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Exercises · Q4

Q.Express A=(351−2)A=\begin{pmatrix}3&5\\1&-2\end{pmatrix} as the sum of a symmetric matrix and a skew-symmetric matrix.

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First find the transpose: A′=(315−2)A'=\begin{pmatrix}3&1\\5&-2\end{pmatrix}.

Symmetric part:

12(A+A′)=12[(351−2)+(315−2)]=12(666−4)=(333−2)\tfrac12(A+A')=\tfrac12\left[\begin{pmatrix}3&5\\1&-2\end{pmatrix}+\begin{pmatrix}3&1\\5&-2\end{pmatrix}\right]=\tfrac12\begin{pmatrix}6&6\\6&-4\end{pmatrix}=\begin{pmatrix}3&3\\3&-2\end{pmatrix}

Call this PP. Check P′=PP'=P: transposing (333−2)\begin{pmatrix}3&3\\3&-2\end{pmatrix} swaps the off-diagonal 33's with each other, giving back the same matrix — so PP is genuinely symmetric.

Skew-symmetric part:

12(A−A′)=12[(351−2)−(315−2)]=12(04−40)=(02−20)\tfrac12(A-A')=\tfrac12\left[\begin{pmatrix}3&5\\1&-2\end{pmatrix}-\begin{pmatrix}3&1\\5&-2\end{pmatrix}\right]=\tfrac12\begin{pmatrix}0&4\\-4&0\end{pmatrix}=\begin{pmatrix}0&2\\-2&0\end{pmatrix}

Call this QQ. Check Q′=−QQ'=-Q: transposing (02−20)\begin{pmatrix}0&2\\-2&0\end{pmatrix} gives (0−220)\begin{pmatrix}0&-2\\2&0\end{pmatrix}, which is exactly −Q-Q — confirming QQ is genuinely skew-symmetric, and its diagonal entries are 00 as required.

Independent check: add PP and QQ back together — (333−2)+(02−20)=(351−2)=A\begin{pmatrix}3&3\\3&-2\end{pmatrix}+\begin{pmatrix}0&2\\-2&0\end{pmatrix}=\begin{pmatrix}3&5\\1&-2\end{pmatrix}=A ✓ — recovering the original matrix exactly, which independently confirms both parts were computed correctly.

✓Final answer

A=(333−2)+(02−20)A=\begin{pmatrix}3&3\\3&-2\end{pmatrix}+\begin{pmatrix}0&2\\-2&0\end{pmatrix} (symmetric ++ skew-symmetric).

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