Inverse Matrices: Solve a Two-Variable System and Verify Both Equations

The matrix method for simultaneous equations depends on three decisions: keep the variables in a fixed order, check that the coefficient matrix has an inverse, and multiply on the correct side. These worked examples using inverse matrices concentrate on that algorithm. Every answer is then checked in both original equations, so an attractive-looking matrix cannot hide an incorrect solution.

We use real numbers and two linear equations in two unknowns. The equations are already supplied; building equations from a story is a separate skill. No geometric transformation interpretation is needed. In the compact notation below, a semicolon separates matrix rows: [a, b; c, d] has first row a, b and second row c, d. A column [x; y] has x above y.

Build AX = B without changing the variable order

For equations ax + by = e and cx + dy = f, choose X = [x; y]. The coefficient matrix is A = [a, b; c, d], and the right-hand column is B = [e; f]. Then AX = B. Each row of the matrix equation reproduces one original equation: the first row gives ax + by, while the second gives cx + dy.

Rearrange both equations into the same order before copying coefficients. A missing variable has coefficient zero. A minus sign belongs to its coefficient, and an expression such as y − 3x must be read as −3x + y if your column order is x, then y. Changing only one row to y, then x creates a different system.

Check the determinant before finding an inverse

For A = [a, b; c, d], the determinant is det A = ad − bc. If this number is nonzero, A⁻¹ = (1/det A)[d, −b; −c, a]. Swap the two main-diagonal entries, change the signs of the other two entries, and multiply every entry by the reciprocal determinant. The superscript −1 means inverse matrix, not taking the reciprocal of each original entry.

Multiplying AX = B on the left by A⁻¹ gives A⁻¹AX = A⁻¹B. Because A⁻¹A is the identity matrix I = [1, 0; 0, 1], the result is X = A⁻¹B. The inverse operation cancels the coefficient matrix while leaving the unknown column.

If det A = 0, stop the inverse method. Division by zero is undefined. The equations might describe the same line, giving infinitely many solutions, or parallel distinct lines, giving no solution. A zero determinant by itself does not distinguish these cases.

Worked case 1: integer solution with a negative determinant

Solve 2x + y = 11 and x − y = 1. With X = [x; y], A = [2, 1; 1, −1] and B = [11; 1]. The determinant is 2(−1) − 1(1) = −3. Therefore A⁻¹ = (−1/3)[−1, −1; −1, 2].

Multiply each inverse row by the right-hand column:

X = (−1/3)[(−1)(11) + (−1)(1); (−1)(11) + 2(1)] = (−1/3)[−12; −9] = [4; 3].

Thus x = 4 and y = 3. Substitution gives 2(4) + 3 = 11 and 4 − 3 = 1. Both checks pass; checking just the first equation would not establish that the pair solves the system.

We can also verify the inverse itself. The product [2, 1; 1, −1][−1, −1; −1, 2] is [−3, 0; 0, −3]: its entries are −2 − 1, −2 + 2, −1 + 1, and −1 − 2. Multiplication by −1/3 gives I. This is an independent check on the sign and scale of the inverse.

Worked case 2: a minus sign in a coefficient

Solve 3x − 2y = 4 and 2x + y = 5. The matrices are A = [3, −2; 2, 1] and B = [4; 5]. Calculate det A = 3(1) − (−2)(2) = 7. Notice that subtracting the negative product adds four.

A⁻¹ = (1/7)[1, 2; −2, 3]. Therefore X = (1/7)[1(4) + 2(5); (−2)(4) + 3(5)] = (1/7)[14; 7] = [2; 1].

The solution is x = 2, y = 1. The first equation gives 6 − 2 = 4; the second gives 4 + 1 = 5. As another check, A[1, 2; −2, 3] = [3 + 4, 6 − 6; 2 − 2, 4 + 3] = [7, 0; 0, 7]. Dividing by seven produces the identity.

Worked case 3: accept a fractional solution

Solve 4x + 3y = 7 and −2x + 5y = −10. Here A = [4, 3; −2, 5], B = [7; −10], and det A = 4(5) − 3(−2) = 26.

A⁻¹ = (1/26)[5, −3; 2, 4]. Thus X = (1/26)[5(7) + (−3)(−10); 2(7) + 4(−10)] = (1/26)[65; −26] = [5/2; −1].

So x = 5/2 and y = −1. Keep the fraction exact until checking: 4(5/2) + 3(−1) = 10 − 3 = 7, and −2(5/2) + 5(−1) = −5 − 5 = −10. A fractional answer is not evidence of an error. The equations, rather than an expectation of whole numbers, determine the solution.

Why multiplication order matters

Let P = [1, 1; 0, 1] and Q = [2, 0; 0, 3]. Row-by-column multiplication gives PQ = [2, 3; 0, 3], whereas QP = [2, 2; 0, 3]. The top-right entries differ. Matrix multiplication is therefore not generally commutative.

For AX = B, A⁻¹B is a valid 2 × 1 product. BA⁻¹ is not even defined: B has one column, but A⁻¹ has two rows. The inner dimensions must match. Think “inverse on the left of both sides,” rather than “move A across and divide.” There is no ordinary matrix division operation that ignores order.

Three practice tasks with full solutions

Practice 1: solve x + 2y = 8 and 3x − y = 3

A = [1, 2; 3, −1] has determinant −1 − 6 = −7. Its inverse is (−1/7)[−1, −2; −3, 1]. Multiplication gives X = (−1/7)[−8 − 6; −24 + 3] = [2; 3]. Check: 2 + 2(3) = 8 and 3(2) − 3 = 3.

Practice 2: solve 2x + 3y = 1 and 4x − y = 9

The determinant of [2, 3; 4, −1] is −2 − 12 = −14. Its inverse is (−1/14)[−1, −3; −4, 2]. Hence X = (−1/14)[−1 − 27; −4 + 18] = [2; −1]. Substitution gives 4 − 3 = 1 and 8 + 1 = 9.

Practice 3: diagnose 2x − y = 4 and 6x − 3y = 13

The determinant is 2(−3) − (−1)(6) = 0. Multiplying the first equation by three would give 6x − 3y = 12, contradicting 13. There is no solution. If the second right-hand side were 12 instead, every pair with y = 2x − 4 would work, so there would be infinitely many solutions.

Frequently asked questions

Can a negative determinant be used?

Yes. Only zero prevents inversion. Retain the negative sign when multiplying by its reciprocal.

Do I swap every entry to find the inverse?

No. Swap the main-diagonal entries, negate the other entries, and divide the entire resulting matrix by the determinant.

Can I use y before x?

Yes, if every coefficient row and the unknown column use that same order. Label it explicitly.

Does determinant zero mean x and y are zero?

No. It means the coefficient matrix has no inverse. Inspect the original equations for inconsistency or dependence.

Why verify both equations?

One equation usually has many solutions. Only a pair satisfying both belongs to the system’s solution set.

Should I turn fractions into decimals?

Not unless useful or requested. Exact fractions avoid rounding errors and make substitution easier to audit.

What is a quick inverse check?

Multiply the proposed inverse by the original matrix. The result must be I, including zeros in both off-diagonal positions.

Choose your next matrix practice

For each new system, write the variable order, determinant, inverse, product and two substitutions. The existing Mathematics topical revision collection provides navigation to further resources. Its contents may require membership; this lesson does not promise a particular matrix worksheet or download.