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Determinants
Introduction
Determinant
Practice: Evaluating Determinants
Area of a Triangle
Practice: Area and Collinearity
Minors and Cofactors
Practice: Minors and Cofactors
Adjoint and Inverse of a Matrix
Practice: Adjoint and Inverse
Applications of Determinants and Matrices
Practice: Solving Systems with Matrices
Miscellaneous Examples
Chapter Review: Mixed Practice
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Glossary

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DeterminantsDeterminant

Reading time: ~30 min

Here is the general idea behind the number you just met. To every square matrix A = [a_{ij}] of order n we attach one number (real or complex), called the determinant of A. Think of it as a machine that takes in a square matrix and gives out a number: if M is the set of square matrices and K the set of numbers, the determinant is a function f : M \to K with f(A) = k.

The determinant of A is written |A|, \det A or \Delta. For a matrix written out in full, the square brackets turn into straight bars:

A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \;\Rightarrow\; |A| = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}

For Aman's first pair of clues, A = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} and |A| = 1 \times (-1) - 2 \times 2 = -5.

Tip

|A| is read as "determinant of A", not "modulus of A". A determinant can be negative, as −5 shows.

A matrix of order 2 × 3 , because only matrices have determinants.

Determinant of a matrix of order one

The smallest case first. A matrix of order one holds a single number, A = [a], and its determinant is that number itself: |A| = a.

So for A = [-7], the determinant is |A| = . The bars here do not strip off the minus sign.

Determinant of a matrix of order two

For a matrix of order 2,

A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}

the determinant is

|A| = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{21}a_{12}

Multiply along the leading diagonal, then subtract the product along the other diagonal. It is exactly the number from the opening, where a_{11}a_{22} - a_{21}a_{12} was a_1b_2 - a_2b_1.

The number has a picture too. Read the two columns of A as arrows from the origin — the blue arm is column 1, the green arm is column 2 — and complete them into a parallelogram.

Drag the tips of the two arms and watch the parallelogram and |A|. Complete all three missions.

So the size of |A| is the area of the parallelogram made by the two columns, and its sign records the turn from column 1 to column 2: anticlockwise gives a positive determinant, clockwise a negative one. When one column is a multiple of the other, the parallelogram is flat and |A| = 0 — the same thing that happened to the parallel lines in the introduction.

Two Worked Examples

Example 1 : Evaluate \begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix}.

Solution : Leading diagonal minus the other diagonal:

\begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix} = 2 \times 2 - 4 \times (-1) = 4 + 4 =

Example 2 : Evaluate \begin{vmatrix} x & x + 1 \\ x - 1 & x \end{vmatrix}.

Solution : The determinant is x \cdot x - (x + 1)(x - 1), and (x + 1)(x - 1) = .

So the determinant is x^2 - (x^2 - 1) = — the same for every value of x.

Note: multiplying every entry by k

Take A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}, so |A| = 2 - 8 = -6. Then 2A = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix} and |2A| = 8 - 32 = -24 = 4|A|. Each product in the formula has two entries, and each entry picked up a factor 2 — so for any 2 × 2 matrix, |kA| = k^2|A|.

Speed Round: Cross-Multiply

Leading diagonal product minus the other diagonal product — and watch the signs of negative entries. After each answer you'll see the working.

Get five right to move on.

Determinant of a matrix of order 3 × 3

A 3 × 3 determinant is worked out by breaking it into 2 × 2 determinants. This is called expanding the determinant along a row or a column. Take

|A| = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}

Expansion along the first row (R1). Walk along R1. For each element, strike out its row and its column and multiply the element by the 2 × 2 determinant that is left. The element a_{1j} also gets the sign (-1)^{1+j}.

Step 1: a_{11} sits in row 1 and column 1, so its sign is (-1)^{1+1} = , and its part is a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix}.

Step 2: a_{12} gets the sign (-1)^{1+2} = , so its part is -a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix}.

Step 3: a_{13} gets the sign (-1)^{1+3} = +1, so its part is a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}.

Adding the three parts:

\begin{aligned} |A| = {} & a_{11}(a_{22}a_{33} - a_{32}a_{23}) \\ & - a_{12}(a_{21}a_{33} - a_{31}a_{23}) \\ & + a_{13}(a_{21}a_{32} - a_{31}a_{22}) \end{aligned}

Multiplied out, this is six products, and each one takes exactly one element from every row and every column:

\begin{aligned} |A| = {} & a_{11}a_{22}a_{33} - a_{11}a_{32}a_{23} - a_{12}a_{21}a_{33} \\ & + a_{12}a_{31}a_{23} + a_{13}a_{21}a_{32} - a_{13}a_{31}a_{22} \end{aligned}

Expansion along the second row (R2) works the same way, with the signs −, +, −:

\begin{aligned} |A| = {} & -a_{21}(a_{12}a_{33} - a_{32}a_{13}) \\ & + a_{22}(a_{11}a_{33} - a_{31}a_{13}) \\ & - a_{23}(a_{11}a_{32} - a_{31}a_{12}) \end{aligned}

Multiply it out and rearrange: you get the very same six products. Expanding along the first column (C1) gives them again. So it does not matter which row or column you expand along — the value of the determinant is always the same.

The Sign Chessboard

Remembering (-1)^{i+j} for every position is easier as a picture. The element a_{ij} gets + when i + j is even and − when i + j is odd.

Build the sign pattern of a 3 × 3 determinant: tap a square to switch its sign until every square shows the sign (-1)^{i+j}.

Read the board along R2 and you get the signs of the R2 expansion, ; down C3 you get .

Tip: let the zeros do the work

  1. Expand along the row or column that contains the most zeros.
  2. An element that is 0 contributes 0, so you never need to work out the 2 × 2 determinant beside it.

Example 3 : Evaluate the determinant \Delta = \begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix}.

Choose a row or a column to expand along; the lab writes out every term with its sign. Complete all three missions.

Solution : The third column has two zeros, so expand along C3. Only a_{13} = 4 survives, and its sign is (-1)^{1+3} = +1:

\Delta = 4\begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} - 0 + 0 = 4(-1 - 12) =

Example 4 : Evaluate \Delta = \begin{vmatrix} 0 & \sin\alpha & -\cos\alpha \\ -\sin\alpha & 0 & \sin\beta \\ \cos\alpha & -\sin\beta & 0 \end{vmatrix}.

Solution : Expand along R1. The first element is 0, so only two 2 × 2 determinants are needed:

\Delta = 0 - \sin\alpha \begin{vmatrix} -\sin\alpha & \sin\beta \\ \cos\alpha & 0 \end{vmatrix} + (-\cos\alpha)\begin{vmatrix} -\sin\alpha & 0 \\ \cos\alpha & -\sin\beta \end{vmatrix}

The first of them is 0 - \sin\beta\cos\alpha, that is .

The second is \sin\alpha\sin\beta - 0, that is .

So \Delta = -\sin\alpha(-\sin\beta\cos\alpha) - \cos\alpha(\sin\alpha\sin\beta) = \sin\alpha\sin\beta\cos\alpha - \cos\alpha\sin\alpha\sin\beta =

Example 5 : Find the values of x for which \begin{vmatrix} 3 & x \\ x & 1 \end{vmatrix} = \begin{vmatrix} 3 & 2 \\ 4 & 1 \end{vmatrix}.

Solution : The left-hand side is 3 \times 1 - x \times x = , and the right-hand side is 3 - 8 = .

So 3 - x^2 = -5, that is x^2 = 8, and x = .

Note: |kA| for a 3 × 3 matrix

Each of the six products in a 3 × 3 determinant has three entries. Multiply every entry of A by k and each product picks up k^3, so |kA| = k^3|A|. For example, if |A| = 4 then |3A| = 27 \times 4 = 108.

Sina