Determinants and Matrices — Maths STD 11 Science — Question

$\therefore \quad A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{52}\end{array}\right]$

$a_{i j}=\frac{(i-j)^2}{5-i}$

$\therefore \quad a_{11}=\frac{(1-1)^2}{5-1}=0, a_{12}=\frac{(1-2)^2}{5-1}=\frac{(-1)^2}{4}=\frac{1}{4}$,

$a_{21}=\frac{(2-1)^2}{5-2}=\frac{1}{3}, a_{22}=\frac{(2-2)^2}{5-2}=0$

$a_{31}=\frac{(3-1)^2}{5-3}=\frac{2^2}{2}=2, a_{32}=\frac{(3-2)^2}{5-3}=\frac{1}{2}$

$\therefore \quad A=\left[\begin{array}{ll}0 & \frac{1}{4} \\ \frac{1}{3} & 0 \\ 2 & \frac{1}{2}\end{array}\right]$

[Note: Answer given in the textbook is $A=\left[\begin{array}{cc}0 & \frac{1}{4} \\ \frac{1}{2} & 0 \\ 2 & \frac{1}{2}\end{array}\right]$

However, as per our calculation it is $\left[\begin{array}{cc}0 & \frac{1}{4} \\ \frac{1}{3} & 0 \\ 2 & \frac{1}{2}\end{array}\right]$.