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# Determinants

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Practice Determinants
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Finding Determinants of Matrices

A mathematical theorem states that a matrix is singular if and only if its determinant is zero. Is the following matrix singular?

$\begin{bmatrix}2 & 1 & 3\\0 & 2 & 1\\-1 & 3 & 0\end{bmatrix}$

### Watch This

Watch the first part of this video, until about 4:00.

### Guidance

Each square matrix has a real number value associated with it called its determinant. This value is denoted by $det \ A$ or $|A|$ .

Finding the Determinant of a $2 \times 2$ matrix:

$det\begin{bmatrix}a & b\\c & d\end{bmatrix} = \begin{vmatrix}{\color{red}a} & {\color{blue}b}\\{\color{blue}c} & {\color{red}d}\end{vmatrix} = {\color{red}ad}- {\color{blue}bc}$

Finding the Determinant of a $3 \times 3$ matrix: To begin, we will repeat the first two columns after matrix. Next, calculate the products and sums as shown below and find the difference. The result is the determinant of the $3 \times 3$ matrix.

Using the determinant to find the Area of a Triangle in the coordinate plane:

We can find the area of a triangle with vertices $(x_1,y_1), (x_2,y_2)$ and $(x_3,y_3)$ using the formula below

$A = \pm\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1\end{vmatrix},$ where the $\pm$ accounts for the possibility that the determinant could be negative but area should always be positive.

Using the calculator to find the determinant of a matrix: If you are using a TI-83 or TI-84, access the Matrix menu by either pressing MATRIX or ( $2^{nd} \ x^{-1}$ MATRIX). Now you can choose to EDIT matrix $A$ . Change the dimensions as needed and enter the data values. Now return to the home screen ( $2^{nd}$ MODE QUIT) and return to the MATRIX menu. Arrow over to MATH and select 1:det( by pressing ENTER. Go into the MATRIX menu once more to select 1: $[A]$ under the NAMES column. Press ENTER. Your screen should show $det([A]$ at this time. Press ENTER once more and the result will be your determinant. These directions work for square matrices of any size.

#### Example A

Find the $det\begin{bmatrix}3 & -4\\1 & 5\end{bmatrix}.$

Solution: Using the rule above for a $2 \times 2$ matrix, the determinant can be found as shown:

$det\begin{bmatrix}3 & -4\\1 & 5\end{bmatrix} = \begin{vmatrix}{\color{red}3} & {\color{blue}-4}\\{\color{blue}1} & {\color{red}5}\end{vmatrix} = {\color{red}(3)(5)}-{\color{blue}(-4)(1)} = {\color{red}15}-{\color{blue}(-4)} = 19$

#### Example B

Find the $det\begin{bmatrix}2 & -3 & 5\\-4 & 7 & 1\\3 & 8 & 6\end{bmatrix}.$

Solution:

First, we need to repeat the first two columns. Then we can find the diagonal products as shown:

$\begin{bmatrix}2 & -3 & 5\\-4 & 7 & 1\\3 & 8 & 6\end{bmatrix}\begin{matrix}2 & -3\\-4 & 7\\3 & 8\end{matrix}\\{\color{red}(2 \cdot 7 \cdot 6)+(-3 \cdot 1 \cdot 3)+(5\cdot-4\cdot8)} & \ {\color{red}= 84+-9+-160 = -85}\\= \ {\color{blue}(3\cdot7\cdot5)+(8\cdot 1\cdot 2)+(6\cdot-4\cdot-3)} & \ {\color{blue}= 105+16+72 = 193}\\{\color{red}-85}-{\color{blue}193} &= -278$

#### Example C

Find the area of the triangle with vertices (2, -1), (4, 5) and (8, 1)

Solution:

The first step is to set up our matrix and find the determinant as shown:

$\begin{bmatrix}2 & -1 & 1\\4 & 5 & 1\\8 & 1 & 1\end{bmatrix}\begin{matrix}2 & -1\\4 & 5\\8 & 1\end{matrix}\\{\color{red}(2\cdot5\cdot1)+(-1\cdot1\cdot8)+(1\cdot4\cdot1)} & \ {\color{red}=10+-8+4=6}\\= \ {\color{blue}(8\cdot5\cdot1)+(1\cdot1\cdot2)+(1\cdot4\cdot-1)} & \ {\color{blue}= 40+2-4=38}\\{\color{red}6}-{\color{blue}38} &= -32$

Now we can multiply this determinant, -32, by $-\frac{1}{2}$ (we will multiply by the negative in order to have a positive result) to get 16. So the area of the triangle is 16 $u^{2}$ .

Intro Problem Revisit To find the determinant, we first need to repeat the first two columns. Then we can find the diagonal products as shown:

$\begin{bmatrix}2 & 1 & 3\\0 & 2 & 1\\-1 & 3 & 0\end{bmatrix}\begin{matrix}2 & 1\\0 & 2\\-1 & 3\end{matrix}\\{\color{red}(2 \cdot 2 \cdot 0)+(1 \cdot 1 \cdot -1)+(3 \cdot 0 \cdot 3)} & \ {\color{red}= 0+ -1 + 0 = -1}\\= \ {\color{blue}(3 \cdot 2 \cdot -1)+(2 \cdot 1 \cdot 3)+(1 \cdot 0 \cdot 0)} & \ {\color{blue}= -6 + 6 + 0 = 0}\\{\color{red}-1}-{\color{blue}0} &= -1$

The determinant is not zero and therefore the matrix is not singular.

### Guided Practice

Find the determinants of the matrices below.

1. $\begin{bmatrix}-1 & 8\\2 & -9\end{bmatrix}$

2. $\begin{bmatrix}-2 & 4 & -3\\5 & -6 & 1\\-4 & 1 & -2\end{bmatrix}$

3. Find the area of the triangle with vertices (-5, 2), (8, -1) and (3, 9)

1. $\begin{vmatrix}{\color{red}-1} & {\color{blue}8}\\{\color{blue}2} & {\color{red}-9}\end{vmatrix} = {\color{red}(-1)(-9)}-{\color{blue}(8)(2)} = {\color{red}9}-{\color{blue}16} = -7$

2. $\begin{vmatrix}-2 & 4 & -3\\5 & -6 & 1\\-4 & 1 & -2\end{vmatrix}\begin{matrix}-2 & 4\\5 & -6\\-4 & 1\end{matrix}\\{\color{red}(-2\cdot-6\cdot-2)+(4\cdot-1\cdot-4)+(-3\cdot5\cdot1)} & \ {\color{red}= -24+16-15=-23}\\= \ {\color{blue}(-4\cdot-6\cdot-3)+(1\cdot1\cdot-2)+(-2\cdot5\cdot4)} & \ {\color{blue}= -72-2-40=-114}\\{\color{red}-23}-{\color{blue}(-114)} &= 91$

3. $\begin{vmatrix}-5 & 2 & 1\\8 & -1 & 1\\3 & 9 & 1\end{vmatrix}\begin{matrix}-5 & 2\\8 & -1\\3 & 9\end{matrix}\\= \ {\color{red}(5+6+72)}-{\color{blue}(-3 + -45 + 16)} = {\color{red}83}-{\color{blue}(-32)} = 115$

So the area is $\frac{1}{2}(115) = 57.5 \ u^{2}$ .

### Practice

1. .
$\begin{bmatrix}2 & -1\\3 & 5\end{bmatrix}$
1. .
$\begin{bmatrix}-3 & -2\\6 & 4\end{bmatrix}$
1. .
$\begin{bmatrix}5 & 10\\-3 & -7\end{bmatrix}$
1. .
$\begin{bmatrix}-4 & 8\\3 & 5\end{bmatrix}$
1. .
$\begin{bmatrix}11 & 3\\7 & 2\end{bmatrix}$
1. .
$\begin{bmatrix}9 & 3\\2 & -1\end{bmatrix}$
1. .
$\begin{bmatrix}1 & -1 & 3\\5 & 0 & 6\\-4 & 8 & 2\end{bmatrix}$
1. .
$\begin{bmatrix}5 & -2 & 1\\6 & 1 & 0\\-3 & 2 & 4\end{bmatrix}$
1. .
$\begin{bmatrix}4 & -1 & 2\\3 & 0 & 1\\-2 & 5 & 6\end{bmatrix}$

Find the area of each triangle with vertices given below.

1. (2, -1), (-5, 2) and (0, 6)
2. (-8, 12), (10, 5) and (1, -4)
3. (-7, 2), (8, 0) and (3, -4)

Find the value of $a$ in the matrices below.

1. .
$\begin{vmatrix}a & 3\\8 & 2\end{vmatrix} = -10$
1. .
$\begin{vmatrix}4 & a\\3 & 5\end{vmatrix} = -1$
1. .
$\begin{vmatrix}2 & -1 & 3\\4 & 5 & 2\\-3 & 0 & a\end{vmatrix} = 23$

### Vocabulary Language: English

determinant

determinant

The determinant is a single number descriptor of a square matrix. The determinant is computed from the entries of the matrix, and has many properties and interpretations explored in linear algebra.
Sarrus’ rule

Sarrus’ rule

Sarrus’ rule is a memorization technique that enables you to compute the determinant of matrices efficiently.
Square matrix

Square matrix

A square matrix is a matrix in which the number of rows equals the number of columns.