Matrices Notes FE Sem 1 Applied Maths-I | Mumbai University

Matrices give a compact way to store and manipulate systems of linear equations, which arise in circuit analysis (KVL and KCL), structures and signal processing. This module introduces Hermitian, skew-Hermitian, unitary and orthogonal matrices, the rank of a matrix found by echelon form, normal form and PAQ form, and uses rank to decide whether a system of linear equations has no solution, a unique solution or infinitely many.

Summary

Pre-requisites: addition, multiplication and transpose of matrices, the inverse \( A^{-1} = \frac{\text{adj}\,A}{|A|} \) of a non-singular matrix, and symmetric (\( A^T = A \)) and skew-symmetric (\( A^T = -A \)) matrices. For a complex matrix A, \( \bar A \) is obtained by conjugating each element and \( A^\theta = (\bar A)^T \) is its conjugate transpose (also written \( A^* \)).

Types of matrices (5.1): A square matrix A is Hermitian if \( A^\theta = A \) (its diagonal elements are real), skew-Hermitian if \( A^\theta = -A \) (its diagonal elements are zero or purely imaginary), unitary if \( A^\theta A = AA^\theta = I \), and orthogonal (real A) if \( A^TA = AA^T = I \). A real symmetric matrix is Hermitian and a real orthogonal matrix is unitary.

Properties (without proof): every square matrix can be written uniquely as the sum of a Hermitian and a skew-Hermitian matrix, \( A = \frac12(A + A^\theta) + \frac12(A - A^\theta) \); if A is Hermitian then iA is skew-Hermitian; the inverse and transpose of an orthogonal matrix are orthogonal, \( A^{-1} = A^T \), and \( |A| = \pm1 \); for a unitary matrix \( A^{-1} = A^\theta \) and \( |\,|A|\,| = 1 \); the product of two orthogonal (unitary) matrices is orthogonal (unitary). The rows (and columns) of an orthogonal matrix form an orthonormal set.

Rank of a matrix (5.1): the rank ρ(A) is the order of the largest non-zero minor of A, which equals the number of linearly independent rows. Elementary row and column transformations (interchange, multiplying by a non-zero constant, adding a multiple of one row or column to another) do not change the rank.

Echelon form: reduce A by row transformations so that every all-zero row is at the bottom and the number of zeros before the first non-zero element increases row by row. The rank equals the number of non-zero rows. Normal form: using both row and column transformations, reduce A to \( \begin{bmatrix} I_r & 0 \\ 0 & 0 \end{bmatrix} \); then ρ(A) = r.

PAQ form (3 × 3 matrices): write A = I A I. Apply every row transformation to A and to the pre-factor I, and every column transformation to A and to the post-factor I, until A becomes the normal form N. The pre-factor becomes P and the post-factor Q, giving PAQ = N with P and Q non-singular. The rank is the order of the identity block in N.

Non-homogeneous system AX = B (5.2): form the augmented matrix [A | B] and reduce to echelon form. If ρ(A) ≠ ρ[A | B] the system is inconsistent (no solution). If ρ(A) = ρ[A | B] = n (the number of unknowns) there is a unique solution; if ρ(A) = ρ[A | B] = r < n there are infinitely many solutions with n − r arbitrary parameters.

Homogeneous system AX = 0 (5.2): it is always consistent since X = 0 is a solution. If ρ(A) = n only the trivial solution exists; if ρ(A) = r < n there are infinitely many non-trivial solutions with n − r free parameters. For a square system, non-trivial solutions exist if and only if |A| = 0. Self-learning topics: application of the inverse of a matrix to coding theory (encoding a message by multiplying with an invertible matrix and decoding with its inverse), and reduction of m × n matrices to normal and PAQ form.

Mind map

Method: testing consistency and solving AX = B

Important definitions

Formulas

Key points to remember

• Hermitian: diagonal real; skew-Hermitian: diagonal zero or purely imaginary.
• For checking unitary or orthogonal, compute A^θA (or A^TA) and show it equals I; showing rows are orthonormal is equivalent.
• Rank of an m × n matrix is at most min(m, n); rank of the null matrix is 0.
• Use only row transformations on the augmented matrix [A | B]; column transformations would mix up the unknowns.
• In PAQ form, row operations go to P (the pre-factor) and column operations go to Q (the post-factor).
• A homogeneous system always has the trivial solution; it has non-trivial solutions only when ρ(A) < n.
• For parameter problems (λ, μ), reduce to echelon form and examine when the last row of A becomes zero.
• Number of free parameters = number of unknowns − rank.

Solved examples

Example 1. Express \( A = \begin{bmatrix} 2 + 3i & 1 - i \\ 3 + i & 4i \end{bmatrix} \) as the sum of a Hermitian and a skew-Hermitian matrix.

\( A^\theta = \begin{bmatrix} 2 - 3i & 3 - i \\ 1 + i & -4i \end{bmatrix} \).

\( P = \frac12(A + A^\theta) = \begin{bmatrix} 2 & 2 - i \\ 2 + i & 0 \end{bmatrix} \); \( P^\theta = P \), so P is Hermitian.

\( Q = \frac12(A - A^\theta) = \begin{bmatrix} 3i & -1 \\ 1 & 4i \end{bmatrix} \); \( Q^\theta = \begin{bmatrix} -3i & 1 \\ -1 & -4i \end{bmatrix} = -Q \), so Q is skew-Hermitian.

A = P + Q.

Example 2. Show that \( A = \frac12\begin{bmatrix} 1 + i & -1 + i \\ 1 + i & 1 - i \end{bmatrix} \) is unitary.

\( A^\theta = \frac12\begin{bmatrix} 1 - i & 1 - i \\ -1 - i & 1 + i \end{bmatrix} \).

\( A^\theta A = \frac14\begin{bmatrix} (1 - i)(1 + i) + (1 - i)(1 + i) & (1 - i)(-1 + i) + (1 - i)(1 - i) \\ (-1 - i)(1 + i) + (1 + i)(1 + i) & (-1 - i)(-1 + i) + (1 + i)(1 - i) \end{bmatrix} \).

Entries: 2 + 2 = 4; \( -(1 - i)^2 + (1 - i)^2 = 0 \); \( -(1 + i)^2 + (1 + i)^2 = 0 \); 2 + 2 = 4.

\( A^\theta A = \frac14\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} = I \), hence A is unitary.

Example 3. Find a, b, c if \( A = \frac13\begin{bmatrix} 1 & 2 & a \\ 2 & 1 & b \\ 2 & -2 & c \end{bmatrix} \) is orthogonal.

For an orthogonal matrix the columns are orthonormal. Columns 1 and 2, (1, 2, 2)/3 and (2, 1, −2)/3, already have unit length and are orthogonal (2 + 2 − 4 = 0).

Column 3 must be orthogonal to both: a + 2b + 2c = 0 and 2a + b − 2c = 0.

Solving (cross-multiplication): \( \frac{a}{-4 - 2} = \frac{b}{4 + 2} = \frac{c}{1 - 4} \Rightarrow a : b : c = 2 : -2 : 1 \).

Unit length: \( \frac{a^2 + b^2 + c^2}{9} = 1 \Rightarrow (a, b, c) = k(2, -2, 1) \) with \( 9k^2 = 9 \), k = ±1.

So a = 2, b = −2, c = 1 or a = −2, b = 2, c = −1.

Example 4. Find the rank of \( A = \begin{bmatrix} 1 & 2 & 3 & 2 \\ 2 & 3 & 5 & 1 \\ 1 & 3 & 4 & 5 \end{bmatrix} \) by reducing to echelon form.

R₂ → R₂ − 2R₁, R₃ → R₃ − R₁: \( \begin{bmatrix} 1 & 2 & 3 & 2 \\ 0 & -1 & -1 & -3 \\ 0 & 1 & 1 & 3 \end{bmatrix} \).

R₃ → R₃ + R₂: \( \begin{bmatrix} 1 & 2 & 3 & 2 \\ 0 & -1 & -1 & -3 \\ 0 & 0 & 0 & 0 \end{bmatrix} \).

This is echelon form with two non-zero rows, so ρ(A) = 2.

Example 5. Find non-singular matrices P and Q such that PAQ is in normal form, for \( A = \begin{bmatrix} 1 & 1 & 2 \\ 1 & 2 & 3 \\ 0 & -1 & -1 \end{bmatrix} \). Hence find the rank.

Write \( A = I_3\,A\,I_3 \).

R₂ → R₂ − R₁ (apply to A and pre-factor): A becomes [[1, 1, 2], [0, 1, 1], [0, −1, −1]], pre-factor [[1, 0, 0], [−1, 1, 0], [0, 0, 1]].

R₃ → R₃ + R₂: A becomes [[1, 1, 2], [0, 1, 1], [0, 0, 0]], pre-factor [[1, 0, 0], [−1, 1, 0], [−1, 1, 1]].

C₂ → C₂ − C₁, C₃ → C₃ − 2C₁ (apply to A and post-factor): A becomes [[1, 0, 0], [0, 1, 1], [0, 0, 0]], post-factor [[1, −1, −2], [0, 1, 0], [0, 0, 1]].

C₃ → C₃ − C₂: A becomes [[1, 0, 0], [0, 1, 0], [0, 0, 0]], post-factor [[1, −1, −1], [0, 1, −1], [0, 0, 1]].

So \( P = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -1 & 1 & 1 \end{bmatrix} \), \( Q = \begin{bmatrix} 1 & -1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix} \), \( PAQ = \begin{bmatrix} I_2 & 0 \\ 0 & 0 \end{bmatrix} \), and ρ(A) = 2.

Example 6. Test for consistency and solve: x + y + z = 6, x − y + 2z = 5, 3x + y + z = 8.

\( [A|B] = \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 1 & -1 & 2 & | & 5 \\ 3 & 1 & 1 & | & 8 \end{bmatrix} \).

R₂ → R₂ − R₁, R₃ → R₃ − 3R₁: rows (0, −2, 1 | −1) and (0, −2, −2 | −10).

R₃ → R₃ − R₂: (0, 0, −3 | −9).

ρ(A) = ρ[A|B] = 3 = number of unknowns, so the system is consistent with a unique solution.

Back substitution: −3z = −9 ⇒ z = 3; −2y + z = −1 ⇒ y = 2; x + y + z = 6 ⇒ x = 1.

Solution x = 1, y = 2, z = 3.

Example 7. For what values of λ and μ does the system x + y + z = 6, x + 2y + 3z = 10, x + 2y + λz = μ have (i) no solution, (ii) a unique solution, (iii) infinitely many solutions?

\( [A|B] = \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 1 & 2 & 3 & | & 10 \\ 1 & 2 & \lambda & | & \mu \end{bmatrix} \).

R₂ → R₂ − R₁, R₃ → R₃ − R₁: (0, 1, 2 | 4), (0, 1, λ − 1 | μ − 6). R₃ → R₃ − R₂: (0, 0, λ − 3 | μ − 10).

(i) λ = 3, μ ≠ 10: ρ(A) = 2, ρ[A|B] = 3, no solution.

(ii) λ ≠ 3 (any μ): ρ(A) = ρ[A|B] = 3, unique solution.

(iii) λ = 3, μ = 10: ρ(A) = ρ[A|B] = 2 < 3, infinitely many solutions: with z = k, y = 4 − 2k, x = 2 + k.

Example 8. Find the values of λ for which 3x + y − λz = 0, 4x − 2y − 3z = 0, 2λx + 4y + λz = 0 has a non-trivial solution, and find the solutions.

Non-trivial solutions require \( |A| = \begin{vmatrix} 3 & 1 & -\lambda \\ 4 & -2 & -3 \\ 2\lambda & 4 & \lambda \end{vmatrix} = 0 \).

\( |A| = 3(-2\lambda + 12) - 1(4\lambda + 6\lambda) - \lambda(16 + 4\lambda) = -4\lambda^2 - 32\lambda + 36 = -4(\lambda + 9)(\lambda - 1) \).

So λ = 1 or λ = −9.

λ = 1: from 3x + y − z = 0 and 4x − 2y − 3z = 0, \( \frac{x}{-3 - 2} = \frac{y}{-4 + 9} = \frac{z}{-6 - 4} \Rightarrow x : y : z = 1 : -1 : 2 \); solution x = k, y = −k, z = 2k (it satisfies 2x + 4y + z = 0).

λ = −9: from 3x + y + 9z = 0 and 4x − 2y − 3z = 0, \( \frac{x}{-3 + 18} = \frac{y}{36 + 9} = \frac{z}{-6 - 4} \Rightarrow x : y : z = 3 : 9 : -2 \); solution x = 3k, y = 9k, z = −2k (check: −54 + 36 + 18 = 0).

Important questions and answers

Q1. Define Hermitian and skew-Hermitian matrices and show that the diagonal elements of a Hermitian matrix are real. (3 marks)

A is Hermitian if \( A^\theta = A \) and skew-Hermitian if \( A^\theta = -A \).

If A is Hermitian, \( a_{ij} = \overline{a_{ji}} \). For i = j, \( a_{ii} = \overline{a_{ii}} \), so each diagonal element equals its own conjugate and is therefore real.

Q2. Show that \( A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \) is orthogonal and find \( A^{-1} \). (3 marks)

\( A^TA = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} = \begin{bmatrix} \cos^2\theta + \sin^2\theta & 0 \\ 0 & \sin^2\theta + \cos^2\theta \end{bmatrix} = I \).

Hence A is orthogonal and \( A^{-1} = A^T = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \).

Q3. If A is Hermitian, show that iA is skew-Hermitian. (3 marks)

\( (iA)^\theta = \bar i\,A^\theta = -iA^\theta = -iA \) since \( A^\theta = A \).

Hence iA is skew-Hermitian.

Q4. Reduce \( A = \begin{bmatrix} 2 & 3 & 4 \\ 3 & 1 & 2 \\ -1 & 2 & 2 \end{bmatrix} \) to normal form and find its rank. (5 marks)

R₁ ↔ R₃ and R₁ → −R₁: [[1, −2, −2], [3, 1, 2], [2, 3, 4]].

R₂ → R₂ − 3R₁, R₃ → R₃ − 2R₁: [[1, −2, −2], [0, 7, 8], [0, 7, 8]].

R₃ → R₃ − R₂: [[1, −2, −2], [0, 7, 8], [0, 0, 0]].

C₂ → C₂ + 2C₁, C₃ → C₃ + 2C₁: [[1, 0, 0], [0, 7, 8], [0, 0, 0]]. C₂ → C₂/7, then C₃ → C₃ − 8C₂: [[1, 0, 0], [0, 1, 0], [0, 0, 0]].

Normal form \( \begin{bmatrix} I_2 & 0 \\ 0 & 0 \end{bmatrix} \), so ρ(A) = 2. (Check: |A| = 2(2 − 4) − 3(6 + 2) + 4(6 + 1) = −4 − 24 + 28 = 0.)

Q5. Show that the system x + y + z = 1, x + 2y + 3z = 2, 2x + 3y + 4z = 4 is inconsistent. (3 marks)

R₂ → R₂ − R₁, R₃ → R₃ − 2R₁ on [A|B]: (0, 1, 2 | 1), (0, 1, 2 | 2).

R₃ → R₃ − R₂: (0, 0, 0 | 1).

ρ(A) = 2 but ρ[A|B] = 3, so the system is inconsistent.

Q6. Solve x + y + z = 3, x + 2y + 3z = 4, x + 4y + 9z = 6. (4 marks)

R₂ → R₂ − R₁, R₃ → R₃ − R₁: (0, 1, 2 | 1), (0, 3, 8 | 3). R₃ → R₃ − 3R₂: (0, 0, 2 | 0).

ρ(A) = ρ[A|B] = 3: unique solution.

2z = 0 ⇒ z = 0; y + 2z = 1 ⇒ y = 1; x = 3 − 1 − 0 = 2.

Q7. Show that the only solution of x + 2y + 3z = 0, 3x + 4y + 4z = 0, 7x + 10y + 12z = 0 is the trivial one. (3 marks)

\( |A| = 1(48 - 40) - 2(36 - 28) + 3(30 - 28) = 8 - 16 + 6 = -2 \neq 0 \).

So ρ(A) = 3 = number of unknowns, and only x = y = z = 0 satisfies the system.

Q8. Solve the homogeneous system x + y − 2z = 0, 2x − y + z = 0, 4x + y − 3z = 0. (5 marks)

R₂ → R₂ − 2R₁, R₃ → R₃ − 4R₁: (0, −3, 5), (0, −3, 5). R₃ → R₃ − R₂: (0, 0, 0).

ρ(A) = 2 < 3, so there are infinitely many non-trivial solutions with one parameter.

Let z = 3k: −3y + 15k = 0 ⇒ y = 5k; x = −y + 2z = −5k + 6k = k.

Solution x = k, y = 5k, z = 3k. Check in third equation: 4k + 5k − 9k = 0.

Q9. State the conditions for consistency of AX = B in terms of rank. (3 marks)

AX = B in n unknowns is consistent if and only if ρ(A) = ρ[A|B].

If ρ(A) = ρ[A|B] = n, the solution is unique; if ρ(A) = ρ[A|B] = r < n, there are infinitely many solutions with n − r arbitrary constants; if ρ(A) ≠ ρ[A|B], the system has no solution.

Q10. Prove that \( A = \frac13\begin{bmatrix} -1 & 2 & 2 \\ 2 & -1 & 2 \\ 2 & 2 & -1 \end{bmatrix} \) is orthogonal. (3 marks)

A is symmetric, so \( A^T = A \). Row-by-row products: row1·row1 = 1 + 4 + 4 = 9, row1·row2 = −2 − 2 + 4 = 0, row1·row3 = −2 + 4 − 2 = 0, row2·row2 = 4 + 1 + 4 = 9, row2·row3 = 4 − 2 − 2 = 0, row3·row3 = 4 + 4 + 1 = 9.

So \( AA^T = \frac19\begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} = I \). Hence A is orthogonal.

Exam tips

• Write every elementary transformation (R₂ → R₂ − 2R₁ etc.) beside each step; MU awards step marks for them.
• In PAQ problems keep three matrices side by side (pre-factor, A, post-factor) and update them together.
• For consistency questions always state the ranks of A and [A|B] explicitly before concluding.
• In λ, μ problems give all three cases (no solution, unique, infinite) with clear conditions.
• Verify your final solution by substituting into one original equation.
• For homogeneous square systems, finding |A| first quickly tells you whether non-trivial solutions exist.

FAQs

Is every unitary matrix orthogonal?

No. A real unitary matrix is orthogonal, but a unitary matrix with complex entries is not orthogonal in general. Orthogonal is the real special case of unitary.

Can I use column transformations to find rank?

Yes, for the rank of a matrix both row and column transformations are allowed. But when solving AX = B with the augmented matrix, use row transformations only.

Is the normal form of a matrix unique?

The normal form itself is unique because it depends only on the rank, but the matrices P and Q in PAQ = N are not unique; different sequences of operations give different P and Q.

How are matrices used for KVL and KCL?

Applying Kirchhoff's laws to a network gives linear equations in the unknown currents or voltages; writing them as AX = B lets rank and elimination methods decide solvability and find the solution.

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Applied Mathematics-I) and were prepared with AI help and checked.