What are matrices? A matrix is a group of numbers arranged in rows and columns. For example, below is a matrix with $2$ rows and $3$ columns:
$$\boldsymbol{A}=\begin{pmatrix}
4&6&5\\
5&9&1\\
\end{pmatrix}$$
The general convention is to use bold uppercase letters such as $\boldsymbol{A}$ to denote a matrix. We refer to a matrix with $2$ rows and $3$ columns as a $2\times3$ matrix where $\times$ is read as "by" . To emphasize that the entries are real numbers, we also sometimes denote the shape like $\mathbb{R}^{2\times3}$ .
Some textbooks use square brackets instead of circular brackets:
$$\boldsymbol{A}=\begin{bmatrix}
4&6&5\\
5&9&1\\
\end{bmatrix}$$
We use the following notation to represent a generic $m\times{n}$ matrix:
$$\boldsymbol{A}=\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix}$$
The convention is to denote each number in the matrix using a non-bold lowercase letter with subscripts. For instance, $a_{12}$ represents the number in the first row second column. We typically match the notation used for the matrix and the numbers it contains - for instance, the numbers in matrix $\boldsymbol{B}$ are denoted as $b$ .
Relationship between vectors and matrices Recall that vectors
$$\boldsymbol{v}=
\begin{pmatrix}
3\\5
\end{pmatrix}$$
This can be treated as a matrix with $2$ rows and $1$ column. In fact, matrices can be thought of as a generalization of vectors that allow for multiple columns!
Special matrices
Definition.
Square matrix
A square matrix contains the same number of rows and columns. Below is an example of a $3\times3$ square matrix:
$$\begin{pmatrix}
4&1&5\\8&1&2\\2&1&2
\end{pmatrix}$$
Definition.
Zero matrix
The zero matrix contains all zeros. Below is an example of a $2\times3$ zero matrix:
$$\begin{pmatrix}
0&0&0\\0&0&0
\end{pmatrix}$$
Definition.
Identity matrix
The identity matrix is a square matrix denoted by $\boldsymbol{I}$ and has the main diagonals filled with $1$ s and others filled with $0$ s. Below is an example of a $3\times3$ identity matrix:
$$\boldsymbol{I}_3=\begin{pmatrix}
1&0&0\\0&1&0\\0&0&1
\end{pmatrix}$$
As we did here, we sometimes use the notation $\boldsymbol{I}_n$ to denote an $n\times{n}$ identity matrix. We will later explore how the identity matrix plays a critical role in linear algebra.
Matrix addition and subtraction Suppose we have two matrices of the same shape $\boldsymbol{A}$ and $\boldsymbol{B}$ . To compute $\boldsymbol{A}+\boldsymbol{B}$ , we add the corresponding pair of numbers in the matrices. For instance, consider the following matrices:
$$\boldsymbol{A}=\begin{pmatrix}
3&6&5\\
5&9&1\\
\end{pmatrix},\;\;\;\;\;
\boldsymbol{B}=\begin{pmatrix}
0&2&1\\
1&3&2\\
\end{pmatrix}$$
Their sum is:
$$\begin{align*}
\boldsymbol{A}+\boldsymbol{B}&=\begin{pmatrix}
3&6&5\\
5&9&1\\
\end{pmatrix}+\begin{pmatrix}
0&2&1\\
1&3&2\\
\end{pmatrix}\\&=
\begin{pmatrix}
3+0&6+2&5+1\\
5+1&9+3&1+2\\
\end{pmatrix}\\
&=
\begin{pmatrix}
3&8&6\\
6&12&3\\
\end{pmatrix}
\end{align*}$$
Similarly, to compute $\boldsymbol{A}-\boldsymbol{B}$ , we subtract the numbers in $\boldsymbol{B}$ from the corresponding numbers in $\boldsymbol{A}$ like so:
$$\begin{align*}
\boldsymbol{A}-\boldsymbol{B}&=\begin{pmatrix}
3&6&5\\
5&9&1\\
\end{pmatrix}-\begin{pmatrix}
0&2&1\\
1&3&2\\
\end{pmatrix}\\&=
\begin{pmatrix}
3-0&6-2&5-1\\
5-1&9-3&1-2\\
\end{pmatrix}\\
&=
\begin{pmatrix}
3&4&4\\
4&6&-1\\
\end{pmatrix}
\end{align*}$$
Note that matrix addition and subtraction are only defined for the case when the two matrices have the same number of rows and columns.
Scalar-matrix multiplication Scalar-matrix multiplication works just like scalar-vector multiplication. For instance, multiplying a scalar $k$ to a $2\times3$ matrix:
$$k
\begin{pmatrix}
a_{11}&a_{12}&a_{13}\\
a_{21}&a_{22}&a_{23}\\
\end{pmatrix}=
\begin{pmatrix}
ka_{11}&ka_{12}&ka_{13}\\
ka_{21}&ka_{22}&ka_{23}\\
\end{pmatrix}$$
Here's a more concrete example:
$$3
\begin{pmatrix}
1&4&2\\
0&2&1\\
\end{pmatrix}=
\begin{pmatrix}
3\times1&3\times4&3\times2\\
3\times0&3\times2&3\times1\\
\end{pmatrix}
=
\begin{pmatrix}
3&12&6\\
0&6&3\\
\end{pmatrix}$$
Matrix-matrix multiplication Unfortunately, matrix multiplication is not as straightforward as matrix addition. Let's go through an example - consider the following matrices:
$$\boldsymbol{A}=\begin{pmatrix}
3&6&5\\
5&0&1\\
\end{pmatrix},\;\;\;\;\;
\boldsymbol{B}=
\begin{pmatrix}2&4&1&0\\1&3&3&1\\0&5&1&2\\\end{pmatrix}$$
Matrix $\boldsymbol{A}$ is a $2\times3$ while matrix $\boldsymbol{B}$ is $3\times4$ . The matrix product $\boldsymbol{AB}$ is another matrix whose shape is determined by the following rule:
$$\large{\underset{({\color{green}2}\times{\color{red}3})}{\boldsymbol{A}}\;\;
\underset{({\color{red}3}\times{\color{blue}4})}{\boldsymbol{B}}\;\;
=\;\;
\underset{{(\color{green}2}\times{\color{blue}4})}{\boldsymbol{AB}}}$$
In words, matrix $\boldsymbol{AB}$ contains:
Note that the number of columns and the number of rows of $\boldsymbol{A}$ and $\boldsymbol{B}$ must match - otherwise matrix multiplication is not defined! We will show why this rule always holds later, but for now, please accept that $\boldsymbol{AB}$ will be a $2\times4$ matrix in this case.
WARNING
We write the product of matrices $\boldsymbol{A}$ and $\boldsymbol{B}$ as $\boldsymbol{AB}$ instead of $\boldsymbol{A}\times\boldsymbol{B}$ . The notation $\boldsymbol{A}\times\boldsymbol{B}$ is not correct and should be avoided.
For notational convenience, let's define $\boldsymbol{C}=\boldsymbol{AB}$ . Our goal now is to compute the entries of $\boldsymbol{C}$ below:
$$\boldsymbol{C}=\begin{pmatrix}
c_{11}&c_{12}&c_{13}&c_{14}\\
c_{21}&c_{22}&c_{23}&c_{24}\\
\end{pmatrix}$$
To compute $c_{11}$ , we take the dot product $1$ st row of $\boldsymbol{A}$ and $1$ st column of $\boldsymbol{B}$ below:
$$\boldsymbol{AB}=\begin{pmatrix}
\color{green}3&\color{green}6&\color{green}5\\5&0&1\\\end{pmatrix}
\begin{pmatrix}\color{green}2&4&1&0\\\color{green}1&3&3&1\\\color{green}0&5&1&2\\\end{pmatrix}$$
Therefore, $c_{11}$ is:
$$\begin{align*}
c_{11}
&=(3)(2)+(6)(1)+(5)(0)\\
&=12
\end{align*}$$
Next, to compute $c_{12}$ , we take the dot product of the $1$ st row of $\boldsymbol{A}$ and the $2$ nd column of $\boldsymbol{B}$ below:
$$\boldsymbol{AB}=\begin{pmatrix}
\color{green}3&\color{green}6&\color{green}5\\5&0&1\\\end{pmatrix}
\begin{pmatrix}2&\color{green}4&1&0\\1&\color{green}3&3&1\\0&\color{green}5&1&2\\\end{pmatrix}$$
Therefore, $c_{12}$ is:
$$\begin{align*}
c_{12}
&=(3)(4)+(6)(3)+(5)(5)\\
&=55
\end{align*}$$
Can you see how there is a pattern in how the entries of $\boldsymbol{AB}$ are computed? $c_{13}$ is computed by taking the dot product of the $1$ st row of $\boldsymbol{A}$ and the $3$ rd column of $\boldsymbol{B}$ .
Let's now move on to the entries of the $2$ nd row of $\boldsymbol{AB}$ . To compute $c_{21}$ , we take the dot product of the $2$ nd row of $\boldsymbol{A}$ and the $1$ st column of $\boldsymbol{B}$ below:
$$\boldsymbol{AB}=\begin{pmatrix}
3&6&5\\\color{green}5&\color{green}0&\color{green}1\\\end{pmatrix}
\begin{pmatrix}\color{green}2&4&1&0\\\color{green}1&3&3&1\\\color{green}0&5&1&2\\\end{pmatrix}$$
Therefore, $c_{21}$ is:
$$\begin{align*}
c_{21}
&=(5)(2)+(0)(1)+(1)(0)\\
&=10
\end{align*}$$
To compute $c_{22}$ , we take the dot product of the $2$ nd row of $\boldsymbol{A}$ and the $2$ nd column of $\boldsymbol{B}$ below:
$$\boldsymbol{AB}=\begin{pmatrix}
3&6&5\\\color{green}5&\color{green}0&\color{green}1\\\end{pmatrix}
\begin{pmatrix}2&\color{green}4&1&0\\1&\color{green}3&3&1\\0&\color{green}5&1&2\\\end{pmatrix}$$
Therefore, $c_{22}$ is:
$$\begin{align*}
c_{22}
&=(5)(4)+(0)(3)+(1)(5)\\
&=25
\end{align*}$$
As you would expect, we can compute $c_{23}$ by taking the dot product of the $2$ nd row of $\boldsymbol{A}$ and the $3$ rd column of $\boldsymbol{B}$ . To generalize, the entry $c_{ij}$ is computed by taking the dot product of the $i$ -th row of $\boldsymbol{A}$ and the $j$ -th column of $\boldsymbol{B}$ . In this way, finding the entries of the matrix product involves taking the dot product of the rows of the first matrix and the columns of the second matrix!
Example.
Computing matrix product of 2x2 matrices
Let's go through a simpler example - consider the following matrices:
$$\boldsymbol{A}=\begin{pmatrix}
3&2\\
1&4\\
\end{pmatrix},\;\;\;\;\;
\boldsymbol{B}=
\begin{pmatrix}5&0\\6&7\end{pmatrix}$$
Solution . Firstly, the shape of $\boldsymbol{AB}$ is:
$$\large{\underset{({\color{green}2}\times{\color{red}2})}{\boldsymbol{A}}\;\;
\underset{({\color{red}2}\times{\color{blue}2})}{\boldsymbol{B}}\;\;
=\;\;
\underset{{(\color{green}2}\times{\color{blue}2})}{\boldsymbol{AB}}}$$
Note that the product $\boldsymbol{AB}$ is defined because the number of columns of $\boldsymbol{A}$ and the number of rows of $\boldsymbol{B}$ match. Let's now compute the entries of $\boldsymbol{AB}$ by taking the dot products of the rows of $\boldsymbol{A}$ and the columns of $\boldsymbol{B}$ like so:
$$\begin{align*}
\boldsymbol{AB}&=\begin{pmatrix}
\color{green}3&\color{green}2\\\color{green}1&\color{green}4\\\end{pmatrix}
\begin{pmatrix}\color{red}5&\color{red}0\\\color{red}6&\color{red}7\end{pmatrix}\\
&=\begin{pmatrix}
{\color{green}(3)}{\color{red}(5)}+{\color{green}(2)}{\color{red}(6)}&{\color{green}(3){\color{red}(0)}}+{\color{green}(2)}{\color{red}(7)}\\
{\color{green}(1)}{\color{red}(5)}+{\color{green}(4)}{\color{red}(6)}&{\color{green}(1){\color{red}(0)}}+{\color{green}(4)}{\color{red}(7)}
\end{pmatrix}\\
&=\begin{pmatrix}
27&14\\
29&28
\end{pmatrix}
\end{align*}$$
Theorem.
Deducing the resulting shape of a matrix product
If $\boldsymbol{A}$ is an $m\times{n}$ matrix and $\boldsymbol{B}$ is an $n\times{r}$ matrix, then the shape of their product is $m\times{r}$ , that is:
$$\large{\underset{({\color{green}m}\times{\color{red}n})}{\boldsymbol{A}}\;\;
\underset{({\color{red}n}\times{\color{blue}r})}{\boldsymbol{B}}\;\;
=\;\;
\underset{{(\color{green}m}\times{\color{blue}r})}{\boldsymbol{AB}}}$$
Note that if the number of columns of $\boldsymbol{A}$ and the number of rows of $\boldsymbol{B}$ do not match, then the product is not defined.
Proof . Suppose $\boldsymbol{A}$ is a $2\times3$ matrix represented below:
$$\boldsymbol{A}=\begin{pmatrix}
*&*&*\\
*&*&*\end{pmatrix}$$
Suppose we have another matrix $\boldsymbol{B}$ and we take product $\boldsymbol{AB}$ . What must the shape of $\boldsymbol{B}$ be for the matrix multiplication to work? Let's consider the following case:
$$\boldsymbol{AB}=\begin{pmatrix}
*&*&*\\
*&*&*\end{pmatrix}
\begin{pmatrix}
\bullet&\bullet\\
\bullet&\bullet\\
\bullet&\bullet\\
\bullet&\bullet\\
\end{pmatrix}$$
Does this matrix multiplication work? Recall that the top-left entry of $\boldsymbol{AB}$ is computed by taking the dot product of the first row of $\boldsymbol{A}$ and the first column of $\boldsymbol{B}$ like so:
$$\boldsymbol{AB}=\begin{pmatrix}
\color{green}*&\color{green}*&\color{green}*\\
*&*&*\end{pmatrix}
\begin{pmatrix}
\color{green}\bullet&\bullet\\
\color{green}\bullet&\bullet\\
\color{green}\bullet&\bullet\\
\color{green}\bullet&\bullet\\
\end{pmatrix}$$
Here, the dot product is not defined because the first row of $\boldsymbol{A}$ contains $3$ numbers whereas the first column of $\boldsymbol{B}$ contains $4$ numbers. The only way for the dot product to work is if the number of rows of $\boldsymbol{B}$ is the same as the number of columns of $\boldsymbol{A}$ like so:
$$\begin{equation}\label{eq:iL6ajgdqb1tLrVa09rS}
\boldsymbol{AB}=\begin{pmatrix}
\color{green}*&\color{green}*&\color{green}*\\
*&*&*\end{pmatrix}
\begin{pmatrix}
\color{green}\bullet&\bullet\\
\color{green}\bullet&\bullet\\
\color{green}\bullet&\bullet\\
\end{pmatrix}
\end{equation}$$
Now that we know how many rows $\boldsymbol{B}$ must have, how about the number of columns? Recall that the entry $c_{ij}$ of $\boldsymbol{C}=\boldsymbol{AB}$ is computed by taking the dot product of the $i$ -th row of $\boldsymbol{A}$ and the $j$ -th column of $\boldsymbol{B}$ . This means that in the case of \eqref{eq:iL6ajgdqb1tLrVa09rS} , because $\boldsymbol{A}$ has $2$ rows and $\boldsymbol{B}$ has $2$ columns, the product $\boldsymbol{C}$ would take on the following shape:
$$\begin{pmatrix}
c_{11}&c_{12}\\
c_{21}&c_{22}\\
\end{pmatrix}$$
What if the shape of $\boldsymbol{B}$ was as follows:
$$\boldsymbol{C}=\boldsymbol{AB}=\begin{pmatrix}
*&*&*\\
*&*&*\end{pmatrix}
\begin{pmatrix}
\bullet&\bullet&\bullet\\
\bullet&\bullet&\bullet\\
\bullet&\bullet&\bullet\\
\end{pmatrix}$$
Since $\boldsymbol{A}$ has $2$ rows and $\boldsymbol{B}$ has $3$ columns, $\boldsymbol{C}$ would also have $2$ rows and $3$ columns:
$$\begin{pmatrix}
c_{11}&c_{12}&c_{13}\\
c_{21}&c_{22}&c_{23}\\
\end{pmatrix}$$
In general, if $\boldsymbol{A}$ has $m$ rows and $\boldsymbol{B}$ has $r$ columns, then $\boldsymbol{C}$ must have $m$ rows and $r$ columns. The way to remember this rule is to first write down the shapes of the two matrices $\boldsymbol{A}$ and $\boldsymbol{B}$ like so:
$$\large{\underset{({\color{green}m}\times{\color{red}n})}{\boldsymbol{A}}\;\;
\underset{({\color{red}n}\times{\color{blue}r})}{\boldsymbol{B}}\;\;
=\;\;
\underset{{(\color{green}m}\times{\color{blue}r})}{\boldsymbol{AB}}}$$
If the inner numbers ($\color{red}n$ in this case) are the same, then the product is defined - otherwise, the product is not defined. The shape of the resulting matrix is equal to the outer numbers ($\color{green}m$ and $\color{blue}r$ in this case).
Example.
Shape of a matrix-vector product
The shape of a matrix-vector product is:
$$\large{\underset{({\color{green}m}\times{\color{red}n})}{\boldsymbol{A}}\;\;
\underset{({\color{red}n}\times{\color{blue}1})}{\boldsymbol{v}}\;\;
=\;\;
\underset{{(\color{green}m}\times{\color{blue}1})}{\boldsymbol{Av}}}$$
This means that a matrix-vector product results in another vector!
Theorem.
Suppose we have an $m\times{n}$ matrix $\boldsymbol{A}$ and $n\times{r}$ matrix $\boldsymbol{B}$ . Each entry of the matrix product $\boldsymbol{AB}$ can be computed like so:
$$\big[\boldsymbol{AB}\big]_{ij}=
\sum_{t=1}^{n}a_{it}b_{tj}$$
Here, $[\boldsymbol{AB}]_{ij}$ represents the value of the $i$ -th row $j$ -th column in $\boldsymbol{AB}$ .
Proof . Suppose we have the following matrices:
$$\boldsymbol{A}=\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\smash\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix},\;\;\;\;\;\;
\boldsymbol{B}=
\begin{pmatrix}
b_{11}&b_{12}&\cdots&b_{1r}\\
b_{21}&b_{22}&\cdots&b_{2r}\\
\vdots&\vdots&\smash\ddots&\vdots\\
b_{n1}&b_{n2}&\cdots&b_{nr}
\end{pmatrix}$$
By the rule of matrix multiplication, the entry at the 1st row 1st column is:
$$\begin{align*}
[\boldsymbol{AB}]_{11}&=
a_{11}b_{11}+
a_{12}b_{21}+
\cdots+
a_{1n}b_{n1}
\\&=
\sum^n_{t=1}a_{1t}b_{t1}
\end{align*}$$
Similarly, the entry at the 1st row 2nd column is:
$$\begin{align*}
[\boldsymbol{AB}]_{12}&=
a_{11}b_{12}+
a_{12}b_{22}+
\cdots+
a_{1n}b_{n2}
\\&=
\sum^n_{t=1}a_{1t}b_{t2}
\end{align*}$$
Similarly, the entry at the 2nd row 1st column is:
$$\begin{align*}
[\boldsymbol{AB}]_{21}&=
a_{21}b_{11}+
a_{22}b_{21}+
\cdots+
a_{2n}b_{n1}
\\&=
\sum^n_{t=1}a_{2t}b_{t1}
\end{align*}$$
Therefore, the general formula for matrix multiplication is:
$$\big[\boldsymbol{AB}\big]_{ij}=
\sum_{t=1}^{n}a_{it}b_{tj}$$
This completes the proof.
Theorem.
Matrix multiplication is not commutative
In general, matrix multiplication is not commutative, that is:
$$\boldsymbol{AB}\ne
\boldsymbol{BA}$$
Where $\boldsymbol{A}$ and $\boldsymbol{B}$ are matrices.
Example . Consider the following matrices:
$$\boldsymbol{A}=\begin{pmatrix}
3&1\\
0&2\\
\end{pmatrix},\;\;\;\;\;
\boldsymbol{B}=\begin{pmatrix}
4&6\\
5&7\\
\end{pmatrix}$$
The product $\boldsymbol{AB}$ and $\boldsymbol{BA}$ are:
$$\boldsymbol{AB}=\begin{pmatrix}
17&25\\
10&14\\
\end{pmatrix},\;\;\;\;\;
\boldsymbol{BA}=\begin{pmatrix}
12&16\\
15&19\\
\end{pmatrix}$$
Notice that $\boldsymbol{AB}\ne{\boldsymbol{BA}}$ . Therefore, unlike scalar multiplication, the ordering of the multiplication is important!
Theorem.
Multiplying a matrix with an identity matrix
One of the key properties of identity matrices $\boldsymbol{I}$ is that multiplying them with another matrix $\boldsymbol{A}$ will yield the matrix itself, that is:
$$\boldsymbol{IA}=\boldsymbol{AI}=\boldsymbol{A}$$
Proof . Consider the $m\times{n}$ matrix $\boldsymbol{A}$ and $n\times{n}$ identity matrix $\boldsymbol{I}_n$ below:
$$\boldsymbol{A}=\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix},\;\;\;\;\;\;
\boldsymbol{I}_n=
\begin{pmatrix}
1&0&\cdots&0\\
0&1&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&1\\
\end{pmatrix}$$
The product $\boldsymbol{A}\boldsymbol{I}_n$ is:
$$\begin{align*}
\boldsymbol{A}\boldsymbol{I}_n
&=\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix}
\begin{pmatrix}
1&0&\cdots&0\\
0&1&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&1\\
\end{pmatrix}\\
&=\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix}\\
&=\boldsymbol{A}
\end{align*}$$
Similarly, computing the product $\boldsymbol{I}_n\boldsymbol{A}$ will yield $\boldsymbol{A}$ . This completes the proof.
Example.
Computing the product of a matrix and an identity matrix
Compute the following product:
$$\begin{pmatrix}
3&1&5\\8&1&2\\2&1&2
\end{pmatrix}
\begin{pmatrix}
1&0&0\\0&1&0\\0&0&1
\end{pmatrix}$$
Solution . Notice how the right matrix is the identity matrix $\boldsymbol{I}_3$ . The product will therefore return the left matrix:
$$\begin{pmatrix}
3&1&5\\8&1&2\\2&1&2
\end{pmatrix}
\begin{pmatrix}
1&0&0\\0&1&0\\0&0&1
\end{pmatrix}=
\begin{pmatrix}
3&1&5\\8&1&2\\2&1&2
\end{pmatrix}$$
Theorem.
Matrix product as a column operation
Let $\boldsymbol{A}$ be any $m\times{n}$ matrix and $\boldsymbol{B}$ be any $n\times{r}$ matrix represented below:
$$\boldsymbol{B}=
\begin{pmatrix}
\vert&\vert&\cdots&\vert\\
\boldsymbol{b_1}&\boldsymbol{b_2}&\cdots&\boldsymbol{b_n}\\
\vert&\vert&\cdots&\vert
\end{pmatrix}\\$$
Here, the columns of $\boldsymbol{B}$ are represented as vectors $\boldsymbol{b}_1$ , $\boldsymbol{b}_2$ , $\cdots$ , $\boldsymbol{b}_n$ .
The product $\boldsymbol{AB}$ is:
$$\boldsymbol{AB}=
\begin{pmatrix}
\vert&\vert&\cdots&\vert\\
\boldsymbol{A}\boldsymbol{b}_1&\boldsymbol{A}\boldsymbol{b}_2&\cdots&\boldsymbol{A}\boldsymbol{b}_n\\
\vert&\vert&\cdots&\vert
\end{pmatrix}$$
Note that $\boldsymbol{A}\boldsymbol{b}_1$ is a vector as explained above link
Proof . Let matrices $\boldsymbol{A}$ and $\boldsymbol{B}$ be represented as follows:
$$\boldsymbol{A}=\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\smash\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix},\;\;\;\;\;\;
\boldsymbol{B}=
\begin{pmatrix}
b_{11}&b_{12}&\cdots&b_{1r}\\
b_{21}&b_{22}&\cdots&b_{2r}\\
\vdots&\vdots&\smash\ddots&\vdots\\
b_{n1}&b_{n2}&\cdots&b_{nr}
\end{pmatrix}$$
The first column of the matrix product $\boldsymbol{AB}$ is:
$$\begin{align*}
\begin{pmatrix}
a_{11}b_{11}+a_{12}b_{21}+\cdots+a_{1n}b_{n1}\\
a_{21}b_{11}+a_{22}b_{21}+\cdots+a_{2n}b_{n1}\\
\vdots\\
a_{m1}b_{11}+a_{m2}b_{21}+\cdots+a_{mn}b_{n1}\\
\end{pmatrix}&=
\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}\\
\end{pmatrix}
\begin{pmatrix}
b_{11}\\b_{21}\\\vdots\\b_{n1}
\end{pmatrix}
=\boldsymbol{A}\boldsymbol{b}_1
\end{align*}$$
The second column of matrix product $\boldsymbol{AB}$ is:
$$\begin{align*}
\begin{pmatrix}
a_{11}b_{12}+a_{12}b_{22}+\cdots+a_{1n}b_{n2}\\
a_{21}b_{12}+a_{22}b_{22}+\cdots+a_{2n}b_{n2}\\
\vdots\\
a_{m1}b_{12}+a_{m2}b_{22}+\cdots+a_{mn}b_{n2}\\
\end{pmatrix}&=
\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}\\
\end{pmatrix}
\begin{pmatrix}
b_{12}\\b_{22}\\\vdots\\b_{n2}
\end{pmatrix}
=\boldsymbol{A}\boldsymbol{b}_2
\end{align*}$$
We can therefore infer that the columns of $\boldsymbol{AB}$ are:
$$\boldsymbol{AB}=
\begin{pmatrix}
\vert&\vert&\cdots&\vert\\
\boldsymbol{A}\boldsymbol{b}_1&\boldsymbol{A}\boldsymbol{b}_2&\cdots&\boldsymbol{A}\boldsymbol{b}_n\\
\vert&\vert&\cdots&\vert
\end{pmatrix}$$
This completes the proof.
Example.
Computing matrix product column by column
Consider the following matrices:
$$\boldsymbol{A}=\begin{pmatrix}
4&3\\
1&2
\end{pmatrix},\;\;\;\;\;\;\boldsymbol{B}=
\begin{pmatrix}
0&3\\5&5
\end{pmatrix}$$
Find $\boldsymbol{AB}$ using theorem link
Solution . The first column of $\boldsymbol{AB}$ is:
$$\begin{pmatrix}
4&3\\1&2
\end{pmatrix}
\begin{pmatrix}0\\5\end{pmatrix}
=
\begin{pmatrix}15\\10\end{pmatrix}$$
The second column of $\boldsymbol{AB}$ is:
$$\begin{pmatrix}
4&3\\1&2
\end{pmatrix}
\begin{pmatrix}3\\5\end{pmatrix}
=
\begin{pmatrix}27\\13\end{pmatrix}$$
Therefore, $\boldsymbol{AB}$ is:
$$\boldsymbol{AB}=\begin{pmatrix}
15&27\\
10&13
\end{pmatrix}$$
Theorem.
Column-row expansion of matrix multiplication
If $\boldsymbol{A}$ is an $m\times{n}$ matrix composed of column vectors and $\boldsymbol{B}$ is an $n\times{k}$ matrix composed of row vectors, then their product $\boldsymbol{AB}$ can be expressed as:
$$\begin{pmatrix}
\vert&\vert&\cdots&\vert\\
\boldsymbol{a}_1&\boldsymbol{a}_2&
\cdots&\boldsymbol{a}_n\\\vert&\vert&\cdots&\vert
\end{pmatrix}\begin{pmatrix}
-&\boldsymbol{r}_1&-\\
-&\boldsymbol{r}_2&-\\
\vdots&\vdots&\vdots\\
-&\boldsymbol{r}_n&-
\end{pmatrix}=
\boldsymbol{a}_1\boldsymbol{r}_1+
\boldsymbol{a}_2\boldsymbol{r}_2+
\cdots+\boldsymbol{a}_n\boldsymbol{r}_n$$
Proof . Suppose matrices $\boldsymbol{A}$ and $\boldsymbol{B}$ are:
$$\boldsymbol{A}=\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\smash\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix},\;\;\;\;\;\;
\boldsymbol{B}=
\begin{pmatrix}
b_{11}&b_{12}&\cdots&b_{1k}\\
b_{21}&b_{22}&\cdots&b_{2k}\\
\vdots&\vdots&\smash\ddots&\vdots\\
b_{n1}&b_{n2}&\cdots&b_{nk}
\end{pmatrix}$$
By theorem link $i$ -th row $j$ -th column of $\boldsymbol{AB}$ is:
$$\begin{equation}\label{eq:aZeFUBZvJv5kR9HV8dI}
\big[\boldsymbol{AB}\big]_{ij}=
\sum_{t=1}^{n}a_{it}b_{tj}
\end{equation}$$
Now, the product $\boldsymbol{a}_1\boldsymbol{r}_1$ is:
$$\boldsymbol{a}_1\boldsymbol{r}_1=
\begin{pmatrix}
a_{11}\\a_{21}\\\vdots\\a_{m1}
\end{pmatrix}\begin{pmatrix}
b_{11}&b_{12}&\cdots&b_{1k}
\end{pmatrix}=
\begin{pmatrix}
a_{11}b_{11}&a_{11}b_{12}&\cdots&a_{11}b_{1k}\\
a_{21}b_{11}&a_{21}b_{12}&\cdots&a_{21}b_{1k}\\
\vdots&\vdots&\smash\ddots&\vdots\\
a_{m1}b_{11}&a_{m1}b_{12}&\cdots&a_{m1}b_{1k}\\
\end{pmatrix}$$
The product $\boldsymbol{a}_2\boldsymbol{r}_2$ is:
$$\boldsymbol{a}_2\boldsymbol{r}_2=
\begin{pmatrix}
a_{12}\\a_{22}\\\vdots\\a_{m2}
\end{pmatrix}\begin{pmatrix}
b_{21}&b_{22}&\cdots&b_{2k}
\end{pmatrix}=
\begin{pmatrix}
a_{12}b_{21}&a_{12}b_{22}&\cdots&a_{12}b_{2k}\\
a_{22}b_{21}&a_{22}b_{22}&\cdots&a_{22}b_{2k}\\
\vdots&\vdots&\smash\ddots&\vdots\\
a_{m2}b_{21}&a_{m2}b_{22}&\cdots&a_{m2}b_{2k}\\
\end{pmatrix}$$
The product $\boldsymbol{a}_n
\boldsymbol{r}_n$ is:
$$\boldsymbol{a}_n\boldsymbol{r}_n=
\begin{pmatrix}
a_{1n}\\a_{2n}\\\vdots\\a_{mn}
\end{pmatrix}\begin{pmatrix}
b_{n1}&b_{n2}&\cdots&b_{nk}
\end{pmatrix}=
\begin{pmatrix}
a_{1n}b_{n1}&a_{1n}b_{n2}&\cdots&a_{1n}b_{nk}\\
a_{2n}b_{n1}&a_{2n}b_{n2}&\cdots&a_{2n}b_{nk}\\
\vdots&\vdots&\smash\ddots&\vdots\\
a_{mn}b_{n1}&a_{mn}b_{n2}&\cdots&a_{mn}b_{nk}\\
\end{pmatrix}$$
Let's now denote $\boldsymbol{C}$ as the sum of all these matrices:
$$\boldsymbol{C}=\boldsymbol{a}_1\boldsymbol{r}_1+
\boldsymbol{a}_2\boldsymbol{r}_2+
\cdots+
\boldsymbol{a}_n\boldsymbol{r}_n$$
By inspection, the entry at the $i$ -th row $j$ -th column of $\boldsymbol{C}$ is:
$$\begin{equation}\label{eq:Ym8pBCLVIPoDcS0Z49d}
\big[\boldsymbol{C}\big]_{ij}=
\sum_{t=1}^{n}a_{it}b_{tj}
\end{equation}$$
Note that \eqref{eq:aZeFUBZvJv5kR9HV8dI} matches \eqref{eq:Ym8pBCLVIPoDcS0Z49d} , which means that every entry of $\boldsymbol{AB}$ and $\boldsymbol{C}$ are equal. Moreover, both $\boldsymbol{AB}$ and $\boldsymbol{C}$ have the shape $m\times{k}$ . Therefore, $\boldsymbol{AB}=\boldsymbol{C}$ and hence:
$$\boldsymbol{AB}=\boldsymbol{a}_1\boldsymbol{r}_1+
\boldsymbol{a}_2\boldsymbol{r}_2+
\cdots+
\boldsymbol{a}_n\boldsymbol{r}_n$$
This completes the proof.
Example.
Finding the column-row expansion of a matrix product
Find the column-row expansion of the following matrix product:
$$\begin{pmatrix}
1&2&3\\
4&5&6
\end{pmatrix}
\begin{pmatrix}
-1&-2\\
-3&-4\\
-5&-6\\
\end{pmatrix}$$
Solution . We apply the expansion rule to get:
$$\begin{pmatrix}
1&2&3\\4&5&6
\end{pmatrix}
\begin{pmatrix}
-1&-2\\-3&-4\\-5&-6\\
\end{pmatrix}=
\begin{pmatrix}1\\4\end{pmatrix}
\begin{pmatrix}-1&-2\end{pmatrix}+
\begin{pmatrix}2\\5\end{pmatrix}
\begin{pmatrix}-3&-4\end{pmatrix}+
\begin{pmatrix}3\\6\end{pmatrix}
\begin{pmatrix}-5&-6\end{pmatrix}
$$
Theorem.
Expressing a sum of scalar-vector products using a matrix-vector product
Consider the following sum:
$$x_1\boldsymbol{a}_1+
x_2\boldsymbol{a}_2+
\cdots+
x_n\boldsymbol{a}_n$$
Where $x_i\in\mathbb{R}$ and $\boldsymbol{a}_i\in\mathbb{R}^m$ for $i=1,2,\cdots,n$ . This can be expressed as a matrix-vector product:
$$\begin{equation}\label{eq:y0B5oUWXjxeqQX1HUIN}
x_1\boldsymbol{a}_1+
x_2\boldsymbol{a}_2+
\cdots+
x_n\boldsymbol{a}_n=
\begin{pmatrix}
\vert&\vert&\cdots&\vert\\
\boldsymbol{a}_1&\boldsymbol{a}_2&\cdots&\boldsymbol{a}_n\\
\vert&\vert&\cdots&\vert
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
\vdots\\
x_n\\
\end{pmatrix}=
\boldsymbol{A}\boldsymbol{x}
\end{equation}$$
Here, $\boldsymbol{A}$ is a matrix whose columns are composed of vectors $\boldsymbol{a}_i$ .
Proof . Suppose matrix $\boldsymbol{A}\in\mathbb{R}^{m\times{n}}$ and $\boldsymbol{x}\in\mathbb{R}^{n}$ are as follows:
$$\boldsymbol{A}=\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix},\;\;\;\;\;\;
\boldsymbol{x}=
\begin{pmatrix}
x_1\\x_2\\\vdots\\x_n
\end{pmatrix}$$
Taking their product:
$$\begin{align*}
\boldsymbol{A}\boldsymbol{x}&=\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix}
\begin{pmatrix}
x_1\\x_2\\\vdots\\x_n
\end{pmatrix}\\
&=\begin{pmatrix}
a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n\\
a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n\\
\vdots\\
a_{m1}x_1+a_{m2}x_2+\cdots+a_{mn}x_n
\end{pmatrix}\\
&=x_1\begin{pmatrix}
a_{11}\\a_{21}\\\vdots\\a_{m1}
\end{pmatrix}+x_2\begin{pmatrix}
a_{12}\\a_{22}\\\vdots\\a_{m2}
\end{pmatrix}
+\cdots
+x_n\begin{pmatrix}
a_{1n}\\a_{2n}\\\vdots\\a_{mn}
\end{pmatrix}\\
&=x_1\boldsymbol{a}_1+x_2\boldsymbol{a}_2+\cdots+x_n\boldsymbol{a}_n
\end{align*}$$
Where $\boldsymbol{a}_1$ , $\boldsymbol{a}_2$ , $\cdots$ , $\boldsymbol{a}_n$ are the columns of matrix $\boldsymbol{A}$ . This completes the proof.
Example.
Expressing a sum of three vectors
Express the following sum as a matrix-vector product:
$$2\boldsymbol{a}_1+
3\boldsymbol{a}_2+
5\boldsymbol{a}_3$$
Where $\boldsymbol{a}_1$ , $\boldsymbol{a}_2$ and $\boldsymbol{a}_3$ are the following vectors:
$$\boldsymbol{a}_1=\begin{pmatrix}
4\\3\\2
\end{pmatrix},\;\;\;\;
\boldsymbol{a}_2=\begin{pmatrix}
1\\8\\7
\end{pmatrix},\;\;\;\;
\boldsymbol{a}_3=\begin{pmatrix}
6\\4\\2
\end{pmatrix}$$
Solution . By theorem link
$$\begin{align*}
2\boldsymbol{a}_1+
3\boldsymbol{a}_2+
5\boldsymbol{a}_3\;\;{\color{blue}=}\;
\begin{pmatrix}
\vert&\vert&\vert\\
\boldsymbol{a}_1&\boldsymbol{a}_2&\boldsymbol{a}_3\\
\vert&\vert&\vert
\end{pmatrix}
\begin{pmatrix}
2\\3\\5
\end{pmatrix}\;{\color{blue}=}\;
\begin{pmatrix}
4&1&6\\3&8&4\\2&7&2\\
\end{pmatrix}
\begin{pmatrix}
2\\3\\5
\end{pmatrix}
\end{align*}$$
Theorem.
Important properties of matrix-vector products
If $\boldsymbol{A}$ is an $m\times{n}$ matrix, $\boldsymbol{v}$ and $\boldsymbol{w}$ are vectors in $\mathbb{R}^n$ , and $c$ is a scalar in $\mathbb{R}$ , then:
$\boldsymbol{A}(\boldsymbol{v}+\boldsymbol{w})=
\boldsymbol{A}\boldsymbol{v}+\boldsymbol{A}\boldsymbol{w}$ .
$\boldsymbol{A}(c\boldsymbol{v})
=c(\boldsymbol{A}\boldsymbol{v})$ .
$(\boldsymbol{A}+\boldsymbol{B})\boldsymbol{v}=
\boldsymbol{A}\boldsymbol{v}+\boldsymbol{B}\boldsymbol{v}$ .
Proof . We will prove these properties for the simple case when $n=3$ but the proofs can easily be generalized. Suppose we have matrices $\boldsymbol{A},\boldsymbol{B}
\in\mathbb{R}^{m\times3}$ and vectors $\boldsymbol{u},\boldsymbol{v}\in\mathbb{R}^3$ like so:
$$\boldsymbol{v}=
\begin{pmatrix}
v_1\\v_2\\v_3\end{pmatrix},\;\;\;\;\;\;
\boldsymbol{w}=\begin{pmatrix}
w_1\\w_2\\w_3
\end{pmatrix},\;\;\;\;\;\;
\boldsymbol{A}=\begin{pmatrix}
\vert&\vert&\vert\\
\boldsymbol{a}_1&\boldsymbol{a}_2&\boldsymbol{a}_3\\
\vert&\vert&\vert\end{pmatrix},\;\;\;\;\;\;
\boldsymbol{B}=\begin{pmatrix}
\vert&\vert&\vert\\
\boldsymbol{b}_1&\boldsymbol{b}_2&\boldsymbol{b}_3\\
\vert&\vert&\vert\end{pmatrix}$$
We start by proving the first property:
$$\begin{align*}
\boldsymbol{A}(\boldsymbol{v}+\boldsymbol{w})&=
\begin{pmatrix}
\vert&\vert&\vert\\
\boldsymbol{a}_1&\boldsymbol{a}_2&\boldsymbol{a}_3\\
\vert&\vert&\vert\\
\end{pmatrix}\left[
\begin{pmatrix}
v_1\\v_2\\v_3
\end{pmatrix}+
\begin{pmatrix}
w_1\\w_2\\w_3
\end{pmatrix}\right]\\
&=
\begin{pmatrix}
\vert&\vert&\vert\\
\boldsymbol{a}_1&\boldsymbol{a}_2&\boldsymbol{a}_3\\
\vert&\vert&\vert\\
\end{pmatrix}\begin{pmatrix}
v_1+w_1\\v_2+w_2\\v_3+w_3
\end{pmatrix}\\
&=(v_1+w_1)\boldsymbol{a}_1+(v_2+w_2)\boldsymbol{a}_2+(v_3+w_3)\boldsymbol{a}_3\\
&=v_1\boldsymbol{a}_1+w_1\boldsymbol{a}_1+v_2\boldsymbol{a}_2+w_2\boldsymbol{a}_2+v_3\boldsymbol{a}_3+w_3\boldsymbol{a}_3\\
&=(v_1\boldsymbol{a}_1+v_2\boldsymbol{a}_2+v_3\boldsymbol{a}_3)+(w_1\boldsymbol{a}_1+w_2\boldsymbol{a}_2+w_3\boldsymbol{a}_3)\\
&=\begin{pmatrix}
\vert&\vert&\vert\\
\boldsymbol{a}_1&\boldsymbol{a}_2&\boldsymbol{a}_3\\
\vert&\vert&\vert
\end{pmatrix}\begin{pmatrix}
u_1\\u_2\\u_3
\end{pmatrix}+
\begin{pmatrix}
\vert&\vert&\vert\\
\boldsymbol{a}_1&\boldsymbol{a}_2&\boldsymbol{a}_3\\
\vert&\vert&\vert
\end{pmatrix}\begin{pmatrix}
v_1\\v_2\\v_3
\end{pmatrix}\\
&=\boldsymbol{Au}+\boldsymbol{Av}
\end{align*}$$
Here, for the third and sixth equality, we used theorem link
Next, we prove the second property:
$$\begin{align*}
\boldsymbol{A}(c\boldsymbol{v})
&=\boldsymbol{A}\left[c\begin{pmatrix}
v_1\\v_2\\v_3
\end{pmatrix}\right]\\
&=\boldsymbol{A}\begin{pmatrix}
cv_1\\cv_2\\cv_3\end{pmatrix}\\
&=\begin{pmatrix}
\vert&\vert&\vert\\
\boldsymbol{a}_1&\boldsymbol{a}_2&\boldsymbol{a}_3\\
\vert&\vert&\vert\\\end{pmatrix}\begin{pmatrix}cv_1\\cv_2\\cv_3\end{pmatrix}\\
&=cv_1\boldsymbol{a}_1+cv_2\boldsymbol{a}_2+cv_3\boldsymbol{a}_3\\
&=c(v_1\boldsymbol{a}_1+v_2\boldsymbol{a}_2+v_3\boldsymbol{a}_3)\\&=c\left[\begin{pmatrix}
\vert&\vert&\vert\\\boldsymbol{a}_1&\boldsymbol{a}_2&\boldsymbol{a}_3\\
\vert&\vert&\vert\\\end{pmatrix}\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}\right]\\
&=c\boldsymbol{A}\boldsymbol{v}
\end{align*}$$
Here, we used theorem link
Finally, we prove the last property:
$$\begin{align*}
(\boldsymbol{A}+\boldsymbol{B})\boldsymbol{v}
&=\left[\begin{pmatrix}
\vert&\vert&\vert\\
\boldsymbol{a}_1&\boldsymbol{a}_2&\boldsymbol{a}_3\\
\vert&\vert&\vert\end{pmatrix}+
\begin{pmatrix}
\vert&\vert&\vert\\
\boldsymbol{b}_1&\boldsymbol{b}_2&\boldsymbol{b}_3\\
\vert&\vert&\vert\end{pmatrix}\right]\boldsymbol{v}\\
&=\begin{pmatrix}
\vert&\vert&\vert\\
\boldsymbol{a}_1+\boldsymbol{b}_1&\boldsymbol{a}_2+\boldsymbol{b}_2&\boldsymbol{a}_3+\boldsymbol{b}_3\\
\vert&\vert&\vert\end{pmatrix}\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}\\
&=(\boldsymbol{a}_1+\boldsymbol{b}_1)v_1+
(\boldsymbol{a}_2+\boldsymbol{b}_2)v_2+
(\boldsymbol{a}_3+\boldsymbol{b}_3)v_3\\
&=v_1\boldsymbol{a}_1+v_1\boldsymbol{b}_1+v_2\boldsymbol{a}_2+v_2\boldsymbol{b}_2+v_3\boldsymbol{a}_3+v_3\boldsymbol{b}_3\\
&=(v_1\boldsymbol{a}_1+v_2\boldsymbol{a}_2+v_3\boldsymbol{a}_3)+(v_1\boldsymbol{b}_1+v_2\boldsymbol{b}_2+v_3\boldsymbol{b}_3)\\
&=\begin{pmatrix}\vert&\vert&\vert\\\boldsymbol{a}_1&\boldsymbol{a}_2&\boldsymbol{a}_3\\
\vert&\vert&\vert\end{pmatrix}\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}+
\begin{pmatrix}\vert&\vert&\vert\\\boldsymbol{b}_1&\boldsymbol{b}_2&\boldsymbol{b}_3\\
\vert&\vert&\vert\end{pmatrix}\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}\\
&=\boldsymbol{Av}+\boldsymbol{Bv}
\end{align*}$$
This completes the proofs of the three properties.
Practice problems What is the shape of the following matrix:
$$\boldsymbol{A}=\begin{pmatrix}
1&6&5&1\\
5&9&1&3\\
\end{pmatrix}$$
Since the matrix has 2 rows and 4 columns, the shape is $2\times4$ .
Consider the following matrices:
$$\boldsymbol{A}=\begin{pmatrix}
2&3\\
4&2
\end{pmatrix},\;\;\;\;\;\;\boldsymbol{B}=
\begin{pmatrix}
1&3\\0&5
\end{pmatrix}$$
Compute $\boldsymbol{A}+\boldsymbol{B}$ .
Your answer
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The matrix addition is defined here because $\boldsymbol{A}$ and $\boldsymbol{B}$ both have the same shape. Matrix addition works by adding the values in the same entry like so:
$$\begin{align*}
\boldsymbol{A}+\boldsymbol{B}
&=\begin{pmatrix}2&3\\4&2\end{pmatrix}+
\begin{pmatrix}1&3\\0&5\end{pmatrix}\\
&=\begin{pmatrix}2+1&3+3\\4+0&2+5\end{pmatrix}\\
&=\begin{pmatrix}3&6\\4&7\end{pmatrix}
\end{align*}$$
Suppose $\boldsymbol{A}$ is a $2\times3$ matrix and $\boldsymbol{B}$ is a $3\times4$ matrix. Infer the shape of their product $\boldsymbol{AB}$ .
By theorem link $\boldsymbol{AB}$ would have the same number of rows as $\boldsymbol{A}$ and the same number of columns as $\boldsymbol{B}$ . This means $\boldsymbol{AB}$ would have shape $2\times4$ .
Consider the following matrices:
$$\boldsymbol{A}=\begin{pmatrix}
2&3\\
4&2
\end{pmatrix},\;\;\;\;\;\;\boldsymbol{B}=
\begin{pmatrix}
1&3\\0&5
\end{pmatrix}$$
Compute $\boldsymbol{AB}$ .
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$$\begin{align*}
\boldsymbol{AB}&=
\begin{pmatrix}2&3\\4&2\end{pmatrix}
\begin{pmatrix}1&3\\0&5\end{pmatrix}\\
&=\begin{pmatrix}(2)(1)+(3)(0)&(2)(3)+(3)(5)\\
(4)(1)+(2)(0)&(4)(3)+(2)(5)\end{pmatrix}\\
&=\begin{pmatrix}2&21\\
4&22\end{pmatrix}
\end{align*}$$
Compute the following:
$$2\begin{pmatrix}
1&3\\
4&1
\end{pmatrix}$$
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$$2\begin{pmatrix}
1&3\\4&1
\end{pmatrix}=
\begin{pmatrix}
2(1)&2(3)\\2(4)&2(1)
\end{pmatrix}=
\begin{pmatrix}
2&6\\8&2
\end{pmatrix}$$
Consider the following matrices:
$$\boldsymbol{A}=\begin{pmatrix}
0&1&1\\
3&2&3\\
1&1&2\\
\end{pmatrix},\;\;\;\;\;\;\boldsymbol{B}=
\begin{pmatrix}
1&0&1\\0&3&2\\1&2&1
\end{pmatrix}$$
Compute $\boldsymbol{AB}$ .
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$$\begin{align*}
\boldsymbol{AB}&=\begin{pmatrix}
0&1&1\\3&2&3\\1&1&2\\
\end{pmatrix}
\begin{pmatrix}
1&0&1\\0&3&2\\1&2&1
\end{pmatrix}\\
&=\begin{pmatrix}
(0)(1)+(1)(0)+(1)(1)&
(0)(0)+(1)(3)+(1)(2)&
(0)(1)+(1)(2)+(1)(1)\\
(3)(1)+(2)(0)+(3)(1)&
(3)(0)+(2)(3)+(3)(2)&
(3)(1)+(2)(2)+(3)(1)\\
(1)(1)+(1)(0)+(2)(1)&
(1)(0)+(1)(3)+(2)(2)&
(1)(1)+(1)(2)+(2)(1)\\
\end{pmatrix}\\
&=\begin{pmatrix}
1&5&3\\6&12&10\\3&7&5
\end{pmatrix}
\end{align*}$$
Compute the following:
$$\begin{pmatrix}
0&1&1\\
3&2&3\\
1&1&2\\
\end{pmatrix}
\begin{pmatrix}
1&0&0\\0&1&0\\0&0&1
\end{pmatrix}$$
Your answer
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Since the matrix on the right is an identity matrix, the resulting product would simply be the matrix on the left.
Rewrite the following as a sum of scalar-vector products:
$$\begin{pmatrix}
5&8&6\\
1&5&7
\end{pmatrix}
\begin{pmatrix}
2\\3\\4\\
\end{pmatrix}$$
By theorem link
$$2\begin{pmatrix}5\\1\end{pmatrix}+
3\begin{pmatrix}8\\5\end{pmatrix}+
4\begin{pmatrix}6\\7\end{pmatrix}$$
