# Finding a Matrix to make an equation work

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?

• 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 $B= A^{-1}C$ – user247327 Dec 1 '16 at 16:28

$AB=C$ 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}$$• Awesome! Thanks to everyone who posted a response. Some of this matrix stuff really confuses me. – cparks10 Dec 2 '16 at 16:29 Hint Note that A is invertible because its determinant is \det(A)=2. So:$$ AB=C \Rightarrow A^{-1}AB=A^{-1}C \Rightarrow B=A^{-1}C$$and you need$A^{-1}$. 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\$.