← Explore exam analytics

Matrices and Determinants · TOPIC LESSON

Matrix Algebra

Track dimensions and multiplication order, then count matrices by their genuinely independent choices.

JEE Main · JEE Advanced75 min study guide10 indexed sections

BY THE END, YOU CAN

  • Add, multiply and transpose matrices with correct dimensions and entry rules.
  • Use inverses on the correct side and recognize failures of scalar-style cancellation.
  • Work with symmetric, skew-symmetric, idempotent and nilpotent matrices.
  • Count finite-entry matrices under symmetry, triangularity and invertibility restrictions.
  • State the underlying number system when an algebraic constraint affects counting.

BEFORE YOU START

  • Arithmetic with arrays, finite sums and elementary algebra.
  • Factorials and basic multiplication principle for counting.
  • Two-by-two determinants for inverse and invertibility checks.

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 2016. This is a limited lesson sample, not a complete exam-question bank.

1 sourced sample question available across the listed exams and years.

Matched topics: Matrix Algebra.

JEE Advanced · 2016 · Paper 2 · Code 1 · Mathematics · printed Q37 (subject Q1)

Paraphrased exam question

Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.

Let P have rows (1,0,0), (4,1,0), (16,4,1), and let I be the 3 × 3 identity matrix. Define Q = P⁵⁰ − I = (qᵢⱼ). Find (q₃₁ + q₃₂)/q₂₁.

Reveal answer & worked solution

Answer: 103; independently derived and consistent with archived option B.

  1. Write P=I+N. The only nonzero entries of N are N₂₁=4, N₃₁=16 and N₃₂=4. Multiplication gives (N²)₃₁=16, with all other entries zero; N³=0.
  2. The binomial expansion truncates: Q=50N+binom(50,2)N²=50N+1225N².
  3. Therefore q₂₁=q₃₂=200 and q₃₁=50(16)+1225(16)=20400. The requested ratio is (20400+200)/200=103.

Source: JEE (Advanced) 2016, Paper 2, Code 1, Q37; solution independently derived; archived-key finality unverified. Solution written for this lesson.

01 / UNDERSTAND & APPLY

1. Shape is part of the mathematical object

A matrix is a rectangular array with a specified number of rows and columns. An m×n matrix has m rows and n columns, and its entry aᵢⱼ lies in row i, column j. Two matrices are equal only when their shapes match and every corresponding entry is equal. An ordered arrangement matters: swapping entries usually produces a different matrix.

Matrices can store data or represent linear transformations. A matrix A of shape m×n maps a column vector with n coordinates to a column vector with m coordinates by multiplication. This input-output interpretation explains the dimension rule before any arithmetic: the input must have the number of entries the transformation expects.

Addition and subtraction are entrywise and require equal shapes. Scalar multiplication multiplies every entry by the same scalar. The zero matrix has every entry zero and a shape appropriate to the context. The identity matrix Iₙ is square, with ones on its main diagonal and zeros elsewhere; it leaves n-coordinate vectors unchanged.

We use real matrices unless a different number system is stated. A binary matrix here means its entries are chosen from {0,1} but operations are ordinary real arithmetic. It does not automatically mean arithmetic modulo two. This distinction becomes important when determinant or invertibility conditions are imposed.

02 / UNDERSTAND & APPLY

2. Row by column is composition, not entrywise multiplication

If A is m×n and B is n×p, the product AB is m×p. Its entry in row i, column j is Σₖ₌₁ⁿaᵢₖbₖⱼ. Take row i of A, column j of B, multiply matching entries and add. The matching inner dimension n is the length of each such dot product.

If column vectors are used, AB means first apply B and then apply A: ABx = A(Bx). This explains why order matters. Even when AB and BA both exist and have the same shape, they need not be equal. A horizontal shear followed by a vertical shear generally differs from doing them in reverse order.

Multiplication is associative where the dimensions allow it: (AB)C = A(BC). It distributes over addition. However, the usual scalar expansion (A + B)² = A² + 2AB + B² needs AB = BA. Without commutation, the middle terms are AB + BA. Likewise, (A − B)(A + B) need not equal A² − B².

A product can be zero even when neither factor is zero. This is a major difference from real scalar multiplication. Matrix cancellation from AB = AC requires additional information such as invertibility of A; merely knowing A ≠ 0 is not enough.

ORIGINAL WORKED EXAMPLE

A rectangular product

For A = [[1,2,0],[0,−1,3]] and B = [[2,1],[1,0],[0,2]], compute AB.

  1. A has shape 2×3 and B has shape 3×2, so AB exists and has shape 2×2.
  2. The first row entries are 1·2 + 2·1 + 0·0 = 4 and 1·1 + 2·0 + 0·2 = 1.
  3. The second row entries are 0·2 + (−1)·1 + 3·0 = −1 and 0·1 + (−1)·0 + 3·2 = 6.
  4. Thus AB = [[4,1],[−1,6]]. The reverse product BA also exists but has shape 3×3, so it cannot equal AB.

Answer AB = [[4,1],[−1,6]].

ORIGINAL WORKED EXAMPLE

A concrete order check

For A = [[1,1],[0,1]] and B = [[1,0],[1,1]], compare AB and BA.

  1. Row-by-column multiplication gives AB = [[2,1],[1,1]].
  2. Reversing the order gives BA = [[1,1],[1,2]].
  3. The products differ in their diagonal entries, so these matrices do not commute.
  4. Therefore replacing AB + BA by 2AB in the square of A + B would be invalid for this pair.

Answer AB = [[2,1],[1,1]]; BA = [[1,1],[1,2]].

03 / UNDERSTAND & APPLY

3. Transposition swaps rows and columns and reverses products

The transpose Aᵀ has entry (Aᵀ)ᵢⱼ = aⱼᵢ. An m×n matrix becomes n×m. Transposition respects sums and scalar multiples, but a product reverses order: (AB)ᵀ = BᵀAᵀ. The dimensions themselves suggest this reversal, and the row-column entry formula proves it.

A square real matrix is symmetric if Aᵀ = A and skew-symmetric if Aᵀ = −A. For a skew-symmetric real matrix, aᵢᵢ = −aᵢᵢ forces every diagonal entry to be zero. Its lower-triangular entries are the negatives of the corresponding upper-triangular entries. This is why skew-symmetric matrices have fewer independent choices.

Every real square matrix has a unique decomposition A = S + K with S symmetric and K skew-symmetric: S = (A + Aᵀ)/2 and K = (A − Aᵀ)/2. Adding gives A, while transposing checks the two required properties. Uniqueness follows because a real matrix that is both symmetric and skew-symmetric must be zero.

A product of symmetric matrices need not be symmetric. If A and B are symmetric, then (AB)ᵀ = BA, so AB is symmetric exactly when AB = BA. In contrast, AᵀA is always symmetric and has xᵀAᵀAx = |Ax|² ≥ 0 for every real column x. The latter identity often provides a useful structural check.

ORIGINAL WORKED EXAMPLE

Separate symmetric and skew-symmetric parts

Decompose A = [[1,4],[−2,3]] into symmetric and skew-symmetric matrices.

  1. The transpose is Aᵀ = [[1,−2],[4,3]].
  2. Compute S = (A + Aᵀ)/2 = [[1,1],[1,3]]. Its off-diagonal entries agree.
  3. Compute K = (A − Aᵀ)/2 = [[0,3],[−3,0]]. Its diagonal is zero and transposition negates it.
  4. Adding S and K returns [[1,4],[−2,3]], so the decomposition is verified entry by entry.

Answer S = [[1,1],[1,3]], K = [[0,3],[−3,0]].

04 / UNDERSTAND & APPLY

4. An inverse undoes a transformation on the correct side

A square matrix A is invertible if there is a matrix A⁻¹ satisfying AA⁻¹ = A⁻¹A = I. For A = [[a,b],[c,d]], the inverse exists exactly when ad − bc ≠ 0, and equals (1/(ad−bc))[[d,−b],[−c,a]]. The denominator is the determinant, not the sum of diagonal entries.

To solve AX = B with invertible A, multiply on the left: X = A⁻¹B. To solve XA = B, multiply on the right: X = BA⁻¹. The side is determined by where A appears. A casual instruction to “divide by A” is ambiguous and should be replaced by a correctly ordered multiplication.

For invertible square matrices, (AB)⁻¹ = B⁻¹A⁻¹. The reversal is necessary because AB B⁻¹A⁻¹ = I. Also (Aᵀ)⁻¹ = (A⁻¹)ᵀ. An inverse of a sum generally has no simple sum formula: (A + B)⁻¹ is not A⁻¹ + B⁻¹, even when all three inverses exist.

A singular matrix may still allow solutions of AX = B for some B, but the inverse shortcut is unavailable and uniqueness can fail. Solving the associated linear system is then the appropriate method. Singularity does not mean every equation involving A is inconsistent.

ORIGINAL WORKED EXAMPLE

Solve a matrix equation and verify by multiplication

Let A = [[2,1],[1,1]] and b = [[5],[3]]. Solve Ax = b.

  1. The determinant is 2·1 − 1·1 = 1, so A⁻¹ = [[1,−1],[−1,2]].
  2. Multiply on the left: x = A⁻¹b = [[5 − 3],[−5 + 6]] = [[2],[1]].
  3. Check Ax = [[2·2 + 1],[2 + 1]] = [[5],[3]], which is the original right-hand side.
  4. The invertibility ensures this solution is unique. Multiplying b by A⁻¹ on the right is not even dimensionally defined for this column b.

Answer x = [[2],[1]].

05 / UNDERSTAND & APPLY

5. Polynomial identities can simplify high powers

A matrix is idempotent when P² = P, as happens for many projection matrices. A matrix is nilpotent when Nᵏ = 0 for some positive integer k. Neither condition permits ordinary scalar factor cancellation. From P(P − I) = 0 one cannot conclude P = 0 or P = I; for example diag(1,0) is a third possibility.

For N² = 0, powers of I + N simplify because I commutes with N and the ordinary binomial expansion is valid for this pair. Every term containing N² vanishes, leaving (I + N)ᵐ = I + mN for nonnegative integer m. Its inverse is I − N because (I + N)(I − N) = I − N² = I.

More generally, a polynomial relation satisfied by A can reduce high powers to lower powers. If A² = αA + βI, repeated substitution expresses every higher power as a linear combination of A and I. The scalar coefficients follow a recurrence. Verify the starting matrix identity directly before using it as a shortcut.

An orthogonal real matrix satisfies QᵀQ = I and preserves Euclidean lengths: |Qx|² = xᵀQᵀQx = |x|². Its inverse is Qᵀ. A symmetric matrix is not automatically orthogonal, and an orthogonal matrix need not be symmetric. Similar-looking names describe different equations.

ORIGINAL WORKED EXAMPLE

A high power without repeated multiplication

For N = [[0,1],[0,0]], find (I + N)²⁰ and (I + N)⁻¹.

  1. Direct multiplication gives N² = 0 although N is not the zero matrix.
  2. Since I commutes with N, the binomial theorem gives (I + N)²⁰ = I + 20N; every higher power of N vanishes.
  3. Thus the twentieth power is [[1,20],[0,1]].
  4. The inverse is I − N = [[1,−1],[0,1]], and multiplying it by I + N gives I on either side.

Answer (I + N)²⁰ = [[1,20],[0,1]]; inverse [[1,−1],[0,1]].

06 / UNDERSTAND & APPLY

6. Count independent entries rather than visible boxes

If every entry of an m×n matrix can independently take one of q values, there are q^(mn) matrices. The word independently is doing the work. A condition such as symmetry ties pairs of entries together, a diagonal condition may fix entries, and an invertibility condition couples several choices through a determinant.

For an n×n symmetric matrix with entries in a fixed q-element set, choose its n diagonal entries and its n(n−1)/2 entries above the diagonal. The entries below are then forced by reflection. The count is q^[n(n+1)/2]. No ordering of the chosen values is forgotten: each chosen position remains distinct.

For a skew-symmetric real matrix, the diagonal is zero and each lower entry is the negative of an upper entry. If the allowed set S is closed under negation and contains zero, with |S| = q, the count is q^[n(n−1)/2]. If S lacks some negatives, not every upper choice is allowed. For S = {0,1}, skew-symmetry forces every off-diagonal entry to be zero too, so only the zero matrix remains.

An upper-triangular matrix has n(n+1)/2 potentially free entries, including the diagonal. With a common q-choice alphabet containing zero, its count is q^[n(n+1)/2]. If the diagonal is fixed in advance, only n(n−1)/2 positions remain free. A diagonal matrix has only n independent positions, and a scalar matrix λI has only one.

ORIGINAL WORKED EXAMPLE

Symmetry versus skew-symmetry over the same alphabet

Count 3×3 symmetric and 3×3 skew-symmetric real matrices with entries from {−1,0,1}.

  1. A symmetric 3×3 matrix has three free diagonal entries and three free entries above the diagonal, for six independent positions.
  2. Each has three choices, giving 3⁶ = 729 symmetric matrices.
  3. A skew-symmetric real matrix has its diagonal forced to zero. Its three upper-triangular entries have three choices each, and the lower entries are their negatives, all still in the alphabet.
  4. Therefore the skew-symmetric count is 3³ = 27. Counting all nine displayed positions independently would ignore the defining constraints.

Answer 729 symmetric matrices and 27 skew-symmetric matrices.

07 / UNDERSTAND & APPLY

7. Structural restrictions can change a count completely

A triangular matrix is invertible exactly when every diagonal entry is nonzero, since its determinant is their product. For upper-triangular binary matrices under real arithmetic, every diagonal must therefore be 1 and the entries above the diagonal remain arbitrary. The count is 2^[n(n−1)/2], not the count of all triangular matrices.

A permutation matrix has exactly one 1 in each row and each column, with other entries zero. Choose the column for the first row in n ways, then a different column for the second in n − 1 ways, and continue. The count is n!, and all such matrices are invertible. Requiring exactly one 1 in each row alone would give nⁿ because repeated columns would be allowed.

For small matrices, case splits on a strategic entry or row are often safer than an unsupported general formula. A 2×2 binary matrix is invertible over the reals if its two rows are nonzero and not scalar multiples. Among the three nonzero binary row vectors, no two distinct ones are scalar multiples, so there are 3·2 = 6 invertible matrices.

The field of arithmetic must be stated. A real binary 3×3 matrix can have determinant −2 and be invertible over the reals, while the same entries represent a singular matrix under arithmetic modulo two because −2 becomes zero. Formulas counting invertible matrices over finite fields cannot be transferred to real matrices merely because their entries are 0 and 1.

ORIGINAL WORKED EXAMPLE

Invertible symmetric binary matrices

Count the real invertible symmetric 2×2 matrices with entries in {0,1}.

  1. Write A = [[a,b],[b,c]] with a,b,c independently in {0,1}. Its determinant is ac − b² = ac − b.
  2. If b = 0, invertibility requires ac = 1, so a = c = 1. This gives one matrix.
  3. If b = 1, invertibility requires ac ≠ 1. Of the four ordered choices (a,c), only (1,1) is excluded, leaving three.
  4. The disjoint cases total 1 + 3 = 4. The symmetry constraint means there were only eight candidates to begin with, not sixteen.

Answer 4 matrices.

08 / UNDERSTAND & APPLY

8. Original practice: dimensions, order and structure

Before multiplying, write the shapes. Before cancelling, identify an inverse or another justified condition. These habits are more reliable than checking the final entries alone.

ORIGINAL PRACTICE · 1

If A is 2×4 and B is 4×3, what is the shape of AB, and does BA exist?

Reveal answer & reasoning

AB is 2×3; BA does not exist.

The inner dimensions 4 match in AB. For BA, the inner dimensions would be 3 and 2, which do not match.

ORIGINAL PRACTICE · 2

Compute [[1,2],[3,4]] + 2[[0,1],[−1,2]].

Reveal answer & reasoning

[[1,4],[1,8]].

Multiply every entry in the second matrix by 2, then add corresponding entries: [[1+0,2+2],[3−2,4+4]].

ORIGINAL PRACTICE · 3

Find the transpose of [[1,2,3],[4,5,6]].

Reveal answer & reasoning

[[1,4],[2,5],[3,6]].

The original rows become columns, changing the shape from 2×3 to 3×2.

ORIGINAL PRACTICE · 4

Give a nonzero 2×2 real matrix whose square is zero.

Reveal answer & reasoning

[[0,1],[0,0]].

Its only possible nonzero product entry would require the first row’s 1 to meet a nonzero entry in the second row, but that row is zero. Direct multiplication gives the zero matrix.

ORIGINAL PRACTICE · 5

Does P² = P force P to equal 0 or I?

Reveal answer & reasoning

No.

P = [[1,0],[0,0]] is idempotent but is neither the zero matrix nor the identity. Matrix zero products do not permit scalar-style factor cancellation.

ORIGINAL PRACTICE · 6

Find the inverse of [[1,2],[0,1]].

Reveal answer & reasoning

[[1,−2],[0,1]].

The determinant is 1. The two-by-two inverse formula gives the stated matrix, and multiplying the pair in either order yields I.

ORIGINAL PRACTICE · 7

If A and B are symmetric, under what condition is AB symmetric?

Reveal answer & reasoning

Exactly when AB = BA.

Since (AB)ᵀ = BᵀAᵀ = BA, the equation (AB)ᵀ = AB is equivalent to commutation.

09 / UNDERSTAND & APPLY

9. Original practice: state the free choices explicitly

Every count should identify a set of independent choices and explain why each admissible matrix is counted exactly once. If the rule couples entries, classify that coupling before using a power.

ORIGINAL PRACTICE · 1

How many 2×3 binary matrices are there?

Reveal answer & reasoning

64.

There are six independently chosen entries, each with two choices, giving 2⁶.

ORIGINAL PRACTICE · 2

How many symmetric 4×4 binary matrices are there?

Reveal answer & reasoning

1024.

There are 4 diagonal and 6 upper-triangular independent entries, totaling 10. The lower entries are forced, so the count is 2¹⁰.

ORIGINAL PRACTICE · 3

How many upper-triangular 3×3 binary matrices are invertible over the reals?

Reveal answer & reasoning

8.

All three diagonal entries must be 1. The three entries above the diagonal are free, giving 2³ = 8.

ORIGINAL PRACTICE · 4

How many 3×3 binary matrices have exactly one 1 in each row, with no column restriction?

Reveal answer & reasoning

27.

Each of three rows independently chooses one of three columns, so the count is 3³. Repeated columns are allowed under the stated condition.

ORIGINAL PRACTICE · 5

How many 3×3 permutation matrices are there?

Reveal answer & reasoning

6.

The occupied columns must all be different. Choose them in 3·2·1 = 3! ways.

ORIGINAL PRACTICE · 6

How many skew-symmetric real 3×3 matrices have entries in {0,1}?

Reveal answer & reasoning

1.

The diagonal must be zero. An off-diagonal 1 would force a matching −1, outside the allowed set. Thus every entry must be zero.

ORIGINAL PRACTICE · 7

For A = [[1,1,0],[1,0,1],[0,1,1]], compare invertibility over ℝ and modulo two.

Reveal answer & reasoning

Invertible over ℝ, singular modulo two.

Its ordinary determinant is −2, nonzero over the reals. Modulo two this determinant is zero. The alphabet alone does not specify the arithmetic system.

10 / UNDERSTAND & APPLY

10. A matrix answer needs both algebra and a shape check

Dimensions decide whether an expression exists before arithmetic decides its value. Multiplication order records transformation order. Transposition and inversion reverse products. Singular matrices block cancellation, while special identities such as N² = 0 can make powers easy. These structural facts explain the rules rather than leaving them as isolated exceptions.

For counting, begin with unconstrained positions and then identify which entries are fixed, mirrored, negated or coupled by invertibility. Positive counts such as q^(mn) rely on independence; they are not universal formulas for every matrix condition. For small cases, a direct case split or enumeration is a valuable check on a proposed general argument.

The most common errors are treating multiplication as entrywise, assuming AB = BA, putting the inverse on the wrong side, counting symmetric off-diagonal partners independently, allowing nonzero skew-symmetric diagonal entries over the reals, and importing a finite-field count into real arithmetic.

Verify a matrix equation by multiplication, a symmetry claim by transposition, an inverse by both-sided identity checks, and a count by explaining a one-to-one correspondence between choices and matrices. These are short checks that protect against different kinds of error and make a solution independently readable.

KEEP THE BIG IDEAS

Quick revision

  • Check dimensions and multiplication order before computing entries.
  • Nonzero matrices can have zero products, so cancellation needs a justified inverse.
  • Transpose and inverse reverse the order of products.
  • Count independent entries, then enforce global conditions such as invertibility.
  • Binary entries do not imply arithmetic modulo two unless explicitly stated.
Back to the top ↑

READ WITH CONTEXT

References & scope

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.

Explore this topic in context

Analytics use a separately reviewed, incomplete historical corpus. Opening a topic does not change its counts or denominator.

KEEP EXPLORING

Your next idea is waiting.

Back to Mathematics index →Browse all topic notes