Answer 41

Oh I don't care any more. Thank you.

Answer 42

Thank you for answering, not for making me not care anymore :P