Kerala KEAM 5 June 2024 Solved Paper

Solution:Given that $A$ is a symmetric matrix and $B$ is a skew-symmetric matrix, we know the following properties:- A symmetric matrix satisfies $A^T = A$.- A skew-symmetric matrix satisfies $B^T = -B$.We are given that: $$A + B = \begin{pmatrix} 1 & 3-2 & 5 \end{pmatrix}$$We need to find $A - B$.Let's define: $$A = \begin{pmatrix} a & bb & d \end{pmatrix}, \quad B = \begin{pmatrix} 0 & e-e & 0 \end{pmatrix}$$Since $A$ is symmetric, the off-diagonal elements are equal, and for $B$, the diagonal elements are zero, and the off-diagonal elements are negatives of each other.Now, from the equation $A + B = \begin{pmatrix} 1 & 3 < br>-2 & 5 \end{pmatrix}$, we can write: $$\begin{pmatrix} a & b < br>b & d \end{pmatrix} + \begin{pmatrix} 0 & e < br>-e & 0 \end{pmatrix} = \begin{pmatrix} 1 & 3 < br>-2 & 5 \end{pmatrix}$$This gives us the following system of equations:1. $a = 1$2. $b + e = 3$3. $b - e = -2$4. $d = 5$From equations 2 and 3, we can solve for $b$ and $e$: $$b + e = 3 \quad and \quad b - e = -2$$Adding these two equations gives: $$2b = 1 \quad \Rightarrow \quad b = \frac{1}{2}$$Substitute $b = \frac{1}{2}$ into $b + e = 3$: $$\frac{1}{2} + e = 3 \quad \Rightarrow \quad e = \frac{5}{2}$$Now, we can find $A - B$: $$A - B = \begin{pmatrix} a & b < br>b & d \end{pmatrix} - \begin{pmatrix} 0 & e < br>-e & 0 \end{pmatrix} = \begin{pmatrix} 1 & \frac{1}{2} < br>\frac{1}{2} & 5 \end{pmatrix} - \begin{pmatrix} 0 & \frac{5}{2} < br>-\frac{5}{2} & 0 \end{pmatrix}$$This gives: $$A - B = \begin{pmatrix} 1 & -2 < br>3 & 5 \end{pmatrix}$$Thus, the correct answer is option (D), $\begin{pmatrix} 1 & -2 < br>3 & 5 \end{pmatrix}$. Quick Tip: To find $A - B$ when $A + B$ is given, simply subtract the matrices using the known properties of symmetric and skew-symmetric matrices. Use the relationship $A + B + A - B = 2A$ to help.