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Lesson 5 of 6

Determinants · Lesson 5 of 6

Solving Systems with Matrix Inverses

“Represent equations as AX=B, test what their coefficient matrix tells you, and solve consistent systems with a unique solution.”

Learning Objectives

• Build coefficient, unknown, and constant matrices with a consistent variable order. • Distinguish consistent and inconsistent systems and recognise when a solution is unique. • Derive and apply X=A⁻¹B using multiplication on the left. • Interpret the determinant and adjoint tests without overclaiming in the singular case. • Translate verbal conditions into equations and check solutions in the original setting.

A whole system in one matrix equation

A linear equation links unknown quantities by coefficients. A system asks for values that satisfy all its equations at once. Matrices organise the coefficients, variables, and right-hand constants separately so that the whole system can be written as a single equation.

Choose and keep a variable order, such as x,y,z. Each equation gives one row of the coefficient matrix A, even when a coefficient is zero. Put the variables in a column X in that same order, and put the right-hand constants in a column B in equation order. The row-column multiplication AX then rebuilds exactly the original left-hand expressions.

Representing a three-variable systemLaTeX
A has order 3 × 3 and X and B have order 3 × 1. A two-equation system in x and y uses a 2 × 2 coefficient matrix and two-entry columns.

For x+z=2y, first rearrange to x−2y+z=0; its coefficient row is [1,−2,1]. For y+3z=11, the coefficient row is [0,1,3], because no x term occurs. Moving a term to the other side changes its sign, while an absent variable contributes a zero. Rechecking the product AX is a useful way to detect a transcription mistake before finding an inverse.

Existence and uniqueness are different questions

A system can have one solution, more than one solution, or none. The word consistent concerns whether at least one solution exists. The determinant then identifies the case in which the square coefficient matrix guarantees exactly one solution.

Definition
Consistent system

A system is consistent if at least one set of values satisfies every equation. A system with a unique solution and a system with infinitely many solutions are both consistent.

Definition
Inconsistent system

A system is inconsistent if no set of values satisfies all its equations simultaneously.

When det A ≠ 0, A⁻¹ exists. Starting from AX=B, multiply both sides on the left by A⁻¹: A⁻¹(AX)=A⁻¹B. Associativity gives (A⁻¹A)X=A⁻¹B, so IX=A⁻¹B and finally X=A⁻¹B. We did not interchange the order of matrices. In a three-variable system, BA⁻¹ would not even have compatible dimensions, since B has one column.

Unique solution by the matrix methodLaTeX
Every solution must equal this one computed column. Conversely, substituting it gives A(A⁻¹B)=B, so it really is a solution. These two statements establish existence and uniqueness.
A determinant test guides the next stepSystem AX = Bdet A ≠ 0det A = 0Unique solutionX = A⁻¹BNo inverse formulaCheck consistency furtherA singular coefficient matrix does not by itself mean “no solution”.
What the determinant test decides— A nonzero determinant gives one solution. A zero determinant needs further reasoning about compatibility.
Example — Solve two equations by an inverse

Problem
Solve 2x+5y=1 and 3x+2y=7 by the matrix method.

  1. 1.Use A=[2,5;3,2], X=[x;y], and B=[1;7]. Check that AX reproduces both equations.
  2. 2.The determinant is 2·2−5·3=−11, so the inverse exists. The adjoint is [2,−5;−3,2], giving A⁻¹=(−1/11)[2,−5;−3,2].
  3. 3.Multiply by the constants: (adj A)B=[2·1−5·7;−3·1+2·7]=[−33;11].
  4. 4.Hence X=(−1/11)[−33;11]=[3;−1], so x=3 and y=−1.
  5. 5.Substitute in the originals: 2·3+5(−1)=1 and 3·3+2(−1)=7. Both are satisfied.
Example — Solve three equations and check every row

Problem
Solve 3x−2y+3z=8, 2x+y−z=1, and 4x−3y+2z=4.

  1. 1.The coefficient matrix is A=[3,−2,3;2,1,−1;4,−3,2] and B=[8;1;4]. Expansion gives det A=3(2−3)+2(4+4)+3(−6−4)=−17.
  2. 2.Calculate the cofactor rows [−1,−8,−10], [−5,−6,1], and [−1,9,7]. Transpose them to get adj A=[−1,−5,−1;−8,−6,9;−10,1,7].
  3. 3.Multiply the adjoint by B. The first entry is −8−5−4=−17; the second is −64−6+36=−34; the third is −80+1+28=−51.
  4. 4.Divide this column by det A=−17 to get X=[1;2;3]. Thus x=1, y=2, z=3.
  5. 5.The original left sides become 3−4+9=8, 2+2−3=1, and 4−6+6=4. Every equation, not just the first, checks.

What happens when the determinant is zero

A zero determinant tells us the inverse method cannot produce a unique solution. It does not decide whether there are infinitely many solutions or none. A necessary compatibility condition follows by multiplying AX=B by adj A on the left, using an identity that remains valid for singular matrices.

If det A=0, then (adj A)A=O. Therefore any solution would satisfy (adj A)AX=(adj A)B, or O=(adj A)B. If the right side is not the zero column, we have a contradiction, so there can be no solution. If it is zero, this necessary test has passed, but it may have discarded information: the original equations must still be examined.

The singular-case testLaTeX
O here is the zero column with the same size as B. In a general three-variable system, a zero result in the second line can occur for a consistent or an inconsistent system.
Example — A contradiction detected by the adjoint

Problem
Test x+3y=5 and 2x+6y=8 for consistency.

  1. 1.Here A=[1,3;2,6] has determinant 6−6=0. Its adjoint is [6,−3;−2,1].
  2. 2.With B=[5;8], calculate (adj A)B=[30−24;−10+8]=[6;−2], which is not O. Thus the system is inconsistent.
  3. 3.The original equations explain the contradiction directly: twice the first would require 2x+6y=10, but the second requires the same expression to be 8.
  4. 4.The determinant sign was not the problem. It was zero, and the right-hand constants were incompatible with the dependent left-hand sides.
Example — Why a zero adjoint test may leave the answer open

Problem
Take A=[1,1,1;2,2,2;3,3,3]. Compare AX=[1;2;3] with AX=[1;2;4].

  1. 1.Every order-two minor of A is zero, so adj A=O and det A=0. Therefore (adj A)B=O for either constant column.
  2. 2.For B=[1;2;3], all equations reduce to x+y+z=1. Choose y and z freely and put x=1−y−z. There are infinitely many solutions, so this system is consistent.
  3. 3.For B=[1;2;4], the first equation says x+y+z=1 while the third says 3(x+y+z)=4. They cannot both hold, so this system is inconsistent.
  4. 4.The same zero test occurred in both systems. It was necessary for a solution, but in this setting it was not sufficient to guarantee one.
Singular does not mean inconsistent

A singular coefficient matrix has no inverse. The system may still have infinitely many solutions. Also, (adj A)B=O does not by itself settle the general three-variable case. Return to the original equations when the test leaves compatibility undecided.

Build equations from a verbal problem

In an application, the mathematical work begins before the inverse. Define the unknowns clearly, translate each statement into a linear equation, and retain the units and meanings of the unknowns. After solving the matrix equation, interpret the values in the original setting.

Example — Find three numbers from three conditions

Problem
Three numbers have sum 6. The second plus three times the third is 11. The first plus the third equals twice the second. Find the numbers.

  1. 1.Let the numbers be x,y,z in that order. The conditions give x+y+z=6, y+3z=11, and x−2y+z=0.
  2. 2.Thus A=[1,1,1;0,1,3;1,−2,1], X=[x;y;z], B=[6;11;0]. The coefficient −2 in the last row comes from moving 2y to the left.
  3. 3.Expanding gives det A=7+3−1=9. The cofactor rows are [7,3,−1], [−3,0,3], and [2,−3,1], so adj A=[7,−3,2;3,0,−3;−1,3,1].
  4. 4.Calculate (adj A)B=[42−33+0;18+0+0;−6+33+0]=[9;18;27]. Dividing by 9 gives [1;2;3].
  5. 5.The numbers are 1,2,3. Check all verbal conditions: their sum is 6, 2+3·3=11, and 1+3=2·2.
Example — Interpret prices per kilogram

Problem
Four kilograms of onion, three of wheat, and two of rice cost ₹60. Two, four, and six kilograms respectively cost ₹90; six, two, and three kilograms respectively cost ₹70. Find the three prices per kilogram.

  1. 1.Let x,y,z be the onion, wheat, and rice prices in rupees per kilogram. Then A=[4,3,2;2,4,6;6,2,3] and B=[60;90;70]. Each coefficient is a purchased mass in kilograms.
  2. 2.The determinant is 4(12−12)−3(6−36)+2(4−24)=50, so the price system has a unique solution.
  3. 3.The cofactor calculation gives adj A=[0,−5,10;30,0,−20;−20,10,10]. Therefore (adj A)B=[−450+700;1800−1400;−1200+900+700]=[250;400;400].
  4. 4.Divide by 50 to obtain X=[5;8;8]. The prices are ₹5, ₹8, and ₹8 per kilogram in the stated order.
  5. 5.Check the totals: 4·5+3·8+2·8=60, 2·5+4·8+6·8=90, and 6·5+2·8+3·8=70. The units kilograms × rupees per kilogram give rupees.
Coefficient testConclusionNext action
det A ≠ 0One solution; consistentCalculate A⁻¹B and check
det A = 0 and (adj A)B ≠ ONo solution; inconsistentExplain the contradiction
det A = 0 and (adj A)B = ONot decided in generalExamine the original equations

Quiz

Quick check

Which coefficient row represents x+z=2y in variable order x,y,z?

Quick check

A consistent system is one that has which property?

Quick check

Why is the matrix-method solution A⁻¹B rather than BA⁻¹?

Quick check

If det A ≠ 0 for a square coefficient matrix, what follows?

Quick check

If det A=0 and (adj A)B≠O, what follows?

Quick check

For a general three-variable system, det A=0 and (adj A)B=O imply what?

Quick check

In a price problem, x is rupees per kilogram and a coefficient is kilograms. What unit does their product have?

Practice Problems

Practice Problems
  1. Check consistency for x+2y=2, 2x+3y=3; and for 2x−y=5, x+y=4. State the determinant in each case.
  2. Check consistency for x+3y=5 and 2x+6y=8. Verify your conclusion directly from the equations as well as the adjoint test.
  3. Discuss consistency for all real a in x+y+z=1, 2x+3y+2z=2, ax+ay+2az=4. Separate the values for which the coefficient determinant is zero.
  4. Check consistency for 3x−y−2z=2, 2y−z=−1, 3x−5y=3. If the determinant is zero, examine a combination of the equations.
  5. Check consistency for 5x−y+4z=5, 2x+3y+5z=2, 5x−2y+6z=−1. Show how your test reaches its conclusion.
  6. Use the matrix method to solve each system: 5x+2y=4, 7x+3y=5; and 2x−y=−2, 3x+4y=3. Check both original equations.
  7. Use the matrix method to solve each system: 4x−3y=3, 3x−5y=7; and 5x+2y=3, 3x+2y=5.
  8. Solve 2x+y+z=1, x−2y−z=3/2, 3y−5z=9 using X=A⁻¹B. Keep the fractional constant exact.
  9. Solve x−y+z=4, 2x+y−3z=0, x+y+z=2 by the matrix method.
  10. Solve 2x+3y+3z=5, x−2y+z=−4, 3x−y−2z=3 by the matrix method.
  11. Solve x−y+2z=7, 3x+4y−5z=−5, 2x−y+3z=12 by the matrix method.
  12. Find A⁻¹ for A=[2,−3,5;3,2,−4;1,1,−2]. Use it to solve 2x−3y+5z=11, 3x+2y−4z=−5, x+y−2z=−3.
  13. Set up and solve the onion, wheat, and rice price problem from this lesson independently. Label every row, column, and unit, then check the three bills.
  14. Give two systems with the same singular coefficient matrix, one consistent and one inconsistent. Explain why the determinant alone cannot distinguish them.

Key Takeaways

Key Takeaways

• Write AX=B with a fixed variable order, explicit zero coefficients, and correct rearranged signs. • Consistent means at least one solution; inconsistent means none. • A nonzero coefficient determinant gives the unique solution X=A⁻¹B. • Premultiply by A⁻¹; do not interchange matrix factors. • For det A=0, a nonzero (adj A)B proves inconsistency, while a zero value may require further checking. • Translate and interpret applications carefully, and check every original equation and its units.

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Adjoints and Inverse Matrices

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Chapter Summary and Practice