Course 8.2Lesson 1 of 3

Evaluating a determinant

A determinant is a single number that belongs to a square matrix. For a $2 \times 2$ matrix it is $ad - bc$ . For larger matrices it is found by expanding along a row or a column, using minors and cofactors with the sign pattern $+\ -\ +$ . Any row or column gives the same value, so choose the one with the most zeros. A triangular matrix has the product of its diagonal entries as its determinant.

about 25 min1 simulation1 practice question1 Python program5 questions to check yourself

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What you will be able to do

find the determinant of a $2 \times 2$ matrix and understand it as an area;
find the minor and the cofactor of any entry;
expand a $3 \times 3$ determinant along any row or column;
pick the row or column that makes the work shortest.

A number from a square matrix

Every square matrix has a determinant, a single number. We write it $\det A$ or $|A|$ , with straight bars. The bars do not mean a positive number here. A determinant can be negative. A matrix that is not square has no determinant.

For a $2 \times 2$ matrix the rule is

\displaystyle \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc.

Multiply along the main diagonal, then subtract the product along the other diagonal. For example $

\displaystyle \begin{vmatrix} 3 & 5 \\ 2 & 4 \end{vmatrix}

= 12 - 10 = 2$.

The determinant as an area

Take the two columns of the matrix as arrows from the origin. They are the sides of a parallelogram. The absolute value of the determinant is the area of that parallelogram. The sign tells the turning direction: it is positive when the second arrow is anticlockwise from the first.

A parallelogram on a grid, drawn from the origin with two sides given by the arrows (3, 1) in blue and (1, 2) in red. The fourth corner is at (4, 3). The parallelogram is shaded, and the label inside says area equals 5. A caption says the determinant with columns 3, 1 and 1, 2 is 3 times 2 minus 1 times 1, which is 5.

With columns $(3, 1)$ and $(1, 2)$ , the determinant is $3 \cdot 2 - 1 \cdot 1 = 5$ , and the parallelogram has area $5$ . If the two arrows lie along one line, there is no parallelogram, the area is $0$ , and so is the determinant.

Minors and cofactors

For a larger matrix, we reduce the work to smaller determinants.

The minor $M_{ij}$ of the entry $a_{ij}$ is the determinant left when row $i$ and column $j$ are deleted. The cofactor is the minor with a sign:

\displaystyle C_{ij} = (-1)^{i + j} M_{ij}.

The signs form a chessboard pattern, with $+$ in the top left corner:

\displaystyle \begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}.

For $A =

\displaystyle \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{bmatrix}

$, the minor$ M_{23} $is$

\displaystyle \begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix}

= 8 - 14 = -6 $. The cofactor is$ C_{23} = (-1)^{2+3}(-6) = 6$.

Expanding a $3 \times 3$ determinant

Expand along row $i$ : multiply each entry of the row by its own cofactor, and add.

\displaystyle |A| = a_{i1}C_{i1} + a_{i2}C_{i2} + a_{i3}C_{i3}.

The same works down any column. All six choices (three rows and three columns) give the same number. Along the first row,

\displaystyle \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}.

Take the matrix $D =

\displaystyle \begin{bmatrix} 2 & 1 & 3 \\ 0 & 4 & -1 \\ 5 & 2 & 1 \end{bmatrix}

$. Along the first row:

\displaystyle |D| = 2(4 + 2) - 1(0 + 5) + 3(0 - 20) = 12 - 5 - 60 = -53.

A shortcut called Sarrus' rule exists for $3 \times 3$ determinants, but it does not work for larger ones. The expansion works for every size, so we use it.

Choose the row or column with the most zeros

A zero entry removes a whole term. Look at $

\displaystyle \begin{bmatrix} 3 & 0 & 2 \\ 1 & 5 & 0 \\ 4 & 2 & 1 \end{bmatrix}

$. Column$ 3 $contains a$ 0$. Expanding down it:

Two panels. On the left a 3 by 3 grid of signs: the top row is plus, minus, plus, the middle row is minus, plus, minus, and the bottom row is plus, minus, plus, like a chessboard. On the right the matrix with rows 3 0 2, then 1 5 0, then 4 2 1 has its third column circled. The caption says to expand along the column with the most zeros.

\displaystyle 2 \cdot C_{13} + 0 \cdot C_{23} + 1 \cdot C_{33} = 2 \cdot (+1)\begin{vmatrix} 1 & 5 \\ 4 & 2 \end{vmatrix} + 1 \cdot (+1)\begin{vmatrix} 3 & 0 \\ 1 & 5 \end{vmatrix} = 2(-18) + 15 = -21.

For a triangular matrix, expand again and again along the column or row of zeros. The result is the product of the diagonal entries. For example $

\displaystyle \begin{vmatrix} 1 & 0 & 0 \\ 5 & 2 & 0 \\ 7 & 4 & 3 \end{vmatrix}

= 1 \cdot 2 \cdot 3 = 6$.

A worked example

Evaluate $

\displaystyle \begin{vmatrix} 2 & 0 & 1 \\ 3 & 4 & -2 \\ 1 & 5 & 6 \end{vmatrix}

$ along its first row.

The entries. $a_{11} = 2$ , $a_{12} = 0$ , $a_{13} = 1$ .
The cofactors. $C_{11} = +

\displaystyle \begin{vmatrix} 4 & -2 \\ 5 & 6 \end{vmatrix}

= 24 + 10 = 34 $.$ C_{13} = +

\displaystyle \begin{vmatrix} 3 & 4 \\ 1 & 5 \end{vmatrix}

= 15 - 4 = 11 $. The term for$ a_{12} $is$ 0 $, so$ C_{12}$ is not needed.

The sum. $2 \cdot 34 + 0 + 1 \cdot 11 = 79$ .

Practise with new numbers

Try it in Python

The program finds a determinant by expanding along the first row. It calls itself on the smaller matrices that remain, and it checks the answer against numpy.

Common mistakes

Forgetting the sign pattern. The middle term of the first-row expansion carries a minus sign.
Adding the products in a $2 \times 2$ determinant. It is $ad - bc$ .
Mixing up minor and cofactor. The cofactor is the minor with the sign $(-1)^{i+j}$ .
Multiplying a row of $A$ by the cofactors of a different row. That gives $0$ , not $|A|$ .
Thinking $|A|$ is always positive. It can be negative or zero.

Remember these

\displaystyle \begin{vmatrix} a & b \\ c & d \end{vmatrix}

= ad - bc$. It is the area of the parallelogram on the columns, with a sign.

$C_{ij} = (-1)^{i+j}M_{ij}$ , and $|A|$ is the sum of (entry times cofactor) along any row or column.
Expand along the row or column with the most zeros.
The determinant of a triangular matrix is the product of the diagonal.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

Find $

\displaystyle \begin{vmatrix} 4 & 7 \\ 2 & 5 \end{vmatrix}

Find the minor $M_{23}$ and the cofactor $C_{23}$ of $

\displaystyle \begin{bmatrix} 2 & 1 & 4 \\ 0 & 3 & 5 \\ 1 & 6 & 7 \end{bmatrix}

\displaystyle \begin{vmatrix} 2 & 1 \\ 1 & 6 \end{vmatrix}

= 11 $, and$ C_{23} = (-1)⁵ \cdot 11 = -11$.

Find $

\displaystyle \begin{vmatrix} 2 & 3 \\ 4 & 6 \end{vmatrix}

$. What do you notice about the rows?

Find $

\displaystyle \begin{vmatrix} 3 & 0 & 0 \\ 1 & 2 & 0 \\ 4 & 5 & 6 \end{vmatrix}

Which row or column would you expand $

\displaystyle \begin{vmatrix} 5 & 2 & 0 \\ 1 & 3 & 0 \\ 4 & 7 & 2 \end{vmatrix}

$ along, and what is its value?

\displaystyle \begin{vmatrix} 5 & 2 \\ 1 & 3 \end{vmatrix}

= 2 \cdot 13 = 26$.

References

NCERT, Mathematics Textbook for Class XII, Part I, Chapter 4 (Determinants), the sections on the determinant, minors and cofactors.
OpenStax, Precalculus 2e, Section 9.8 (Solving Systems with Cramer's Rule), which evaluates $2 \times 2$ and $3 \times 3$ determinants.

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Problems (3)

Evaluate the determinant below by expanding along the first row, and give the cofactors C₁₁ and C₁₃…Easy · High School (11–12) · by StemPy Lab

Let A = 1, 2, 3; 0, 4, 5; 1, 0, 6. Find |A|…Medium · High School (11–12) · by StemPy Lab

For which values of x is x + 1, 3, 5; 2, x + 2, 5; 2, 3, x + 4 = 0? …Medium · High School (11–12) · by StemPy Lab

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