Find the inverse of a matrix if it exists and use it to solve systems of linear equations

Learning goal: Find the inverse of a 2 × 2 matrix when it exists, recognize from the determinant when it does not, use the inverse to solve a system written as AX = B by computing X = A⁻¹B, and use technology to find the inverse of a matrix of dimension 3 × 3 or greater and then verify the resulting solution by hand.

Scope note. A.REI.8 and A.REI.9 are marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate standards "for students who are planning on taking advanced courses". They are taught here in full, as extension content for students on that path.

The identity matrix and what an inverse is. The n × n identity matrix I has 1 in every position on the main diagonal and 0 everywhere else; for 2 × 2 it is [[1, 0], [0, 1]]. It behaves like the number 1: IA = A and AI = A for any n × n matrix A, and IX = X for any n × 1 vector. A square matrix A is invertible if there is a matrix A⁻¹ of the same size with A⁻¹A = I and AA⁻¹ = I; A⁻¹ is called the inverse of A. The inverse plays the role of a reciprocal. Just as 0 has no reciprocal, some matrices have no inverse, and the phrase "if it exists" in the goal is not decoration: deciding existence is the first step.

The 2 × 2 formula and the determinant. For A = [[a, b], [c, d]], the number ad − bc is the determinant of A, written det A. If det A ≠ 0, then A⁻¹ = (1/(ad − bc)) [[d, −b], [−c, a]]: swap the diagonal entries, negate the other two, and divide every entry by the determinant. If det A = 0, no inverse exists. The formula can be checked by multiplying: [[a, b], [c, d]] times [[d, −b], [−c, a]] gives [[ad − bc, −ab + ba], [cd − dc, −cb + ad]] = [[ad − bc, 0], [0, ad − bc]], which is (ad − bc) I, so dividing by ad − bc yields I exactly when that division is allowed.

A fully worked model. Solve 2x + 3y = 8 and x − y = −1 with an inverse. From the previous lesson, A = [[2, 3], [1, −1]], X = [x, y], B = [8, −1]. First, det A = 2(−1) − 3(1) = −2 − 3 = −5, which is not zero, so A⁻¹ exists. Then A⁻¹ = (1/(−5)) [[−1, −3], [−1, 2]] = [[1/5, 3/5], [1/5, −2/5]]. Confirm it is the inverse: A⁻¹A has first row (1/5)(2) + (3/5)(1) = 1 and (1/5)(3) + (3/5)(−1) = 0, and second row (1/5)(2) + (−2/5)(1) = 0 and (1/5)(3) + (−2/5)(−1) = 3/5 + 2/5 = 1, so A⁻¹A = I. Now multiply both sides of AX = B on the left by A⁻¹: A⁻¹AX = A⁻¹B, so IX = A⁻¹B, so X = A⁻¹B. Compute: A⁻¹B = [(1/5)(8) + (3/5)(−1), (1/5)(8) + (−2/5)(−1)] = [(8 − 3)/5, (8 + 2)/5] = [1, 2]. So x = 1 and y = 2. Substitute back into the original equations: 2(1) + 3(2) = 8 and 1 − 2 = −1. The solution matches the one found by substitution, as it must.

Why the multiplication is on the left, and why the solution is unique. Matrix multiplication is not commutative, so "multiply both sides by A⁻¹" must specify a side; because X sits to the right of A in AX, the inverse goes on the left of both sides so that A⁻¹A collapses to I next to X. Writing BA⁻¹ is not even defined when B is a column vector. The argument also shows uniqueness: if AX₁ = B and AX₂ = B, then X₁ = A⁻¹B = X₂. So when A⁻¹ exists, the system has exactly one solution. Conversely, when det A = 0 the system has either no solution or infinitely many, and the inverse method cannot be used; elimination still can.

When the inverse does not exist. Take A = [[2, 4], [1, 2]]. Then det A = 4 − 4 = 0, and the formula would require dividing by zero, so there is no inverse. The rows are proportional, the second row being half the first, and that is the geometric story: the two equations describe lines with the same slope. The system 2x + 4y = 6, x + 2y = 3 has infinitely many solutions (the second equation is half the first), while 2x + 4y = 6, x + 2y = 1 has none. A zero determinant means the system cannot have exactly one solution, so no single vector A⁻¹B could be the answer.

Exact fractions are the answer; decimals are the check. Solve 5x + 3y = 4 and 2x − 7y = 9. A = [[5, 3], [2, −7]], det A = −35 − 6 = −41. A⁻¹ = (1/(−41)) [[−7, −3], [−2, 5]] = (1/41) [[7, 3], [2, −5]]. Then X = A⁻¹B = (1/41) [7(4) + 3(9), 2(4) − 5(9)] = (1/41) [55, −37] = [55/41, −37/41], approximately (1.3415, −0.9024). Check the second equation: 2(55/41) − 7(−37/41) = (110 + 259)/41 = 369/41 = 9. This matches the elimination result from the systems lesson.

Matrices of dimension 3 × 3 or greater: use technology, then verify by hand. A 3 × 3 inverse can be computed by hand, but the arithmetic is long and error-prone, so a graphing calculator, spreadsheet or computer algebra system is the right tool for finding it. The student's job is to set up A and B correctly, obtain A⁻¹, compute X = A⁻¹B, and then verify by hand that AX = B. For x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2, the coefficient matrix is A = [[1, 1, 1], [2, −1, 1], [1, 2, −1]] and B = [6, 3, 2]. Technology reports det A = 7 and A⁻¹ = (1/7) [[−1, 3, 2], [3, −2, 1], [5, −1, −3]]. Then X = A⁻¹B = (1/7) [−6 + 9 + 4, 18 − 6 + 2, 30 − 3 − 6] = (1/7) [7, 14, 21] = [1, 2, 3]. Verify by hand: AX = [1 + 2 + 3, 2 − 2 + 3, 1 + 4 − 3] = [6, 3, 2] = B. A tool can also be asked to confirm A⁻¹A = I. If the tool reports that the matrix is singular, that is the technology's way of saying det A = 0 and no inverse exists.

How to get the inverse from a tool, step by step. The keystrokes differ from tool to tool, but the steps are the same. In a spreadsheet: type the nine entries of A into a 3 × 3 block of cells, row by row (say A1 to C3), and the three entries of B into a column (E1 to E3). In an empty cell type =MDETERM(A1:C3); it shows 7 (sometimes 7.000000000000001, because the tool works in decimals, which is read as 7). Then enter =MINVERSE(A1:C3) so that its result fills a 3 × 3 block; the block shows −0.142857, 0.428571, 0.285714 in the first row, 0.428571, −0.285714, 0.142857 in the second and 0.714286, −0.142857, −0.428571 in the third. Multiplying every entry by the determinant 7 turns these decimals back into the integers −1, 3, 2 / 3, −2, 1 / 5, −1, −3, which is how the fraction form (1/7)[[−1, 3, 2], [3, −2, 1], [5, −1, −3]] above was read off. Finally =MMULT(MINVERSE(A1:C3), E1:E3) gives the column 1, 2, 3. On a graphing calculator: open the matrix editor, set the dimensions of a matrix to 3 × 3, type the entries row by row, set a second matrix to 3 × 1 for B, then on the main screen compute the first matrix raised to the power −1 for A⁻¹ and that inverse times the second matrix for X. Reading an error: enter A = [[1, 2, 3], [2, 4, 6], [1, 0, 1]], whose second row is twice the first. The determinant is 0, and the inverse command returns an error (a spreadsheet shows an error value such as #NUM!; a calculator reports a singular matrix) instead of a matrix. Because the tool works in decimals, a matrix whose determinant is 0 can occasionally come back with enormous entries instead of an error, so always check the determinant and the product A⁻¹A as well.

The misconception and why it is wrong. The most frequent 2 × 2 error is swapping the diagonal entries but forgetting to negate the off-diagonal ones, writing (1/det) [[d, b], [c, a]]. Multiplying that by A does not give I; for A = [[2, 5], [1, 3]] with det 1, the wrong candidate [[3, 5], [1, 2]] times A gives [[11, 30], [4, 11]], nowhere near I, while the correct A⁻¹ = [[3, −5], [−1, 2]] times A gives [[6 − 5, 15 − 15], [−2 + 2, −5 + 6]] = I. Checking A⁻¹A = I takes one multiplication and catches the error every time. A second misconception is to divide B by A entry by entry; there is no such operation, and X = A⁻¹B is a matrix product, not a division.

Worked 2 × 2 inverse and solution: Solve 3x − 4y = 5 and 2x + y = −3. A = [[3, −4], [2, 1]], B = [5, −3]. det A = 3(1) − (−4)(2) = 3 + 8 = 11 ≠ 0, so A⁻¹ exists: A⁻¹ = (1/11) [[1, 4], [−2, 3]]. Check A⁻¹A: (1/11) [[3 + 8, −4 + 4], [−6 + 6, 8 + 3]] = (1/11) [[11, 0], [0, 11]] = I. Then X = A⁻¹B = (1/11) [5 − 12, −10 − 9] = [−7/11, −19/11] ≈ (−0.6364, −1.7273). Substitute into the originals: 3(−7/11) − 4(−19/11) = (−21 + 76)/11 = 55/11 = 5, and 2(−7/11) − 19/11 = −33/11 = −3.

Worked singular matrix: Does A = [[3, 6], [2, 4]] have an inverse? det A = 12 − 12 = 0, so no. The rows are proportional (the first is 1.5 times the second). A system 3x + 6y = p, 2x + 4y = q therefore has either no solution (if p ≠ 1.5q) or infinitely many (if p = 1.5q), never exactly one, which is why no inverse can produce a single solution vector.

Worked 3 × 3 with technology: Solve 2x − z = 1, y + 3z = 4, x + y = 2. A = [[2, 0, −1], [0, 1, 3], [1, 1, 0]], B = [1, 4, 2]. Technology gives det A = −5 and A⁻¹ = (1/5) [[3, 1, −1], [−3, −1, 6], [1, 2, −2]]. Then X = A⁻¹B = (1/5) [3 + 4 − 2, −3 − 4 + 12, 1 + 8 − 4] = (1/5) [5, 5, 5] = [1, 1, 1]. Verify by hand: AX = [2 − 1, 1 + 3, 1 + 1] = [1, 4, 2] = B. Also spot-check one entry of A⁻¹A: row 1 of A⁻¹ times column 1 of A is (1/5)(3·2 + 1·0 + (−1)·1) = (1/5)(5) = 1, the correct diagonal entry.

Worked 3 × 3 with technology, second example: Solve 2x + y − z = −6, x − 3y + 2z = 1, 3x + 2y + z = −1. A = [[2, 1, −1], [1, −3, 2], [3, 2, 1]], B = [−6, 1, −1]. Technology gives det A = −20 and A⁻¹ = (1/20) [[7, 3, 1], [−5, −5, 5], [−11, 1, 7]]. X = A⁻¹B = (1/20) [−42 + 3 − 1, 30 − 5 − 5, 66 + 1 − 7] = (1/20) [−40, 20, 60] = [−2, 1, 3]. Verify: AX = [−4 + 1 − 3, −2 − 3 + 6, −6 + 2 + 3] = [−6, 1, −1] = B, matching the elimination result from the systems lesson.

Inverse record: For each of [[4, 7], [1, 2]], [[2, 4], [1, 2]], [[5, 3], [2, −7]] and [[2, 5], [1, 3]], compute the determinant, state whether an inverse exists, write the inverse when it does, and show the product A⁻¹A with every entry's arithmetic. For the matrix without an inverse, write two systems with that coefficient matrix, one with no solution and one with infinitely many, and explain why neither could have been solved by an inverse.

Independent performance: Write two systems of two linear equations in two unknowns, one whose coefficient matrix has a nonzero determinant and one whose determinant is zero. For the first, represent it as AX = B, find A⁻¹ by the 2 × 2 formula, confirm A⁻¹A = I, compute X = A⁻¹B, give the solution exactly and as decimals, and substitute it into both original equations. For the second, show that the determinant is zero, explain what that means for the number of solutions, and solve or classify the system by elimination instead. Then write one system of three linear equations in three unknowns by choosing the solution first, represent it as AX = B, and use technology to check that det A ≠ 0 (if it is 0, your equations are dependent and the system does not have exactly one solution, so change one equation and check again); then use technology to find A⁻¹ (record which tool you used and the inverse it reported), compute X = A⁻¹B by hand from that inverse, and verify AX = B by hand. Success criteria: the existence of each inverse is decided by the determinant before any inverse is written; each inverse is confirmed by a product equal to I; the vector variable is recovered by left-multiplying by A⁻¹ and the reason for the side is stated; and every solution is substituted back into the original equations.