Skip to content
Worked Examples · Example 35

Q.Find A−1A^{-1}, if A=[10−2−51]A = \begin{bmatrix} 10 & -2 \\ -5 & 1 \end{bmatrix}.

CBSENCERTSubjective· 2mImportance★★★★★
44% · 35/80 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 →

det⁡A=0\det A = 0, so AA is singular and has no inverse.

A−1=1det⁡A adj(A)A^{-1}=\dfrac{1}{\det A}\,\mathrm{adj}(A) exists iff det⁡A≠0\det A \neq 0. For [abcd]\begin{bmatrix}a&b\\c&d\end{bmatrix}, det⁡=ad−bc\det = ad-bc.

  1. Given

A=[10−2−51].A = \begin{bmatrix} 10 & -2 \\ -5 & 1 \end{bmatrix}.

  1. Compute the determinant: det⁡A=(10)(1)−(−2)(−5)=10−10=0.\det A = (10)(1) - (-2)(-5) = 10 - 10 = 0. …

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.