GATE 2016 | MATHS | Q-48

Question

Let \( M= \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \) and \( e^M = Id + M + \frac{M^2}{2!} + \frac{M^3}{3!} + \frac{M^4}{4!} + \ldots \). If \( e^M =  [b_{ij}]\). then

\[\frac{1}{e} \sum_{i=1}^{3} \sum_{j=1}^{3} b_{ij}
\]

is equal to ________________________

Answer 1

notice that $$\frac{1}{e} \sum_{i=1}^3\sum_{j=1}^3 b_{ij} = \text{sum of entries of }e^{M-1}$$

so $$M’ = M-I = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$

$$(M’)^2 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\  0& 0 & 0 \end{bmatrix}$$

$$(M’)^3 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\  0& 0 & 0 \end{bmatrix} \ so \ (M’)^n = 0 \ where \ n\geq3 $$

so $$e^{M-1} = e^{M’} = I + M’ + (M’)^2/2  + 0  = \begin{bmatrix} 1 & 1 & 0.5 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} $$

so $$ans = 1+1+1+1+1+0.5=5.5$$