"Let $$A=\left(\begin{array}{rr}4 & -2 \\ \alpha & \beta\end{array}\right)$$. If $\mathrm{A}^{2}+\gamma \mathrm{A}+18 \mathrm{I}=\mathrm{O}$, then $\operatorname{det}(\mathrm{A})$ is equal to _____________."

Question

Accepted Answer

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Characteristic equation of A is given by

$\left| {A - \lambda I} \right| = 0$

$$\left| {\matrix{ {4 - \lambda } & { - 2} \cr \alpha & {\beta - \lambda } \cr } } \right| = 0$$

$\Rightarrow {\lambda ^2} - (4 + \beta )\lambda + (4\beta + 2\alpha ) = 0$

So, ${A^2} - (4 + \beta )A + (4\beta + 2\alpha )I = 0$

$|A| = 4\beta + 2\alpha = 18$

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Let $$A=\left(\begin{array}{rr}4 & -2 \\ \alpha & \beta\end{array}\right)$$.

If $\mathrm{A}^{2}+\gamma \mathrm{A}+18 \mathrm{I}=\mathrm{O}$, then $\operatorname{det}(\mathrm{A})$ is equal to _____________.

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Let $$A=\left(\begin{array}{rr}4 & -2 \\ \alpha & \beta\end{array}\right)$$. If $\mathrm{A}^{2}+\gamma \mathrm{A}+18 \mathrm{I}=\mathrm{O}$, then $\operatorname{det}(\mathrm{A})$ is equal to _____________.

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More Matrices and Determinants questions

Drill 25 more like these. Every day. Free.

Let $$A=\left(\begin{array}{rr}4 & -2 \\ \alpha & \beta\end{array}\right)$$.

If $\mathrm{A}^{2}+\gamma \mathrm{A}+18 \mathrm{I}=\mathrm{O}$, then $\operatorname{det}(\mathrm{A})$ is equal to _____________.