Matrix multiplication

HelpNeeder

New member
If A, B, C and D are arbitrary matrices (with the correct amount of rown and columns to be multiplied)

and if

A * B = C

can

A * D = C ?

lex

Junior Member
There exist matrices A, B, C, D: B ≠ D and AB = C = AD
(if that's what you mean).

All you need is that $$\displaystyle \text{A}\boldsymbol{x}=\boldsymbol{c_j}$$ for some column $$\displaystyle \boldsymbol{c_j}$$ of C, to have non-unique solutions for $$\displaystyle \boldsymbol{x}$$.

E.g.
$$\displaystyle \left(\begin{matrix}2&1&3\\0&1&2\\\end{matrix}\right)\left(\begin{matrix}1\\2\\3\\\end{matrix}\right)=\left(\begin{matrix}13\\8\\\end{matrix}\right)$$
$$\displaystyle \left(\begin{matrix}2&1&3\\0&1&2\\\end{matrix}\right)\left(\begin{matrix}2\\6\\1\\\end{matrix}\right)=\left(\begin{matrix}13\\8\\\end{matrix}\right)$$

HelpNeeder

New member
Thank you! The thing is that I do not understand why in this case the solution is unique:

Here is the proof:

Becayse if I multiply A-1 to another b, then I can still have the same u?

HelpNeeder

New member
And the statement
If there is a b in Rn such that the equation Ax = b is consistent, then the solution is unique.
is false.
So I do not understand why they are talking about uniqueness here.

HelpNeeder

New member
And the statement
If the equation Ax = b has at least one solution for each b in Rn, then the solution is unique for each b.
is true...

lex

Junior Member
Thank you! The thing is that I do not understand why in this case the solution is unique:
View attachment 26285

Here is the proof:
View attachment 26286
Becayse if I multiply A-1 to another b, then I can still have the same u?
Can there exist matrices A, B, C, D:
B ≠ D and AB = C and AD = C
As I showed, such things do exist.
However it does not mean that for all matrices A, C (of correct sizes), there always exist B ≠ D: AB = C and AD = C

In particular, if A is invertible: then for matrix C, there is a unique matrix B: AB = C (namely A-1C)
(and so there is a unique solution to Ax=b (namely A-1b)).

HelpNeeder

New member
Thank you! But how can we be sure that A-1C is unique?

lex

Junior Member
Thank you! But how can we be sure that A-1C is unique?
AB = C $$\displaystyle \quad$$(1)
As A is invertible, $$\displaystyle \text{A}^{-1}$$ exists.
You can then left multiply both sides of equation (1) by $$\displaystyle \text{A}^{-1}$$
$$\displaystyle \text{A}^{-1}\text{AB} = \text{A}^{-1} \text{C}$$
$$\displaystyle \text{IB}$$ = $$\displaystyle \text{A}^{-1}\text{C}$$
$$\displaystyle \text{B=}\text{A}^{-1}\text{C}$$

$$\displaystyle \text{A}^{-1}\text{C}$$ is the product of two matrices. You only get one unique answer when you multiply two matrices together.

Thank you!

Jomo

Elite Member
Thank you! But how can we be sure that A-1C is unique?
I know that Lex has already told you why A-1C is unique but I have to repeat the answer. Whenever you multiply two matrices of proper sizes you only get one answer!!!!