GATE 2018 EE Q.17: solution and answer

Official answer: 5.5Source: Official answer key, GATE 2018 EE (IIT Guwahati)

1. Let the eigenvalues of $\mathbf{A}$ be ${\lambda }_1,{\lambda }_2$, counted over the complex numbers. The trace is the sum of the eigenvalues: ${\lambda }_1+{\lambda }_2=\text{tr}(\mathbf{A})=4$.

2. The eigenvalues of ${\mathbf{A}}^2$ are ${\lambda }_1^2,{\lambda }_2^2$, so ${\lambda }_1^2+{\lambda }_2^2=\text{tr}({\mathbf{A}}^2)=5$.

3. The determinant is the product of the eigenvalues: $\det (\mathbf{A})={\lambda }_1{\lambda }_2$.

4. Use the identity $({\lambda }_1+{\lambda }_2{)}^2={\lambda }_1^2+{\lambda }_2^2+2{\lambda }_1{\lambda }_2$: $$16=5+2{\lambda }_1{\lambda }_2\Rightarrow {\lambda }_1{\lambda }_2=\frac{11}{2}=5.5$$

5. The same result without eigenvalues: by the Cayley–Hamilton theorem every $2\times 2$ matrix satisfies ${\mathbf{A}}^2-\text{tr}(\mathbf{A})\mathbf{A}+\det (\mathbf{A})\mathbf{I}=0$. Taking the trace of both sides (with $\text{tr}(\mathbf{I})=2$) gives $$\det (\mathbf{A})=\frac{[\text{tr}(\mathbf{A}){]}^2-\text{tr}({\mathbf{A}}^2)}{2}=\frac{16-5}{2}=5.5$$

6. Check: the eigenvalues solve ${\lambda }^2-4\lambda +5.5=0$, so they are the complex pair $2\pm j\sqrt{1.5}$. That is allowed, because a real matrix can have complex eigenvalues, and the sum and product relations still hold. For example, $\mathbf{A}=\left[\begin{array}{cc}2& -1.5\\ 1& 2\end{array}\right]$ has trace $4$, ${\mathbf{A}}^2=\left[\begin{array}{cc}2.5& -6\\ 4& 2.5\end{array}\right]$ has trace $5$, and $\det (\mathbf{A})=4+1.5=5.5$. The determinant is non-zero, as a non-singular matrix requires, so $\det (\mathbf{A})=5.5$.