Maharashtra Board Class 12 Maths Part 1 Chapter 2 Matrices 2.4 Solutions

Question 1. Find A^T, if

(i) A = \[ \begin{bmatrix} 1 & 3 \\ -4 & 5 \end{bmatrix} \]
(ii) A = \[ \begin{bmatrix} 2 & -6 & 1 \\ -4 & 0 & 5 \end{bmatrix} \]
Answer: Solution:
(i) A = \[ \begin{bmatrix} 1 & 3 \\ -4 & 5 \end{bmatrix} \]
\( \implies \) A^T = \[ \begin{bmatrix} 1 & -4 \\ 3 & 5 \end{bmatrix} \]
(ii) A = \[ \begin{bmatrix} 2 & -6 & 1 \\ -4 & 0 & 5 \end{bmatrix} \]
\( \implies \) A^T = \[ \begin{bmatrix} 2 & -4 \\ -6 & 0 \\ 1 & 5 \end{bmatrix} \]
In simple words: Transposing a matrix involves swapping its rows and columns. This operation changes the dimensions of the matrix unless it's a square matrix.

🎯 Exam Tip: Pay close attention to the order of elements when converting rows to columns (or vice-versa) to avoid errors. This is a fundamental operation in matrix algebra.

Question 2. If A = [a_ij]_3×3 where a_ij = 2(i - j). Find A and A^T. State whether A and A^T both are symmetric or skew-symmetric matrices.
Answer: Solution:
A=[a_ij]_3×3 = \[ \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \]
Given: \( a_{ij} = 2(i-j) \)
\( \implies a_{11} = 2(1-1) = 0, a_{12} = 2(1-2) = -2, \)
\( a_{13} = 2(1-3) = -4, a_{21} = 2(2-1) = 2, \)
\( a_{22} = 2(2-2) = 0, a_{23} = 2(2-3) = -2, \)
\( a_{31} = 2(3-1) = 4, a_{32} = 2(3-2) = 2, \)
\( a_{33} = 2(3-3) = 0 \)
\( \implies \) A = \[ \begin{bmatrix} 0 & -2 & -4 \\ 2 & 0 & -2 \\ 4 & 2 & 0 \end{bmatrix} \]
\( \implies \) A^T = \[ \begin{bmatrix} 0 & 2 & 4 \\ -2 & 0 & 2 \\ -4 & -2 & 0 \end{bmatrix} \]
\( \implies -A = \[ \begin{bmatrix} 0 & 2 & 4 \\ -2 & 0 & 2 \\ -4 & -2 & 0 \end{bmatrix} \] \)
\( \implies \) A = -A^T and A^T = -A
Hence, A and A^T are both skew-symmetric matrices.
In simple words: A matrix is skew-symmetric if its transpose is equal to its negative. This means all diagonal elements must be zero, and off-diagonal elements \(a_{ij}\) and \(a_{ji}\) must be opposites.

🎯 Exam Tip: When checking for skew-symmetry, ensure all diagonal elements are zero and corresponding off-diagonal elements have opposite signs. This is a key characteristic to quickly identify such matrices.

Question 3. If A = \[ \begin{bmatrix} 5 & -3 \\ 4 & -3 \\ -2 & 1 \end{bmatrix} \] , prove that \( (A^T)^T = A \).
Answer: Solution:
A = \[ \begin{bmatrix} 5 & -3 \\ 4 & -3 \\ -2 & 1 \end{bmatrix} \]
\( \implies A^T = \[ \begin{bmatrix} 5 & 4 & -2 \\ -3 & -3 & 1 \end{bmatrix} \] \)
\( \implies (A^T)^T = \[ \begin{bmatrix} 5 & -3 \\ 4 & -3 \\ -2 & 1 \end{bmatrix} \] = A \)
In simple words: The operation of transposing a matrix twice returns the original matrix, effectively undoing the first transpose.

🎯 Exam Tip: This property is fundamental and often used to simplify expressions or proofs involving multiple transpositions. It's a quick check for matrix identity.

Question 4. If A = \[ \begin{bmatrix} 1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9 \end{bmatrix} \] , prove that \( A^T = A \).
Answer: Solution:
A = \[ \begin{bmatrix} 1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9 \end{bmatrix} \] ... (1)
\( \implies A^T = \[ \begin{bmatrix} 1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9 \end{bmatrix} \] ... (2) \)
From (1) and (2), \( A^T = A \).
In simple words: A matrix is symmetric if it remains unchanged after being transposed, meaning its elements are mirrored across the main diagonal.

🎯 Exam Tip: For a symmetric matrix, elements a_ij must be equal to a_ji for all i and j. Check corresponding off-diagonal elements carefully to verify symmetry.

Question 5. If A = \[ \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} \] , B = \[ \begin{bmatrix} 2 & 1 \\ 4 & -1 \\ -3 & 3 \end{bmatrix} \] , C = \[ \begin{bmatrix} 1 & 2 \\ -1 & 4 \\ -2 & 3 \end{bmatrix} \] , then show that
(i) \( (A + B)^T = A^T + B^T \)
(ii) \( (A - C)^T = A^T - C^T \)
Answer: Solution:
(i) \( A+B = \[ \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} + \begin{bmatrix} 2 & 1 \\ 4 & -1 \\ -3 & 3 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 2+2 & -3+1 \\ 5+4 & -4-1 \\ -6-3 & 1+3 \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ 9 & -5 \\ -9 & 4 \end{bmatrix} \] \)
\( \implies (A + B)^T = \[ \begin{bmatrix} 4 & 9 & -9 \\ -2 & -5 & 4 \end{bmatrix} \] ... (1) \)
\( A^T = \[ \begin{bmatrix} 2 & 5 & -6 \\ -3 & -4 & 1 \end{bmatrix} \] \)
\( B^T = \[ \begin{bmatrix} 2 & 4 & -3 \\ 1 & -1 & 3 \end{bmatrix} \] \)
\( \implies A^T + B^T = \[ \begin{bmatrix} 2 & 5 & -6 \\ -3 & -4 & 1 \end{bmatrix} + \begin{bmatrix} 2 & 4 & -3 \\ 1 & -1 & 3 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 2+2 & 5+4 & -6-3 \\ -3+1 & -4-1 & 1+3 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 4 & 9 & -9 \\ -2 & -5 & 4 \end{bmatrix} \] ... (2) \)
From (1) and (2),
\( (A + B)^T = A^T + B^T \).
(ii) \( A-C = \[ \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ -1 & 4 \\ -2 & 3 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 2-1 & -3-2 \\ 5-(-1) & -4-4 \\ -6-(-2) & 1-3 \end{bmatrix} = \begin{bmatrix} 1 & -5 \\ 6 & -8 \\ -4 & -2 \end{bmatrix} \] \)
\( \implies (A-C)^T = \[ \begin{bmatrix} 1 & 6 & -4 \\ -5 & -8 & -2 \end{bmatrix} \] ... (1) \)
\( A^T = \[ \begin{bmatrix} 2 & 5 & -6 \\ -3 & -4 & 1 \end{bmatrix} \] \)
\( C^T = \[ \begin{bmatrix} 1 & -1 & -2 \\ 2 & 4 & 3 \end{bmatrix} \] \)
\( \implies A^T - C^T = \[ \begin{bmatrix} 2 & 5 & -6 \\ -3 & -4 & 1 \end{bmatrix} - \begin{bmatrix} 1 & -1 & -2 \\ 2 & 4 & 3 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 2-1 & 5-(-1) & -6-(-2) \\ -3-2 & -4-4 & 1-3 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 1 & 6 & -4 \\ -5 & -8 & -2 \end{bmatrix} \] ... (2) \)
From (1) and (2), \( (A-C)^T = A^T - C^T \).
In simple words: The transpose operation distributes over matrix addition and subtraction, meaning you can take the transpose of each matrix first and then add or subtract them, or vice-versa.

🎯 Exam Tip: This property is fundamental for simplifying complex matrix expressions. Remember that matrix addition/subtraction and transposition operations can be interchanged without affecting the result.

Question 6. If A = \[ \begin{bmatrix} 5 & 4 \\ -2 & 3 \end{bmatrix} \] and B = \[ \begin{bmatrix} -1 & 3 \\ 4 & -1 \end{bmatrix} \] , then find C^T, such that \( 3A - 2B + C = I \), where I is the unit matrix of order 2.
Answer: Solution:
\( 3A-2B+C=I \)
\( \implies C = I - 3A + 2B \)
\( = \[ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] - 3 \[ \begin{bmatrix} 5 & 4 \\ -2 & 3 \end{bmatrix} \] + 2 \[ \begin{bmatrix} -1 & 3 \\ 4 & -1 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] - \[ \begin{bmatrix} 15 & 12 \\ -6 & 9 \end{bmatrix} \] + \[ \begin{bmatrix} -2 & 6 \\ 8 & -2 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 1-15+(-2) & 0-12+6 \\ 0-(-6)+8 & 1-9-2 \end{bmatrix} \] \)
\( \implies C = \[ \begin{bmatrix} -16 & -6 \\ 14 & -10 \end{bmatrix} \] \)
\( \implies C^T = \[ \begin{bmatrix} -16 & 14 \\ -6 & -10 \end{bmatrix} \] \)
In simple words: To solve for an unknown matrix in an equation, treat matrices like algebraic variables but follow matrix operation rules. Isolate the unknown matrix and then perform the required operations.

🎯 Exam Tip: Be careful with scalar multiplication and matrix addition/subtraction. Ensure the correct matrix operations are performed in order and keep track of signs and element positions.

Question 7. If A = \[ \begin{bmatrix} 7 & 3 & 0 \\ 0 & 4 & -2 \end{bmatrix} \] , B = \[ \begin{bmatrix} 0 & -2 & 3 \\ 2 & 1 & -4 \end{bmatrix} \] , then find
(i) \( A^T + 4B^T \)
(ii) \( 5A^T - 5B^T \)
Answer: Solution:
A = \[ \begin{bmatrix} 7 & 3 & 0 \\ 0 & 4 & -2 \end{bmatrix} \]
\( \implies A^T = \[ \begin{bmatrix} 7 & 0 \\ 3 & 4 \\ 0 & -2 \end{bmatrix} \] \)
B = \[ \begin{bmatrix} 0 & -2 & 3 \\ 2 & 1 & -4 \end{bmatrix} \]
\( \implies B^T = \[ \begin{bmatrix} 0 & 2 \\ -2 & 1 \\ 3 & -4 \end{bmatrix} \] \)
(i) \( A^T + 4B^T = \[ \begin{bmatrix} 7 & 0 \\ 3 & 4 \\ 0 & -2 \end{bmatrix} \] + 4 \[ \begin{bmatrix} 0 & 2 \\ -2 & 1 \\ 3 & -4 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 7 & 0 \\ 3 & 4 \\ 0 & -2 \end{bmatrix} \] + \[ \begin{bmatrix} 0 & 8 \\ -8 & 4 \\ 12 & -16 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 7+0 & 0+8 \\ 3-8 & 4+4 \\ 0+12 & -2-16 \end{bmatrix} = \begin{bmatrix} 7 & 8 \\ -5 & 8 \\ 12 & -18 \end{bmatrix} \] \)
(ii) \( 5A^T - 5B^T = 5 \[ \begin{bmatrix} 7 & 0 \\ 3 & 4 \\ 0 & -2 \end{bmatrix} \] - 5 \[ \begin{bmatrix} 0 & 2 \\ -2 & 1 \\ 3 & -4 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 35 & 0 \\ 15 & 20 \\ 0 & -10 \end{bmatrix} \] - \[ \begin{bmatrix} 0 & 10 \\ -10 & 5 \\ 15 & -20 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 35-0 & 0-10 \\ 15-(-10) & 20-5 \\ 0-15 & -10-(-20) \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 35 & -10 \\ 25 & 15 \\ -15 & 10 \end{bmatrix} \] \)
In simple words: This problem demonstrates scalar multiplication applied to transposed matrices, followed by matrix addition or subtraction. Each element of a matrix is multiplied by the scalar before other operations.

🎯 Exam Tip: Multiply each element of the transposed matrix by the scalar before performing the final addition or subtraction. Ensure correct sign handling, especially with negative numbers.

Question 8. If A = \[ \begin{bmatrix} 1 & 0 & 1 \\ 3 & 1 & 2 \end{bmatrix} \] , B = \[ \begin{bmatrix} 2 & 1 & -4 \\ 3 & 5 & -2 \end{bmatrix} \] and C = \[ \begin{bmatrix} 0 & 2 & 3 \\ -1 & -1 & 0 \end{bmatrix} \] , verify that \( (A + 2B + 3C)^T = A^T + 2B^T + 3C^T \)
Answer: Solution:
\( A+2B+3C \)
\( = \[ \begin{bmatrix} 1 & 0 & 1 \\ 3 & 1 & 2 \end{bmatrix} \] + 2 \[ \begin{bmatrix} 2 & 1 & -4 \\ 3 & 5 & -2 \end{bmatrix} \] + 3 \[ \begin{bmatrix} 0 & 2 & 3 \\ -1 & -1 & 0 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 1 & 0 & 1 \\ 3 & 1 & 2 \end{bmatrix} \] + \[ \begin{bmatrix} 4 & 2 & -8 \\ 6 & 10 & -4 \end{bmatrix} \] + \[ \begin{bmatrix} 0 & 6 & 9 \\ -3 & -3 & 0 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 1+4+0 & 0+2+6 & 1-8+9 \\ 3+6-3 & 1+10-3 & 2-4+0 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 5 & 8 & 2 \\ 6 & 8 & -2 \end{bmatrix} \] \)
\( \implies (A + 2B + 3C)^T = \[ \begin{bmatrix} 5 & 6 \\ 8 & 8 \\ 2 & -2 \end{bmatrix} \] ... (1) \)
\( A^T = \[ \begin{bmatrix} 1 & 3 \\ 0 & 1 \\ 1 & 2 \end{bmatrix} \] \)
\( B^T = \[ \begin{bmatrix} 2 & 3 \\ 1 & 5 \\ -4 & -2 \end{bmatrix} \] \)
\( C^T = \[ \begin{bmatrix} 0 & -1 \\ 2 & -1 \\ 3 & 0 \end{bmatrix} \] \)
\( \implies A^T + 2B^T + 3C^T \)
\( = \[ \begin{bmatrix} 1 & 3 \\ 0 & 1 \\ 1 & 2 \end{bmatrix} \] + 2 \[ \begin{bmatrix} 2 & 3 \\ 1 & 5 \\ -4 & -2 \end{bmatrix} \] + 3 \[ \begin{bmatrix} 0 & -1 \\ 2 & -1 \\ 3 & 0 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 1 & 3 \\ 0 & 1 \\ 1 & 2 \end{bmatrix} \] + \[ \begin{bmatrix} 4 & 6 \\ 2 & 10 \\ -8 & -4 \end{bmatrix} \] + \[ \begin{bmatrix} 0 & -3 \\ 6 & -3 \\ 9 & 0 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 1+4+0 & 3+6-3 \\ 0+2+6 & 1+10-3 \\ 1-8+9 & 2-4+0 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 5 & 6 \\ 8 & 8 \\ 2 & -2 \end{bmatrix} \] ... (2) \)
From (1) and (2),
\( (A + 2B + 3C)^T = A^T + 2B^T + 3C^T \).
In simple words: This property shows that the transpose of a linear combination of matrices is the same as the linear combination of their transposes.

🎯 Exam Tip: To verify this property, calculate both sides of the equation independently, ensuring all matrix additions, scalar multiplications, and transpositions are performed accurately, then compare the final matrices.

Question 9. If A = \[ \begin{bmatrix} -1 & 2 & 1 \\ -3 & 2 & -3 \end{bmatrix} \] and B = \[ \begin{bmatrix} 2 & 1 \\ -3 & 2 \\ -1 & 3 \end{bmatrix} \] , prove that \( (A + B^T)^T = A^T + B \).
Answer: Solution:
A = \[ \begin{bmatrix} -1 & 2 & 1 \\ -3 & 2 & -3 \end{bmatrix} \]
B = \[ \begin{bmatrix} 2 & 1 \\ -3 & 2 \\ -1 & 3 \end{bmatrix} \]
\( \implies A^T = \[ \begin{bmatrix} -1 & -3 \\ 2 & 2 \\ 1 & -3 \end{bmatrix} \] \)
\( B^T = \[ \begin{bmatrix} 2 & -3 & -1 \\ 1 & 2 & 3 \end{bmatrix} \] \)
\( \implies A + B^T = \[ \begin{bmatrix} -1 & 2 & 1 \\ -3 & 2 & -3 \end{bmatrix} + \begin{bmatrix} 2 & -3 & -1 \\ 1 & 2 & 3 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} -1+2 & 2-3 & 1-1 \\ -3+1 & 2+2 & -3+3 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 1 & -1 & 0 \\ -2 & 4 & 0 \end{bmatrix} \] \)
\( \implies (A+B^T)^T = \[ \begin{bmatrix} 1 & -2 \\ -1 & 4 \\ 0 & 0 \end{bmatrix} \] ... (1) \)
\( A^T + B = \[ \begin{bmatrix} -1 & -3 \\ 2 & 2 \\ 1 & -3 \end{bmatrix} + \begin{bmatrix} 2 & 1 \\ -3 & 2 \\ -1 & 3 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} -1+2 & -3+1 \\ 2-3 & 2+2 \\ 1-1 & -3+3 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 1 & -2 \\ -1 & 4 \\ 0 & 0 \end{bmatrix} \] ... (2) \)
From (1) and (2),
\( (A + B^T)^T = A^T + B \).
In simple words: This problem verifies a matrix property involving the transpose of a sum, where one of the matrices in the sum is already transposed. It shows how transposition applies to combined matrix operations.

🎯 Exam Tip: Correctly identify and compute the transpose of each matrix component before performing addition and the final transpose. Pay close attention to matrix dimensions for valid operations.

Question 10. Prove that \( A + A^T \) is symmetric and \( A - A^T \) is a skew-symmetric matrix, where
(i) A = \[ \begin{bmatrix} 1 & 2 & 4 \\ 3 & 2 & 1 \\ -2 & -3 & 2 \end{bmatrix} \]
(ii) A = \[ \begin{bmatrix} 5 & 2 & -4 \\ 3 & -7 & 2 \\ 4 & -5 & -3 \end{bmatrix} \]
Answer: Solution:
(i) \( A = \[ \begin{bmatrix} 1 & 2 & 4 \\ 3 & 2 & 1 \\ -2 & -3 & 2 \end{bmatrix} \] \)
\( \implies A^T = \[ \begin{bmatrix} 1 & 3 & -2 \\ 2 & 2 & -3 \\ 4 & 1 & 2 \end{bmatrix} \] \)
\( \implies A + A^T = \[ \begin{bmatrix} 1 & 2 & 4 \\ 3 & 2 & 1 \\ -2 & -3 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 3 & -2 \\ 2 & 2 & -3 \\ 4 & 1 & 2 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 1+1 & 2+3 & 4-2 \\ 3+2 & 2+2 & 1-3 \\ -2+4 & -3+1 & 2+2 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 2 & 5 & 2 \\ 5 & 4 & -2 \\ 2 & -2 & 4 \end{bmatrix} \] \)
This is a symmetric matrix (by definition).
Also, \( A - A^T = \[ \begin{bmatrix} 1 & 2 & 4 \\ 3 & 2 & 1 \\ -2 & -3 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 3 & -2 \\ 2 & 2 & -3 \\ 4 & 1 & 2 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 1-1 & 2-3 & 4-(-2) \\ 3-2 & 2-2 & 1-(-3) \\ -2-4 & -3-1 & 2-2 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 0 & -1 & 6 \\ 1 & 0 & 4 \\ -6 & -4 & 0 \end{bmatrix} \] \)
This is a skew-symmetric matrix (by definition).
(ii) \( A = \[ \begin{bmatrix} 5 & 2 & -4 \\ 3 & -7 & 2 \\ 4 & -5 & -3 \end{bmatrix} \] \)
\( \implies A^T = \[ \begin{bmatrix} 5 & 3 & 4 \\ 2 & -7 & -5 \\ -4 & 2 & -3 \end{bmatrix} \] \)
\( \implies A + A^T = \[ \begin{bmatrix} 5 & 2 & -4 \\ 3 & -7 & 2 \\ 4 & -5 & -3 \end{bmatrix} + \begin{bmatrix} 5 & 3 & 4 \\ 2 & -7 & -5 \\ -4 & 2 & -3 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 5+5 & 2+3 & -4+4 \\ 3+2 & -7-7 & 2-5 \\ 4-4 & -5+2 & -3-3 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 10 & 5 & 0 \\ 5 & -14 & -3 \\ 0 & -3 & -6 \end{bmatrix} \] \)
\( \implies (A + A^T)^T = A^T + (A^T)^T = A^T + A \) i.e., \( A + A^T = (A + A^T)^T \)
\( \implies A + A^T \) is a symmetric matrix.
Also, \( A - A^T = \[ \begin{bmatrix} 5 & 2 & -4 \\ 3 & -7 & 2 \\ 4 & -5 & -3 \end{bmatrix} - \begin{bmatrix} 5 & 3 & 4 \\ 2 & -7 & -5 \\ -4 & 2 & -3 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 5-5 & 2-3 & -4-4 \\ 3-2 & -7+7 & 2+5 \\ 4+4 & -5-2 & -3+3 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 0 & -1 & -8 \\ 1 & 0 & 7 \\ 8 & -7 & 0 \end{bmatrix} \] \)
\( \implies (A - A^T)^T = \[ \begin{bmatrix} 0 & 1 & 8 \\ -1 & 0 & -7 \\ -8 & 7 & 0 \end{bmatrix} \] \)
\( = - \[ \begin{bmatrix} 0 & -1 & -8 \\ 1 & 0 & 7 \\ 8 & -7 & 0 \end{bmatrix} \] \)
\( \implies (A - A^T)^T = - (A - A^T) \)
\( \implies A - A^T \) is a skew-symmetric matrix.
In simple words: For any square matrix A, the sum \(A+A^T\) always results in a symmetric matrix, while the difference \(A-A^T\) always results in a skew-symmetric matrix.

🎯 Exam Tip: To prove a matrix is symmetric, show it equals its transpose. To prove it's skew-symmetric, show it equals the negative of its transpose. These are essential definitions for matrix classification.

Question 11. Express each of the following matrix as the sum of a symmetric and a skew-symmetric matrix:
(i) \[ \begin{bmatrix} 4 & -2 \\ 3 & -5 \end{bmatrix} \]
(ii) \[ \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \]
Answer: Solution:
(i) Let \( A = \[ \begin{bmatrix} 4 & -2 \\ 3 & -5 \end{bmatrix} \] \)
Then \( A^T = \[ \begin{bmatrix} 4 & 3 \\ -2 & -5 \end{bmatrix} \] \)
\( \implies A + A^T = \[ \begin{bmatrix} 4 & -2 \\ 3 & -5 \end{bmatrix} + \begin{bmatrix} 4 & 3 \\ -2 & -5 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 4+4 & -2+3 \\ 3-2 & -5-5 \end{bmatrix} = \begin{bmatrix} 8 & 1 \\ 1 & -10 \end{bmatrix} \] \)
This is a symmetric matrix.
Also, \( A - A^T = \[ \begin{bmatrix} 4 & -2 \\ 3 & -5 \end{bmatrix} - \begin{bmatrix} 4 & 3 \\ -2 & -5 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 4-4 & -2-3 \\ 3-(-2) & -5-(-5) \end{bmatrix} = \begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix} \] \)
This is a skew-symmetric matrix.
Now, \( A = \frac{1}{2} (A+A^T) + \frac{1}{2} (A-A^T) \)
\( = \frac{1}{2} \[ \begin{bmatrix} 8 & 1 \\ 1 & -10 \end{bmatrix} \] + \frac{1}{2} \[ \begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix} \] \)
\( \implies A = \[ \begin{bmatrix} \frac{1}{2}(8) & \frac{1}{2}(1) \\ \frac{1}{2}(1) & \frac{1}{2}(-10) \end{bmatrix} \] + \[ \begin{bmatrix} \frac{1}{2}(0) & \frac{1}{2}(-5) \\ \frac{1}{2}(5) & \frac{1}{2}(0) \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 4 & \frac{1}{2} \\ \frac{1}{2} & -5 \end{bmatrix} \] + \[ \begin{bmatrix} 0 & -\frac{5}{2} \\ \frac{5}{2} & 0 \end{bmatrix} \] \)
(ii) Let \( A = \[ \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \] \)
Then \( A^T = \[ \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \] \)
\( \implies A + A^T = \[ \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} + \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 3+3 & 3-2 & -1-4 \\ -2+3 & -2-2 & 1-5 \\ -4-1 & -5+1 & 2+2 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} \] \)
This is a symmetric matrix.
Also, \( A - A^T \)
\( = \[ \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} - \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 3-3 & 3-(-2) & -1-(-4) \\ -2-3 & -2-(-2) & 1-(-5) \\ -4-(-1) & -5-1 & 2-2 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix} \] \)
This is a skew-symmetric matrix.
Now, \( A = \frac{1}{2} (A+A^T) + \frac{1}{2} (A-A^T) \)
\( \implies A = \frac{1}{2} \[ \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} \] + \frac{1}{2} \[ \begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix} \] \)
In simple words: Any square matrix can be uniquely decomposed into the sum of a symmetric matrix and a skew-symmetric matrix using specific formulas involving the matrix and its transpose.

🎯 Exam Tip: Remember the formulas: \( P = \frac{1}{2}(A+A^T) \) for the symmetric part and \( Q = \frac{1}{2}(A-A^T) \) for the skew-symmetric part. Accuracy in calculating \( A+A^T \) and \( A-A^T \) is key.

Question 12. If A = \[ \begin{bmatrix} 2 & -1 \\ 3 & -2 \\ 4 & 1 \end{bmatrix} \] and B = \[ \begin{bmatrix} 0 & 3 & -4 \\ 2 & -1 & 1 \end{bmatrix} \] , verify that
(i) \( (AB)^T = B^T A^T \)
(ii) \( (BA)^T = A^T B^T \)
Answer: Solution:
\( A = \[ \begin{bmatrix} 2 & -1 \\ 3 & -2 \\ 4 & 1 \end{bmatrix} \] \)
\( B = \[ \begin{bmatrix} 0 & 3 & -4 \\ 2 & -1 & 1 \end{bmatrix} \] \)
\( \implies A^T = \[ \begin{bmatrix} 2 & 3 & 4 \\ -1 & -2 & 1 \end{bmatrix} \] \)
\( B^T = \[ \begin{bmatrix} 0 & 2 \\ 3 & -1 \\ -4 & 1 \end{bmatrix} \] \)
(i) \( \implies AB = \[ \begin{bmatrix} 2 & -1 \\ 3 & -2 \\ 4 & 1 \end{bmatrix} \] \[ \begin{bmatrix} 0 & 3 & -4 \\ 2 & -1 & 1 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 2(0)+(-1)(2) & 2(3)+(-1)(-1) & 2(-4)+(-1)(1) \\ 3(0)+(-2)(2) & 3(3)+(-2)(-1) & 3(-4)+(-2)(1) \\ 4(0)+1(2) & 4(3)+1(-1) & 4(-4)+1(1) \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 0-2 & 6+1 & -8-1 \\ 0-4 & 9+2 & -12-2 \\ 0+2 & 12-1 & -16+1 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} -2 & 7 & -9 \\ -4 & 11 & -14 \\ 2 & 11 & -15 \end{bmatrix} \] \)
\( \implies (AB)^T = \[ \begin{bmatrix} -2 & -4 & 2 \\ 7 & 11 & 11 \\ -9 & -14 & -15 \end{bmatrix} \] ... (1) \)
\( B^T A^T = \[ \begin{bmatrix} 0 & 2 \\ 3 & -1 \\ -4 & 1 \end{bmatrix} \] \[ \begin{bmatrix} 2 & 3 & 4 \\ -1 & -2 & 1 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 0(2)+2(-1) & 0(3)+2(-2) & 0(4)+2(1) \\ 3(2)+(-1)(-1) & 3(3)+(-1)(-2) & 3(4)+(-1)(1) \\ -4(2)+1(-1) & -4(3)+1(-2) & -4(4)+1(1) \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 0-2 & 0-4 & 0+2 \\ 6+1 & 9+2 & 12-1 \\ -8-1 & -12-2 & -16+1 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} -2 & -4 & 2 \\ 7 & 11 & 11 \\ -9 & -14 & -15 \end{bmatrix} \] ... (2) \)
From (1) and (2),
\( (AB)^T = B^T A^T \).
(ii) \( BA = \[ \begin{bmatrix} 0 & 3 & -4 \\ 2 & -1 & 1 \end{bmatrix} \] \[ \begin{bmatrix} 2 & -1 \\ 3 & -2 \\ 4 & 1 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 0(2)+3(3)+(-4)(4) & 0(-1)+3(-2)+(-4)(1) \\ 2(2)+(-1)(3)+1(4) & 2(-1)+(-1)(-2)+1(1) \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 0+9-16 & 0-6-4 \\ 4-3+4 & -2+2+1 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} -7 & -10 \\ 5 & 1 \end{bmatrix} \] \)
\( \implies (BA)^T = \[ \begin{bmatrix} -7 & 5 \\ -10 & 1 \end{bmatrix} \] ... (1) \)
\( A^T B^T = \[ \begin{bmatrix} 2 & 3 & 4 \\ -1 & -2 & 1 \end{bmatrix} \] \[ \begin{bmatrix} 0 & 2 \\ 3 & -1 \\ -4 & 1 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 2(0)+3(3)+4(-4) & 2(2)+3(-1)+4(1) \\ (-1)(0)+(-2)(3)+1(-4) & (-1)(2)+(-2)(-1)+1(1) \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} 0+9-16 & 4-3+4 \\ 0-6-4 & -2+2+1 \end{bmatrix} \] \)
\( = \[ \begin{bmatrix} -7 & 5 \\ -10 & 1 \end{bmatrix} \] ... (2) \)
From (1) and (2), \( (BA)^T = A^T B^T \).
In simple words: The reverse order law for transposes states that the transpose of a product of matrices is the product of their transposes, but in reverse order. This means \((AB)^T = B^T A^T\).

🎯 Exam Tip: This is a critical property of matrix transposes. Always remember to reverse the order of matrices when taking the transpose of a product; failure to do so is a common error.

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