If A = {bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 {bmatrix} then A^{-1} =

Correct Answer is : A2A²A2 - A - 3I

Correct Answer is : A2A²A2 - A - 3I

Test Index

AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 5 of 160

Marks: +1, -0

If A =

\begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}

A^{-1}

Solution:

The characteristic equation of the square matrix is written as,

|A-\lambda I|=0

Then,

\begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}

-\lambda \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=0

\begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}

-\lambda \begin{bmatrix}\lambda ,0,0 \\ 0,\lambda ,0 \\ 0,0,\lambda\end{bmatrix}=0

\begin{bmatrix}1-\lambda & 0 & -2 \\ -2 & -1-\lambda & 2 \\ 3 & 4 & 1-\lambda\end{bmatrix}=0

Simplifying the above matrix, we get,

(1-\lambda)[-(1+\lambda)-8]-2(-8+3(\lambda+1))=0

(1-\lambda) \left[ - (1-\lambda^2) - 8 \right] -2(-8+3\lambda+3)=0

(1-\lambda) (\lambda^2-9 )

-2(-5+3\lambda)=0

\lambda^2-9-\lambda^3+9\lambda

-6\lambda+10=0

Further simplify the above,

-\lambda^3+\lambda^2+3\lambda+1=0

\lambda^3-\lambda^2-3\lambda-1=0

From the Caley Hamilton’s theorem,

A^{3}-A^{2}-3 A-I=0

A^{-1} (A^{3}-A^{2}-3 A-I )=0

A^{-1}=A^{2}-A-3 I

Go to Question:

Prev Question

Next Question