If A satisfies the equation \[x^3 - 5 x^2 + 4x + \lambda = 0\] then A-1 exists if _____________ .

\[\lambda eq 0\] A satisfies \[ x^3 - 5 x^2 + 4x + \lambda = 0 . \] \[ \Rightarrow A^3 - 5 A^2 + 4A = - \lambda\] \[\text{ Assuming }A^{- 1}\text{ exists, we get }\] \[ A^{- 1} \left( A^3 - 5 A^2 + 4A \right) = - \lambda A^{- 1} \] \[ \Rightarrow A^2 - 5A + 4 = - A^{- 1} \lambda \] \[ \Rightarrow A^{- 1} = \frac{- \left( A^2 - 5A + 4 \right)}{\lambda}\] \[\text{ Thus, }A^{- 1}\text{ exists if } \lambda eq 0 .\]

If a Satisfies the Equation X 3 − 5 X 2 + 4 X + λ = 0 Then A-1 Exists If - Mathematics

Question

If A satisfies the equation $x^{3} - 5 x^{2} + 4 x + λ = 0$ then A^-1 exists if _____________ .

Options

$λ = 1$
$λ \neq 2$
$λ \neq - 1$
$λ \neq 0$

MCQ

Solution

$λ \neq 0$

A satisfies $x^{3} - 5 x^{2} + 4 x + λ = 0 .$

$\Rightarrow A^{3} - 5 A^{2} + 4 A = - λ$

$Assuming A^{- 1} exists, we get$

$A^{- 1} (A^{3} - 5 A^{2} + 4 A) = - λ A^{- 1}$

$\Rightarrow A^{2} - 5 A + 4 = - A^{- 1} λ$

$\Rightarrow A^{- 1} = \frac{- (A^{2} - 5 A + 4)}{λ}$

$Thus, A^{- 1} exists if λ \neq 0 .$

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Inverse of Matrix - Inverse of a Square Matrix by the Adjoint Method

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Q 12 Q 11 Q 13

Chapter 7: Adjoint and Inverse of a Matrix - Exercise 7.4 [Page 37]

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RD Sharma Mathematics [English] Class 12

Chapter 7 Adjoint and Inverse of a Matrix
Exercise 7.4 | Q 12 | Page 37

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