Lesson 14: Finding Determinants

New Concept

Every square matrix is associated with one real number called the determinant of the matrix.

What’s next

Next, you’ll master the technique for calculating the determinant of $2 \times 2$ and $3 \times 3$ matrices and use it to solve problems.

The determinant of a $2 \times 2$ square matrix is found by subtracting the product of the entries on one diagonal from the product of the entries on the other diagonal. The formula is:

Evaluate $\begin{vmatrix} -5 & 2 \\ 4 & 6 \end{vmatrix} = (-5)(6) - (4)(2) = -30 - 8 = -38$.
Find $x$ if $\begin{vmatrix} x+1 & 2 \\ 5 & 3 \end{vmatrix} = 5$. Solution: $(3)(x+1) - (5)(2) = 5 \implies 3x + 3 - 10 = 5 \implies 3x = 12 \implies x=4$.

Think of it as a diagonal duel! To find the magic number for a 2x2 matrix, you multiply the numbers on the main diagonal from top-left to bottom-right. Then, from that result, you subtract the product of the other diagonal, from bottom-left to top-right. This simple ad - cb formula is your key to unlocking the determinant's value!

To find the determinant of a $3 \times 3$ matrix, multiply each element in one row by its minor (the determinant of the matrix that remains when you cover the element's row and column), and combine using alternating signs.

Evaluate $\begin{vmatrix} 2 & 1 & 0 \\ 3 & -1 & 4 \\ -2 & 5 & 6 \end{vmatrix} = 2\begin{vmatrix} -1 & 4 \\ 5 & 6 \end{vmatrix} - 1\begin{vmatrix} 3 & 4 \\ -2 & 6 \end{vmatrix} + 0\begin{vmatrix} 3 & -1 \\ -2 & 5 \end{vmatrix} = 2(-26) - 1(26) + 0 = -78$.
Evaluate $\begin{vmatrix} 1 & 3 & 5 \\ 2 & 4 & 6 \\ 0 & 1 & 2 \end{vmatrix} = 1\begin{vmatrix} 4 & 6 \\ 1 & 2 \end{vmatrix} - 3\begin{vmatrix} 2 & 6 \\ 0 & 2 \end{vmatrix} + 5\begin{vmatrix} 2 & 4 \\ 0 & 1 \end{vmatrix} = 1(2) - 3(4) + 5(2) = 0$.

To crack a 3x3 matrix, break it into smaller 2x2 puzzles. Choose a row, typically the first one. For each element, cover its row and column to find its 'minor'—the determinant of the 2x2 grid that remains. Multiply the element by its minor, then combine the results using an alternating plus-minus-plus pattern. It's that simple!

A second method for a $3 \times 3$ determinant involves repeating the first two columns to the right of the matrix. Add the products of the three downward-sloping diagonals, and subtract the sum of the products of the three upward-sloping diagonals.

Evaluate $\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$. Rewrite: $\begin{array}{ccccc} 1 & 2 & 3 & 1 & 2 \\ 4 & 5 & 6 & 4 & 5 \\ 7 & 8 & 9 & 7 & 8 \end{array}$. Result: $((45)+(84)+(96)) - ((105)+(48)+(72)) = 225 - 225 = 0$.
Evaluate $\begin{vmatrix} 3 & -1 & 2 \\ 0 & 5 & -3 \\ 1 & 4 & -2 \end{vmatrix}$. Rewrite: $\begin{array}{ccccc} 3 & -1 & 2 & 3 & -1 \\ 0 & 5 & -3 & 0 & 5 \\ 1 & 4 & -2 & 1 & 4 \end{array}$. Result: $((-30)+(-3)+0) - (10+(-36)+0) = -33 - (-26) = -7$.

This is a fantastic visual shortcut for 3x3 determinants, no minors needed! First, copy the first two columns and place them next to the matrix. Then, multiply along the three downward diagonals and add their products. Do the same for the three upward diagonals. Finally, subtract the total of the 'up' products from the 'down' products.

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ can be found using the formula:

Find the area of a triangle with vertices $(2,4), (5,1), (-1,-2)$. Area = $|\frac{1}{2} \begin{vmatrix} 2 & 5 & -1 \\ 4 & 1 & -2 \\ 1 & 1 & 1 \end{vmatrix}| = |\frac{1}{2}(2(3) - 5(6) + (-1)(3))| = |\frac{-27}{2}| = 13.5$ square units.

Believe it or not, determinants can find a triangle's area! Just take the coordinates of its three vertices. Plug the x-values into the top row of a 3x3 matrix and the y-values into the middle row. Fill the bottom row with 1s, find the determinant, multiply by 1/2, and take the absolute value since area can't be negative.