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Determinants

Number calculated from the entries in a matrix primarily through multiplication and subtraction.

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

Determinant of a Matrix

Each square matrix has a real number value associated with it called its determinant. This value is denoted by \begin{align*}det \ A\end{align*} or \begin{align*}|A|\end{align*}.

Finding the Determinant of a \begin{align*}2 \times 2\end{align*} matrix:

\begin{align*}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}\end{align*}

Finding the Determinant of a \begin{align*}3 \times 3\end{align*} 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 \begin{align*}3 \times 3\end{align*} 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 \begin{align*}(x_1,y_1), (x_2,y_2)\end{align*} and \begin{align*}(x_3,y_3)\end{align*} using the formula below

\begin{align*}A = \pm\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix},\end{align*} where the \begin{align*}\pm\end{align*} 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 (\begin{align*}2^{nd} \ x^{-1}\end{align*} MATRIX). Now you can choose to EDIT matrix \begin{align*}A\end{align*}. Change the dimensions as needed and enter the data values. Now return to the home screen (\begin{align*}2^{nd}\end{align*} 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:\begin{align*}[A]\end{align*} under the NAMES column. Press ENTER. Your screen should show \begin{align*}det([A]\end{align*} at this time. Press ENTER once more and the result will be your determinant. These directions work for square matrices of any size.

Let's solve the following problems relating to determinants of matrices.

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

Using the rule above for a \begin{align*}2 \times 2\end{align*} matrix, the determinant can be found as shown:

\begin{align*}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\end{align*}

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

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

\begin{align*}\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\end{align*}

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

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

\begin{align*}\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\end{align*}

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

Examples

Example 1

Earlier, you were asked to determine whether the following matrix is singular.

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

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

\begin{align*}\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\end{align*}

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

Example 2

Find the determinant of the matrix below:

\begin{align*}\begin{bmatrix} -1 & 8\\ 2 & -9\end{bmatrix}\end{align*}

\begin{align*}\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\end{align*}

Example 3

Find the determinant of the matrix below:
\begin{align*}\begin{bmatrix} -2 & 4 & -3\\ 5 & -6 & 1\\ -4 & 1 & -2\end{bmatrix}\end{align*}

\begin{align*}\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\end{align*}

Example 4

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

\begin{align*}\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\end{align*}

So the area is \begin{align*}\frac{1}{2}(115) = 57.5 \ u^{2}\end{align*}.

Review

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

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 \begin{align*}a\end{align*} in the matrices below.

1. .
\begin{align*}\begin{vmatrix} a & 3\\ 8 & 2\end{vmatrix} = -10\end{align*}
1. .
\begin{align*}\begin{vmatrix} 4 & a\\ 3 & 5\end{vmatrix} = -1\end{align*}
1. .

\begin{align*}\begin{vmatrix} 2 & -1 & 3\\ 4 & 5 & 2\\ -3 & 0 & a\end{vmatrix} = 23\end{align*}

To see the Review answers, open this PDF file and look for section 4.7.

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Vocabulary Language: English

TermDefinition
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 is a memorization technique that enables you to compute the determinant of matrices efficiently.
Square matrix A square matrix is a matrix in which the number of rows equals the number of columns.