Unit 7: Systems of Equations and Matrices · ~25 min
Objective
Compute the determinant of 2×2 and 3×3 matrices and apply Cramer's Rule to solve a system of linear equations.
Gaussian elimination in Lesson 7.4 solves a system by working through it step by step. The determinant takes a completely different approach: it's a single number, computed directly from a matrix's entries, that answers the yes/or/no question "does this system have exactly one solution?" — and, through Cramer's Rule, hands you that solution in one formula.
The determinant of a 2×2 matrix
For a matrix A =
a
b
c
d
the determinant, written det(A) or |A|, is
det(A) = ad − bc — the product of the main diagonal minus the product of the other diagonal.
Geometrically, |det(A)| is the area of the parallelogram formed by treating the matrix's two columns as vectors starting at the origin — it measures how much the matrix stretches or shrinks area.
The unit square maps to a parallelogram whose area is |det(A)|
The determinant of a 3×3 matrix
Expand along the first row, using each entry's minor — the 2×2 determinant left after deleting that entry's row and column — with alternating signs:
det = a(ei − fh) − b(di − fg) + c(dh − eg) for the matrix with rows (a, b, c), (d, e, f), (g, h, i).
Cramer's Rule
For a system of two equations in x and y, let D be the determinant of the coefficient matrix, \(D_{x}\) the determinant with the x-column replaced by the constants, and \(D_{y}\) the determinant with the y-column replaced by the constants:
x = \(\frac{D_{x}}{D}\) and y = \(\frac{D_{y}}{D}\) — valid whenever D ≠ 0.
D = 0 — Cramer's Rule fails; the system has either no solution or infinitely many, and matrices alone can't distinguish which without further work.
The same pattern extends to three variables: \(D_{z}\) replaces the z-column with the constants, and z = \(\frac{D_{z}}{D.}\)
Worked Example 1 · Cramer's Rule for a 2×2 system
ProblemSolve 3x + 2y = 12 and x − 2y = −4 using Cramer's Rule.
1D =
3
2
1
−2
= 3(−2) − 2(1) = −6 − 2 = −8.
2\(D_{x}\) (replace the x-column with the constants) =
12
2
−4
−2
= 12(−2) − 2(−4) = −24 + 8 = −16.
3\(D_{y}\) (replace the y-column with the constants) =
1.Find the determinant of [[4, 3], [2, 5]].
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4(5) − 3(2) = 20 − 6 = 14.
2.Find the determinant of the identity matrix [[1, 0], [0, 1]].
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1(1) − 0(0) = 1 — the identity matrix always has determinant 1, since it leaves area unchanged.
3.Use Cramer's Rule to solve 2x + y = 7 and x − y = 2.
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D = 2(−1) − 1(1) = −3. \(D_{x}\) = 7(−1) − 1(2) = −9, so x = \(\frac{-9}{-3}\) = 3. \(D_{y}\) = 2(2) − 7(1) = −3, so y = \(\frac{-3}{-3}\) = 1. Solution: (3, 1).
4.What does D = 0 tell you about a system, and what can't Cramer's Rule tell you in that case?
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D = 0 means the system does not have exactly one solution — it has either no solution or infinitely many. Cramer's Rule can't distinguish which; you'd need elimination (Lesson 7.1) or row reduction (Lesson 7.4) to determine that.
5.Find the determinant of [[2, 0, 1], [3, 1, 0], [1, 2, 1]].
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