Question

How many 3x3 matrices 'A' can you find such that
x x+y
A y = x-y
z 0
for all choices of x, y and z?

Answer 1

\[A \left(\begin{matrix}x \\ y\ \\z\end{matrix}\right) =\left(\begin{matrix}x+y \\ x-y\\0\end{matrix}\right)\]

Answer 2

just let a11, a12, a13..... as entries of A
then you have a11x +a12 y +a13 z = x +y

Answer 3

I can only think of one so far

\[\large \left(\begin{matrix} 1&1&0\\1&-1&0\\0&0&0\end{matrix} \right) \left(\begin{matrix}x \\ y\ \\z\end{matrix}\right) =\left(\begin{matrix}x+y \\ x-y\\0\end{matrix}\right)\]

Answer 4

make a comparison to get a11 = 1, a12 = 1 a13 =0

Answer 5

@rational , your answer is correct, but how did you arrive to that.
@Loser66 i tried, but there's too many unknowns

Answer 6

@rational the last line can't be 0, it makes A is not invertible.

Answer 7

it has to be 0 for arbitrary x y z

Answer 8

I took linear combinations of rows of (x,y,z) since the matrix A was left multiplied to the vector (x,y,z) @mickey4691

Answer 9

ok, let me show my work
for the last one, we have a31 x +a32y +a33z =0, one of a can be 0 and others is 1 and -1
the permutation of it is 6, so that we have 6 matrices

Answer 10

but you are assuming that one of them is zero which would defy the purpose of proving? @Loser66
@rational what do you mean by linear combinations?

Answer 11

that won't result in 0 for z component right ? because a31 = 0, a32=1,a33=-1 gives y-z, not 0

Answer 12

left multiplying a vector by a matrix is same as taking linear combinations of rows of vector

Answer 13

OH yeah!!!! My bad. I am sooooo sorry.

Answer 14

\[\large \left(\begin{matrix} a&b&c \end{matrix} \right) \left(\begin{matrix}x \\ y\ \\z\end{matrix}\right) =\left(\begin{matrix}ax +by+cz\end{matrix}\right)\]

Answer 15

we can interpret this multiplication as taking linear combinations of "rows" of second matrix:

take "a" of first row
take "b" of second row
take "c" of third row

and add them

Answer 16

So if you want \(ax-by\) as result, you would do :

\[\large \left(\begin{matrix} a&-b&0 \end{matrix} \right) \left(\begin{matrix}x \\ y\ \\z\end{matrix}\right) =\left(\begin{matrix}ax -by\end{matrix}\right)\]

right ?

Answer 17

I see what you're doing there but this method is more of a inspection and comparison? correct?

Answer 18

nope,
this is one of the four ways of multiplying two matrices

Answer 19

and the result of matrix multiplication is unique
so we won't be missing out any solutions i hope

Answer 20

hmm, i will attach a photo of my working. Give me a few. Thanks. I understand

Answer 21

Okay...

Answer 22

yup

Answer 23

Looks good !

Answer 24

Thank you @rational and @Loser66

Answer 25

yw :)