Determinants
A single number that decides whether a matrix is invertible, a system is solvable, and a triangle has area
- What a determinant is, and how to evaluate \(2\times2\) and \(3\times3\) ones.
- Minors and cofactors, and expansion along any row or column.
- The properties that let you simplify a determinant before computing it.
- The area of a triangle and the test for collinearity.
- The adjoint, the inverse \(A^{-1}=\tfrac{1}{\det A}\operatorname{adj}A\), and Cramer's rule.
- How the determinant decides the consistency of a system, including the homogeneous case.
What Is a Determinant?
To every square matrix we attach a single number, its determinant, written \(\det A\) or \(|A|\). Small as it is, this number answers big questions: whether the matrix can be inverted, whether a linear system has a unique solution, and — geometrically — how much the matrix scales area or volume. The previous chapter left invertibility hanging on this one quantity; here we finally pin it down.
"Down-product minus up-product." Everything larger is built from this by the cofactor expansion below. A determinant is only defined for square matrices.
Minors & Cofactors
To break a large determinant into smaller ones we delete a row and a column and look at what remains.
The minor \(M_{ij}\) is the determinant left after deleting row \(i\) and column \(j\). The cofactor attaches the sign \((-1)^{i+j}\) — a checkerboard of \(+,-,+\) starting from the top-left.
Expansion of a 3×3 Determinant
A determinant equals the sum of each entry of any one row (or column) times its cofactor. You may expand along whichever line has the most zeros — the answer is the same.
Properties of Determinants
The properties are where determinants become powerful: they let you transform a determinant into a simpler equivalent one before expanding, and they prove identities almost by inspection.
| Property | Effect on \(\det\) |
|---|---|
| Transpose: \(\det A^{T}=\det A\) | rows and columns are interchangeable |
| Swap two rows/columns | sign changes |
| Two identical rows/columns | determinant is \(0\) |
| Multiply a row by \(k\) | determinant multiplied by \(k\) |
| \(R_i\to R_i+kR_j\) | determinant unchanged |
| Product: \(\det(AB)=\det A\,\det B\) | multiplicative |
| Triangular matrix | product of diagonal entries |
Multiplying a single row by \(k\) multiplies the determinant by \(k\). But multiplying the entire \(n\times n\) matrix by \(k\) scales every one of its \(n\) rows, so \(\det(kA)=k^{n}\det A\). Confusing these two is a frequent slip.
Area of a Triangle
Determinants carry geometric meaning. The signed area swept by coordinate vectors is exactly a determinant — which gives a clean formula for the area of a triangle from its vertices, and an instant test for three points lying on a line.
The three points are collinear exactly when this determinant is zero — no area, no triangle. The same determinant set to zero is the equation of the line through two given points.
The Adjoint & the Inverse
Cofactors do more than evaluate a determinant — assembled correctly they build the inverse. The adjoint is the transpose of the matrix of cofactors.
So \(A\) is invertible iff \(\det A\neq0\). Two useful corollaries for an \(n\times n\) matrix: \(\det(\operatorname{adj}A)=(\det A)^{\,n-1}\) and \(\det(A^{-1})=\dfrac{1}{\det A}\).
Cramer's Rule
For a square system \(AX=B\) with \(\det A\neq0\), each unknown is a ratio of determinants — no matrix inversion required.
Here \(A_i\) is \(A\) with its \(i\)-th column replaced by the column \(B\). The rule applies only when \(\Delta=\det A\neq0\) — precisely the case of a unique solution.
Consistency of a System
The single number \(\Delta=\det A\) classifies the whole behaviour of a non-homogeneous system \(AX=B\).
| Condition | Conclusion |
|---|---|
| \(\Delta\neq0\) | consistent — a unique solution (Cramer) |
| \(\Delta=0\) and \((\operatorname{adj}A)B\neq O\) | inconsistent — no solution |
| \(\Delta=0\) and \((\operatorname{adj}A)B=O\) | consistent — infinitely many solutions |
Homogeneous Systems
A system \(AX=O\) (all constants zero) is always consistent — the zero vector solves it. The only question is whether other solutions exist.
If \(\det A\neq0\), the only solution is the trivial one \(X=O\). The condition \(\det A=0\) is therefore the gateway to non-trivial solutions — a fact used constantly to find unknown parameters.
Strategies & Standard Results
A few habits make determinants fast and reliable.
Use \(R_i\to R_i+kR_j\) to create zeros, then expand along the new line of zeros.
If every entry of a row shares a factor, pull it out front — it scales the whole determinant.
If two rows or columns become identical (or proportional), the determinant is zero on sight.
Putting It to Work
Problem. Evaluate \(\begin{vmatrix}2&-1&3\\0&4&1\\5&2&0\end{vmatrix}\).
Solution. Expand along the first column (it has a zero):
Problem. Show \(\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}=(b-a)(c-a)(c-b)\).
Solution. Apply \(R_2\to R_2-R_1\) and \(R_3\to R_3-R_1\), factor \((b-a)\) and \((c-a)\) from the new rows:
Problem. Are the points \((1,2),(3,8),(5,14)\) collinear?
Solution. Test the area determinant:
The determinant is zero, so the three points are collinear.
Problem. Find \(A^{-1}\) for \(A=\begin{bmatrix}1&2\\3&4\end{bmatrix}\).
Solution. \(\det A=-2\); the adjoint swaps the diagonal and negates the off-diagonal:
Problem. Solve \(2x+y=5,\ x-3y=-1\).
Solution. \(\Delta=\begin{vmatrix}2&1\\1&-3\end{vmatrix}=-7\). Replace columns by the constants:
Problem. For which \(\lambda\) does \(x+y+z=0,\ x+\lambda y+z=0,\ x+y+\lambda z=0\) have a non-trivial solution?
Solution. Set the coefficient determinant to zero:
Only at \(\lambda=1\) (where the equations coincide) are non-trivial solutions possible.
Chapter Summary
\(2\times2\) is \(ad-bc\); larger ones expand by cofactors along any row or column.
\(C_{ij}=(-1)^{i+j}M_{ij}\); \(\det A=\sum_j a_{ij}C_{ij}\).
Swap flips sign, equal rows give 0, \(R_i\to R_i+kR_j\) is free, \(\det(kA)=k^n\det A\).
Area \(=\tfrac12|\Delta|\); \(\Delta=0\) means the three points are collinear.
\(A^{-1}=\tfrac{1}{\det A}\operatorname{adj}A\); \(\det(\operatorname{adj}A)=(\det A)^{n-1}\).
Cramer when \(\Delta\neq0\); homogeneous has non-trivial solutions iff \(\Delta=0\).
Problems
Simplify with the properties before expanding wherever you can. Difficulty rises down the list.
- Evaluate \(\begin{vmatrix}3&1&2\\-1&2&4\\0&5&1\end{vmatrix}\).
- Find \(x\) if \(\begin{vmatrix}x&2\\3&x\end{vmatrix}=10\).
- Find the area of the triangle with vertices \((1,1),(4,2),(2,5)\).
- Show that \(\begin{vmatrix}b+c&a&a\\b&c+a&b\\c&c&a+b\end{vmatrix}=4abc\).
- Without expanding, prove \(\begin{vmatrix}1&a&b+c\\1&b&c+a\\1&c&a+b\end{vmatrix}=0\).
- If \(A\) is a \(3\times3\) matrix with \(\det A=4\), find \(\det(2A)\) and \(\det(\operatorname{adj}A)\).
- Solve by Cramer's rule: \(x+y+z=6,\ x-y+z=2,\ 2x+y-z=1\).
- For which \(k\) is the system \(x+ky+3z=0,\ 4x+3y+kz=0,\ 2x+y+2z=0\) having a non-trivial solution?
- Examine the consistency of \(2x+3y=5,\ 4x+6y=11\) using the determinant test.
- Prove the product property \(\det(AB)=\det A\,\det B\) for two \(2\times2\) matrices.
- If \(\alpha,\beta,\gamma\) are the roots of \(x^3+px+q=0\), evaluate \(\begin{vmatrix}\alpha&\beta&\gamma\\\beta&\gamma&\alpha\\\gamma&\alpha&\beta\end{vmatrix}\).
- Find all \(\lambda\) for which the system \(x+y+z=1,\ x+2y+4z=\lambda,\ x+4y+10z=\lambda^2\) is consistent.