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Matrices and Matrix Operations

Find the sum and difference of two matrices.

Matrices and Matrix Operations

Write the augmented matrix for a system of equations (IA 4.5.1)
Add, subtract matrices and multiply a matrix by a scalar

A matrix is a rectangular array of numbers arranged in rows and columns.

A matrix with m rows and n columns has dimension m×n.

Each number in the matrix is called an element or entry in the matrix.

The matrix on the left below has 2 rows and 3 columns and so it has order 2×3. We say it is a 2 by 3 matrix.

We will use a matrix to represent systems of equations.

Each column then would be the coefficients of one of the variables in the system or the constants.

A vertical line replaces the equal signs.

We call the resulting matrix the augmented matrix for the system of equations.

Example

Write each system of linear equations as an augmented matrix

Try it.

ⓐ $\left\{\begin{array}{l}3x-y=-1 \\ 2y=2x+5\end{array}\right$

ⓑ $\left\{\begin{array}{l}4x+3y=-2 \\ x-2y-3z=7 \\ 2x-y+2z=-6\end{array}\right$

Solution

ⓐ We first rewrite the second equation in standard form

$\left\{\begin{array}{l}3x-y=-1 \\ -2x+2y=5\end{array}\right$

Next we write the augmented matrix

$\left\{\begin{array}{l}3x-y=-1 \\ -2x+2y=5\end{array}\right⇒\overset{x\ y\}{[\begin{array}{ll}3 & -1 \\ -2 & 2\end{array}\ |\ \begin{array}{l}-1 \\ 5\end{array}]}$

ⓑ Each equation is in standard form

Write the augmented matrix

$\left\{\begin{array}{l}4x+3y=-2 \\ x-2y-3z=7 \\ 2x-y+2z=-6\end{array}\right⇒\overset{x\ y\ z\}{[\begin{array}{lll}4 & 3 & 0 \\ 1 & -2 & -3 \\ 2 & -1 & 2\end{array}\ |\ \begin{array}{l}-2 \\ 7 \\ -6\end{array}]}$

Write each system of linear equations as an augmented matrix

Try it.

$\left\{\begin{array}{l}2x-5y=-3 \\ 4x=3y-1\end{array}\right$

Try it.

$\left\{\begin{array}{l}4x+3y-2z=-3 \\ -2x+y-3z=4 \\ -x-4y+5z=-2\end{array}\right$

Condensed — the full section is in OpenStax College Algebra 2e.

Finding the Sum and Difference of Two Matrices

To solve a problem like the one described for the soccer teams, we can use a matrix, which is a rectangular array of numbers. A row in a matrix is a set of numbers that are aligned horizontally. A column in a matrix is a set of numbers that are aligned vertically. Each number is an entry, sometimes called an element, of the matrix. Matrices (plural) are enclosed in [ ] or ( ), and are usually named with capital letters. For example, three matrices named $A,B,$ and $C$ are shown below.

\[A=[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}],B=[\begin{array}{lll}1 & 2 & 7 \\ 0 & -5 & 6 \\ 7 & 8 & 2\end{array}],C=[\begin{array}{l}-1 \\ \ 0 \\ \ 3\end{array}\ \ \begin{array}{l}3 \\ 2 \\ 1\end{array}]\]

Condensed — the full section is in OpenStax College Algebra 2e.

Finding Scalar Multiples of a Matrix

Besides adding and subtracting whole matrices, there are many situations in which we need to multiply a matrix by a constant called a scalar. Recall that a scalar is a real number quantity that has magnitude, but not direction. For example, time, temperature, and distance are scalar quantities. The process of scalar multiplication involves multiplying each entry in a matrix by a scalar. A scalar multiple is any entry of a matrix that results from scalar multiplication.

Consider a real-world scenario in which a university needs to add to its inventory of computers, computer tables, and chairs in two of the campus labs due to increased enrollment. They estimate that 15% more equipment is needed in both labs. The school’s current inventory is displayed in .

	Lab A	Lab B
Computers	15	27
Computer Tables	16	34
Chairs	16	34

Converting the data to a matrix, we have

\[{C}_{2013}=[\begin{array}{l}15 \\ 16 \\ 16\end{array}\ \ \ \ \begin{array}{l}27 \\ 34 \\ 34\end{array}]\]

To calculate how much computer equipment will be needed, we multiply all entries in matrix $C$ by 0.15.

\[(0.15){C}_{2013}=[\begin{array}{l}(0.15)15 \\ (0.15)16 \\ (0.15)16\end{array}\ \ \ \ \begin{array}{l}(0.15)27 \\ (0.15)34 \\ (0.15)34\end{array}]=[\begin{array}{l}2.25 \\ 2.4 \\ 2.4\end{array}\ \ \ \begin{array}{l}4.05 \\ 5.1 \\ 5.1\end{array}]\]

We must round up to the next integer, so the amount of new equipment needed is

\[[\begin{array}{l}3 \\ 3 \\ 3\end{array}\ \ \ \begin{array}{l}5 \\ 6 \\ 6\end{array}]\]

Adding the two matrices as shown below, we see the new inventory amounts.

\[[\begin{array}{l}15 \\ 16 \\ 16\end{array}\ \ \ \ \begin{array}{l}27 \\ 34 \\ 34\end{array}]+[\begin{array}{l}3 \\ 3 \\ 3\end{array}\ \ \ \begin{array}{l}5 \\ 6 \\ 6\end{array}]=[\begin{array}{l}18 \\ 19 \\ 19\end{array}\ \ \ \begin{array}{l}32 \\ 40 \\ 40\end{array}]\]

This means

\[{C}_{2014}=[\begin{array}{l}18 \\ 19 \\ 19\end{array}\ \ \ \begin{array}{l}32 \\ 40 \\ 40\end{array}]\]

Condensed — the full section is in OpenStax College Algebra 2e.

Finding the Product of Two Matrices

In addition to multiplying a matrix by a scalar, we can multiply two matrices. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. If $A$ is an $\ m\times r$ matrix and $B$ is an $\ r\times n$ matrix, then the product matrix $AB$ is an $\ m\times n$ matrix. For example, the product $AB$ is possible because the number of columns in $A$ is the same as the number of rows in $B.$ If the inner dimensions do not match, the product is not defined.

We multiply entries of $A$ with entries of $B$ according to a specific pattern as outlined below. The process of matrix multiplication becomes clearer when working a problem with real numbers.

To obtain the entries in row $i$ of $AB,$ we multiply the entries in row $i$ of $A$ by column $j$ in $B$ and add. For example, given matrices $A$ and $B,$ where the dimensions of $A$ are $2\times 3$ and the dimensions of $B$ are $3\times 3,$ the product of $AB$ will be a $2\times 3$ matrix.

\[A=[\begin{array}{lll}{a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23}\end{array}]\ \text{and }B=[\begin{array}{lll}{b}_{11} & {b}_{12} & {b}_{13} \\ {b}_{21} & {b}_{22} & {b}_{23} \\ {b}_{31} & {b}_{32} & {b}_{33}\end{array}]\]

Multiply and add as follows to obtain the first entry of the product matrix $AB.$

To obtain the entry in row 1, column 1 of $AB,$ multiply the first row in $A$ by the first column in $B,$ and add. \[[\begin{array}{lll}{a}_{11} & {a}_{12} & {a}_{13}\end{array}][\begin{array}{l}{b}_{11} \\ {b}_{21} \\ {b}_{31}\end{array}]={a}_{11}⋅{b}_{11}+{a}_{12}⋅{b}_{21}+{a}_{13}⋅{b}_{31}\]
To obtain the entry in row 1, column 2 of $AB,$ multiply the first row of $A$ by the second column in $B,$ and add. \[[\begin{array}{lll}{a}_{11} & {a}_{12} & {a}_{13}\end{array}][\begin{array}{l}{b}_{12} \\ {b}_{22} \\ {b}_{32}\end{array}]={a}_{11}⋅{b}_{12}+{a}_{12}⋅{b}_{22}+{a}_{13}⋅{b}_{32}\]
To obtain the entry in row 1, column 3 of $AB,$ multiply the first row of $A$ by the third column in $B,$ and add. \[[\begin{array}{lll}{a}_{11} & {a}_{12} & {a}_{13}\end{array}][\begin{array}{l}{b}_{13} \\ {b}_{23} \\ {b}_{33}\end{array}]={a}_{11}⋅{b}_{13}+{a}_{12}⋅{b}_{23}+{a}_{13}⋅{b}_{33}\]

We proceed the same way to obtain the second row of $AB.$ In other words, row 2 of $A$ times column 1 of $B;$ row 2 of $A$ times column 2 of $B;$ row 2 of $A$ times column 3 of $B.$ When complete, the product matrix will be

Condensed — the full section is in OpenStax College Algebra 2e.

Key Concepts

A matrix is a rectangular array of numbers. Entries are arranged in rows and columns.
The dimensions of a matrix refer to the number of rows and the number of columns. A $3\times 2$ matrix has three rows and two columns. See .
We add and subtract matrices of equal dimensions by adding and subtracting corresponding entries of each matrix. See , , , and .
Scalar multiplication involves multiplying each entry in a matrix by a constant. See .
Scalar multiplication is often required before addition or subtraction can occur. See .
Multiplying matrices is possible when inner dimensions are the same—the number of columns in the first matrix must match the number of rows in the second.
The product of two matrices, $A$ and $B,$ is obtained by multiplying each entry in row 1 of $A$ by each entry in column 1 of $B;$ then multiply each entry of row 1 of $A$ by each entry in columns 2 of $B,$ and so on. See and .
Many real-world problems can often be solved using matrices. See .
We can use a calculator to perform matrix operations after saving each matrix as a matrix variable. See .

Matrices and Matrix Operations

Write the augmented matrix for a system of equations (IA 4.5.1)
Add, subtract matrices and multiply a matrix by a scalar

A matrix is a rectangular array of numbers arranged in rows and columns.

A matrix with m rows and n columns has dimension m×n.

Each number in the matrix is called an element or entry in the matrix.

The matrix on the left below has 2 rows and 3 columns and so it has order 2×3. We say it is a 2 by 3 matrix.

We will use a matrix to represent systems of equations.

Each column then would be the coefficients of one of the variables in the system or the constants.

A vertical line replaces the equal signs.

We call the resulting matrix the augmented matrix for the system of equations.

Example

Write each system of linear equations as an augmented matrix

Try it.

ⓐ $\left\{\begin{array}{l}3x-y=-1 \\ 2y=2x+5\end{array}\right$

ⓑ $\left\{\begin{array}{l}4x+3y=-2 \\ x-2y-3z=7 \\ 2x-y+2z=-6\end{array}\right$

Solution

ⓐ We first rewrite the second equation in standard form

$\left\{\begin{array}{l}3x-y=-1 \\ -2x+2y=5\end{array}\right$

Next we write the augmented matrix

$\left\{\begin{array}{l}3x-y=-1 \\ -2x+2y=5\end{array}\right⇒\overset{x\ y\}{[\begin{array}{ll}3 & -1 \\ -2 & 2\end{array}\ |\ \begin{array}{l}-1 \\ 5\end{array}]}$

ⓑ Each equation is in standard form

Write the augmented matrix

Write each system of linear equations as an augmented matrix

Try it.

$\left\{\begin{array}{l}2x-5y=-3 \\ 4x=3y-1\end{array}\right$

Try it.

$\left\{\begin{array}{l}4x+3y-2z=-3 \\ -2x+y-3z=4 \\ -x-4y+5z=-2\end{array}\right$

Condensed — the full section is in OpenStax Precalculus 2e.

Finding the Sum and Difference of Two Matrices

Condensed — the full section is in OpenStax Precalculus 2e.

Finding Scalar Multiples of a Matrix

	Lab A	Lab B
Computers	15	27
Computer Tables	16	34
Chairs	16	34

Converting the data to a matrix, we have

\[{C}_{2013}=[\begin{array}{l}15 \\ 16 \\ 16\end{array}\ \ \ \ \begin{array}{l}27 \\ 34 \\ 34\end{array}]\]

To calculate how much computer equipment will be needed, we multiply all entries in matrix $C$ by 0.15.

We must round up to the next integer, so the amount of new equipment needed is

\[[\begin{array}{l}3 \\ 3 \\ 3\end{array}\ \ \ \begin{array}{l}5 \\ 6 \\ 6\end{array}]\]

Adding the two matrices as shown below, we see the new inventory amounts.

This means

\[{C}_{2014}=[\begin{array}{l}18 \\ 19 \\ 19\end{array}\ \ \ \begin{array}{l}32 \\ 40 \\ 40\end{array}]\]

Condensed — the full section is in OpenStax Precalculus 2e.

Finding the Product of Two Matrices

Multiply and add as follows to obtain the first entry of the product matrix $AB.$

To obtain the entry in row 1, column 1 of $AB,$ multiply the first row in $A$ by the first column in $B,$ and add. \[[\begin{array}{lll}{a}_{11} & {a}_{12} & {a}_{13}\end{array}][\begin{array}{l}{b}_{11} \\ {b}_{21} \\ {b}_{31}\end{array}]={a}_{11}⋅{b}_{11}+{a}_{12}⋅{b}_{21}+{a}_{13}⋅{b}_{31}\]
To obtain the entry in row 1, column 2 of $AB,$ multiply the first row of $A$ by the second column in $B,$ and add. \[[\begin{array}{lll}{a}_{11} & {a}_{12} & {a}_{13}\end{array}][\begin{array}{l}{b}_{12} \\ {b}_{22} \\ {b}_{32}\end{array}]={a}_{11}⋅{b}_{12}+{a}_{12}⋅{b}_{22}+{a}_{13}⋅{b}_{32}\]
To obtain the entry in row 1, column 3 of $AB,$ multiply the first row of $A$ by the third column in $B,$ and add. \[[\begin{array}{lll}{a}_{11} & {a}_{12} & {a}_{13}\end{array}][\begin{array}{l}{b}_{13} \\ {b}_{23} \\ {b}_{33}\end{array}]={a}_{11}⋅{b}_{13}+{a}_{12}⋅{b}_{23}+{a}_{13}⋅{b}_{33}\]

Condensed — the full section is in OpenStax Precalculus 2e.

Key Concepts

A matrix is a rectangular array of numbers. Entries are arranged in rows and columns.
The dimensions of a matrix refer to the number of rows and the number of columns. A $3\times 2$ matrix has three rows and two columns. See .
We add and subtract matrices of equal dimensions by adding and subtracting corresponding entries of each matrix. See , , , and .
Scalar multiplication involves multiplying each entry in a matrix by a constant. See .
Scalar multiplication is often required before addition or subtraction can occur. See .
Multiplying matrices is possible when inner dimensions are the same—the number of columns in the first matrix must match the number of rows in the second.
The product of two matrices, $A$ and $B,$ is obtained by multiplying each entry in row 1 of $A$ by each entry in column 1 of $B;$ then multiply each entry of row 1 of $A$ by each entry in columns 2 of $B,$ and so on. See and .
Many real-world problems can often be solved using matrices. See .
We can use a calculator to perform matrix operations after saving each matrix as a matrix variable. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

ⓐ $\left\{\begin{array}{l}3x-y=-1 \\ 2y=2x+5\end{array}\right$
ⓑ $\left\{\begin{array}{l}4x+3y=-2 \\ x-2y-3z=7 \\ 2x-y+2z=-6\end{array}\right$

விடை தெரியப்படுத்து

ⓐ We first rewrite the second equation in standard form
$\left\{\begin{array}{l}3x-y=-1 \\ -2x+2y=5\end{array}\right$
Next we write the augmented matrix
$\left\{\begin{array}{l}3x-y=-1 \\ -2x+2y=5\end{array}\right⇒\overset{x\ y\}{[\begin{array}{ll}3 & -1 \\ -2 & 2\end{array}\ |\ \begin{array}{l}-1 \\ 5\end{array}]}$
ⓑ Each equation is in standard form
Write the augmented matrix
$\left\{\begin{array}{l}4x+3y=-2 \\ x-2y-3z=7 \\ 2x-y+2z=-6\end{array}\right⇒\overset{x\ y\ z\}{[\begin{array}{lll}4 & 3 & 0 \\ 1 & -2 & -3 \\ 2 & -1 & 2\end{array}\ |\ \begin{array}{l}-2 \\ 7 \\ -6\end{array}]}$
$\left\{\begin{array}{l}2x-5y=-3 \\ 4x=3y-1\end{array}\right$
$\left\{\begin{array}{l}4x+3y-2z=-3 \\ -2x+y-3z=4 \\ -x-4y+5z=-2\end{array}\right$
ⓐ Add the two matrices $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]B=\left[\begin{array}{ll}e & f \\ g & h\end{array}\right]$
ⓑ Subtract the two matrices $A=\left[\begin{array}{ll}2 & -4 \\ 5 & 3\end{array}\right]B=\left[\begin{array}{ll}6 & 9 \\ 7 & 8\end{array}\right]$
ⓒ Multiply the matrix $A=\left[\begin{array}{ll}2 & -4 \\ 5 & 3\end{array}\right]$ by 5.

விடை தெரியப்படுத்து

ⓐ $A+B=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]+\left[\begin{array}{ll}e & f \\ g & h\end{array}\right]=\left[\begin{array}{ll}a+e & b+f \\ c+g & d+h\end{array}\right]$
ⓑ $A-B=\left[\begin{array}{ll}2 & -4 \\ 5 & 3\end{array}\right]-\left[\begin{array}{ll}6 & 9 \\ 7 & 8\end{array}\right]=\left[\begin{array}{ll}2-6 & -4-9 \\ 5-7 & 3-8\end{array}\right]=\left[\begin{array}{ll}-4 & -13 \\ -2 & -5\end{array}\right]$
ⓒ $5A=\left[\begin{array}{ll}5(2) & 5(-4) \\ 5(5) & 5(3)\end{array}\right]=\left[\begin{array}{ll}10 & -20 \\ 25 & 15\end{array}\right]$
Add the two matrices $A=\left[\begin{array}{ll}l & m \\ n & p\end{array}\right]B=\left[\begin{array}{ll}q & r \\ s & t\end{array}\right]$
Subtract the two matrices $A=\left[\begin{array}{ll}-3 & 2 \\ 1 & 0\end{array}\right]B=\left[\begin{array}{ll}-5 & 4 \\ 1 & 5\end{array}\right]$
Multiply the matrix $A=\left[\begin{array}{ll}2 & -4 \\ 5 & 3\end{array}\right]$ by –2
Find $2A+3B$ when $A=\left[\begin{array}{ll}1 & -6 \\ 4 & 8\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & -5 \\ 3 & -1\end{array}\right]$
Given matrix $A:$
1. ⓐWhat are the dimensions of matrix $A?$
2. ⓑWhat are the entries at ${a}_{31}$ and ${a}_{22}?$ \[A=[\begin{array}{llll}2 & & 1 & 0 \\ 2 & & 4 & 7 \\ 3 & & 1 & -2\end{array}]\]
விடை தெரியப்படுத்து
1. ⓐThe dimensions are $\ 3\times 3$ because there are three rows and three columns.
2. ⓑEntry ${a}_{31}$ is the number at row 3, column 1, which is 3. The entry ${a}_{22}$ is the number at row 2, column 2, which is 4. Remember, the row comes first, then the column.
Find the sum of $A$ and $B,$ given
\[A=[\begin{array}{ll}a & b \\ c & d\end{array}]\ \text{ and }B=[\begin{array}{ll}e & f \\ g & h\end{array}]\]

விடை தெரியப்படுத்து

Add corresponding entries.
\[\begin{array}{lllll}A+B=[\begin{array}{ll}a & b \\ c & d\end{array}]+[\begin{array}{ll}e & f \\ g & h\end{array}] \\ \ =[\begin{array}{lll}a+e & & b+f \\ c+g & & d+h\end{array}]\end{array}\]
Find the sum of $A$ and $B.$
\[A=[\begin{array}{ll}4 & 1 \\ 3 & 2\end{array}]\ \text{ and }B=[\begin{array}{ll}5 & 9 \\ 0 & 7\end{array}]\]

விடை தெரியப்படுத்து

Add corresponding entries. Add the entry in row 1, column 1, ${a}_{11},$ of matrix $A$ to the entry in row 1, column 1, ${b}_{11},$ of $B.$ Continue the pattern until all entries have been added.
\[\begin{array}{lllll}A+B=[\begin{array}{ll}4 & 1 \\ 3 & 2\end{array}]+[\begin{array}{ll}5 & 9 \\ 0 & 7\end{array}] \\ \ =[\begin{array}{lll}4+5 & & 1+9 \\ 3+0 & & 2+7\end{array}] \\ \ =[\begin{array}{ll}9 & 10 \\ 3 & 9\end{array}]\end{array}\]
Find the difference of $A$ and $B.$
\[A=[\begin{array}{ll}-2 & 3 \\ 0 & 1\end{array}]\ \text{ and }B=[\begin{array}{ll}8 & 1 \\ 5 & 4\end{array}]\]

விடை தெரியப்படுத்து

We subtract the corresponding entries of each matrix.
\[\begin{array}{lllll}A-B=[\begin{array}{ll}-2 & 3 \\ 0 & 1\end{array}]-[\begin{array}{ll}8 & 1 \\ 5 & 4\end{array}] \\ \ =[\begin{array}{lll}-2-8 & & 3-1 \\ 0-5 & & 1-4\end{array}] \\ \ =[\begin{array}{lll}-10 & & 2 \\ -5 & & -3\end{array}]\end{array}\]
Given $A$ and $B:$
1. ⓐFind the sum.
2. ⓑFind the difference.
\[A=[\begin{array}{lll}2 & -10 & -2 \\ 14 & 12 & 10 \\ 4 & -2 & 2\end{array}]\ \text{and }B=[\begin{array}{lll}6 & 10 & -2 \\ 0 & -12 & -4 \\ -5 & 2 & -2\end{array}]\]
விடை தெரியப்படுத்து
1. ⓐAdd the corresponding entries. \[\begin{array}{lllllllllllll} \\ A+B=[\begin{array}{lll}2 & \ -10 & \ -2 \\ 14 & \ 12 & \ 10 \\ 4 & \ -2 & \ 2\end{array}]+[\begin{array}{lll}6 & \ 10 & \ -2 \\ 0 & \ -12 & \ -4 \\ -5 & \ 2 & \ -2\end{array}] \\ \ \ \ \ \ \ \ =[\begin{array}{lll}2+6 & \ -10+10 & \ -2-2 \\ 14+0 & \ 12-12 & \ 10-4 \\ 4-5 & \ -2+2 & \ 2-2\end{array}] \\ \ \ \ \ \ \ \ =[\begin{array}{lll}8 & \ 0 & \ -4 \\ 14 & \ 0 & \ 6 \\ -1 & \ 0 & \ 0\end{array}]\end{array}\]
2. ⓑSubtract the corresponding entries. \[\begin{array}{lllllllllllll} \\ A-B=[\begin{array}{lll}2 & -10 & -2 \\ 14 & 12 & 10 \\ 4 & -2 & 2\end{array}]-[\begin{array}{lll}6 & 10 & -2 \\ 0 & -12 & -4 \\ -5 & 2 & -2\end{array}] \\ \ \ \ \ \ \ \ =[\begin{array}{lll}2-6 & \ -10-10 & \ -2+2 \\ 14-0 & \ 12+12 & \ 10+4 \\ 4+5 & \ -2-2 & \ 2+2\end{array}] \\ \ \ \ \ \ \ \ =[\begin{array}{lll}-4 & \ -20 & \ 0 \\ 14 & \ 24 & \ 14 \\ 9 & \ -4 & \ 4\end{array}]\end{array}\]
Add matrix $A$ and matrix $B.$
\[A=[\begin{array}{ll}2 & 6 \\ 1 & 0 \\ 1 & -3\end{array}]\ \text{ and }B=[\begin{array}{ll}3 & -2 \\ 1 & 5 \\ -4 & 3\end{array}]\]

விடை தெரியப்படுத்து

$A+B=[\begin{array}{l}2 \\ 1 \\ 1\end{array}\begin{array}{l}\ \ 6 \\ \text{}\text{}\text{}\ \ \ 0 \\ \ -3\end{array}]+[\ \begin{array}{l}\ 3 \\ 1 \\ -4\end{array}\begin{array}{l}\ -2 \\ \ \ 5 \\ \ \ \ 3\end{array}]$
$=[\begin{array}{l}2\ +\ 3 \\ 1\ \ +\ \ 1 \\ 1+(-4)\end{array}\ \ \ \begin{array}{l}6+(-2) \\ 0\ +\ 5 \\ -3\ \ +\ \ 3\end{array}]=[\begin{array}{l}\ 5 \\ \ 2 \\ -3\end{array}\ \ \ \begin{array}{l}4 \\ 5 \\ 0\end{array}]$
Multiply matrix $A$ by the scalar 3.
\[A=[\begin{array}{ll}8 & 1 \\ 5 & 4\end{array}]\]

விடை தெரியப்படுத்து

Multiply each entry in $A$ by the scalar 3.
\[\begin{array}{lll}3A=3[\begin{array}{ll}8 & \ 1 \\ 5 & \ 4\end{array}] \\ \ \ \ \ =[\begin{array}{ll}3⋅8 & \ 3⋅1 \\ 3⋅5 & \ 3⋅4\end{array}] \\ \ \ \ \ =[\begin{array}{ll}24 & 3 \\ 15 & 12\end{array}]\end{array}\]
Given matrix $B,$ find $-2B$ where
\[B=[\begin{array}{ll}4 & 1 \\ 3 & 2\end{array}]\]

விடை தெரியப்படுத்து

$-2B=[\begin{array}{ll}-8 & -2 \\ -6 & -4\end{array}]$
Find the sum $3A+2B.$
\[A=[\begin{array}{lll}1 & -2 & 0 \\ 0 & -1 & 2 \\ 4 & 3 & -6\end{array}]\ \text{and }B=[\begin{array}{lll}-1 & 2 & 1 \\ 0 & -3 & 2 \\ 0 & 1 & -4\end{array}]\]

விடை தெரியப்படுத்து

First, find $3A,$ then $2B.$
\[\begin{array}{lllllll}\begin{array}{lllllll} \\ \\ 3A=[\begin{array}{lll}3⋅1 & \ 3(-2) & \ 3⋅0 \\ 3⋅0 & \ 3(-1) & \ 3⋅2 \\ 3⋅4 & \ 3⋅3 & \ 3(-6)\end{array}]\end{array} \\ \ \ \ =[\begin{array}{lll}3 & \ -6 & \ 0 \\ 0 & \ -3 & \ 6 \\ 12 & \ 9 & \ -18\end{array}]\end{array}\]\[\begin{array}{lllllll}\begin{array}{lllllll} \\ \\ 2B=[\begin{array}{lll}2(-1) & \ 2⋅2 & \ 2⋅1 \\ 2⋅0 & \ 2(-3) & \ 2⋅2 \\ 2⋅0 & \ 2⋅1 & \ 2(-4)\end{array}]\end{array} \\ \ \ \ =[\begin{array}{lll}-2 & 4 & 2 \\ 0 & -6 & 4 \\ 0 & 2 & -8\end{array}]\end{array}\]
Now, add $3A+2B.$
\[\begin{array}{lllllllllllll} \\ \\ 3A+2B=[\begin{array}{lll}3 & -6 & 0 \\ 0 & -3 & 6 \\ 12 & 9 & -18\end{array}]+[\begin{array}{lll}-2 & 4 & 2 \\ 0 & -6 & 4 \\ 0 & 2 & -8\end{array}] \\ \ =[\begin{array}{lll}3-2 & \ -6+4 & 0+2 \\ 0+0 & \ -3-6 & 6+4 \\ 12+0 & \ 9+2 & -18-8\end{array}] \\ \ =[\begin{array}{lll}1 & \ -2 & 2 \\ 0 & \ -9 & 10 \\ 12 & \ 11 & -26\end{array}]\end{array}\]
Multiply matrix $A$ and matrix $B.$
\[A=[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}]\ \text{ and }B=[\begin{array}{ll}5 & 6 \\ 7 & 8\end{array}]\]

விடை தெரியப்படுத்து

First, we check the dimensions of the matrices. Matrix $A$ has dimensions $2\times 2$ and matrix $B$ has dimensions $2\times 2.$ The inner dimensions are the same so we can perform the multiplication. The product will have the dimensions $2\times 2.$
We perform the operations outlined previously.
Given $A$ and $B:$
1. ⓐ Find $AB.$
2. ⓑ Find $BA.$
\[A=[\begin{array}{lll}\begin{array}{lll}-1 & 2 & 3\end{array} \\ \begin{array}{lll}\ \ 4 & 0 & 5\end{array}\end{array}]\ \text{and }B=[\begin{array}{l}\ 5 \\ -4 \\ \ 2\end{array}\ \ \begin{array}{l}-1 \\ \ 0 \\ \ 3\end{array}]\]
விடை தெரியப்படுத்து
1. ⓐAs the dimensions of $A$ are $2\times 3$ and the dimensions of $B$ are $3\times 2,$ these matrices can be multiplied together because the number of columns in $A$ matches the number of rows in $B.$ The resulting product will be a $2\times 2$ matrix, the number of rows in $A$ by the number of columns in $B.$ \[\begin{array}{llllllll} \\ AB=[\begin{array}{lll}-1 & 2 & 3 \\ 4 & 0 & 5\end{array}]\ [\begin{array}{ll}5 & -1 \\ -4 & 0 \\ 2 & 3\end{array}] \\ =[\begin{array}{ll}-1(5)+2(-4)+3(2) & \ -1(-1)+2(0)+3(3) \\ 4(5)+0(-4)+5(2) & \ 4(-1)+0(0)+5(3)\end{array}] \\ =[\begin{array}{ll}-7 & 10 \\ 30 & 11\end{array}]\end{array}\]
2. ⓑThe dimensions of $B$ are $3\times 2$ and the dimensions of $A$ are $2\times 3.$ The inner dimensions match so the product is defined and will be a $3\times 3$ matrix. \[\begin{array}{llllllll} \\ BA=[\begin{array}{ll}5 & -1 \\ -4 & 0 \\ 2 & 3\end{array}]\ [\begin{array}{lll}-1 & 2 & 3 \\ 4 & 0 & 5\end{array}] \\ =[\begin{array}{lll}5(-1)+-1(4) & \ 5(2)+-1(0) & \ 5(3)+-1(5) \\ -4(-1)+0(4) & \ -4(2)+0(0) & \ -4(3)+0(5) \\ 2(-1)+3(4) & \ 2(2)+3(0) & \ 2(3)+3(5)\end{array}] \\ =[\begin{array}{lll}-9 & 10 & 10 \\ 4 & -8 & -12 \\ 10 & 4 & 21\end{array}]\end{array}\]
Let’s return to the problem presented at the opening of this section. We have , representing the equipment needs of two soccer teams.
Wildcats Mud Cats
Goals 6 10
Balls 30 24
Jerseys 14 20
We are also given the prices of the equipment, as shown in .
Goal $300
Ball $10
Jersey $30
We will convert the data to matrices. Thus, the equipment need matrix is written as
\[E=[\begin{array}{l}6 \\ 30 \\ 14\end{array}\ \ \ \begin{array}{l}10 \\ 24 \\ 20\end{array}]\]
The cost matrix is written as
\[C=[\begin{array}{lll}300 & 10 & 30\end{array}]\]
We perform matrix multiplication to obtain costs for the equipment.
\[\begin{array}{llllll} \\ \\ CE=[\begin{array}{lll}300 & 10 & 30\end{array}][\begin{array}{ll}6 & 10 \\ 30 & 24 \\ 14 & 20\end{array}] \\ =[\begin{array}{ll}300(6)+10(30)+30(14) & 300(10)+10(24)+30(20)\end{array}] \\ =[\begin{array}{ll}2,520 & 3,840\end{array}]\end{array}\]
The total cost for equipment for the Wildcats is $2,520, and the total cost for equipment for the Mud Cats is $3,840.
Find $AB-C$ given
\[A=[\begin{array}{lll}-15 & 25 & 32 \\ 41 & -7 & -28 \\ 10 & 34 & -2\end{array}],B=[\begin{array}{lll}45 & 21 & -37 \\ -24 & 52 & 19 \\ 6 & -48 & -31\end{array}],\text{and }C=[\begin{array}{lll}-100 & -89 & -98 \\ 25 & -56 & 74 \\ -67 & 42 & -75\end{array}].\]

விடை தெரியப்படுத்து

On the matrix page of the calculator, we enter matrix $A$ above as the matrix variable $[A],$ matrix $B$ above as the matrix variable $[B],$ and matrix $C$ above as the matrix variable $[C].$
On the home screen of the calculator, we type in the problem and call up each matrix variable as needed.
\[[A][B]-[C]\]
The calculator gives us the following matrix.
\[[\begin{array}{lll}-983 & \ -462 & \ 136 \\ 1,820 & \ 1,897 & \ -856 \\ -311 & \ 2,032 & \ 413\end{array}]\]
Can we add any two matrices together? If so, explain why; if not, explain why not and give an example of two matrices that cannot be added together.

விடை தெரியப்படுத்து

No, they must have the same dimensions. An example would include two matrices of different dimensions. One cannot add the following two matrices because the first is a $2\times 2$ matrix and the second is a $2\times 3$ matrix. $[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}]+[\begin{array}{lll}6 & 5 & 4 \\ 3 & 2 & 1\end{array}]$ has no sum.
Can we multiply any column matrix by any row matrix? Explain why or why not.
Can both the products $AB$ and $BA$ be defined? If so, explain how; if not, explain why.

விடை தெரியப்படுத்து

Yes, if the dimensions of $A$ are $m\times n$ and the dimensions of $B$ are $n\times m,$ both products will be defined.
Can any two matrices of the same size be multiplied? If so, explain why, and if not, explain why not and give an example of two matrices of the same size that cannot be multiplied together.
Does matrix multiplication commute? That is, does $AB=BA?$ If so, prove why it does. If not, explain why it does not.

விடை தெரியப்படுத்து

Not necessarily. To find $AB,$ we multiply the first row of $A$ by the first column of $B$ to get the first entry of $AB.$ To find $BA,$ we multiply the first row of $B$ by the first column of $A$ to get the first entry of $BA.$ Thus, if those are unequal, then the matrix multiplication does not commute.
$\frac{1}{2}C$
$C-0.5D$
$100D-10E$

விடை தெரியப்படுத்து

$[\begin{array}{lll}-840 & 650 & -530 \\ 330 & 360 & 250 \\ -10 & 900 & 110\end{array}]$
${A}^{2}$
${B}^{2}$

விடை தெரியப்படுத்து

$[\begin{array}{ll}1,400 & 700 \\ -1,400 & 700\end{array}]$
${C}^{2}$
${B}^{2}{A}^{2}$

விடை தெரியப்படுத்து

$[\begin{array}{ll}332,500 & 927,500 \\ -227,500 & 87,500\end{array}]$
${A}^{2}{B}^{2}$
${(AB)}^{2}$

விடை தெரியப்படுத்து

$[\begin{array}{ll}490,000 & 0 \\ 0 & 490,000\end{array}]$
${(BA)}^{2}$
${D}^{2}$

விடை தெரியப்படுத்து

$[\begin{array}{lll}-3 & -2 & -2 \\ -28 & 59 & 46 \\ -4 & 16 & 7\end{array}]$
${A}^{2}$
${D}^{3}$

விடை தெரியப்படுத்து

$[\begin{array}{lll}1 & -18 & -9 \\ -198 & 505 & 369 \\ -72 & 126 & 91\end{array}]$
${B}^{2}$

விடை தெரியப்படுத்து

$[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}]$

Symbols used here

imaginary unit
i² = −1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

matrix
A rectangular array of numbers; a linear map.

\det A,\ |A|

determinant
Scaling factor of area/volume under A; zero means singular.

A^{-1},\ A^{T}

inverse, transpose
The matrix that undoes A; A with rows and columns swapped.

\lambda

lambda (eigenvalue)
The factor by which an eigenvector is stretched: Av = λv.

\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|

dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: Matrices and Matrix Operations

Find the sum and difference of two matrices.
Find scalar multiples of a matrix.
Find the product of two matrices.
Write the augmented matrix for a system of equations (IA 4.5.1)
Add, subtract matrices and multiply a matrix by a scalar
To obtain the entry in row 1, column 1 of
To obtain the entry in row 1, column 2 of
To obtain the entry in row 1, column 3 of

Questions people ask

What does a determinant mean geometrically?

It is the factor by which the matrix scales area (2×2) or volume (3×3), with a negative sign if orientation flips. Zero means the matrix flattens space and cannot be undone.

What is an eigenvector?

A direction the matrix does not turn — it only stretches it by the eigenvalue. Along eigenvectors a complicated matrix acts like multiplication by a number.

Why is matrix multiplication not commutative?

Because a matrix is a transformation and AB means "do B, then A". Rotating then reflecting is not the same as reflecting then rotating.

உங்களை முயற்சிக்கவும்

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

மேலும் Linear Algebra

Determinants Matrix inverse Eigenvalues and eigenvectors Matrix multiplication Row reduction Vector spaces, span and linear independence Orthogonality, projections and least squares Diagonalisation and matrix powers Linear transformations and change of basis

	Wildcats	Mud Cats
Goals	6	10
Balls	30	24
Jerseys	14	20

Goal	$300
Ball	$10
Jersey	$30