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If \(A=\left[\begin{array}{ccc}1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1\end{array}\right]\), then \(A^{-1}=\)
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Verified Answer
The correct answer is:
\(A^2-A-3\)
Characteristic equation of any square matrix \(A\) is given by
\(\begin{aligned}
& |A-\lambda I|=0 \\
& \Rightarrow \quad\left[\begin{array}{ccc}
1 & 0 & -2 \\
-2 & -1 & 2 \\
3 & 4 & 1
\end{array}\right]-\lambda\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=0 \\
& \Rightarrow \quad\left[\begin{array}{ccc}
1 & 0 & -2 \\
-2 & -1 & 2 \\
3 & 4 & 1
\end{array}\right]-\left[\begin{array}{ccc}
\lambda & 0 & 0 \\
0 & \lambda & 0 \\
0 & 0 & \lambda
\end{array}\right]=0 \\
& \Rightarrow \quad\left|\begin{array}{ccc}
1-\lambda & 0 & -2 \\
-2 & -1-\lambda & 2 \\
3 & 4 & 1-\lambda
\end{array}\right|=0 \\
& \Rightarrow(1-\lambda)[-(1+\lambda)(1-\lambda)-8]-2[-8+3(\lambda+1)]=0 \\
& \Rightarrow \quad(1-\lambda)\left[-\left(1-\lambda^2\right)-8\right]-2(-8+3 \lambda+3)=0 \\
& \Rightarrow \quad(1-\lambda)\left(\lambda^2-9\right)-2(-5+3 \lambda)=0 \\
& \Rightarrow \quad \lambda^2-9-\lambda^3+9 \lambda-6 \lambda+10=0 \\
& \Rightarrow \quad-\lambda^3+\lambda^2+3 \lambda+1=0 \\
& \Rightarrow \quad \lambda^3-\lambda^2-3 \lambda-1=0
\end{aligned}\)
According to Caley Hamilton, every square matrix satisfies its characteristics equations.
\(\begin{aligned}
& \therefore \quad A^3-A^2-3 A-I=0 \\
& \Rightarrow \quad A^{-1}\left(A^3-A^2-3 A-I\right)=0 \\
& \Rightarrow \quad A^2-A-3 I-A^{-1}=0 \\
& \Rightarrow \quad A^{-1}=A^2-A-3 I
\end{aligned}\)
\(\begin{aligned}
& |A-\lambda I|=0 \\
& \Rightarrow \quad\left[\begin{array}{ccc}
1 & 0 & -2 \\
-2 & -1 & 2 \\
3 & 4 & 1
\end{array}\right]-\lambda\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=0 \\
& \Rightarrow \quad\left[\begin{array}{ccc}
1 & 0 & -2 \\
-2 & -1 & 2 \\
3 & 4 & 1
\end{array}\right]-\left[\begin{array}{ccc}
\lambda & 0 & 0 \\
0 & \lambda & 0 \\
0 & 0 & \lambda
\end{array}\right]=0 \\
& \Rightarrow \quad\left|\begin{array}{ccc}
1-\lambda & 0 & -2 \\
-2 & -1-\lambda & 2 \\
3 & 4 & 1-\lambda
\end{array}\right|=0 \\
& \Rightarrow(1-\lambda)[-(1+\lambda)(1-\lambda)-8]-2[-8+3(\lambda+1)]=0 \\
& \Rightarrow \quad(1-\lambda)\left[-\left(1-\lambda^2\right)-8\right]-2(-8+3 \lambda+3)=0 \\
& \Rightarrow \quad(1-\lambda)\left(\lambda^2-9\right)-2(-5+3 \lambda)=0 \\
& \Rightarrow \quad \lambda^2-9-\lambda^3+9 \lambda-6 \lambda+10=0 \\
& \Rightarrow \quad-\lambda^3+\lambda^2+3 \lambda+1=0 \\
& \Rightarrow \quad \lambda^3-\lambda^2-3 \lambda-1=0
\end{aligned}\)
According to Caley Hamilton, every square matrix satisfies its characteristics equations.
\(\begin{aligned}
& \therefore \quad A^3-A^2-3 A-I=0 \\
& \Rightarrow \quad A^{-1}\left(A^3-A^2-3 A-I\right)=0 \\
& \Rightarrow \quad A^2-A-3 I-A^{-1}=0 \\
& \Rightarrow \quad A^{-1}=A^2-A-3 I
\end{aligned}\)
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