Determinants · Lesson 6 of 6
Chapter Summary and Practice
“Connect determinant calculations, geometry, adjoints, inverses, and systems through complete mixed worked problems.”
• Choose a suitable determinant calculation or geometric interpretation for a mixed problem. • Connect minors and cofactors to adjoints, invertibility, and matrix-method solutions. • Apply determinant and inverse identities with their correct conditions and factor order. • Use substitutions to turn a reciprocal-variable system into a linear system and recover valid original variables. • Reason about symbolic and trigonometric determinants, including independence and ranges. • Check results using signs, dimensions, units, and substitution in the original problem.
See the chapter as a connected chain
We began with a rule that turns a square matrix into one number. We then used that number to measure triangle area and test collinearity. Minors and cofactors explained expansion, and their cancellation patterns built an adjoint. Dividing that adjoint by a nonzero determinant produced an inverse, which then solved a whole system of linear equations.
These connections help us choose a method. A question asking for a triangle’s area calls for a coordinate determinant and a final magnitude. A question asking whether a matrix has an inverse calls for its determinant before any lengthy cofactor work. A question asking for unknown quantities first needs a correctly organised matrix equation. The same mathematical tool can serve different purposes, so the meaning of the final number or matrix must remain clear.
| Idea | Relationship or method | What to watch |
|---|---|---|
| Determinant and order | A square array produces one scalar; order one returns its entry | Determinant bars do not mean modulus |
| Order two | det [a,b;c,d] = ad−bc | Subtract the complete second product |
| Order three | Expand along any row or column | Use deletion and the correct position signs |
| Scaling | det(kA) = kⁿdet A | n is the order, not the number of entries |
| Triangle area | T = |D|/2 for rows [xᵢ,yᵢ,1] | Keep original coordinate signs until the end |
| Collinearity and lines | D=0; make the third point (x,y) for a line equation | Two fixed points must be distinct to define one line |
| Minor and cofactor | Delete row i and column j; Aᵢⱼ=(−1)^(i+j)Mᵢⱼ | A cofactor already contains its position sign |
| Cofactor sums | Matching row or column gives det A; different ones give 0 | Match the subscripts, not just the entry values |
| Adjoint | Transpose the cofactor matrix; A(adj A)=(det A)I | Do not confuse an adjoint with an ordinary transpose |
| Inverse | det A≠0; A⁻¹=(adj A)/det A | A singular matrix still has an adjoint |
| Product and derived identities | det(AB)=det A·det B; (AB)⁻¹=B⁻¹A⁻¹ | Square orders must match and inverse factors must exist |
| Determinants of derived matrices | det(adj A)=(det A)^(n−1); det(A⁻¹)=1/det A | Use n≥2 for the stated adjoint relationship and det A≠0 for the inverse |
| Linear systems | AX=B; if det A≠0, X=A⁻¹B | Premultiply and check all original equations |
| Singular systems | Nonzero (adj A)B proves inconsistency | A zero result does not settle the general case |
Choose an expansion that reveals structure
A determinant containing variables need not lead to complicated algebra. A zero-rich row or column can remove terms, and identities may make surviving terms cancel. Independence from a parameter means that the final expression has no remaining dependence on that parameter; it does not mean that the determinant must be zero.
Problem
Show that det [x,sin θ,cos θ;−sin θ,−x,1;cos θ,1,x] does not depend on θ.
- 1.Write s=sin θ and c=cos θ to keep the expansion readable. Expand row 1: D=x(−x²−1)−s(−sx−c)+c(−s+xc).
- 2.Distribute every term: D=−x³−x+s²x+sc−sc+xc².
- 3.The mixed terms sc cancel. The remaining angle terms combine to x(s²+c²)=x because sin²θ+cos²θ=1.
- 4.Thus D=−x³−x+x=−x³. It may depend on x, but it is independent of θ for every fixed x.
Problem
Evaluate det [1,x,y;1,x+y,y;1,x,x+y].
- 1.Expand along row 1: D=[(x+y)(x+y)−xy]−x[(x+y)−y]+y[x−(x+y)].
- 2.The first bracket is x²+xy+y²; the second is x; the third is −y.
- 3.Hence D=x²+xy+y²−x²−y²=xy. Complete the products before trying to combine terms.
- 4.A quick check at x=2,y=3 gives D=6. If either variable is zero, the expression predicts zero, which agrees with the resulting repeated rows.
For a range question, do not stop when the determinant has been expanded. Express it using a quantity whose range is known and check that the endpoint values can actually occur. For A=[1,sin θ,1;−sin θ,1,sin θ;−1,−sin θ,1], expansion gives det A=2+2sin²θ. As θ varies from 0 to 2π, sin²θ covers [0,1], so det A covers [2,4], including both endpoints. No inverse calculation is required because the determinant is already positive throughout.
Move between geometry and algebra with care
The determinant itself is signed, while geometric area is nonnegative. That distinction should guide both calculation and reverse problems. A simple independent check through a horizontal base or a known line can confirm that the algebra has retained every possible point.
Problem
The triangle with vertices (2,−6), (5,4), and (k,4) has area 35. Find k.
- 1.Form D=det [2,−6,1;5,4,1;k,4,1]. Expansion gives D=2(4−4)+6(5−k)+(20−4k)=50−10k.
- 2.Set |D|/2=35, so |50−10k|=70. The positive branch 50−10k=70 gives k=−2.
- 3.The negative branch 50−10k=−70 gives k=12. Both are valid.
- 4.The two points (5,4) and (k,4) form a horizontal base of length |k−5|; the height from (2,−6) is 10. The area is 5|k−5|, which is 35 for both k values.
A zero coordinate determinant means the three recorded points are collinear. A zero coefficient determinant means the coefficient matrix is singular and does not give a unique solution by an inverse. It does not automatically prove that the system has no solution. Interpret the zero in the problem where it arose.
Use an identity to reduce unnecessary work
A supplied product can identify an inverse directly. Likewise, the adjoint formula can prove a relationship between derived matrices without recalculating nine cofactors on each side. Such shortcuts are valid only when we state the order and nonzero conditions they need.
Problem
Let A=[1,−1,2;0,2,−3;3,−2,4] and C=[−2,0,1;9,2,−3;6,1,−2]. Use their product to solve x−y+2z=1, 2y−3z=1, 3x−2y+4z=2.
- 1.Multiply AC. Its first row is [−2−9+12,0−2+2,1+3−4]=[1,0,0]. Its second is [0+18−18,0+4−3,0−6+6]=[0,1,0].
- 2.Its third row is [−6−18+24,0−4+4,3+6−8]=[0,0,1], so AC=I.
- 3.The product determinant rule gives (det A)(det C)=1, proving A nonsingular. Therefore C=A⁻¹; its reverse product is also I.
- 4.The constant column is B=[1;1;2]. Multiply X=CB=[−2+0+2;9+2−6;6+1−4]=[0;5;3].
- 5.Thus x=0,y=5,z=3. The original left sides are 0−5+6=1, 10−9=1, and 0−10+12=2. The supplied product has replaced a fresh cofactor calculation.
Problem
For an invertible square matrix A of order at least two, explain why (adj A)⁻¹=adj(A⁻¹) and (A⁻¹)⁻¹=A.
- 1.From A⁻¹=(adj A)/det A, rearrange to adj A=(det A)A⁻¹. Since det A is nonzero, adj A is invertible and its inverse is A/(det A).
- 2.Apply the adjoint relationship to A⁻¹: adj(A⁻¹)=det(A⁻¹)(A⁻¹)⁻¹. The determinant factor is 1/det A.
- 3.The equations AA⁻¹=A⁻¹A=I mean that A itself is the two-sided inverse of A⁻¹. Thus (A⁻¹)⁻¹=A.
- 4.Consequently adj(A⁻¹)=A/(det A), equal to (adj A)⁻¹. Both sides are matrices, and every reciprocal used was justified by invertibility.
For a product AB, write (AB)⁻¹=B⁻¹A⁻¹ before substituting any known inverse matrices. For a diagonal matrix [x,0,0;0,y,0;0,0,z] with nonzero x,y,z, each diagonal entry can be undone separately: the inverse is [1/x,0,0;0,1/y,0;0,0,1/z]. Multiplying gives ones on the diagonal and zeros elsewhere. This special structure does not justify taking reciprocals entry by entry in a general matrix, where off-diagonal sums also matter.
A substitution can reveal a linear system
Not every problem displays its linear structure in the original variables. If unknowns occur only through reciprocals, we may treat those reciprocals as new variables. We then use the matrix method on the transformed system and finally return to the original variables, checking all nonzero requirements.
Problem
Solve 2/x+3/y+10/z=4, 4/x−6/y+5/z=1, and 6/x+9/y−20/z=2.
- 1.The original equations require x,y,z to be nonzero. Put u=1/x, v=1/y, w=1/z. The equations become 2u+3v+10w=4, 4u−6v+5w=1, 6u+9v−20w=2.
- 2.Use coefficient matrix H=[2,3,10;4,−6,5;6,9,−20] and constant column b=[4;1;2]. Its determinant is 1200, so the transformed system has a unique solution.
- 3.Calculate adj H=[75,150,75;110,−100,30;72,0,−24]. Multiplying by b gives [300+150+150;440−100+60;288+0−48]=[600;400;240].
- 4.Divide by 1200 to obtain u=1/2, v=1/3, w=1/5. These are all nonzero, so taking reciprocals is allowed: x=2,y=3,z=5.
- 5.Check in the original equations: 1+1+2=4, 2−2+1=1, and 3+3−4=2. Solving for u,v,w was an intermediate step, not the final interpretation of the unknowns.
If a transformed answer had u=0, there would be no finite nonzero real x with 1/x=0. In that situation a solution of the transformed linear system would not yield a solution of the original reciprocal equations. Checking the substitution’s reverse direction is therefore part of the mathematics, not an optional final comment.
Check the object, the conditions, and the result
A reliable mixed solution has three layers: identify what is being calculated, choose a method whose conditions hold, and interpret or verify the result. This habit joins procedural skill with understanding and makes the chapter’s formulas easier to use accurately.
Before calculating, ask whether the requested output is a number, an area, a matrix, or a column of unknowns. During calculation, preserve signs and factor order, and justify division by a determinant. After calculation, use a different check where possible: multiply an inverse in both orders, substitute a system solution, compare an area with a visible base and height, or test a symbolic expression at a convenient numerical input. A single numerical test supports a formula but does not replace its algebraic derivation.
Quiz
An order-three matrix A has det A=2. What is det(2A)?
The determinant for a triangle’s coordinates is zero. Which conclusion follows?
A nonsingular order-three matrix has det A=4. What is det(adj A)?
For fixed x, a trigonometric determinant simplifies to −x³. What does that establish?
If det A=−2 and det B=5, what is det((AB)⁻¹)?
After u=1/x, a transformed system gives u=0. What should be concluded about x?
For 0≤θ≤2π, what is the range of 2+2sin²θ?
If A=[2,0,0;0,−4,0;0,0,5], which diagonal entries does its inverse have?
Practice Problems
- Prove that det [x,sin θ,cos θ;−sin θ,−x,1;cos θ,1,x] is independent of θ. Identify separately the cancellation and the trigonometric identity.
- Evaluate det [cos α cos β,cos α sin β,−sin α;−sin β,cos β,0;sin α cos β,sin α sin β,cos α]. Choose an expansion that uses its zero entry.
- Given A⁻¹=[3,−1,1;−15,6,−5;5,−2,2] and B=[1,2,−2;−1,3,0;0,−2,1], find (AB)⁻¹. Explain the factor order before calculating.
- For A=[1,2,1;2,3,1;1,1,5], verify (adj A)⁻¹=adj(A⁻¹) and (A⁻¹)⁻¹=A. Begin by verifying that the required inverses exist.
- Evaluate det [x,y,x+y;y,x+y,x;x+y,x,y] by direct expansion. Factor the resulting polynomial if useful and check one numerical case.
- Evaluate det [1,x,y;1,x+y,y;1,x,x+y] by expanding along a row. Explain which terms cancel.
- Solve 2/x+3/y+10/z=4, 4/x−6/y+5/z=1, 6/x+9/y−20/z=2 using reciprocal variables and the matrix method. Check the recovered x,y,z.
- Find and verify the inverse of diag(x,y,z) for nonzero real x,y,z. Explain why multiplying diag(1/x,1/y,1/z) by xyz would generally spoil the inverse.
- For A=[1,sin θ,1;−sin θ,1,sin θ;−1,−sin θ,1], determine the complete range of det A for 0≤θ≤2π. Verify that both endpoints occur.
- Use the given inverse product in the worked example to solve the same coefficient system with B=[2;0;1]. Check the resulting three equations.
- Find the area of the triangle (−1,2),(3,2),(0,−2). Derive the equation of the line through its first two vertices, then explain the determinant condition for the third point to lie on that line.
- For A=[2,1;3,2], calculate its determinant, all cofactors, adjoint, and inverse. Verify A(adj A)=(det A)I and AA⁻¹=A⁻¹A=I.
- Give a singular coefficient matrix and choose two constant columns, one producing a consistent system and another an inconsistent one. State which conclusions come from the determinant and which require the original equations.
- A student writes det(3A)=3det A for an order-three matrix, uses the cofactor matrix without transposing to find an inverse, and treats a negative triangle determinant as a negative area. Correct and explain each error.
Key Takeaways
• Determinants connect a square array to a scalar, but the scalar must be interpreted in its geometric or algebraic setting. • Expansion, scaling, and product rules provide different efficient routes; each has its own conditions. • Coordinate determinants give half-magnitude area and a zero test for collinearity. • Minors become signed cofactors, whose transpose is the adjoint; matching and different-row sums explain the adjoint identity. • A nonzero determinant permits an inverse and the unique matrix-method solution; singular systems need compatibility reasoning. • Inverse products reverse factor order, and substitutions must be reversed with their domain restrictions checked. • Mixed practice is complete only when the answer is interpreted and verified in the original problem.
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Solving Systems with Matrix Inverses
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