Use determinant structure to test invertibility, then use the full augmented system to decide consistency.
JEE Main · JEE Advanced80 min study guide11 indexed sections
BY THE END, YOU CAN
Evaluate determinants by cofactors and controlled row operations.
Use determinant identities with the correct matrix size and invertibility conditions.
Solve linear systems by elimination, inverse or Cramer’s rule as appropriate.
Classify parameter-dependent singular systems without dividing by zero.
Connect determinants to signed area, volume and dependence.
BEFORE YOU START
Matrix dimensions, multiplication and inverse concepts.
Solving simultaneous linear equations.
Polynomial factorization and elementary vector geometry.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
Available in this lesson: JEE Advanced 2015; JEE Main 2016. This is a limited lesson sample, not a complete exam-question bank.
2 sourced sample questions available across the listed exams and years.
Answer checked against the official final key. Worked solution written for this lesson.
For the 3 × 3 matrix with entry (i + jα)² in row i, column j (i,j = 1,2,3), its determinant equals −648α. Which offered values satisfy this: −4, 9, −9, 4?
Reveal answer & worked solution
Answer: 9 and −9 (original options B and C).
Write each entry as i² + 2ijα + j²α². Thus the matrix factors as A B, with rows of A equal to (i²,i,1) and columns of B equal to (1,2jα,j²α²).
Direct evaluation gives det A = −2 and det B = 4α³, so the required determinant is −8α³.
Solve −8α³ = −648α: α(α−9)(α+9) = 0. The equation permits 0, 9 and −9, but only 9 and −9 occur among the offered choices.
Source: JEE (Advanced) 2015, Paper 1, Code 1, Q50; official post-feedback answer-key paper. Solution written for this lesson.
Matched topics: Linear Systems.
JEE Main · 2016 · 2016-04-03 · Offline Paper 1 · Code E · Mathematics · printed Q65
Paraphrased exam question
Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.
For how many real λ does x+λy−z=0, λx−y−z=0, x+y−λz=0 have a solution other than (0,0,0)?
Reveal answer & worked solution
Answer: Three values: λ=−1,0,1 (independently derived; source option 4).
A homogeneous square system has a nonzero solution exactly when its coefficient matrix is singular.
Its determinant is det[(1,λ,−1),(λ,−1,−1),(1,1,−λ)]=λ³−λ=λ(λ−1)(λ+1).
Setting the determinant to zero gives the three distinct real values −1,0,1. Singularity guarantees a nontrivial nullspace at each.
Source: JEE (Main) 2016, 2016-04-03, Code E, Q65; original booklet scan archived by Resonance; independently derived solution; final-key status unverified. Solution written for this lesson.
01 / UNDERSTAND & APPLY
1. A scalar attached to a square matrix
A determinant assigns a scalar to a square matrix. It is not another matrix and is not defined by the ordinary determinant rule for a nonsquare array. For A = [[a,b],[c,d]], det A = ad − bc. Geometrically, this is the signed area scale of the transformation represented by A; its magnitude is the area of the parallelogram formed by its two column vectors.
A zero determinant means the square matrix is singular: its columns are linearly dependent, it does not have an inverse, and it collapses at least one direction. A nonzero determinant means invertibility. The sign distinguishes orientation for real matrices, while the magnitude describes area or volume scaling in the corresponding dimension.
A determinant can help solve a linear system, but it does not contain the right-hand side. Thus det A = 0 alone cannot tell whether Ax = b has no solution or infinitely many. The same singular coefficient matrix may produce either outcome for different b. This distinction is central to the second half of the lesson.
We use bracket notation for a matrix and det A or vertical determinant bars for its scalar determinant. The absolute value |det A| is a second operation: it removes orientation when a nonnegative area or volume is wanted. It should not be inserted into algebraic identities where the sign matters.
02 / UNDERSTAND & APPLY
2. Cofactors organize an expansion
The minor Mᵢⱼ is the determinant obtained by deleting row i and column j. The cofactor is Cᵢⱼ = (−1)^(i+j)Mᵢⱼ. The alternating sign pattern begins with plus in the upper-left corner. Expanding along row i gives det A = ΣⱼaᵢⱼCᵢⱼ; expanding along a column gives the analogous sum.
For a 3×3 matrix, expansion along a row reduces the calculation to three 2×2 determinants. Choose a row or column with zeros whenever possible. Each cofactor uses the minor after deleting the chosen row and column; it is not obtained by simply multiplying nearby entries. Keep the middle minus sign visible in a first-row expansion.
The diagonal-product mnemonic sometimes used for 3×3 determinants is specific to order three. Extending the same wraparound pattern to a 4×4 matrix omits terms and is invalid. Cofactor expansion and row reduction work in every finite square size, though row reduction is usually faster for larger matrices.
An independent check can use a different row, a different column or determinant-preserving row operations. Repeating the same expansion is less effective because the same sign mistake may repeat. If a row is a multiple of another, recognize dependence before doing any expansion: the determinant is zero.
ORIGINAL WORKED EXAMPLE
A determinant checked by elimination
Evaluate det [[1,2,1],[2,5,3],[1,0,2]].
Expand along the first row: 1(5·2 − 3·0) − 2(2·2 − 3·1) + 1(2·0 − 5·1).
This is 10 − 2(1) − 5 = 3.
For a separate check, replace row 2 by row 2 − 2 row 1 and row 3 by row 3 − row 1. These operations preserve the determinant.
The resulting matrix is [[1,2,1],[0,1,1],[0,−2,1]]. Expanding down its first column gives det [[1,1],[−2,1]] = 3.
Answer 3.
03 / UNDERSTAND & APPLY
3. Track exactly what an operation changes
Swapping two rows changes the determinant’s sign. Multiplying one row by k multiplies the determinant by k. Adding a multiple of one row to a different row leaves the determinant unchanged. The same rules hold for columns. These are determinant rules; when solving a system, the corresponding operation must also be applied to the right-hand side.
An upper- or lower-triangular determinant is the product of its diagonal entries. Row operations can create a triangular matrix, but account for every row swap and scaling on the way. A common efficient strategy uses only row additions until a pivot requires a swap; that minimizes bookkeeping.
For an n×n matrix, det(kA) = kⁿdet A because every row is scaled. This differs from scaling a single row. Also det(Aᵀ) = det A and det(AB) = det A det B for square matrices of the same order. If A is invertible, det(A⁻¹) = 1/det A.
The determinant is linear in each one row while the other rows are held fixed, but not linear in the entire matrix. In general det(A + B) ≠ det A + det B. For example, in order two, det(2I) = 4 while det I + det I = 2. Distinguish multilinearity from a nonexistent whole-matrix additivity.
ORIGINAL WORKED EXAMPLE
Bookkeeping without entries
A 3×3 matrix A has determinant 5. Matrix B is obtained by swapping two rows, multiplying one row by 4, then adding twice one row to another. Find det B and det(2A).
The row swap changes 5 to −5.
Scaling one row by 4 changes the determinant to −20. The later row addition leaves it unchanged.
Therefore det B = −20.
For 2A, all three rows are scaled by 2, so det(2A) = 2³·5 = 40. A single factor 2 would incorrectly treat whole-matrix scaling as one-row scaling.
Answer det B = −20; det(2A) = 40.
04 / UNDERSTAND & APPLY
4. Factor before expanding a parameter forest
When rows become equal for a particular parameter value, the determinant vanishes at that value, suggesting a polynomial factor. Row or column subtraction can expose that factor explicitly. Since a determinant is linear in each row, a common factor from one row can be pulled out. Repeating this reduces a complicated-looking symbolic determinant to a small remaining calculation.
The Vandermonde determinant with rows (1,a,a²), (1,b,b²), (1,c,c²) equals (b−a)(c−a)(c−b). Subtract the first row from the other two, then factor b−a and c−a from those rows. The remaining 2×2 determinant is (c+a) − (b+a) = c−b. The sign depends on the order of the rows and factors.
For parameter-dependent matrices, factor the determinant completely and record every singular parameter before using an inverse or Cramer’s rule. Those exceptional values are often the main point of the question. A formula valid away from them cannot be extended by simply substituting a zero denominator.
If the determinant polynomial is identically zero, there is a structural dependence for every parameter value, not merely a list of isolated singular values. Examine the rows or columns and then the augmented system if solutions are being requested.
ORIGINAL WORKED EXAMPLE
A symmetric parameter matrix
Find the singular values of t for A(t) = [[t,1,1],[1,t,1],[1,1,t]].
Expand along the first row: det A = t(t² − 1) − (t − 1) + (1 − t).
Simplify to t³ − 3t + 2.
Factor: t³ − 3t + 2 = (t − 1)²(t + 2).
Therefore the matrix is singular at t = 1 and t = −2, and invertible at every other real t. The repeated determinant root at t = 1 does not by itself specify the number of system solutions.
Answer Singular exactly at t = 1 and t = −2.
05 / UNDERSTAND & APPLY
5. The augmented matrix retains the whole system
A linear system can be written Ax = b, where A contains coefficients, x is the unknown column, and b is the right-hand side. The augmented matrix [A|b] keeps both coefficients and constants together. Row operations preserve the solution set when they are reversible: swap equations, multiply an equation by a nonzero scalar, or add a multiple of one equation to another.
Gaussian elimination reduces the system to echelon form. A pivot is the first nonzero coefficient in a nonzero row after earlier columns have been cleared. Back-substitution determines pivot variables in terms of any free variables. If every variable has a pivot and no contradiction occurs, the solution is unique.
A row [0 … 0 | c] with c ≠ 0 states 0 = c and proves inconsistency. A row of all zeros states an identity and contributes no new condition. Free variables in a consistent system produce infinitely many real solutions. Do not read a zero row as a contradiction or as a new equation setting a variable to zero.
In rank language, a system is consistent exactly when rank A = rank[A|b]. If that common rank equals the number of unknowns, the solution is unique; if it is smaller, free variables remain. Rank is the number of pivots, not the number of nonzero entries. These criteria also apply to rectangular systems, where det A is unavailable.
ORIGINAL WORKED EXAMPLE
Elimination with an explicit solution check
Solve x + y + z = 6, 2x + y − z = 1, and x − y + 2z = 5.
Subtract twice the first equation from the second to obtain −y − 3z = −11, or y + 3z = 11.
Subtract the first equation from the third to obtain −2y + z = −1.
From the second reduced equation z = 2y − 1. Substitute into y + 3z = 11 to get 7y − 3 = 11, hence y = 2 and z = 3.
The first original equation gives x = 1. Check the others: 2 + 2 − 3 = 1 and 1 − 2 + 6 = 5.
All three variables are determined, so the system has the unique solution (1,2,3).
Answer x = 1, y = 2, z = 3.
06 / UNDERSTAND & APPLY
6. Cramer’s rule requires a nonzero denominator
For an n×n system Ax = b with det A ≠ 0, Cramer’s rule gives xⱼ = det Aⱼ/det A, where Aⱼ is obtained by replacing column j of A with b. Replace a column, not a row, because column j contains the coefficients multiplying xⱼ across all equations.
Cramer’s rule is a closed-form identity and can be elegant for small symbolic systems or when only one variable is requested. Elimination is usually more efficient numerically for a large system. If det A = 0, the ratios are not defined. Even all replacement determinants being zero is not enough by itself to conclude consistency; inspect the augmented rank or original equations. For example, the 2×2 zero coefficient matrix with a nonzero right-hand side has zero coefficient and replacement determinants but gives an inconsistent system.
The adjugate adj A is the transpose of the cofactor matrix. It satisfies A adj A = adj A A = (det A)I. When det A ≠ 0, divide this identity by det A to obtain A⁻¹ = adj A/det A. Forgetting the transpose confuses the cofactor matrix with the adjugate and generally gives an incorrect inverse.
For a homogeneous system Ax = 0, the zero vector is always a solution. If A is square and invertible, it is the only solution. If A is square and singular, a nonzero solution exists over the reals because at least one free variable remains. The word homogeneous here means zero right-hand side, not a condition on the degree of individual terms.
ORIGINAL WORKED EXAMPLE
Cramer’s rule for one variable
In the system 2x + y = 5 and x − y = 1, find x by Cramer’s rule and then y.
The coefficient determinant is D = det [[2,1],[1,−1]] = −2 − 1 = −3, so the system is uniquely solvable.
Replace the x-column by the constants: Dₓ = det [[5,1],[1,−1]] = −5 − 1 = −6.
Thus x = Dₓ/D = 2. Substitute into x − y = 1 to obtain y = 1.
Checking 2x + y gives 4 + 1 = 5. Replacing a row instead would not represent the variable’s coefficient column.
Answer x = 2, y = 1.
07 / UNDERSTAND & APPLY
7. A zero determinant opens a case analysis
For a parameter-dependent system, first solve the generic nonzero-determinant case, then substitute each singular parameter directly into the original equations. This prevents invalid division and reveals whether equations become redundant or contradictory. The right-hand side is decisive.
In the symmetric matrix A(t) from the earlier example, every row sum is t + 2. With right-hand side (1,1,1), symmetry suggests an equal-coordinate solution x = y = z = 1/(t + 2) whenever the matrix is invertible. At t = 1, all equations collapse to x + y + z = 1. At t = −2, adding all three equations gives 0 = 3, which is impossible.
The generic expression may have a finite value at a singular parameter and still fail to describe the entire solution set. At t = 1, the equal-coordinate point (1/3,1/3,1/3) is one solution, but there are infinitely many others. A removable-looking denominator in one simplified coordinate formula is not proof of uniqueness.
To describe an infinite family, name the free parameters and express the pivot variables in terms of them. For x + y + z = 1, write y = s, z = r and x = 1 − s − r, with s,r arbitrary real. This is a full solution set, whereas one example point only proves existence.
ORIGINAL WORKED EXAMPLE
One matrix, three solution behaviors
Classify and solve t x + y + z = 1, x + t y + z = 1, x + y + t z = 1 for real t.
The coefficient determinant is (t − 1)²(t + 2). For t ≠ 1,−2 the system has a unique solution.
Setting all three coordinates equal gives (t + 2)x = 1, so the unique solution is x = y = z = 1/(t + 2).
At t = 1, all equations are x + y + z = 1. Hence (x,y,z) = (1 − s − r,s,r), for arbitrary real s,r.
At t = −2, adding the three equations gives 0 = 3, so no solution exists.
The two singular values therefore produce different outcomes even though both make the same coefficient determinant zero.
Answer Unique equal-coordinate solution for t ≠ 1,−2; infinitely many for t = 1; none for t = −2.
08 / UNDERSTAND & APPLY
8. Determinants connect algebra to area and dependence
For planar vectors u = (u₁,u₂) and v = (v₁,v₂), the determinant u₁v₂ − u₂v₁ is signed parallelogram area. Its absolute value gives area and half of that gives triangle area. A zero value means the two vectors are parallel or one is zero, so the parallelogram collapses.
For three spatial vectors, the determinant equals their scalar triple product, an oriented parallelepiped volume. A zero determinant means coplanarity of the vectors. When testing points rather than vectors, subtract a common base point first. Otherwise the test may inadvertently refer to a plane or triangle involving the origin.
For a linear transformation A, det A describes how oriented volume scales. Shearing by adding one coordinate multiple to another preserves determinant, swapping axes reverses its sign, and scaling one coordinate multiplies it by that factor. These geometric effects match the row/column operation rules and provide intuition for why those rules look as they do.
The determinant test detects full rank for square matrices, but it does not report the exact smaller rank when it vanishes. A zero matrix and a matrix with one dependent row can both have determinant zero while having very different ranks. Elimination exposes the remaining independent directions.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Both axes use50pixels per coordinate unit. The shaded triangle occupies half the parallelogram. AB then AC is positively oriented in the usual x-right,y-up plane. D is an auxiliary point, not an additional triangle vertex.
ORIGINAL WORKED EXAMPLE
A triangle area from a small determinant
Find the area of the triangle with vertices A = (1,1), B = (4,2), C = (2,5).
Use adjacent displacement vectors AB = (3,1) and AC = (1,4).
Their determinant is 3·4 − 1·1 = 11.
The parallelogram area is |11|, so the triangle area is 11/2.
Reversing the order of the two vectors would give −11, but the absolute-value area would remain 11/2.
Answer 11/2 square units.
09 / UNDERSTAND & APPLY
9. Original practice: evaluate and use structure
Track determinant changes separately from the entries when row operations are involved. Preserve signs until an area or volume explicitly calls for an absolute value.
ORIGINAL PRACTICE · 1
Evaluate det [[3,2],[5,4]].
Reveal answer & reasoning
2.
Use ad − bc: 3·4 − 2·5 = 12 − 10 = 2.
ORIGINAL PRACTICE · 2
Evaluate the determinant of [[2,1,4],[0,−3,5],[0,0,6]].
Reveal answer & reasoning
−36.
The matrix is upper triangular, so multiply the diagonal entries 2(−3)6.
ORIGINAL PRACTICE · 3
If det A = −2 for a 4×4 matrix, find det(3A).
Reveal answer & reasoning
−162.
All four rows are scaled by 3, so det(3A) = 3⁴(−2) = −162.
ORIGINAL PRACTICE · 4
If square A,B have determinants 3 and −4, find det(ABᵀ).
Reveal answer & reasoning
−12.
Transpose preserves determinant and products multiply determinants: det A det Bᵀ = 3(−4).
ORIGINAL PRACTICE · 5
Evaluate det [[1,1,1],[1,3,9],[1,4,16]].
Reveal answer & reasoning
6.
This is the three-row Vandermonde form with a=1,b=3,c=4. Its determinant is (3−1)(4−1)(4−3)=6.
ORIGINAL PRACTICE · 6
A row of a determinant is twice another row. What is the determinant?
Reveal answer & reasoning
0.
Subtracting twice the other row creates a zero row without changing the determinant. Expanding along a zero row gives zero.
10 / UNDERSTAND & APPLY
10. Original practice: unique, infinite or inconsistent
When a system is singular, show the actual compatibility or contradiction. A bare statement that the determinant is zero is not a classification.
ORIGINAL PRACTICE · 1
Solve x + y = 3 and x − y = 1.
Reveal answer & reasoning
x = 2, y = 1.
Adding gives 2x = 4. Substitute x = 2 into the first equation to get y = 1. The coefficient determinant is −2, confirming uniqueness.
ORIGINAL PRACTICE · 2
Classify x + 2y = 3 and 2x + 4y = 6.
Reveal answer & reasoning
Infinitely many: (x,y) = (3 − 2t,t).
The second equation is twice the first, so it adds no restriction. Choose y=t freely and solve for x.
ORIGINAL PRACTICE · 3
Classify x + 2y = 3 and 2x + 4y = 7.
Reveal answer & reasoning
No solution.
Twice the first equation gives left side 2x+4y but right side 6, contradicting the second equation’s right side 7.
ORIGINAL PRACTICE · 4
For which k does x + y = 1, x + ky = 2 have a unique solution, and what is it?
Reveal answer & reasoning
For k ≠ 1: y = 1/(k−1), x = (k−2)/(k−1). For k=1 there is no solution.
Subtract the first equation from the second to get (k−1)y=1. Division is allowed only if k≠1. At k=1 this becomes 0=1, an explicit contradiction.
ORIGINAL PRACTICE · 5
Can a homogeneous square system with determinant zero have only the zero solution?
Reveal answer & reasoning
No over the real numbers.
A singular square coefficient matrix has rank less than the number of unknowns. The homogeneous system is always consistent and has a free variable, so a nonzero solution exists.
ORIGINAL PRACTICE · 6
A three-variable system reduces to x + 2z = 4, y − z = 1 and 0 = 0. Give all solutions.
Reveal answer & reasoning
(x,y,z) = (4 − 2t,1 + t,t), t ∈ ℝ.
The variable z is free. Set z=t and solve the two pivot equations. The zero row is redundant, not a contradiction.
11 / UNDERSTAND & APPLY
11. Determinant information has a precise scope
A nonzero determinant completely settles invertibility of a square matrix and uniqueness for every right-hand side. A zero determinant settles only the failure of invertibility; consistency still belongs to the augmented system. This distinction should remain visible in every parameter problem.
The common determinant errors are losing a cofactor sign, using a 3×3 diagonal trick in higher order, failing to account for a row scaling, assuming det(A+B) is additive, and confusing one-row scaling with scaling the whole matrix. The common system errors are changing coefficients without changing constants, dividing by a parameter before listing its zero case, and using Cramer ratios with a zero denominator.
A reliable solution evaluates or factors the determinant, chooses a justified solving method, checks exceptional parameter values directly, and substitutes the final coordinates into all original equations. For an infinite family, verify the formula for arbitrary free parameters rather than checking only one point.
Geometric interpretations provide another audit: zero determinant means collapsed area or volume, while row additions preserve the corresponding signed measure. These ideas explain the algebra, but do not replace the exact row-reduction information needed when a singular system must be classified.
KEEP THE BIG IDEAS
Quick revision
Determinants apply to square matrices and preserve orientation signs.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.