How to find Transpose of a matrix

How to find Transpose of a matrix , Symmetric & Skew symmetric matrices | Class 12 Matrices Tutorial

Mathematics: Transpose, Symmetric & Skew-Symmetric Matrices

📖 Mathematics Study Notes

Class XII | Chapter: Matrices
✨ Topic: Transpose, Symmetric & Skew-Symmetric Matrices ✨
🔁 Concept 1: Transpose of a Matrix

The transpose of a matrix flips it over its diagonal, switching row and column indices.

📌 Definition

If \(A = [a_{ij}]\) is \(m \times n\), then the transpose \(A^T\) (or \(A’\)) is \(n \times m\) with \((A^T)_{ij} = a_{ji}\).

\(A_{m \times n}\)
\([a_{ij}]\)
⟷
\(A^T_{n \times m}\)
\([a_{ji}]\)
📐 Properties of Transpose
  • \((A^T)^T = A\)
  • \((kA)^T = kA^T\)
  • \((A + B)^T = A^T + B^T\)
  • \((AB)^T = B^T A^T\) (Reversal Law)
✅ Solved Examples (Transpose)
Example 1:

\(A = \begin{bmatrix} 3 & 5 \\ \sqrt{3} & 1 \\ 0 & -1 \end{bmatrix}\), find \(A^T\) and verify \((A^T)^T = A\).

Solution: \(A^T = \begin{bmatrix} 3 & \sqrt{3} & 0 \\ 5 & 1 & -1 \end{bmatrix}\); \((A^T)^T = A\).
Example 2: (Reversal Law)

\(A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}\), \(B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}\). Show \((AB)^T = B^T A^T\).

Solution: \(AB = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}\), \((AB)^T = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}\) = \(B^T A^T\).
Example 3:

\(A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}\) satisfies \(A^T A = I\). Find \(x, y, z\).

Solution: \(A^T A = \operatorname{diag}(2x^2, 6y^2, 3z^2) = I\) ⇒ \(x = \pm \frac{1}{\sqrt{2}}, y = \pm \frac{1}{\sqrt{6}}, z = \pm \frac{1}{\sqrt{3}}\).
Example 4:

\(A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}\), find \(\alpha\) such that \(A + A^T = I\).

Solution: \(A + A^T = \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix} = I\) ⇒ \(\cos\alpha = \frac12\) ⇒ \(\alpha = \frac{\pi}{3}\).
Example 5:

Prove \(\operatorname{Tr}(A) = \operatorname{Tr}(A^T)\).

Solution: Diagonal entries unchanged, so sum of diagonal entries equal.
🪞 Concept 2: Symmetric and Skew-Symmetric Matrices

Square matrices classified by relation between \(A\) and \(A^T\).

🔹 Symmetric Matrix

\(A^T = A\) ⇒ \(a_{ij} = a_{ji}\) for all \(i,j\).

⚡ Skew-Symmetric Matrix

\(A^T = -A\) ⇒ \(a_{ij} = -a_{ji}\). In particular, diagonal entries \(a_{ii} = 0\).

📌 Important Theorems:
• For any square matrix \(A\): \(A + A^T\) is symmetric, \(A – A^T\) is skew-symmetric.
• Every square matrix can be uniquely expressed as sum of symmetric and skew-symmetric parts: \[ A = \underbrace{\frac{1}{2}(A + A^T)}_{\text{symmetric}} + \underbrace{\frac{1}{2}(A – A^T)}_{\text{skew-symmetric}} \]
📘 Solved Examples
Example 1:

Show \(A = \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix}\) is skew-symmetric.

Solution: \(A^T = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix} = -A\).
Example 2:

If \(A, B\) symmetric, prove \(AB – BA\) is skew-symmetric.

Solution: \((AB-BA)^T = (AB)^T – (BA)^T = B^T A^T – A^T B^T = BA – AB = -(AB-BA)\).
Example 3:

Express \(B = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}\) as symmetric + skew-symmetric.

Solution: Symmetric \(P = \frac12(B+B^T)\), Skew \(Q = \frac12(B-B^T)\). Verify \(P+Q = B\).
Example 4:

Prove \(AA^T\) and \(A^T A\) are always symmetric.

Solution: \((AA^T)^T = (A^T)^T A^T = AA^T\), similarly for \(A^T A\).
🏋️ Self Exercise (HOTS)
  1. If \(A^2 = A\), show \((I+A)^3 – 7A = I\).
  2. Prove that every square matrix can be uniquely expressed as sum of symmetric and skew-symmetric matrix.
  3. If \(A = \begin{bmatrix} 0 & x \\ y & 0 \end{bmatrix}\) and \(A^2 = I\), find relation between \(x\) and \(y\).
  4. If \(A\) and \(B\) are symmetric, when is \(AB\) symmetric? (Condition: \(AB = BA\))
  5. If \(A\) is skew-symmetric of odd order, prove \(\det(A)=0\).
  6. Show that if \(A\) is skew-symmetric, then \(A^2\) is symmetric.
  7. If \(AB=BA\), prove by induction \((AB)^n = A^n B^n\).
  8. Find matrix \(X\) such that \(\begin{bmatrix} 2 & -1 \\ 1 & 2 \end{bmatrix}X = \begin{bmatrix} -1 & -8 \\ 1 & -6 \end{bmatrix}\).
  9. If \(A = [a_{ij}]\) is skew-symmetric of order 3, evaluate \(a_{11}+a_{22}+a_{33}+a_{12}+a_{21}\).
  10. If \(A\) is symmetric, prove \(B^T A B\) is symmetric for compatible \(B\).
🔥 Advanced 3×3 / HOTS Challenges
  1. Let \(A = [a_{ij}]\) be \(3\times3\) skew-symmetric with \(a_{12}=2, a_{13}=5, a_{23}=-3\). Write full matrix \(A\) and compute \(S = \sum a_{ii} + (a_{21}\cdot a_{12})\).
  2. If \(A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix}\) satisfies \(AA^T = 9I\), find \(a\) and \(b\).
  3. Let \(A = P+Q\) with \(P\) symmetric, \(Q\) skew-symmetric, \(A = \begin{bmatrix} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -1 & 0 \end{bmatrix}\). Find element \((P)_{23}\) and \((Q)_{12}\).
  4. Given \(A^T = 2A + I\) for \(3\times3\) matrix \(A\), determine \(A\). (Hint: take transpose again)
  5. If \(A\) is \(3\times3\) skew-symmetric with \(a_{12}=1, a_{13}=2, a_{23}=3\) and sum of squares of non-zero elements =30, find \(|A|\) and \(|A+I|\).
💡 Hint for Q14: From \(A^T = 2A + I\) take transpose → \(A = 2A^T + I\). Solve both equations to get \(A = -I\).
📌 Advanced Applications (Board Level)
🎯 Example: Skew-symmetric matrix elements

If \(A\) is \(3\times3\) skew-symmetric and \(a_{12}=2, a_{13}=5, a_{23}=-3\), then \(A = \begin{bmatrix} 0 & 2 & 5 \\ -2 & 0 & -3 \\ -5 & 3 & 0 \end{bmatrix}\). Compute \(S = \sum a_{ii} + a_{21}\cdot a_{12}\).

Solution: Diagonal sum = 0, \(a_{21} = -2, a_{12}=2\) ⇒ product = \(-4\). Hence \(S = -4\).
🎯 Example: Orthogonality condition \(AA^T = 9I\)

Given \(A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix}\) and \(AA^T = 9I\). Solve for \(a,b\).

Solution: \(AA^T = \begin{bmatrix} 1+4+4 & 2+2-4 & a+4+2b \\ \dots \end{bmatrix}\) matching \(9I\) yields equations: \(a+2b=0\) from (1,3) and from (3,1) same, and norm row3: \(a^2+4+b^2=9\). Solving gives \(a=2, b=-1\) or \(a=-2, b=1\).
🎯 Symmetric + Skew-symmetric decomposition

\(A = \begin{bmatrix} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -1 & 0 \end{bmatrix}\). Find \(P_{23}\) (symmetric part) and \(Q_{12}\) (skew part).

Solution: \(P = \frac12(A+A^T)\), \(Q = \frac12(A-A^T)\). Then \(P_{23} = \frac12(5 + (-1)) = 2\). \(Q_{12} = \frac12(4 – 7) = -\frac{3}{2}\).
🎯 Matrix equation \(A^T = 2A + I\)

Find \(3\times3\) matrix \(A\) satisfying \(A^T = 2A + I\).

Solution: Transpose both sides: \((A^T)^T = 2A^T + I \) ⇒ \(A = 2A^T + I\). Substitute \(A^T\) from original: \(A = 2(2A+I) + I = 4A + 3I\) ⇒ \(-3A = 3I\) ⇒ \(A = -I\). Verify: \((-I)^T = -I = 2(-I)+I\) holds.
🎯 Determinant of skew-symmetric matrix (odd order)

If \(A\) is \(3\times3\) skew-symmetric with \(a_{12}=1, a_{13}=2, a_{23}=3\) and sum of squares of nonzero elements = \(1^2+2^2+3^2+(-1)^2+(-2)^2+(-3)^2 = 30\). Find \(|A|\) and \(|A+I|\).

Solution: \(A = \begin{bmatrix} 0 & 1 & 2 \\ -1 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix}\). \(|A| = 0\) (odd order skew-symmetric). \(A+I = \begin{bmatrix} 1 & 1 & 2 \\ -1 & 1 & 3 \\ -2 & -3 & 1 \end{bmatrix}\), \(\det(A+I) = 1(1\cdot1 – 3\cdot(-3)) -1((-1)\cdot1 – 3\cdot(-2)) + 2((-1)(-3)-1\cdot(-2)) = 1(1+9) -1(-1+6) + 2(3+2) = 10 -5 +10 = 15\).
🧠 Quick Visual Memory Aid:
Symmetric: \(A^T = A\)
\(a_{ij} = a_{ji}\)
🟰
Skew-Symmetric: \(A^T = -A\)
\(a_{ij} = -a_{ji}\), diagonal = 0
✨ Any square matrix = Symmetric Part + Skew-Symmetric Part ✨
📐 Master Transpose, Symmetric & Skew-Symmetric | Class XII Mathematics • Matrix Algebra

Matrices Class 12: Transpose, Symmetric & Skew-Symmetric Matrices FAQ | MathStudy.in

Frequently Asked Questions (FAQs) on Matrices: Transpose, Symmetric & Skew-Symmetric Matrices

1. What is the transpose of a matrix and how is it denoted?

The transpose of a matrix is obtained by interchanging its rows and columns. If \(A = [a_{ij}]\) is an \(m \times n\) matrix, then its transpose \(A^T\) (or \(A’\)) is an \(n \times m\) matrix with \((A^T)_{ij} = a_{ji}\). This fundamental operation is the foundation for understanding symmetric and skew-symmetric matrices in Class 12.

2. How does the reversal law \((AB)^T = B^T A^T\) help in matrix algebra?

The reversal law states that the transpose of a product equals the product of transposes in reverse order. For example, if \(A\) is a column matrix and \(B\) is a row matrix, \((AB)^T = B^T A^T\). This property is crucial for proving identities involving symmetric and skew-symmetric matrices, and it appears frequently in board exam problems.

3. What are the key properties of the transpose operation I should memorize?

The essential properties include: \((A^T)^T = A\), \((kA)^T = kA^T\), \((A + B)^T = A^T + B^T\), and the reversal law \((AB)^T = B^T A^T\). These properties are repeatedly used to simplify expressions and prove whether a given matrix is symmetric or skew-symmetric. Mastering them ensures accuracy in higher-order problems.

4. How do you define a symmetric matrix and what are its characteristics?

A square matrix \(A\) is symmetric if \(A^T = A\), meaning \(a_{ij} = a_{ji}\) for all \(i, j\). The diagonal entries can be any real number. Symmetric matrices appear extensively in physics and engineering. In our tutorials, we show that for any square matrix, \(A + A^T\) is always symmetric.

5. What defines a skew-symmetric matrix and what is special about its diagonal?

A square matrix \(A\) is skew-symmetric if \(A^T = -A\), which implies \(a_{ij} = -a_{ji}\) for all \(i, j\). A crucial property is that all diagonal entries must be zero because \(a_{ii} = -a_{ii}\) forces \(a_{ii}=0\). Skew-symmetric matrices of odd order always have zero determinant, a key result often tested in exams.

6. How can any square matrix be expressed as a sum of symmetric and skew-symmetric matrices?

Every square matrix \(A\) can be uniquely decomposed as \(A = P + Q\), where \(P = \frac{1}{2}(A + A^T)\) is symmetric and \(Q = \frac{1}{2}(A – A^T)\) is skew-symmetric. This theorem is fundamental in matrix algebra. For instance, we can decompose a 3×3 matrix into its symmetric and skew-symmetric parts to simplify further calculations.

7. How do you prove that \(AA^T\) and \(A^T A\) are always symmetric matrices?

To prove \(AA^T\) is symmetric, we compute its transpose: \((AA^T)^T = (A^T)^T A^T = A A^T\). Since the transpose equals the original matrix, \(AA^T\) is symmetric. The same proof works for \(A^T A\). This property is frequently used to construct symmetric matrices from any given matrix \(A\).

8. What is the condition for the product of two symmetric matrices to be symmetric?

If \(A\) and \(B\) are symmetric matrices, then \(AB\) is symmetric if and only if \(AB = BA\), i.e., the matrices commute. Because \((AB)^T = B^T A^T = BA\). Therefore, symmetry of the product requires commutativity. This is a typical higher-order thinking skill (HOTS) question in Class 12 board exams.

9. How do you find the symmetric and skew-symmetric parts of a given matrix?

To find the symmetric part \(P\), compute \(\frac{1}{2}(A + A^T)\). For the skew-symmetric part \(Q\), compute \(\frac{1}{2}(A – A^T)\). For example, given \(A = \begin{bmatrix} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -1 & 0 \end{bmatrix}\), the symmetric part gives \(P_{23}=2\) and the skew part gives \(Q_{12}=-\frac{3}{2}\). This decomposition is a standard exam question.

10. Why is the determinant of a skew-symmetric matrix of odd order always zero?

For a skew-symmetric matrix \(A\) of odd order, we have \(A^T = -A\). Taking determinants: \(\det(A^T) = \det(-A)\). Since \(\det(A^T) = \det(A)\) and \(\det(-A) = (-1)^n \det(A)\), we get \(\det(A) = (-1)^n \det(A)\). For odd \(n\), \((-1)^n = -1\), so \(\det(A) = -\det(A)\), implying \(2\det(A)=0\) and thus \(\det(A)=0\). This is an important theoretical result.

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