HSA.REI.C.9: Solving Systems of Linear Equations with Inverse Matrices
In plain English: HSA.REI.C.9 is an advanced (+) Common Core algebra standard that asks students to find the inverse of a matrix when it exists and use it to solve a system of linear equations written as AX = B, so that X = A⁻¹B. Students find 2 × 2 inverses by hand, use technology for 3 × 3 or larger matrices, and recognize that a matrix with determinant 0 has no inverse. It is usually taught in Precalculus.
(+) Find the inverse of a matrix if it exists and use it to solve systems of linear equations (using technology for matrices of dimension 3 × 3 or greater).
Common Core State Standards for Mathematics · Domain: Reasoning with Equations and Inequalities (REI) · Cluster: Solve systems of equations Also written as HSA-REI.C.9 or A-REI.9 · Official standard
Students learn to find the inverse of a square matrix, when it exists, and to use it to solve a system of linear equations. For a 2 × 2 matrix A = [a b; c d], the inverse exists exactly when the determinant ad - bc is not 0, and then A⁻¹ = (1/(ad - bc))[d -b; -c a]. Students confirm an inverse by checking that AA⁻¹ = A⁻¹A = I.
Students then write a system as AX = B, multiply both sides on the left by A⁻¹, and get X = A⁻¹B, the same way they would solve 5x = 35 by multiplying by 1/5. For 3 × 3 and larger systems, students use a graphing calculator or matrix software to find A⁻¹ and the product A⁻¹B, as the standard specifies. The lesson ends with the case where no inverse exists and what that means for the system. Matrices are written row by row, with rows separated by semicolons: [4 7; 2 4] has first row 4, 7 and second row 2, 4, and [x; y] is a column vector.
Learning Objectives
By the end of this lesson, students will be able to:
Compute the determinant of a 2 × 2 matrix and decide whether the matrix has an inverse
Find the inverse of a 2 × 2 matrix by hand and verify it by showing that AA⁻¹ = A⁻¹A = I
Solve a system of two linear equations by writing it as AX = B and computing X = A⁻¹B
Use technology to find the inverse of a 3 × 3 or larger matrix and solve the matching system
Explain why the inverse method fails when the determinant is 0 and what that says about the system
Prior Knowledge Required
Students should already be comfortable with:
Writing a system of linear equations as a matrix equation AX = B HSA.REI.C.8
Multiplying matrices and the identity matrix I (HSN.VM.C.8, HSN.VM.C.10)
Solving systems of two linear equations algebraically and graphically HSA.REI.C.6
"(1) Solve 5x = 35 without dividing: what number can you multiply both sides by? (2) Multiply [2 1; 5 3][3 -1; -5 2]. What do you notice about the result? What would happen if you multiplied in the other order?"
For (1), students multiply by 1/5, the multiplicative inverse of 5, because (1/5)(5) = 1. For (2), the product is [1 0; 0 1] = I in both orders. Name the idea: the two matrices are inverses of each other, and I plays the role of 1. The lesson question is: if we can write a system as AX = B, can we "multiply by 1/A" to get X?
Direct Instruction20 minutes
Part 1: Finding an inverse. Define the inverse of a square matrix A as the matrix A⁻¹ with AA⁻¹ = A⁻¹A = I. For a 2 × 2 matrix, give the procedure:
Compute the determinant: for A = [a b; c d], det(A) = ad - bc.
Decide whether A⁻¹ exists: if det(A) = 0, A has no inverse. Stop.
Swap and negate: swap a and d, and change the signs of b and c to get [d -b; -c a].
Divide by the determinant: A⁻¹ = (1/det(A))[d -b; -c a].
Check: multiply A by your answer. The product must be I.
Part 2: Solving AX = B. Multiply both sides on the left by A⁻¹: A⁻¹AX = A⁻¹B, so IX = A⁻¹B and X = A⁻¹B (Diagram 1). Stress the order: A⁻¹ goes on the left of B, because matrix multiplication is not commutative, and BA⁻¹ is not even defined when B is a column. For 3 × 3 systems, enter A and B on a calculator and compute [A]⁻¹[B]. Work through the examples:
Find a 2 × 2 inverse
Find the inverse of A = [4 7; 2 4]. The determinant is 4(4) - 7(2) = 2.
Solve x + y + z = 6, 2x - y + z = 3 and x + 2y - z = 2. A calculator gives det(A) = 7 and A⁻¹ = (1/7)[-1 3 2; 3 -2 1; 5 -1 -3].
Equation: X = A⁻¹[6; 3; 2] = [1; 2; 3], so (x, y, z) = (1, 2, 3)
Context problem
A student mixes a 10% saline solution with a 30% saline solution to make 20 liters of a 24% solution. With x and y liters: x + y = 20 and 0.1x + 0.3y = 4.8, or x + 3y = 48 after multiplying by 10. A = [1 1; 1 3] has det 2 and A⁻¹ = (1/2)[3 -1; -1 1].
Equation: X = (1/2)[60 - 48; -20 + 48] = [6; 14]: 6 liters of 10% and 14 liters of 30%
After the second example, use Diagram 2. The system 3x + 2y = 7 and 4x + 3y = 11 has a nonzero determinant, and its two lines meet in exactly one point, (-1, 5). The matrix [6 3; 4 2] has determinant 0: the system 6x + 3y = 9 and 4x + 2y = 2 describes two parallel lines with no intersection, and if the second equation were a multiple of the first, such as 4x + 2y = 6, the two lines would be the same, with infinitely many solutions. In both cases there is no single answer X = A⁻¹B, which is why A⁻¹ cannot exist. In the context example, ask students to check that both amounts are positive and that 0.1(6) + 0.3(14) = 4.8.
Guided Practice15 minutes
Pairs work four items and check each inverse by multiplying: (a) find the inverse of [5 2; 7 3] (det 1, inverse [3 -2; -7 5]); (b) find the inverse of [2 -1; 4 6] (det 16, inverse (1/16)[6 1; -4 2]); (c) decide whether [3 -6; -2 4] has an inverse (det 12 - 12 = 0, so no); (d) use the inverse from (a) to solve 5x + 2y = 1 and 7x + 3y = 2 (X = [3 -2; -7 5][1; 2] = [-1; 3]). Listen for these errors: forgetting to divide by the determinant, negating a and d instead of b and c, forgetting that dividing by a negative determinant changes every sign, and multiplying B by A⁻¹ in the wrong order.
Independent Practice15 minutes
Students work alone: (1) find the inverse of [1 3; 2 8] (det 2, inverse [4 -3/2; -1 1/2]); (2) find the value of k for which [k 4; 3 6] has no inverse (6k - 12 = 0, so k = 2); (3) solve 2x + 5y = 4 and x + 3y = 3 with an inverse matrix (A⁻¹ = [3 -5; -1 2], X = [-3; 2]); (4) use a calculator to solve 2x + y - z = 3, x + 3y + 2z = 9 and 3x - y + z = 2 (det 25, X = [1; 2; 1]). For every system, students substitute the solution into the original equations.
Closure5 minutes
Exit ticket: (1) Find the inverse of [3 1; 5 2]. (Answer: det 1, inverse [2 -1; -5 3].) (2) Use it to solve 3x + y = 4 and 5x + 2y = 9. (Answer: X = [2 -1; -5 3][4; 9] = [-1; 7].) (3) Explain in one sentence why [2 8; 1 4] has no inverse. (Its determinant is 8 - 8 = 0.)
Differentiation Strategies
For Struggling Students
Give a fill-in template: det = (__)(__) - (__)(__), then [__ __; __ __] with arrows showing which entries swap and which change sign
Start with matrices whose determinant is 1, so there is no fraction to distribute, before moving to other determinants
Let students check every inverse on a calculator after doing it by hand, so they get immediate feedback
For Advanced Students
Ask students to prove the 2 × 2 inverse formula by multiplying [a b; c d] by (1/(ad - bc))[d -b; -c a]
Ask students to show that (AB)⁻¹ = B⁻¹A⁻¹ for two invertible 2 × 2 matrices of their choice, and explain why the order reverses
Challenge: for a 3 × 3 system whose matrix has determinant 0, use the equations to decide whether there are no solutions or infinitely many
Assessment Guidance
What to Look For
Check that students compute the determinant first and stop when it is 0, instead of writing a formula with division by zero. Every inverse should be verified by a product equal to I. When students solve systems, look for X = A⁻¹B with A⁻¹ on the left, and for a substitution check at the end. For 3 × 3 systems, students should record what they entered in the calculator ([A]⁻¹[B]) and interpret a "singular matrix" message as "no inverse exists".
02
Classroom Activities
3 Activities
1
Inverse or Not? Card Sort
15 minPairs
Pairs sort 8 matrix cards into "has an inverse" and "has no inverse" by computing determinants, then find and check the inverse of every invertible matrix.
The 8 Cards
[3 2; 7 5], [3 9; 1 3], [5 4; 1 1], [4 -2; -6 3]
[2 3; 2 4], [-1 2; 3 -6], [6 5; 1 1], [0 3; -2 1]
Procedure
Partner A computes the determinant of a card; Partner B checks it and places the card in one of two piles
For each invertible matrix, the pair writes the inverse and multiplies to check that the product is I
Answer key: no inverse for [3 9; 1 3], [4 -2; -6 3] and [-1 2; 3 -6]; inverses [5 -2; -7 3], [1 -4; -1 5], [2 -3/2; -1 1], [1 -5; -1 6] and (1/6)[1 -3; 2 0] for the other five
Discussion Questions
Look at the three matrices with no inverse. How is the second row related to the first row in each one?
[0 3; -2 1] has a 0 in it. Why does it still have an inverse?
Modification for Distance Learning
Put the cards on a shared slide with two drop zones. Pairs drag each card and type its determinant and inverse in a text box next to it.
2
Technology Lab: 3 × 3 and 4 × 4 Systems
20 minGroups of 3
Groups model two context problems with 3 × 3 systems and solve them with a graphing calculator or online matrix calculator, as the standard asks for matrices of dimension 3 × 3 or greater.
Scenarios
A bakery tray of cookies uses 2 cups of flour, 1 cup of sugar and 1 cup of butter; a tray of muffins uses 3, 2 and 1; a tray of scones uses 4, 1 and 2. The bakery wants to use exactly 31 cups of flour, 13 of sugar and 14 of butter. How many trays of each? (A = [2 3 4; 1 2 1; 1 1 2], det -1, solution 3 trays of cookies, 3 of muffins, 4 of scones.)
A zoo charged $8 per child, $14 per adult and $10 per senior. On one morning it had 250 visitors, took in $2,860, and had as many adults as children and seniors together. (A = [1 1 1; 8 14 10; -1 1 -1], det -4, solution 70 children, 125 adults, 55 seniors.)
Procedure
The modeler defines the variables and writes the system; the calculator operator enters [A] and [B] and computes [A]⁻¹ and [A]⁻¹[B]; the checker substitutes the answer into every original equation
Groups record the determinant and one row of A⁻¹, and explain why the answer makes sense in the context
Groups rotate roles for the second scenario
Challenge Variation
Give the 4 × 4 system w + x + y + z = 10, 2w + x - y + z = 5, w + 3x + y - 2z = 2 and 3w - x + 2y + z = 11. Groups solve it with technology (the solution is w = 1, x = 2, y = 3, z = 4) and then change one equation so that the calculator reports that the matrix is singular.
3
Crack the Matrix Code
20 minPairs
A secret message was encoded by writing letters as numbers (A = 1, B = 2, ..., Z = 26), grouping them in pairs as column vectors, and multiplying each vector by the encoding matrix E = [2 1; 3 2]. Decoding each pair means solving the system EX = C, which students do with E⁻¹.
Procedure
Find E⁻¹ and check it (det 1, E⁻¹ = [2 -1; -3 2])
Decode the message with coded vectors [27; 41], [58; 96] and [42; 75] by computing E⁻¹ times each one. (The message is [13; 1], [20; 18], [9; 24], which spells MATRIX.)
Each pair encodes a six-letter word of its own with E and trades with another pair to decode
Discussion Questions
Why would the matrix [2 4; 1 2] be a bad encoding matrix?
If you only knew the coded vectors, what would you need to crack the code?
Why must E⁻¹ multiply the coded vector on the left?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Solving AX = B with an Inverse Matrix
Multiplying both sides of AX = B on the left by A⁻¹ turns A into the identity matrix, so X = A⁻¹B. For 3x + 2y = 7 and 4x + 3y = 11, the determinant is 1, A⁻¹ = [3 -2; -4 3], and X = [-1; 5].
Diagram 2: Determinant, Inverse and the Graph of the System
Both graphs are drawn to scale, one grid square per unit. On the left, the determinant is 1, the lines meet at the single point (-1, 5), and X = A⁻¹B gives that point. On the right, the determinant is 0, the lines y = 3 - 2x and y = 1 - 2x are parallel, and the coefficient matrix has no inverse.
04
Homework Assignment
~30 min
HSA.REI.C.9 Homework: Inverse Matrices and Systems
Directions: Show the determinant before every inverse and check every inverse by multiplying. For each system, write it as AX = B, state A⁻¹, compute X = A⁻¹B, and check the solution in the original equations. Use technology for Problem 6 and record what you entered. Matrices are written row by row: [7 3; 2 1] has rows 7, 3 and 2, 1.
Part 1: Finding Inverses (Problems 1-3)
Find the inverse of each matrix, if it exists. If it does not exist, explain why. (a) [7 3; 2 1] (b) [4 6; 2 3] (c) [3 -2; 5 -4]
Show that [5 -3; -3 2] and [2 3; 3 5] are inverses of each other by multiplying them in both orders.
Find every value of k for which the matrix [k 6; 2 k + 1] has no inverse.
Part 2: Solving Systems with Inverses (Problems 4-6)
Solve the system 4x + 5y = 2 and 3x + 4y = 1 using an inverse matrix.
A coffee shop blends Colombian beans that cost $9 per pound with Ethiopian beans that cost $13 per pound to make 40 pounds of a blend worth $10.50 per pound. Write the system as AX = B, find A⁻¹, and use it to find how many pounds of each bean go into the blend.
Use technology to solve x + 2y + z = 9, 2x - y + 3z = 18 and 3x + y - 2z = -3 by computing A⁻¹B. Then replace the third equation with 3x + y + 4z = 27, try to find A⁻¹ again, and explain what happens and why. (Hint: compare the new third equation with the sum of the first two.)
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Determinant and Existence
Determinant correct; existence of the inverse decided and explained
Determinant correct but conclusion missing
Missing or incorrect
Finding the Inverse
Inverse correct and checked by a product equal to I
One sign or scaling error, or no check
Missing or incorrect
Solving with X = A⁻¹B
Correct order A⁻¹B, correct solution, checked in the equations
Method correct with one arithmetic error
Method missing or incorrect
Technology and Interpretation
Calculator work recorded; singular case and context answers explained
Answers correct but explanation incomplete
Missing or incorrect
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again. Matrices are written row by row, with rows separated by semicolons: [4 7; 2 4] has first row 4, 7 and second row 2, 4, and [x; y] is a column vector. A calculator is allowed for questions that say to use technology.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
What is the inverse of [3 5; 1 2]?
Answer: C
The determinant is 3(2) - 5(1) = 1, so the inverse is [2 -5; -1 3]. Check: [3 5; 1 2][2 -5; -1 3] = [1 0; 0 1]. Choice A swaps a and d but forgets to change the signs of b and c. Choice B changes the signs of the wrong entries. Choice D takes the reciprocal of each entry, which is not how matrix inverses work.
Question 2 of 20 · Multiple Choice
Which matrix has no inverse?
Answer: A
det [4 2; 6 3] = 4(3) - 2(6) = 0, so it has no inverse; its second row is 1.5 times the first. The others have determinants 18, -3 and -1, all nonzero. Choice D may look suspicious because of the zeros, but a zero entry does not make the determinant 0.
Question 3 of 20 · Multiple Choice
What is the determinant of [5 -2; 3 4], and does the matrix have an inverse?
Answer: B
det = 5(4) - (-2)(3) = 20 + 6 = 26, which is not 0, so the inverse exists. Choice A subtracts 6 instead of -6: the product bc is -6, and subtracting it adds 6. Choice C reverses the order of the subtraction.
Question 4 of 20 · Multiple Choice
What is the inverse of [2 4; 1 3]?
Answer: D
The determinant is 2(3) - 4(1) = 2, so A⁻¹ = (1/2)[3 -4; -1 2] = [3/2 -2; -1/2 1]. Choice A forgets to divide by the determinant: multiplying it by A gives 2I, not I. Choice B has every sign reversed, which is -A⁻¹. Choice C takes reciprocals of the entries.
Question 5 of 20 · Multiple Choice
How can you tell that a 2 × 2 matrix M is the inverse of A?
Answer: A
By definition, M = A⁻¹ when both products AM and MA equal the identity matrix I. Choice B confuses the multiplicative inverse with an additive idea. Choice C describes multiplying by I, not by an inverse. Choice D is a common misconception: entrywise reciprocals are almost never the inverse.
Question 6 of 20 · Multiple Choice
A system is written as AX = B with B = [5; 2]. You are told that A⁻¹ = [4 -3; -1 1]. What is X?
Answer: C
X = A⁻¹B = [4(5) - 3(2); -1(5) + 1(2)] = [14; -3]. Check: A = [1 3; 1 4], and 14 + 3(-3) = 5 and 14 + 4(-3) = 2. Choice A multiplies B by A instead of A⁻¹. Choice B gets the sign of both products with the entry 2 wrong: 4(5) + 3(2) = 26 and -1(5) - 1(2) = -7. Choice D lists the entries in the wrong order.
Question 7 of 20 · Multiple Choice
What is the first step in solving AX = B with an inverse matrix, if A⁻¹ exists?
Answer: B
A⁻¹AX = A⁻¹B gives IX = A⁻¹B, so X = A⁻¹B. Choice A is not defined: there is no matrix division. Choice C gives AXA⁻¹ = BA⁻¹, which does not isolate X, and BA⁻¹ is not even defined when B is a column vector.
Question 8 of 20 · Multiple Choice
Use an inverse matrix to solve 2x + y = 5 and 7x + 4y = 18.
Answer: D
A = [2 1; 7 4] has determinant 8 - 7 = 1, so A⁻¹ = [4 -1; -7 2] and X = [4(5) - 18; -7(5) + 2(18)] = [2; 1]. Check: 2(2) + 1 = 5 and 7(2) + 4(1) = 18. Choice B multiplies B by A instead of A⁻¹. Choice A swaps x and y.
Question 9 of 20 · Multiple Choice
The coefficient matrix of a system of two linear equations has determinant 0. What can you conclude?
Answer: C
When det(A) = 0, A has no inverse, and the two lines are parallel or the same line. Parallel lines give no solution; the same line gives infinitely many. Choice B describes the case det(A) ≠ 0. Choice D is false: the zero matrix times A is the zero matrix, never I.
Question 10 of 20 · Multiple Choice
On a graphing calculator, the coefficient matrix is stored as [A] and the constants as [B]. Which entry solves the system?
Answer: A
The solution is X = A⁻¹B, so enter [A]⁻¹ times [B]. Choice B multiplies in the wrong order; for a 3 × 1 column [B] and a 3 × 3 matrix the calculator reports a dimension error. Choice C is not a defined operation, and choice D computes AB, not the solution.
Question 11 of 20 · Multiple Choice
Use technology to solve x + y + z = 2, x - y + 2z = 6 and 2x + y - z = -1.
Answer: B
The determinant of A = [1 1 1; 1 -1 2; 2 1 -1] is 7, and A⁻¹B = [1; -1; 2]. Check: 1 - 1 + 2 = 2, 1 + 1 + 4 = 6, 2 - 1 - 2 = -1. Choices A and D satisfy only the first equation, which is why every equation must be checked. Choice C satisfies none of the three equations.
Question 12 of 20 · Multiple Choice
What is the inverse of [1 2; 3 5]?
Answer: D
The determinant is 1(5) - 2(3) = -1, so A⁻¹ = (1/-1)[5 -2; -3 1] = [-5 2; 3 -1]. Choice A forgets to divide by the negative determinant, so every sign is wrong. Choice B negates b and c but does not swap a and d.
Question 13 of 20 · Multiple Choice
A museum sold 150 tickets: adult tickets at $15 and child tickets at $9, for $1,770 in total. Using X = A⁻¹B with A = [1 1; 15 9], how many adult tickets were sold?
Answer: C
det(A) = 9 - 15 = -6, so A⁻¹ = (1/-6)[9 -1; -15 1] and X = (1/-6)[9(150) - 1770; -15(150) + 1770] = (1/-6)[-420; -480] = [70; 80]. So 70 adult and 80 child tickets. Check: 15(70) + 9(80) = 1,770. Choice A is the number of child tickets.
Question 14 of 20 · Multiple Choice
A calculator shows "ERROR: SINGULAR MATRIX" when you ask for the inverse of A = [1 2 3; 4 5 6; 7 8 9]. What does this mean?
Answer: B
A singular matrix is one with determinant 0; it has no inverse. Here the rows are dependent: row 1 + row 3 = 2 × row 2, so det(A) = 0. Choice A is false: the standard expects technology for 3 × 3 and larger matrices. Choice C is false: calculators display fractional inverses.
Question 15 of 20 · Short Answer
Find the inverse of [9 4; 2 1] and check your answer.
For what value of k does [3 k; 2 4] have no inverse? Explain.
The matrix has no inverse when its determinant is 0: 3(4) - 2k = 0, so 12 = 2k and k = 6. Then [3 6; 2 4] has second row equal to 2/3 of the first row.
Question 17 of 20 · Short Answer
Solve 5x + 3y = 1 and 3x + 2y = 0 using an inverse matrix.
A = [5 3; 3 2], det = 10 - 9 = 1, A⁻¹ = [2 -3; -3 5]. X = A⁻¹[1; 0] = [2; -3], so x = 2, y = -3. Check: 5(2) + 3(-3) = 1 and 3(2) + 2(-3) = 0.
Question 18 of 20 · Short Answer
A garden store sells bags of soil for $6 and bags of mulch for $4. On Tuesday it sold 55 bags for $290. Write the system as AX = B, find A⁻¹, and solve.
With s bags of soil and m bags of mulch: s + m = 55 and 6s + 4m = 290, so A = [1 1; 6 4] and B = [55; 290]. det(A) = 4 - 6 = -2, A⁻¹ = (1/-2)[4 -1; -6 1] = [-2 1/2; 3 -1/2]. X = [-110 + 145; 165 - 145] = [35; 20]: 35 bags of soil and 20 bags of mulch. Check: 6(35) + 4(20) = 290.
Question 19 of 20 · Short Answer
Use technology to solve 2x + 3y - z = -2, x - y + 4z = 15 and 3x + 2y + z = 7. State the determinant of the coefficient matrix.
A = [2 3 -1; 1 -1 4; 3 2 1] has determinant 10, so A⁻¹ exists. The calculator gives A⁻¹B = [2; -1; 3], so (x, y, z) = (2, -1, 3). Check: 4 - 3 - 3 = -2, 2 + 1 + 12 = 15, 6 - 2 + 3 = 7.
Question 20 of 20 · Short Answer
Explain why the inverse method cannot solve 2x + 4y = 6 and x + 2y = 5. What are the solutions of this system?
The coefficient matrix [2 4; 1 2] has determinant 4 - 4 = 0, so it has no inverse and X = A⁻¹B cannot be computed. Dividing the first equation by 2 gives x + 2y = 3, which contradicts x + 2y = 5: the lines are parallel, so the system has no solution.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSA.REI.C.9 mean?
HSA.REI.C.9 means students can find the inverse of a square matrix when it has one, and use that inverse to solve a system of linear equations. Written as AX = B, the system has the solution X = A⁻¹B. Students compute 2 × 2 inverses by hand and use technology for 3 × 3 or larger matrices. It follows HSA.REI.C.8, which is about writing the system as a matrix equation.
Is HSA.REI.C.9 Algebra 2 or Precalculus?
It is usually taught in Precalculus, and sometimes in an honors Algebra II course. It is a (+) standard, which Common Core describes as additional mathematics for students who take advanced courses such as calculus, discrete mathematics or advanced statistics. It is taught together with the matrix standards HSN.VM.C.6 to HSN.VM.C.12.
How do you find the inverse of a 2 × 2 matrix?
Compute the determinant ad - bc of A = [a b; c d]. If it is 0, there is no inverse. Otherwise swap a and d, change the signs of b and c, and divide every entry by the determinant: A⁻¹ = (1/(ad - bc))[d -b; -c a]. For example, [4 7; 2 4] has determinant 2 and inverse [2 -7/2; -1 2]. Always check by multiplying: AA⁻¹ must be I.
When does a matrix not have an inverse?
A square matrix has no inverse exactly when its determinant is 0. Such a matrix is called singular. For a 2 × 2 matrix this happens when one row is a multiple of the other, as in [6 3; 4 2]. A matrix that is not square, such as a 2 × 3 matrix, never has an inverse in this sense.
Why is it X = A⁻¹B and not X = BA⁻¹?
Matrix multiplication is not commutative, so the side matters. Multiplying both sides of AX = B on the left by A⁻¹ gives A⁻¹AX = A⁻¹B, and since A⁻¹A = I, X = A⁻¹B. The product BA⁻¹ is not even defined when B is a column vector and A⁻¹ is 2 × 2 or 3 × 3, because the dimensions do not match.
Do students have to find 3 × 3 inverses by hand?
No. The standard says to use technology for matrices of dimension 3 × 3 or greater. Students should know how to enter the matrices on a graphing calculator or in matrix software, compute [A]⁻¹[B], interpret a "singular matrix" message, and check the answer by substituting it into the original equations. Finding 3 × 3 inverses by hand, for example by row reduction, is an extension usually left to linear algebra.
What are common mistakes when using inverse matrices?
Common errors are forgetting to divide by the determinant, negating a and d instead of b and c, losing signs when the determinant is negative, multiplying in the wrong order (BA⁻¹), and taking reciprocals of the entries. Checking that AA⁻¹ = I and substituting the final answer into the system catches each of these.
What does it mean for the system if the determinant is 0?
The inverse method cannot be used, and the system does not have exactly one solution. For two equations in two variables, the lines are parallel (no solution) or the same line (infinitely many solutions). To tell which, compare the equations: 6x + 3y = 9 and 4x + 2y = 2 simplify to 2x + y = 3 and 2x + y = 1, which are parallel lines.
Is the inverse method better than substitution or elimination?
For two equations by hand, elimination is often just as fast. The inverse method pays off when the same coefficient matrix is used with many different constant vectors, as in the matrix code activity, and when technology handles 3 × 3 or larger systems. It also connects solving systems to the idea of an inverse, the same idea as solving 5x = 35 by multiplying by 1/5.
How does this standard connect to later math?
It builds on writing systems as matrix equations (HSA.REI.C.8) and on matrix multiplication and the identity matrix (HSN.VM.C.8 and HSN.VM.C.10). It leads to linear algebra, where inverses, determinants and row reduction are studied for any size of matrix, and to applications in computer graphics, cryptography, economics and engineering.
07
Related Standards
5 standards
These standards connect to HSA.REI.C.9: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSA.REI.C.8Prerequisite
Represent a system of linear equations as a single matrix equation
