Let $$A=\left[\begin{array}{lll}2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b\end{array}\right]$$. If $A^3=4 A^2-A-21 I$, where $I$ is the identity matrix of order $3 \times 3$, then $2 a+3 b$ is equal to
Solution
<p>$$\begin{aligned}
& |A-\lambda I|=0 \\
& \left|\begin{array}{ccc}
2-\lambda & a & 0 \\
1 & 3-\lambda & 1 \\
0 & 5 & b-\lambda
\end{array}\right|=0 \\
& (2-\lambda)[(3-\lambda)(b-\lambda)-5]-a[b-\lambda-0]+0=0 \\
& (2-\lambda)\left[3 b-3 \lambda-b \lambda+\lambda^2-5\right]-a b+a \lambda=0 \\
& \lambda^3-(b+5) \lambda^2+(1-a+5 b) \lambda+(10-6 b+a b)=0 \\
& A^3-(b+5) A^2+(1-a+5 b) A+(10-6 b+a b) I=0 \\
& \Rightarrow \mathrm{b}+5=4,1-a+5 b=1,10-6 b+a b=21 \\
& \Rightarrow a=-5, b=-1 \\
& \Rightarrow 2 a+3 b=-13
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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