Linear Algebra · Book or PDF

A First Course in Linear Algebra

A.G. Hamilton · Study material generated by AI from this edition. Not affiliated with the publisher.

Chapter 7 · pp. 74–89

Chapter 7: Determinants

A determinant is a single number worked out from a square array of numbers. This chapter shows how to compute determinants of different sizes, how row operations change them, and why a non-zero determinant tells you a matrix is invertible. It ends with cofactors, the adjoint matrix, and a rule for the determinant of a product.

Determinants versus matrices, and the 2 x 2 case

A matrix is a rectangular array of numbers. A determinant is one number calculated from a square matrix. Mixing these up is a common mistake. We write $\det A$ for the determinant of the matrix $A$ . (p. 74)

For a 2 x 2 matrix with rows $(a, b)$ and $(c, d)$ :

$\det A = ad - bc$

Multiply down the main diagonal, then subtract the product of the other diagonal. For rows $(1, 2)$ and $(3, 4)$ the value is $1\cdot 4 - 2\cdot 3 = -2$ . The same expression $ad-bc$ turns up when you solve two simultaneous equations in $x$ and $y$ by elimination. (p. 74)

Evaluating 3 x 3 determinants

For a 3 x 3 array with rows $(a_1, a_2, a_3)$ , $(b_1, b_2, b_3)$ , $(c_1, c_2, c_3)$ , the definition is:

$a_1b_2c_3 + a_2b_3c_1 + a_3b_1c_2 - a_3b_2c_1 - a_1b_3c_2 - a_2b_1c_3$

Diagonal method. Write the first two columns again to the right of the array. Add the products along the three left-to-right diagonals. Subtract the products along the three right-to-left diagonals. This reproduces the formula above. (pp. 74–76)

Example: for rows $(1,2,3)$ , $(0,1,2)$ , $(1,0,3)$ , the added products are $3 + 4 + 0$ and the subtracted ones are $3 + 0 + 0$ . The value is $7 - 3 = 4$ . This shortcut works only for 3 x 3 determinants. It does not extend to 4 x 4 or larger. (pp. 74–76)

Expansion by a row or column. This method is the one that extends to larger sizes. Take each entry of a chosen row or column. Multiply it by the smaller determinant left after deleting that entry's row and column. Then combine the terms using alternating signs. For the first row:

$a_1(b_2c_3-b_3c_2) - a_2(b_1c_3-b_3c_1) + a_3(b_1c_2-b_2c_1)$

The signs follow a checkerboard: + - + on the top row, - + - on the next, then + - +. The minus on the middle term is the easiest thing to forget. Any row or column gives the same value. So pick one with zeros, because each zero removes a whole term. (pp. 76–78)

Example: rows $(2,0,1)$ , $(3,1,4)$ , $(1,0,5)$ . The second column has two zeros. Only the entry 1 in the middle survives, with sign +. Its smaller determinant is $2\cdot 5 - 1\cdot 1 = 9$ . So the determinant is 9.

How row operations affect a determinant

There are three kinds of elementary row operation, and each has a simple effect. (pp. 78–79)

Swapping two rows flips the sign of the determinant.
Multiplying one row by $k$ multiplies the determinant by $k$ . A determinant of 5 becomes 10 if one row is doubled.
Adding a multiple of one row to another row leaves the determinant unchanged. Adding 3 times row 1 to row 2 keeps the value at 5.

The book checks these by multiplying out the 3 x 3 case. They hold for all sizes. Matching rules hold for column operations too. (pp. 78–79, 82)

Two identical rows give zero

If two rows of a determinant are the same, its value is 0. To see why, expand along the third row, the one that is not repeated. Each smaller determinant still has two equal rows, and the terms cancel to 0. This same fact is used to prove that adding a multiple of a row changes nothing. The extra expression that appears in that proof is itself a determinant with a repeated row. (pp. 79–80)

Larger determinants

A 4 x 4 determinant is defined by expanding along the first row. Each of the four entries multiplies the 3 x 3 determinant left after deleting its row and column. The signs alternate: + - + -. Expanding along another row or column gives the same value, as long as the alternating signs are kept. Determinants of 5 x 5 and larger are defined the same way, and the row rules still apply. (p. 80)

Triangular matrices and the elimination shortcut

The determinant of a triangular matrix is the product of its main-diagonal entries. For an upper triangular matrix, expand down the first column. Only the top-left entry is non-zero, and it multiplies a smaller triangular determinant. Repeat this and a 4 x 4 matrix gives $a_1b_2c_3d_4$ . Lower triangular matrices work the same way. (p. 81)

This gives a fast method for big determinants. (pp. 81–83)

Apply Gaussian elimination and write down how each row operation changes the determinant.
Read off the determinant of the triangular result.
Undo the recorded changes.

Example: rows $(1,3,-1)$ , $(2,0,1)$ , $(1,1,4)$ . Subtracting multiples of one row from another leaves the determinant unchanged. Dividing a row by $-6$ divides the determinant by $-6$ . The final triangular matrix has determinant 4. So the original is $4 \times (-6) = -24$ . If a row swap had been used, the sign would also flip. (pp. 81–83)

Determinants and invertibility

A $p \times p$ matrix $A$ is invertible exactly when $\det A \neq 0$ . After elimination, the triangular matrix has determinant zero only if its last row is all zeros. That means fewer than $p$ non-zero rows, so the rank is below $p$ . A non-zero determinant therefore matches full rank. This adds a fifth condition to the Equivalence Theorem. The conditions are: (p. 82)

$A$ is invertible.
The rank of $A$ is $p$ .
The columns of $A$ are linearly independent.
$Ax = 0$ has only the solution $x = 0$ .
$\det A \neq 0$ .

Cofactors and the adjoint

Write the entry in row $i$ , column $j$ as $a_{ij}$ . Its cofactor $A_{ij}$ is the smaller determinant found by deleting row $i$ and column $j$ , multiplied by the checkerboard sign for that position. With cofactors, the first-row expansion becomes:

$\det A = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$

To build the adjoint matrix $\mathrm{adj}\,A$ , replace every entry by its cofactor, then transpose the result. By hand this is very long, even for 3 x 3, so it is mostly of theoretical use. (pp. 83–84)

The inverse formula and the product rule

If $A$ is invertible, then:

$A^{-1} = \frac{1}{\det A}\,\mathrm{adj}\,A$

Here is why. Look at the product $A(\mathrm{adj}\,A)$ . Each diagonal entry is a row expansion of $A$ , so it equals $\det A$ . Each off-diagonal entry is an expansion of a determinant with two equal rows, so it is 0. Hence $A(\mathrm{adj}\,A) = (\det A)I$ . Dividing by $\det A$ , which is non-zero because $A$ is invertible, gives the formula. (pp. 84–85)

For $p \times p$ matrices $A$ and $B$ , the determinant of a product is the product of the determinants:

$\det(AB) = (\det A)(\det B)$

The book states this without proof. One consequence is that if either $A$ or $B$ has determinant 0, then $AB$ has determinant 0, so $AB$ is singular. (p. 86)

Check yourself

Find the value of the determinant with rows $(5, 2)$ and $(3, 1)$ . Explain in one sentence how a determinant differs from a matrix.
A 3 x 3 determinant has value 7. What is the new value if you swap two rows, then multiply one row by 3?
Why is it wise to expand a determinant along a row or column that contains zeros?
Use the triangular-matrix rule to find the determinant of the matrix with rows $(2, 5, 1)$ , $(0, 3, 4)$ , $(0, 0, -1)$ .
Elimination on a 4 x 4 matrix ends with a zero row. What does this tell you about its determinant and about whether it is invertible?