What the determinant measures
For a 2 × 2 matrix, \(|A|\) is the (signed) area-scaling factor of the transformation it represents. Zero determinant means the plane is squashed onto a line – which is why such a matrix cannot be undone (no inverse).
Chapter Overview
Every square matrix has a number attached to it – its determinant. The determinant tells us whether a matrix is invertible, gives the area of a triangle from its vertices, and decides whether a system of linear equations has a unique solution. This chapter develops determinants of order up to 3, minors and cofactors, the adjoint and inverse of a matrix, and the matrix method for solving linear equations.
Topics
Key Concepts
- \(|A| \ne 0 \iff A\) is invertible (non-singular).
- \(|kA| = k^n|A|\), \(|AB| = |A||B|\), \(|A^T| = |A|\).
- \(A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|I\) and \(A^{-1} = \dfrac{1}{|A|}\operatorname{adj}A\).
- \(AX = B\) has the unique solution \(X = A^{-1}B\) when \(|A| \ne 0\).
Formulas
2 × 2 determinant
Determinants · Determinant of a Square Matrix\[\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc\]
Main diagonal product minus the other diagonal product.
Scalar multiple
Determinants · Determinant of a Square Matrix\[|kA| = k^{n}|A| \quad (A \text{ of order } n)\]
Each of the \(n\) rows contributes a factor \(k\). Also \(|AB| = |A||B|\) and \(|A^T| = |A|\).
Determinant rules
Determinants · Determinant of a Square Matrix\[|A^T|=|A|,\quad |AB|=|A|\,|B|,\quad |kA|=k^n|A|,\quad |A^{-1}|=\frac{1}{|A|}\]
Area of a triangle
Determinants · Area of a Triangle using Determinants\[\Delta = \frac12\left|\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right|\]
Zero determinant ⇔ the three points are collinear.
Cofactor
Determinants · Minors and Cofactors\[A_{ij} = (-1)^{i+j}M_{ij}\]
\(M_{ij}\): determinant left after deleting row \(i\) and column \(j\).
Adjoint identity
Determinants · Adjoint and Inverse of a Matrix\[A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|\,I\]
Consequences: \(A^{-1} = \dfrac{\operatorname{adj}A}{|A|}\) and \(|\operatorname{adj}A| = |A|^{n-1}\).
Inverse using adjoint
Determinants · Adjoint and Inverse of a Matrix\[A^{-1} = \frac{1}{|A|}\operatorname{adj}A, \quad |A| \ne 0\]
Also \(|A^{-1}| = 1/|A|\).
Adjoint identities
Determinants · Adjoint and Inverse of a Matrix\[A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I,\qquad |\operatorname{adj}A|=|A|^{n-1}\]
Matrix method
Determinants · Solving Systems of Linear Equations\[AX = B \;\Rightarrow\; X = A^{-1}B \quad (|A| \ne 0)\]
Note the order: \(A^{-1}\) multiplies \(B\) from the left.
Key Points
Invertible ⇔ non-zero determinant
Determinants · Determinant of a Square MatrixA square matrix is invertible exactly when \(|A| \ne 0\).
Expand along zeros
Determinants · Determinant of a Square MatrixChoose the row or column with the most zeros – fewer calculations, fewer slips.
Two answers for a given area
Determinants · Area of a Triangle using DeterminantsWhen the area is given, solve \(\det = +2\Delta\) and \(\det = -2\Delta\). Expect two values.
Check |A| first
Determinants · Adjoint and Inverse of a MatrixBefore finding an inverse, evaluate |A|. If it is zero, the inverse does not exist – say so and stop.
Verify by substitution
Determinants · Solving Systems of Linear EquationsAfter solving AX = B, substitute the values back into one or two original equations. It takes 30 seconds and catches most errors.
Consistency test (AX = B)
Determinants · Solving Systems of Linear Equations- \(|A|\neq0\): unique solution \(X=A^{-1}B\).
- \(|A|=0\) and \((\operatorname{adj}A)B\neq O\): no solution (inconsistent).
- \(|A|=0\) and \((\operatorname{adj}A)B=O\): infinitely many solutions or none – check further.
Common Mistakes
Writing |kA| = k|A|
Determinants · Determinant of a Square MatrixFor a 3 × 3 matrix with \(|A| = 4\), claiming \(|2A| = 8\).
Correct: \(|kA| = k^n|A|\), so \(|2A| = 2^3 \times 4 = 32\).
Ignoring the sign pattern
Determinants · Determinant of a Square MatrixExpanding \(a_{11}M_{11} + a_{12}M_{12} + a_{13}M_{13}\) with all plus signs.
Correct: use + − + along row 1 (chessboard pattern), i.e. multiply by cofactors, not minors.
Scalar multiple of a matrix
Determinants · Determinant of a Square MatrixFor a \(3\times3\) matrix, \(|2A|=2^3|A|=8|A|\), not \(2|A|\). A scalar multiplies every row.
Reporting a negative area
Determinants · Area of a Triangle using DeterminantsWriting “Area = −12 square units”.
Correct: Area \(= \tfrac12|\det|\) is always non-negative: 12 square units.
Sign of the cofactor
Determinants · Minors and Cofactors\(A_{ij}=(-1)^{i+j}M_{ij}\). The sign pattern is a chessboard starting with + in the top-left corner.
Forgetting to transpose the cofactor matrix
Determinants · Adjoint and Inverse of a MatrixTaking the cofactor matrix itself as adj A.
Correct: \(\operatorname{adj}A = (\text{cofactor matrix})^T\). The (1,2) entry of adj A is the cofactor \(A_{21}\).
Writing X = BA⁻¹
Determinants · Solving Systems of Linear EquationsMultiplying the constants column on the wrong side.
Correct: from \(AX = B\), pre-multiply by \(A^{-1}\): \(X = A^{-1}B\). \(BA^{-1}\) is not even defined for a 3 × 1 column \(B\).
Solved Examples
Expanding a 3 × 3 determinant
Determinants · Determinant of a Square MatrixEvaluate \(\begin{vmatrix} 2 & 3 & 1 \\ 0 & 1 & 4 \\ 1 & 0 & 2 \end{vmatrix}\).
- Expand along row 1: \(2(2 - 0) - 3(0 - 4) + 1(0 - 1)\).
- \(= 4 + 12 - 1\).
Final Answer: \(15\)
Equation of a line using determinants
Determinants · Area of a Triangle using DeterminantsFind the equation of the line through \((1, 3)\) and \((4, 9)\) using determinants.
- A point \((x, y)\) is on the line iff \(\begin{vmatrix} x & y & 1 \\ 1 & 3 & 1 \\ 4 & 9 & 1 \end{vmatrix} = 0\).
- Expand: \(x(3 - 9) - y(1 - 4) + (9 - 12) = -6x + 3y - 3\).
- \(-6x + 3y - 3 = 0 \iff y = 2x + 1\).
Final Answer: \(y = 2x + 1\)
Solving a 3 × 3 system
Determinants · Solving Systems of Linear EquationsSolve \(x + y + z = 4\), \(x - y + z = 2\), \(2x + y - z = 1\) by the matrix method.
- \(A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{bmatrix}\), \(B = \begin{bmatrix} 4 \\ 2 \\ 1 \end{bmatrix}\); \(|A| = 6\).
- \(\operatorname{adj}A = \begin{bmatrix} 0 & 2 & 2 \\ 3 & -3 & 0 \\ 3 & 1 & -2 \end{bmatrix}\).
- \(X = \frac{1}{6}\operatorname{adj}A\,B = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}\).
- Check in the first equation: 1 + 1 + 2 = 4. ✓
Final Answer: \(x = 1,\ y = 1,\ z = 2\)
Practice & Tests
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