MHB How should this matrix be multiplied

[mathlearn]

$A=\begin{bmatrix}
3&2\\
\end{bmatrix} B=\begin{bmatrix}
1\\
2\end{bmatrix}$

Find the value of the matrix $AB$.

The order of the first matrix is 1*2

The order of the second matrix is 2*1

Matrix AB should be 1*1

I am a bit struggling in determining the way that these two matrices should be multiplied

Many thanks :)

 

[mathmari]

You are correct about the order of the matrices! (Yes)

We have the following:
$$AB=\begin{bmatrix}
3&2\\
\end{bmatrix} \begin{bmatrix}
1\\
2\end{bmatrix}=\begin{bmatrix}
c_{11}
\end{bmatrix}$$

To calculate the element $c_{11}$ we have to multiply the $1$st row of $A$ with the $1$st column of $B$, as an inner product. So we get the following:
$$c_{11}=3\cdot 1+2\cdot 2=3+4=7$$ So, the result is $$AB=\begin{bmatrix}
7
\end{bmatrix}$$

 

[mathlearn]

You are correct about the order of the matrices! (Yes)

We have the following:
$$AB=\begin{bmatrix}
3&2\\
\end{bmatrix} \begin{bmatrix}
1\\
2\end{bmatrix}=\begin{bmatrix}
c_{11}
\end{bmatrix}$$

To calculate the element $c_{11}$ we have to multiply the $1$st row of $A$ with the $1$st column of $B$, as an inner product. So we get the following:
$$c_{11}=3\cdot 1+2\cdot 2=3+4=7$$ So, the result is $$AB=\begin{bmatrix}
7
\end{bmatrix}$$

Thanks :) I can see it in this problem now

But how do we know which row are to be multiplied with what column ?

 

[mathmari]

But how do we know which row are to be multiplied with what column ?

Let $A=\begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{pmatrix}$ and $B=\begin{pmatrix}
1 & 2 \\
3 & 4 \\
5 & 6
\end{pmatrix}$.

We have the following:
$$AB=\begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{pmatrix}\begin{pmatrix}
1 & 2 \\
3 & 4 \\
5 & 6
\end{pmatrix}=\begin{pmatrix}
c_{11} & c_{12} \\
c_{21} & c_{22} \\
c_{31} & c_{32}
\end{pmatrix}$$ To calculate the element $c_{ij}$ we have to multiply the $i$-th row of $A$ with the $j$-th column of $B$, as an inner product.

So, we get the following elements:
$$c_{11}=1\cdot 1+2\cdot 3+3\cdot 5=1+6+15=22 \\
c_{12}=1\cdot 2+2\cdot 4+3\cdot 6=2+8+18=28 \\
c_{21}=4\cdot 1+5\cdot 3+6\cdot 5=4+15+30=49 \\
c_{22}=4\cdot 2+5\cdot 4+6\cdot 6=8+20+36=64 \\
c_{31}=7\cdot 1+8\cdot 3+9\cdot 5=7+24+45=76 \\
c_{32}=7\cdot 2+8\cdot 4+9\cdot 6=14+32+54=100$$ Therefore, the result is the following:
$$AB=\begin{pmatrix}
22 & 28 \\
49 & 64 \\
76 & 100
\end{pmatrix}$$

 

[HallsofIvy]

Thanks :) I can see it in this problem now

But how do we know which row are to be multiplied with what column ?

To multiply AB, all rows of A are multiplied with all columns of B.