GATE Electrical 2021 | Question: 1

Question

Let $p$ and $q$ be real numbers such that $p^{2}+q^{2}=1$ . The eigenvalues of the matrix $\begin{bmatrix} p & q\\ q& -p \end{bmatrix}$are

• A. $1$ and $1$
• B. $1$ and $-1$
• C. $j$ and $-j$
• D. $pq$ and $-pq$

Answer 1

answer -B (1AND-1)

SOLUTION

FORMULA OF EIGEN VALUE

A^2+(diagonal addition)A+(determinant of matrix)=0

where A  is eign value

diagonal addition (p-p)=0

det(matrix)=(-p^2-q^2)

given question (p^2+q^2)=1

then put value

A^2-1=0

the A=1and -1(ans)

Comments

We can follow elimination of options.

Determinant of matrix is -(p^2+q^2) which is -1

Since p and q are real, eigen values must also be real.

pq and -pq cannot be the eigen values because p=1/root(2) and q=1/root(2) is also a possibility for the determinant condition.

Only answer left is 1 and -1