Skip to content
Question of 182

Q.If A=[1201]A=\begin{bmatrix}1&2\\0&1\end{bmatrix} and B=[21−10]B=\begin{bmatrix}2&1\\-1&0\end{bmatrix}, then find 2A+3B2A+3B.

(a) [6502]\begin{bmatrix}6&5\\0&2\end{bmatrix}
(b) [87−32]\begin{bmatrix}8&7\\-3&2\end{bmatrix}
(c) [77−32]\begin{bmatrix}7&7\\-3&2\end{bmatrix}
(d) [87−23]\begin{bmatrix}8&7\\-2&3\end{bmatrix}
Odisha ChseOdisha CHSE +2 Science Board Exam 2022MCQ· 1mImportance★★★★★
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 →

Scale each matrix entrywise, then add corresponding entries.

A=[1201]A=\begin{bmatrix}1&2\\0&1\end{bmatrix}, B=[21−10]B=\begin{bmatrix}2&1\\-1&0\end{bmatrix}.

2A=[2402]2A=\begin{bmatrix}2&4\\0&2\end{bmatrix}, 3B=[63−30]3B=\begin{bmatrix}6&3\\-3&0\end{bmatrix}.

…

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.