Let $\mathrm{A}=\left[\begin{array}{cc}\alpha & -1 \\ 6 & \beta\end{array}\right], \alpha>0$, such that $\operatorname{det}(\mathrm{A})=0$ and $\alpha+\beta=1$. If I denotes $2 \times 2$ identity matrix, then the matrix $(I+A)^8$ is :
Solution
<p>Let $|A|=0 \Rightarrow \alpha \beta-(-6)=0 \Rightarrow \alpha \beta=-6$ and $\alpha+\beta=1 \Rightarrow \alpha \beta$ are roots of the equation</p>
<p>$$\begin{aligned}
& x^2-x-6=0 \Rightarrow x=3,-2 . \text { Since } \alpha>0 \\
& \Rightarrow \quad \alpha=3, \beta=-2 \\
& \Rightarrow \quad A=\left[\begin{array}{ll}
3 & -1 \\
6 & -2
\end{array}\right] \Rightarrow I+A=\left[\begin{array}{ll}
4 & -1 \\
6 & -1
\end{array}\right] \\
& (I+A)^2=\left[\begin{array}{ll}
10 & -3 \\
18 & -5
\end{array}\right] \Rightarrow(I+A)^4=\left[\begin{array}{ll}
46 & -15 \\
90 & -29
\end{array}\right] \\
& \Rightarrow \quad(I+A)^8=\left[\begin{array}{cc}
766 & -255 \\
1530 & -509
\end{array}\right]
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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