Q.(a) If the payment of ₹ 2,000 is made at the end of every quarter for 10 years at the rate of 8% per year, then find the amount of annuity. $[(1.02)^{40} = 2.2080]$

(a) Amount of the annuity $= 2000\cdot\dfrac{(1.02)^{40}-1}{0.02} = ₹1{,}20{,}800$. (b) By matrix inversion ($\det A=50$), onion ₹20/kg, wheat ₹40/kg, rice ₹60/kg.

Part (a): Amount (future value) of an annuity. Payment $P=₹2000$ at the end of every quarter, 10 years, 8% per year.

Step 1 — Quarterly rate and periods. $i = \dfrac{8\%}{4} = 2\% = 0.02$; $n = 10\times4 = 40$.

Step 2 — Amount formula (immediate annuity).

$A = P\cdot\frac{(1+i)^n - 1}{i} = 2000\cdot\frac{(1.02)^{40} - 1}{0.02}.$

Step 3 — Substitute $(1.02)^{40} = 2.2080$.

$A = 2000\cdot\frac{2.2080 - 1}{0.02} = 2000\cdot\frac{1.2080}{0.02} = 2000 \times 60.4 = 120800.$

Part (b): Matrix-inversion method. Let onion $=x$, wheat $=y$, rice $=z$ (₹/kg). The equations are

$4x+3y+2z=320,\quad 2x+4y+6z=560,\quad 6x+2y+3z=380.$

So $A=\begin{pmatrix}4&3&2\\2&4&6\\6&2&3\end{pmatrix},\ B=\begin{pmatrix}320\\560\\380\end{pmatrix}$, and $X=A^{-1}B$.

Step 1 — Determinant.

$\det A = 4(4\cdot3-6\cdot2) - 3(2\cdot3-6\cdot6) + 2(2\cdot2-4\cdot6) = 4(0) - 3(-30) + 2(-20) = 50.$

Step 2 — Adjoint (transpose of the cofactor matrix):

$\text{adj }A = \begin{pmatrix}0 & -5 & 10\\ 30 & 0 & -20\\ -20 & 10 & 10\end{pmatrix}.$

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