UPSC 2017 Maths Optional Paper 1 Q1a — Step-by-Step Solution

10 marks · Section A

Diagonalization via Eigenvectors · Linear Algebra · Read the full method →

Question

Let $A=\begin{pmatrix}2 & 2\\ 1 & 3\end{pmatrix}$ . Find a non-singular matrix $P$ such that $P^{-1}AP$ is a diagonal matrix.

Distinct eigenvalues $\Rightarrow$ diagonalizable; eigenvectors as columns of $P$ , eigenvalues on the diagonal in the same order.

\det(A-\lambda I)=\begin{vmatrix}2-\lambda & 2\\ 1 & 3-\lambda\end{vmatrix}=(2-\lambda)(3-\lambda)-2=\lambda^2-5\lambda+4=(\lambda-1)(\lambda-4).

So the eigenvalues are $\lambda_1=1$ and $\lambda_2=4$ . They are distinct, hence $A$ is diagonalizable.

Solve $(A-I)\mathbf v=\mathbf0$ :

\begin{pmatrix}1 & 2\\ 1 & 2\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\mathbf0\ \Rightarrow\ x+2y=0.

Take $y=1,\ x=-2$ : $\mathbf v_1=\begin{pmatrix}-2\\ 1\end{pmatrix}$ .

Solve $(A-4I)\mathbf v=\mathbf0$ :

\begin{pmatrix}-2 & 2\\ 1 & -1\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\mathbf0\ \Rightarrow\ x=y.

Take $\mathbf v_2=\begin{pmatrix}1\\ 1\end{pmatrix}$ .

Put the eigenvectors as columns of $P$ (order matching the eigenvalues):

P=\begin{pmatrix}-2 & 1\\ 1 & 1\end{pmatrix},\qquad \det P=-2-1=-3\ne 0\ (\text{non-singular}).

Then

\boxed{\;P=\begin{pmatrix}-2 & 1\\ 1 & 1\end{pmatrix},\qquad P^{-1}AP=\begin{pmatrix}1 & 0\\ 0 & 4\end{pmatrix}.\;}