(a) Cramer's rule gives investments ₹250, ₹4000, ₹4250. (b) Least-squares trend $Y=169.43+3.29X$ with origin 1998.

Part (a) — Cramer's rule. Let $x,y,z$ be the amounts (₹) invested at 2%, 3%, 6%.

Equations (interest equation $\times100$):

$x+y+z=8500,$

$2x+3y+6z=38000,$

$x+y-z=0\quad(\text{since }z=x+y).$

Determinant $\Delta$:

$\Delta=\begin{vmatrix}1&1&1\\2&3&6\\1&1&-1\end{vmatrix}=1(-3-6)-1(-2-6)+1(2-3)=-9+8-1=-2.$

$\Delta_x$ (replace column 1):

$\Delta_x=\begin{vmatrix}8500&1&1\\38000&3&6\\0&1&-1\end{vmatrix}=8500(-9)-1(-38000)+1(38000)=-76500+38000+38000=-500.$

$\Delta_y$ (replace column 2):

$\Delta_y=\begin{vmatrix}1&8500&1\\2&38000&6\\1&0&-1\end{vmatrix}=1(-38000)-8500(-8)+1(-38000)=-38000+68000-38000=-8000.$

$\Delta_z$ (replace column 3):

$\Delta_z=\begin{vmatrix}1&1&8500\\2&3&38000\\1&1&0\end{vmatrix}=1(-38000)-1(-38000)+8500(-1)=-8500.$

Solutions:

$x=\frac{\Delta_x}{\Delta}=\frac{-500}{-2}=250,\quad y=\frac{-8000}{-2}=4000,\quad z=\frac{-8500}{-2}=4250.$

Check: $250+4000+4250=8500$ ✓; interest $=0.02(250)+0.03(4000)+0.06(4250)=5+120+255=380$ ✓; $z=x+y$ ✓.

Part (b) — straight-line trend by least squares. Take origin at the middle year 1998, $X=\text{year}-1998$, so $\sum X=0$.

| Year | $X$ | $Y$ | $XY$ | $X^2$ |

| --- | --- | --- | --- | --- | …