Question

If the eigen vectors of the matrix \(\begin{bmatrix}1& 2\\0& 2\end{bmatrix}\) are written in the form \(\begin{bmatrix}1a\end{bmatrix}\) and \(\begin{bmatrix}1b\end{bmatrix}\), then \(a+b=\)

• A. \(0\)
• B. \(\dfrac{1}{2}\)
• C. \(1\)
• D. \(2\)

Hint:

For triangular matrices, eigenvalues are diagonal entries, then solve \((A-\lambda I)X=0\).

Answer 1

The Correct Option is B

Concept:
For an eigenvalue \(\lambda\), eigenvectors are obtained from \[ (A-\lambda I)X=0 \] Given, \[ A=\begin{bmatrix}1& 2\\0& 2\end{bmatrix} \] Since \(A\) is triangular, eigenvalues are the diagonal entries: \[ 1,\quad 2 \]

Step 1: Eigenvector for \(\lambda=1\). \[ A-I= \begin{bmatrix}0& 2\\0& 1\end{bmatrix} \] Let eigenvector be \[ \begin{bmatrix}xy\end{bmatrix} \] Then, \[ 2y=0 \] \[ y=0 \] Taking \(x=1\), \[ \begin{bmatrix}10\end{bmatrix} \] So, \[ a=0 \]

Step 2: Eigenvector for \(\lambda=2\). \[ A-2I= \begin{bmatrix}-1& 20& 0\end{bmatrix} \] So, \[ -x+2y=0 \] \[ x=2y \] Taking \(x=1\), \[ y=\frac{1}{2} \] So, \[ b=\frac{1}{2} \]

Step 3: Find \(a+b\). \[ a+b=0+\frac{1}{2} \] \[ a+b=\frac{1}{2} \]

Step 4: Final answer. \[ \boxed{\frac{1}{2}} \]