If A is a symmetric matrix and B is a skew-symmetric matrix such that A + B = [ \beginmatrix 2 & 3 5 & - 1 \endmatrix ] , then AB is equal to :

[ \beginmatrix 4 & - 2 - 1 & - 4 \endmatrix ]

[ \beginmatrix 4 & - 2 1 & - 4 \endmatrix ]

[ \beginmatrix - 4 & - 2 - 1 & 4 \endmatrix ]

[ \beginmatrix - 4 & 2 1 & 4 \endmatrix ]

Matrices and Determinants PYQ — JEE Main 2019 · 12 Apr · Shift 1 · Q39 with Solution

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Correct answer: D

Since $A$ is symmetric and $B$ is skew-symmetric, for any matrix $M$ we can decompose it as $A=\frac{M+M^T}{2},\qquad B=\frac{M-M^T}{2}.$

=\frac{1}{2}\begin{bmatrix}4&8\\8&-2\end{bmatrix} =\begin{bmatrix}2&4\\4&-1\end{bmatrix}.$$ 4. Compute $B$: $$B=\frac{1}{2}\left(\begin{bmatrix}2&3\\5&-1\end{bmatrix}-\begin{bmatrix}2&5\\3&-1\end{bmatrix}\right) =\frac{1}{2}\begin{bmatrix}0&-2\\2&0\end{bmatrix} =\begin{bmatrix}0&-1\\1&0\end{bmatrix}.$$ 5. Now calculate $AB$: $$AB=\begin{bmatrix}2&4\\4&-1\end{bmatrix}\begin{bmatrix}0&-1\\1&0\end{bmatrix}.$$ Multiply row-wise and column-wise: $$AB= \begin{bmatrix} 2\cdot 0+4\cdot 1 & 2\cdot(-1)+4\cdot 0\\[4pt] 4\cdot 0+(-1)\cdot 1 & 4\cdot(-1)+(-1)\cdot 0 \end{bmatrix} = \begin{bmatrix} 4&-2\\-1&-4 \end{bmatrix}.$$ 6. Compare with the options: - Option A: $\begin{bmatrix}4&-2\\1&-4\end{bmatrix}$ - Option B: $\begin{bmatrix}-4&-2\\-1&4\end{bmatrix}$ - Option C: $\begin{bmatrix}-4&2\\1&4\end{bmatrix}$ - Option D: $\begin{bmatrix}4&-2\\-1&-4\end{bmatrix}$ So the correct answer is **Option D**.

View written solutionFree

Correct answer: D

Since $A$ is symmetric and $B$ is skew-symmetric, for any matrix $M$ we can decompose it as $A=\frac{M+M^T}{2},\qquad B=\frac{M-M^T}{2}.$

=\frac{1}{2}\begin{bmatrix}4&8\\8&-2\end{bmatrix} =\begin{bmatrix}2&4\\4&-1\end{bmatrix}.$$ 4. Compute $B$: $$B=\frac{1}{2}\left(\begin{bmatrix}2&3\\5&-1\end{bmatrix}-\begin{bmatrix}2&5\\3&-1\end{bmatrix}\right) =\frac{1}{2}\begin{bmatrix}0&-2\\2&0\end{bmatrix} =\begin{bmatrix}0&-1\\1&0\end{bmatrix}.$$ 5. Now calculate $AB$: $$AB=\begin{bmatrix}2&4\\4&-1\end{bmatrix}\begin{bmatrix}0&-1\\1&0\end{bmatrix}.$$ Multiply row-wise and column-wise: $$AB= \begin{bmatrix} 2\cdot 0+4\cdot 1 & 2\cdot(-1)+4\cdot 0\\[4pt] 4\cdot 0+(-1)\cdot 1 & 4\cdot(-1)+(-1)\cdot 0 \end{bmatrix} = \begin{bmatrix} 4&-2\\-1&-4 \end{bmatrix}.$$ 6. Compare with the options: - Option A: $\begin{bmatrix}4&-2\\1&-4\end{bmatrix}$ - Option B: $\begin{bmatrix}-4&-2\\-1&4\end{bmatrix}$ - Option C: $\begin{bmatrix}-4&2\\1&4\end{bmatrix}$ - Option D: $\begin{bmatrix}4&-2\\-1&-4\end{bmatrix}$ So the correct answer is **Option D**.