UPSC 2025 Maths Optional Paper 1 Q4c-i — Step-by-Step Solution

12 marks · Section A

Eigenvalues and eigenvectors · Linear Algebra · asked 10× in 14 yrs · Read the full method →

Question

Find the eigenvalues and the corresponding eigenvectors of the matrix

A = \begin{bmatrix} 1 & 2 & 0 \\ 2 & 1 & -6 \\ 2 & -2 & 3 \end{bmatrix}.

Technique

Solve $\det(A-\lambda I)=0$ for the eigenvalues, then solve $(A-\lambda I)\mathbf v=\mathbf 0$ for each eigenvector.

Solution

Step 1 — Characteristic polynomial.

\det(A-\lambda I)=\begin{vmatrix}1-\lambda & 2 & 0\\ 2 & 1-\lambda & -6\\ 2 & -2 & 3-\lambda\end{vmatrix}.

Expand along the first row:

=(1-\lambda)\big[(1-\lambda)(3-\lambda)-(-6)(-2)\big]-2\big[2(3-\lambda)-(-6)(2)\big]+0.

Compute the brackets:

(1-\lambda)(3-\lambda)-12=\lambda^2-4\lambda+3-12=\lambda^2-4\lambda-9,

2(3-\lambda)+12=18-2\lambda.

\det=(1-\lambda)(\lambda^2-4\lambda-9)-2(18-2\lambda).

Expand $(1-\lambda)(\lambda^2-4\lambda-9)=\lambda^2-4\lambda-9-\lambda^3+4\lambda^2+9\lambda=-\lambda^3+5\lambda^2+5\lambda-9.$ Then subtract $2(18-2\lambda)=36-4\lambda$ :

-\lambda^3+5\lambda^2+5\lambda-9-36+4\lambda=-\lambda^3+5\lambda^2+9\lambda-45.

Set $\det(A-\lambda I)=0$ , i.e. (multiplying by $-1$ )

\lambda^3-5\lambda^2-9\lambda+45=0.

Step 2 — Roots. Group: $\lambda^2(\lambda-5)-9(\lambda-5)=(\lambda-5)(\lambda^2-9)=(\lambda-5)(\lambda-3)(\lambda+3)=0.$

\lambda=5,\quad \lambda=3,\quad \lambda=-3.

Step 3 — Eigenvectors.

$\lambda=3$ : solve $(A-3I)\mathbf v=0$ , $A-3I=\begin{bmatrix}-2&2&0\\2&-2&-6\\2&-2&0\end{bmatrix}.$ Row 1: $-2x+2y=0\Rightarrow y=x$ . Row 3: $2x-2y=0$ (same). Row 2: $2x-2y-6z=0\Rightarrow -6z=0\Rightarrow z=0$ . So $\mathbf v=(1,1,0)^T$ .

$\lambda=5$ : $A-5I=\begin{bmatrix}-4&2&0\\2&-4&-6\\2&-2&-2\end{bmatrix}.$ Row 1: $-4x+2y=0\Rightarrow y=2x$ . Row 3: $2x-2y-2z=0\Rightarrow z=x-y=x-2x=-x$ . Check row 2: $2x-4(2x)-6(-x)=2x-8x+6x=0$ ✓. So $\mathbf v=(1,2,-1)^T$ .

$\lambda=-3$ : $A+3I=\begin{bmatrix}4&2&0\\2&4&-6\\2&-2&6\end{bmatrix}.$ Row 1: $4x+2y=0\Rightarrow y=-2x$ . Row 3: $2x-2y+6z=0\Rightarrow 2x+4x+6z=0\Rightarrow z=-x$ . Check row 2: $2x+4(-2x)-6(-x)=2x-8x+6x=0$ ✓. So $\mathbf v=(1,-2,-1)^T$ (equivalently $(-1,2,1)^T$ ).

Answer

\boxed{\;\begin{aligned} \lambda_1&=3,\ &\mathbf v_1&=(1,1,0)^T,\\ \lambda_2&=5,\ &\mathbf v_2&=(1,2,-1)^T,\\ \lambda_3&=-3,\ &\mathbf v_3&=(1,-2,-1)^T\ (\text{or }(-1,2,1)^T). \end{aligned}\;}