Skip to content
Question of 182

Q.Express [15−12]\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix} as the sum of a symmetric and skew symmetric matrix.

Karnataka PUCKarnataka II PUC Board 2023Subjective· 3mImportance★★★★★
0% · 0/182 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Write A=12(A+A′)+12(A−A′)A=\tfrac12(A+A')+\tfrac12(A-A'); the first part is symmetric, the second skew-symmetric.

Step 1 — Find the transpose. Let A=[15−12]A=\begin{bmatrix}1&5\\-1&2\end{bmatrix}. Then

A′=[1−152].A'=\begin{bmatrix}1&-1\\5&2\end{bmatrix}.

Step 2 — Symmetric part P=12(A+A′)P=\tfrac12(A+A').

A+A′=[1+15−1−1+52+2]=[2444],P=12[2444]=[1222].A+A'=\begin{bmatrix}1+1&5-1\\-1+5&2+2\end{bmatrix}=\begin{bmatrix}2&4\\4&4\end{bmatrix},\qquad P=\frac12\begin{bmatrix}2&4\\4&4\end{bmatrix}=\begin{bmatrix}1&2\\2&2\end{bmatrix}.

Since P′=PP'=P, PP is symmetric.

Step 3 — Skew-symmetric part Q=12(A−A′)Q=\tfrac12(A-A').

A−A′=[1−15−(−1)−1−52−2]=[06−60],Q=12[06−60]=[03−30].A-A'=\begin{bmatrix}1-1&5-(-1)\\-1-5&2-2\end{bmatrix}=\begin{bmatrix}0&6\\-6&0\end{bmatrix},\qquad Q=\frac12\begin{bmatrix}0&6\\-6&0\end{bmatrix}=\begin{bmatrix}0&3\\-3&0\end{bmatrix}.

Since Q′=−QQ'=-Q, QQ is skew-symmetric. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.