Matrices Ex 3.3 Class 12 NCERT: Transpose, Symmetric & Skew-Symmetric Matrices

Key Definitions for Matrices Ex 3.3

Transpose of a Matrix (A' or Aᵀ)

If A = [aᵢⱼ] is an m × n matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A. It is denoted by A' or Aᵀ. So, if A = [aᵢⱼ] of order m × n, then A' = [aⱼᵢ] of order n × m.

Symmetric Matrix

A square matrix A = [aᵢⱼ] is said to be symmetric if its transpose is equal to the matrix itself, i.e., A' = A. This means aᵢⱼ = aⱼᵢ for all possible values of i and j.

Skew-Symmetric Matrix

A square matrix A = [aᵢⱼ] is said to be skew-symmetric if its transpose is equal to its negative, i.e., A' = -A. This means aᵢⱼ = -aⱼᵢ for all i and j. A key property is that all diagonal elements of a skew-symmetric matrix are zero (aᵢᵢ = 0).

Properties of Transpose and Their Applications

Understanding the properties of the transpose is crucial for solving problems efficiently. These rules allow us to manipulate matrix expressions involving transposes without calculating the transpose at every step.

1. (A')' = A: Taking the transpose of a transposed matrix returns the original matrix. This is intuitive; if you interchange rows and columns and then do it again, you get back to where you started.
2. (kA)' = kA': The transpose of a matrix multiplied by a scalar 'k' is the same as the scalar multiplied by the transpose of the matrix. The scalar simply scales every element, an operation that isn't affected by the row-column interchange.
3. (A + B)' = A' + B': The transpose of the sum of two matrices is the sum of their transposes. This is because the (i, j)-th element of (A + B) is aᵢⱼ + bᵢⱼ. The (j, i)-th element of (A+B)' is therefore aᵢⱼ + bᵢⱼ. This is the same as the sum of the (j, i)-th elements of A' and B'.
4. (AB)' = B'A' (Reversal Law): This is the most important and often trickiest property. The transpose of the product of two matrices is the product of their transposes in reverse order. This 'reversal law' is fundamental and appears frequently in proofs and problems.

Theorem: Expressing a Matrix as a Sum of Symmetric and Skew-Symmetric Matrices

1. Step 1: Start with any Square Matrix A — This theorem only applies to square matrices (where the number of rows equals the number of columns).
2. Step 2: Find the Transpose A' — Calculate the transpose of A by interchanging its rows and columns.
3. Step 3: Construct the Symmetric Part (P) — Define a new matrix P = ½(A + A'). You can prove that P is always symmetric because P' = [½(A + A')]' = ½(A' + (A')') = ½(A' + A) = P.
4. Step 4: Construct the Skew-Symmetric Part (Q) — Define another matrix Q = ½(A - A'). You can prove that Q is always skew-symmetric because Q' = [½(A - A')]' = ½(A' - (A')') = ½(A' - A) = -½(A - A') = -Q.
5. Step 5: Verify the Sum — Add P and Q together: P + Q = ½(A + A') + ½(A - A') = ½A + ½A' + ½A - ½A' = A. This shows that any square matrix A can be written as the sum of a symmetric matrix P and a skew-symmetric matrix Q.

Worked Examples for Matrices Ex 3.3

• Example 1: Verify the property (A + B)' = A' + B' Let A = [[-1, 2, 3], [5, 7, 9], [-2, 1, 1]] and B = [[-4, 1, -5], [1, 2, 0], [1, 3, 1]]. Step 1: Calculate A + B. A + B = [[-1+(-4), 2+1, 3+(-5)], [5+1, 7+2, 9+0], [-2+1, 1+3, 1+1]] = [[-5, 3, -2], [6, 9, 9], [-1, 4, 2]]. Step 2: Find the transpose of (A + B). (A + B)' = [[-5, 6, -1], [3, 9, 4], [-2, 9, 2]]. This is our L.H.S. Step 3: Find the transposes A' and B' individually. A' = [[-1, 5, -2], [2, 7, 1], [3, 9, 1]] B' = [[-4, 1, 1], [1, 2, 3], [-5, 0, 1]] Step 4: Calculate A' + B'. A' + B' = [[-1+(-4), 5+1, -2+1], [2+1, 7+2, 1+3], [3+(-5), 9+0, 1+1]] = [[-5, 6, -1], [3, 9, 4], [-2, 9, 2]]. This is our R.H.S. Conclusion: Since L.H.S = R.H.S, the property (A + B)' = A' + B' is verified.
• Example 2: Express the matrix A = [[3, 5], [1, -1]] as the sum of a symmetric and a skew-symmetric matrix. Step 1: Find the transpose of A. A' = [[3, 1], [5, -1]]. Step 2: Calculate the symmetric part, P = ½(A + A'). A + A' = [[3+3, 5+1], [1+5, -1+(-1)]] = [[6, 6], [6, -2]]. P = ½ [[6, 6], [6, -2]] = [[3, 3], [3, -1]]. (Check: P' = [[3, 3], [3, -1]] = P. So, P is symmetric.) Step 3: Calculate the skew-symmetric part, Q = ½(A - A'). A - A' = [[3-3, 5-1], [1-5, -1-(-1)]] = [[0, 4], [-4, 0]]. Q = ½ [[0, 4], [-4, 0]] = [[0, 2], [-2, 0]]. (Check: Q' = [[0, -2], [2, 0]] = -Q. So, Q is skew-symmetric.) Step 4: Express A as P + Q. P + Q = [[3, 3], [3, -1]] + [[0, 2], [-2, 0]] = [[3+0, 3+2], [3-2, -1+0]] = [[3, 5], [1, -1]]. Final Answer: This is equal to the original matrix A. Thus, A = P + Q.

Exam Traps and Important Points

Here are some common mistakes students make in exams:

• Reversal Law for Products: The most frequent error is writing (AB)' = A'B'. Remember the correct rule is (AB)' = B'A'. The order is reversed. Always double-check this when working with matrix products.
• Skew-Symmetric Diagonals: For any skew-symmetric matrix A, A' = -A. This implies that for the diagonal elements, aᵢᵢ = -aᵢᵢ, which means 2aᵢᵢ = 0, so aᵢᵢ = 0. All diagonal elements of a skew-symmetric matrix must be zero. If you calculate the skew-symmetric part Q and its diagonal elements are not all zero, you have made a calculation error.
• Forgetting the ½ Factor: When expressing a matrix as the sum P + Q, many students correctly calculate A + A' and A - A' but forget to multiply by ½ to get P and Q. Remember, P = ½(A + A') and Q = ½(A - A').

Practice Questions with Solutions

• Q: If A = [[cos α, sin α], [-sin α, cos α]], then verify that A'A = I (the identity matrix). A: Step 1: Find the transpose of A, which is A'. A' = [[cos α, -sin α], [sin α, cos α]]. Step 2: Multiply A' with A. A'A = [[cos α, -sin α], [sin α, cos α]] * [[cos α, sin α], [-sin α, cos α]] = [[(cos α)(cos α) + (-sin α)(-sin α), (cos α)(sin α) + (-sin α)(cos α)], [(sin α)(cos α) + (cos α)(-sin α), (sin α)(sin α) + (cos α)(cos α)]] Step 3: Simplify the resulting matrix. A'A = [[cos²α + sin²α, cos α sin α - sin α cos α], [sin α cos α - cos α sin α, sin²α + cos²α]] Using the identity sin²α + cos²α = 1, we get: A'A = [[1, 0], [0, 1]] Final answer: A'A = I. Hence, verified.
• Q: If A = [[1], [-4], [3]] and B = [-1, 2, 1], verify that (AB)' = B'A'. A: Step 1: Calculate the product AB. A is 3x1 and B is 1x3, so AB will be 3x3. AB = [[1], [-4], [3]] [-1, 2, 1] = [[(1)(-1), (1)(2), (1)(1)], [(-4)(-1), (-4)(2), (-4)(1)], [(3)(-1), (3)(2), (3)(1)]] = [[-1, 2, 1], [4, -8, -4], [-3, 6, 3]]. Step 2: Find the transpose of AB, i.e., (AB)'. (AB)' = [[-1, 4, -3], [2, -8, 6], [1, -4, 3]]. This is the L.H.S. Step 3: Find the transposes B' and A'. B' = [[-1], [2], [1]] (3x1 matrix) A' = [1, -4, 3] (1x3 matrix) Step 4: Calculate the product B'A'. B'A' = [[-1], [2], [1]] [1, -4, 3] = [[(-1)(1), (-1)(-4), (-1)(3)], [(2)(1), (2)(-4), (2)(3)], [(1)(1), (1)(-4), (1)(3)]] = [[-1, 4, -3], [2, -8, 6], [1, -4, 3]]. This is the R.H.S. Final answer: Since L.H.S. = R.H.S., the reversal law (AB)' = B'A' is verified.
• Q: Find the values of x, y, and z if the matrix A = [[0, 2y, z], [x, y, -z], [x, -y, z]] satisfies the equation A'A = I. A: Step 1: Write down A' and the equation A'A = I. A' = [[0, x, x], [2y, y, -y], [z, -z, z]] A'A = [[0, x, x], [2y, y, -y], [z, -z, z]] * [[0, 2y, z], [x, y, -z], [x, -y, z]] = [[1, 0, 0], [0, 1, 0], [0, 0, 1]] Step 2: Perform the matrix multiplication A'A. Top-left element: (0)(0) + (x)(x) + (x)(x) = 2x² Middle element: (2y)(2y) + (y)(y) + (-y)(-y) = 4y² + y² + y² = 6y² Bottom-right element: (z)(z) + (-z)(-z) + (z)(z) = z² + z² + z² = 3z² Step 3: Equate the diagonal elements of A'A to the diagonal elements of I. 2x² = 1 => x² = 1/2 => x = ±1/√2 6y² = 1 => y² = 1/6 => y = ±1/√6 3z² = 1 => z² = 1/3 => z = ±1/√3 Final answer: x = ±1/√2, y = ±1/√6, z = ±1/√3.
• Q: Express the matrix A = [[6, -2, 2], [-2, 3, -1], [2, -1, 3]] as the sum of a symmetric and a skew-symmetric matrix. A: Step 1: Find the transpose A'. A' = [[6, -2, 2], [-2, 3, -1], [2, -1, 3]]. Notice that A = A', so the matrix A is already symmetric. Step 2: Calculate the symmetric part, P = ½(A + A'). Since A = A', A + A' = A + A = 2A. P = ½(2A) = A = [[6, -2, 2], [-2, 3, -1], [2, -1, 3]]. Step 3: Calculate the skew-symmetric part, Q = ½(A - A'). Since A = A', A - A' = A - A = 0 (the zero matrix). Q = ½(0) = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]. The zero matrix is both symmetric and skew-symmetric. Step 4: Express A as P + Q. A = P + Q = [[6, -2, 2], [-2, 3, -1], [2, -1, 3]] + [[0, 0, 0], [0, 0, 0], [0, 0, 0]] Final answer: The symmetric part is the matrix A itself, and the skew-symmetric part is the zero matrix.

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