Linear Systems with Matrices

1. What Is a Linear System in Matrix Form?

A system of linear equations is a set of equations that share the same unknowns. Matrices give it a compact form: bundle the coefficients into a matrix A, the unknowns into a column vector x, and the right-hand sides into a column vector b, and the whole system becomes one equation:

A x = b

For example, the system 2x + y = 5 and x + 3y = 10 becomes A = [[2, 1], [1, 3]], x = [x, y]ᵀ, b = [5, 10]ᵀ. Solving the system means finding the vector x that satisfies this single matrix equation.

연립일차방정식은 계수를 행렬 A, 미지수를 열벡터 x, 우변을 열벡터 b로 묶으면 하나의 식 Ax = b로 표현됩니다. 예를 들어 2x + y = 5, x + 3y = 10은 A = [[2, 1], [1, 3]], b = [5, 10]ᵀ 입니다. 연립방정식을 푸는 것은 이 식을 만족하는 x를 찾는 것입니다.

2. Three Ways to Solve Ax = b

• Inverse method. When A is square and invertible (det ≠ 0), multiply both sides by A⁻¹: x = A⁻¹b. Clean for small systems, but computing the inverse is costly for large ones.
• Gaussian elimination. Row-reduce the augmented matrix [A | b] to echelon form, then back-substitute. This is the workhorse method and scales well.
• Cramer’s rule. Each unknown is a ratio of determinants: xᵢ = det(Aᵢ) / det(A), where Aᵢ replaces column i with b. Elegant for 2×2 and 3×3, impractical beyond that.

Ax = b는 세 가지로 풉니다: ① 역행렬법 x = A⁻¹b (det ≠ 0일 때), ② 가우스 소거법(첨가행렬 [A | b]를 행 줄임 후 후진대입 — 가장 실용적), ③ 크라메르 공식 xᵢ = det(Aᵢ)/det(A) (2×2·3×3에 적합).

3. Worked Example (Inverse Method)

Solve 2x + y = 5 and x + 3y = 10. Here A = [[2, 1], [1, 3]] and b = [5, 10]ᵀ.

First the determinant: det(A) = 2·3 − 1·1 = 5 (non-zero, so a unique solution exists). The 2×2 inverse is:

A⁻¹ = (1/5) [[3, −1], [−1, 2]]

Then x = A⁻¹b:

• x = (1/5)(3·5 − 1·10) = (1/5)(15 − 10) = 1
• y = (1/5)(−1·5 + 2·10) = (1/5)(−5 + 20) = 3

Check: 2(1) + 3 = 5 ✓ and 1 + 3(3) = 10 ✓.

2x + y = 5, x + 3y = 10에서 A = [[2, 1], [1, 3]], det(A) = 5 (≠ 0 → 유일해). A⁻¹ = (1/5)[[3, −1], [−1, 2]]이고 x = A⁻¹b = [1, 3]ᵀ. 검산: 2·1 + 3 = 5 ✓, 1 + 3·3 = 10 ✓.

4. When Does a Solution Exist?

The determinant of a square system decides everything:

• det(A) ≠ 0 — the matrix is invertible and there is exactly one solution.
• det(A) = 0 — the matrix is singular; the system has either no solution (inconsistent) or infinitely many (dependent equations).

To tell those last two cases apart — or to handle non-square systems — compare the rank of A with the rank of the augmented matrix [A | b]. A solution exists exactly when those two ranks are equal, and it is unique only when that rank also equals the number of unknowns.

정사각 계에서는 det(A) ≠ 0이면 유일해, det(A) = 0이면 해가 없거나 무수히 많습니다. 두 경우의 구분이나 비정사각 계는 A와 첨가행렬 [A | b]의 계수(rank)를 비교합니다 — 두 계수가 같아야 해가 존재하고, 그 값이 미지수 수와 같을 때만 유일합니다.

5. Frequently Asked Questions

How do you write a system of equations as a matrix? Collect the coefficients into A, the unknowns into x, and the constants into b, giving Ax = b.

When does Ax = b have a unique solution? When A is square and det(A) ≠ 0 — equivalently, when A is invertible.

What if det(A) = 0? The system has no unique solution: either no solution at all or infinitely many. Use rank to decide which.

계수를 A, 미지수를 x, 상수를 b로 묶어 Ax = b로 씁니다. A가 정사각이고 det(A) ≠ 0(= 가역)일 때 유일해를 가집니다. det(A) = 0이면 해가 없거나 무수히 많으며, 계수로 구분합니다.

Ready to practice? Drill linear systems and the rest of linear algebra on C:Matrix, or review the full matrix reference and the related determinant, inverse matrix, and rank.

실전 연습은 C:Matrix에서, 전체 개념은 행렬 레퍼런스와 행렬식, 역행렬, 계수(rank) 문서에서 이어집니다.