# 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!!!!