Q.Express the matrix $B = \begin{bmatrix}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{bmatrix}$ as a symmetric matrix (as printed in the paper). OR Find the values of $x$ and $y$ from the following equation: $2\begin{bmatrix}x & 5 \\ 7 & y-3\end{bmatrix} + \begin{bmatrix}3 & -4 \\ 1 & 2\end{bmatrix} = \begin{bmatrix}7 & 6 \\ 15 & 14\end{bmatrix}$.

A single matrix cannot literally be turned into a symmetric one without changing it; the standard NCERT problem on this exact matrix is to express it as the sum of a symmetric part $P=\tfrac12(B+B')$ and a skew-symmetric part $Q=\tfrac12(B-B')$. That standard version is solved below, honestly noting the reading. For the OR part, a direct linear matrix equation is solved.

Part 1 (honest note + standard solution): As literally printed, "express $B$ as a symmetric matrix" is not generally possible unless $B$ already is symmetric (it isn't, since $B\ne B'$ here). This exact matrix is the standard NCERT example for the problem "express $B$ as the sum of a symmetric and a skew-symmetric matrix," so that is solved here.

$B=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix},\qquad B'=\begin{bmatrix}2&-1&1\\-2&3&-2\\-4&4&-3\end{bmatrix}$

Symmetric part $P=\dfrac12(B+B')$:

$B+B' = \begin{bmatrix}4&-3&-3\\-3&6&2\\-3&2&-6\end{bmatrix}\ \Rightarrow\ P = \begin{bmatrix}2&-\tfrac32&-\tfrac32\\-\tfrac32&3&1\\-\tfrac32&1&-3\end{bmatrix}$

Skew-symmetric part $Q=\dfrac12(B-B')$:

$B-B' = \begin{bmatrix}0&-1&-5\\1&0&6\\5&-6&0\end{bmatrix}\ \Rightarrow\ Q = \begin{bmatrix}0&-\tfrac12&-\tfrac52\\ \tfrac12&0&3\\ \tfrac52&-3&0\end{bmatrix}$

Check: $P+Q=B$, $P'=P$ (symmetric), $Q'=-Q$ (skew-symmetric). ✓

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