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Here is my problem:

Suppose $A = \begin{bmatrix}3&2\\5&4\end{bmatrix}$ and $C = \begin{bmatrix}2&0\\1&6\end{bmatrix}$

Find a matrix $B$ such that $AB=C$ or prove that no such matrix exists. Explain your answer.

In order to do this, I am going to need to find the inverse of at least one of the matrices. Will I need to find both? Help?

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  • $\begingroup$ Sigh. You write matrices a and c but then ask about A and C! You want to find B such that AB= C? Yes, there exists a unique B if and only if A is invertible and then $B= A^{-1}C$ $\endgroup$ – user247327 Dec 1 '16 at 16:28
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$AB=C$ therefore $B=A^{-1}C$ if $A$ is invertible

$det(A)=3(4)-5(2)=2$

Therefore $A^{-1}=\begin{bmatrix}2&-1\\-2.5&1.5\end{bmatrix}$

$B=$ $\begin{bmatrix}2&-1\\-2.5&1.5\end{bmatrix}$$\begin{bmatrix}2&0\\1&6\end{bmatrix}$

$$B=\begin{bmatrix}3&-6\\-3.5&9\end{bmatrix}$$

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  • $\begingroup$ Awesome! Thanks to everyone who posted a response. Some of this matrix stuff really confuses me. $\endgroup$ – cparks10 Dec 2 '16 at 16:29
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Hint

Note that $A$ is invertible because its determinant is $\det(A)=2$. So:

$$ AB=C \Rightarrow A^{-1}AB=A^{-1}C \Rightarrow B=A^{-1}C $$

and you need $A^{-1}$.

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While this concrete problem can be solve by inverting $A$, having $A$ invertible is not a necessary condition for the existence of $B$ solving $AB=C$. What that equation does require is that the column span of $C$ is a subspace of the column span of $A$ (since every column of $AB$ will be in the column space of $A$, regardless of what $B$ is). This also leads to the following method to find $B$ provided the condition is met. For each column of $C$, call it $c$, try to solve the equation $Ax=c$. If this fails it proves the column span of $C$ is not contained in the image (column span) of $A$, while if it succeeds, you can use a solution for$~x$ as the corresponding column of$~B$. Repeat for all columns of $C$.

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