Represent a system of linear equations as a single matrix equation AX = B in a vector variable

Learning goal: Represent a system of linear equations as a single matrix equation AX = B, where A is the coefficient matrix, X is a column vector variable holding the unknowns and B is the column vector of constants; translate in both directions between a system and its matrix equation; and verify a proposed solution by matrix multiplication.

Scope note. A.REI.8 is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full, as extension content for students on that path.

Why write a system as one equation. A system such as 2x + 3y = 8 and x − y = −1 is two equations in two unknowns. A matrix equation packs all of it into one statement, AX = B, with one unknown, the vector X. The payoff is conceptual and practical: the whole system becomes a single object that can be manipulated the way a linear equation ax = b is manipulated, and the next lesson uses this form to solve systems with a matrix inverse. The form also scales without change to three, four or more unknowns, where writing out substitutions becomes error-prone.

The three pieces. For a system of m linear equations in n unknowns, written with the unknowns in the same order in every equation and constants on the right: the coefficient matrix A is the m × n matrix whose row i holds the coefficients of equation i, in the order of the unknowns; the vector variable X is the n × 1 column vector whose entries are the unknowns, in that same order; the constant vector B is the m × 1 column vector of the right-hand sides. The matrix product AX is defined because the number of columns of A (n) equals the number of rows of X (n), and the product is m × 1, the same size as B, so the equation AX = B makes sense as an equality of two column vectors.

How matrix multiplication reproduces the system — a full model. Take 2x + 3y = 8 and x − y = −1. Then A = [[2, 3], [1, −1]], writing the matrix row by row, X = [x, y] as a column, and B = [8, −1] as a column. Compute AX by the row-times-column rule: the first entry is (row 1 of A)·X = 2x + 3y, and the second is (row 2 of A)·X = 1x + (−1)y = x − y. So AX is the column [2x + 3y, x − y], and setting it equal to B = [8, −1] says exactly 2x + 3y = 8 and x − y = −1, entry by entry. Nothing has been added or lost; the matrix equation is the system. Solve the system by substitution, x = y − 1, so 2(y − 1) + 3y = 8, 5y = 10, y = 2, x = 1. Now verify in matrix form: with X = [1, 2], AX = [2(1) + 3(2), 1(1) − 1(2)] = [8, −1] = B. A proposed solution is checked by multiplying it by A and comparing with B, one product for the whole system.

Order and zeros matter. The columns of A correspond to the unknowns in a fixed order, so every equation must be arranged with its unknowns in that order before the coefficients are read off. If an unknown is missing from an equation, its coefficient is 0 and a 0 must be written in that position; leaving a gap shifts every later coefficient into the wrong column. For the system 2x − z = 1, y + 3z = 4, x + y = 2, the coefficient matrix is A = [[2, 0, −1], [0, 1, 3], [1, 1, 0]], with X = [x, y, z] and B = [1, 4, 2]. Multiplying, row 1 gives 2x + 0y − z, row 2 gives 0x + y + 3z and row 3 gives x + y + 0z, which are the three left sides. Equations that are not in standard form must be rearranged first: y = 3x + 2 becomes −3x + y = 2, so its row is [−3, 1] with constant 2.

Reading a system from a matrix equation. The translation runs both ways. Given A = [[1, 2], [3, −5]], X = [x, y] and B = [4, −7], multiplying out AX gives [x + 2y, 3x − 5y], so the system is x + 2y = 4 and 3x − 5y = −7. Each row of A together with the matching entry of B is one equation.

Three unknowns. For x + y + z = 6, 2x − y + z = 3 and x + 2y − z = 2, the coefficient matrix is the 3 × 3 matrix A = [[1, 1, 1], [2, −1, 1], [1, 2, −1]], X = [x, y, z] and B = [6, 3, 2]. The earlier lesson found the solution (1, 2, 3) by elimination. Verify by one multiplication: AX = [1 + 2 + 3, 2 − 2 + 3, 1 + 4 − 3] = [6, 3, 2] = B. The check that previously took three separate substitutions is one matrix product.

The equation AX = B as an equation in the vector variable X. In ax = b with a ≠ 0, the solution is x = b/a, or more carefully x = a⁻¹b, multiplying both sides by the reciprocal of a. The matrix equation invites the same thought: if a matrix A⁻¹ exists with A⁻¹A = I, the identity matrix, then multiplying both sides of AX = B on the left by A⁻¹ gives X = A⁻¹B. That is the subject of the next lesson. Two cautions apply now. First, matrices do not always have inverses, so "divide by A" is not always possible. Second, matrix multiplication is not commutative and is defined only when sizes match: for a 2 × 2 matrix A and a 2 × 1 vector X, the product AX is defined and XA is not, since a 2 × 1 times a 2 × 2 has a mismatch between one column and two rows. So the vector variable is written on the right of A, and any multiplication used to solve must be applied on the left of both sides.

The misconception and why it is wrong. Students sometimes write the constants inside A as an extra column and call that AX = B. The matrix with the constants attached, [A | B], is the augmented matrix, which is a bookkeeping device for elimination, not a factor in a product. In the matrix equation, A holds coefficients only, X holds unknowns only, and B holds constants only; that separation is what allows AX to be computed and compared with B. A second error is to write X as a row instead of a column; then AX is not defined at all, because a 2 × 2 matrix cannot be multiplied by a 1 × 2 row on the right.

Worked translation to matrix form: Write 3x − 4y = 5 and 2x + y = −3 as a matrix equation. Both equations are in standard form with x before y. A = [[3, −4], [2, 1]], X = [x, y] (a column), B = [5, −3] (a column). Check by multiplying: AX = [3x − 4y, 2x + y], and AX = B is the original pair. Solving by elimination, add 4 times the second to the first: (3x − 4y) + 4(2x + y) = 5 + 4(−3), which expands to 3x − 4y + 8x + 4y = 5 − 12, so 11x = −7 and x = −7/11; then y = −3 − 2(−7/11) = −33/11 + 14/11 = −19/11. Verify in matrix form: AX = [3(−7/11) − 4(−19/11), 2(−7/11) + (−19/11)] = [(−21 + 76)/11, (−14 − 19)/11] = [55/11, −33/11] = [5, −3] = B.

Worked translation from matrix form with rearrangement: The system y = 3x + 2 and y = −x + 6 is not in standard form. Rearranged: −3x + y = 2 and x + y = 6. So A = [[−3, 1], [1, 1]], X = [x, y], B = [2, 6]. Subtracting the first equation from the second gives 4x = 4, x = 1, y = 5. Check: AX = [−3(1) + 5, 1 + 5] = [2, 6] = B. If the equations had been left as y = 3x + 2, reading off [3, 1] as a row would give the wrong matrix, because the 3x is on the other side of the equals sign.

Worked three-unknown system with missing terms: Represent 2x − z = 1, y + 3z = 4, x + y = 2 as AX = B and verify the solution (1, 1, 1). A = [[2, 0, −1], [0, 1, 3], [1, 1, 0]], X = [x, y, z], B = [1, 4, 2]. The zeros record the missing y in the first equation, the missing x in the second and the missing z in the third. Solving: from the third, y = 2 − x; from the first, z = 2x − 1; substituting into the second, 2 − x + 3(2x − 1) = 4, so 5x − 1 = 4, x = 1, y = 1, z = 1. Verify: AX = [2 − 0 − 1, 0 + 1 + 3, 1 + 1 + 0] = [1, 4, 2] = B.

Worked verification by multiplication only: Is X = [2, 1] a solution of AX = B with A = [[4, −1], [1, 3]] and B = [7, 5]? Multiply: AX = [4(2) − 1(1), 1(2) + 3(1)] = [8 − 1, 2 + 3] = [7, 5] = B. Yes. Is X = [1, 3]? AX = [4 − 3, 1 + 9] = [1, 10] ≠ B, so no; the first entry alone already fails.

Translation record: Make a two-column record. On the left write four systems: two pairs of linear equations in two unknowns (one of them not in standard form), one system of three equations in three unknowns with at least two missing terms, and one system given to you only as A, X and B that you must expand into equations. On the right write the matching matrix equation or system, showing the row-times-column products that prove the two forms say the same thing, and state the dimensions of A, X and B in each case.

Independent performance: Choose a solution first, for instance x = 2, y = −1, z = 3, and write three linear equations in x, y and z that it satisfies, with at least one unknown missing from at least one equation and at least one equation initially written with unknowns on both sides. Rearrange each equation into standard form, represent the system as a single matrix equation AX = B in the vector variable X, and state the dimensions of each piece. Verify your chosen solution by computing the matrix product AX and comparing it with B entry by entry. Then propose a second vector that is not a solution and show by multiplication which entry of AX fails to match B. Finish with a paragraph explaining why AX = B is one equation in the unknown vector X, why X must be a column written on the right of A, and how the matrix product reproduces every equation of the system. Success criteria: coefficients appear in the same variable order in every row, every missing unknown is recorded as 0, the product AX is computed with the row-times-column rule and written out, and the paragraph explains rather than restates.