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If $A$ is a $2 \times 2$ matrix such that $\operatorname{det} A=-21$ and trace of $\mathrm{A}^3$ is 2024 , then the trace of A is
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The correct answer is:
11
Let eigen values of matrices $\mathrm{A}$ are $\lambda_1, \lambda_1 \Rightarrow \lambda_1, \lambda_2=-21$
$\Rightarrow$ Eigen values of matrices $\mathrm{A}^3$ are $\lambda_1{ }^3, \lambda_2{ }^3$
Now $\lambda_1^3, \lambda_2{ }^3=\left(\lambda_1+\lambda_2\right)\left(\lambda_1{ }^2+\lambda_2-\lambda_1 \lambda_2\right)$
$\Rightarrow 2024=\left(\lambda_1+\lambda_2\right)\left[\lambda_1^2+\lambda_2^2+21\right)$
$\Rightarrow 2024=\left(\lambda_1+\lambda_2\right)\left[\left(\lambda_1+\lambda_2\right)^2-2 \lambda_1 \lambda_2+21\right]$
$\Rightarrow 2024=\left(\lambda_1+\lambda_2\right)^3-2 \lambda_1 \lambda_2\left(\lambda_1+\lambda_2\right)+21\left(\lambda_1+\lambda_2\right)$
$\Rightarrow\left(\lambda_1+\lambda_2\right)^3+63\left(\lambda_1+\lambda_2\right)-2024=0$ (A)
Here option (b) satisfied the given $\mathrm{eq}^{\mathrm{n}}$. (A) other options are not satisfied.
$$
\Rightarrow \lambda_1+\lambda_2=\text { trace of } \mathrm{A}=11 \text {. }
$$
$\Rightarrow$ Eigen values of matrices $\mathrm{A}^3$ are $\lambda_1{ }^3, \lambda_2{ }^3$
Now $\lambda_1^3, \lambda_2{ }^3=\left(\lambda_1+\lambda_2\right)\left(\lambda_1{ }^2+\lambda_2-\lambda_1 \lambda_2\right)$
$\Rightarrow 2024=\left(\lambda_1+\lambda_2\right)\left[\lambda_1^2+\lambda_2^2+21\right)$
$\Rightarrow 2024=\left(\lambda_1+\lambda_2\right)\left[\left(\lambda_1+\lambda_2\right)^2-2 \lambda_1 \lambda_2+21\right]$
$\Rightarrow 2024=\left(\lambda_1+\lambda_2\right)^3-2 \lambda_1 \lambda_2\left(\lambda_1+\lambda_2\right)+21\left(\lambda_1+\lambda_2\right)$
$\Rightarrow\left(\lambda_1+\lambda_2\right)^3+63\left(\lambda_1+\lambda_2\right)-2024=0$ (A)
Here option (b) satisfied the given $\mathrm{eq}^{\mathrm{n}}$. (A) other options are not satisfied.
$$
\Rightarrow \lambda_1+\lambda_2=\text { trace of } \mathrm{A}=11 \text {. }
$$
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