Chapter 3 · Matrices

Exercise 3.2 · Q19: Trust fund bonds

Exercise 3.2Multiplication of matrices · Section 3.4

The question

A trust fund has ₹30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

(a) ₹1800 (b) ₹2000

Exam answer

Let ₹ $x$ be invested in the first bond. Then ₹ $(30000 - x)$ is invested in the second.

Total annual interest (₹):

\begin{aligned} &\begin{bmatrix} x & 30000 - x \end{bmatrix}\begin{bmatrix} \frac{5}{100} \\[1mm] \frac{7}{100} \end{bmatrix} \\ &= \begin{bmatrix} \frac{5x}{100} + \frac{7(30000 - x)}{100} \end{bmatrix} \\ &= \begin{bmatrix} \frac{210000 - 2x}{100} \end{bmatrix} \end{aligned}

(a) Equating with $[1800]$ ⁠,

\begin{aligned} \frac{210000 - 2x}{100} &= 1800 \\ \Rightarrow 210000 - 2x &= 180000 \\ \Rightarrow x &= 15000 \end{aligned}

Second bond: $30000 - 15000 = 15000$ ⁠.

$\therefore$ ₹15,000 in the first bond and ₹15,000 in the second bond.

(b) Equating with $[2000]$ ⁠,

\begin{aligned} \frac{210000 - 2x}{100} &= 2000 \\ \Rightarrow 210000 - 2x &= 200000 \\ \Rightarrow x &= 5000 \end{aligned}

Second bond: $30000 - 5000 = 25000$ ⁠.

$\therefore$ ₹5,000 in the first bond and ₹25,000 in the second bond.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Total interest is "amount in bond 1 times its rate, plus amount in bond 2 times its rate". That is exactly a row-times-column product: a $1\times2$ row matrix of amounts times a $2\times1$ column matrix of rates gives a $1\times1$ matrix holding the total interest. With one unknown amount $x$ ⁠, the product gives one linear equation, which we solve.

Given, to find, and the plan

Given: Total ₹30,000; rates 5% and 7% per year; required yearly interest ₹1800 in (a), ₹2000 in (b).
To find: How much goes into each bond, in each case.
Plan: (1) Name the unknown and build the amount matrix and the rate matrix, with labels. (2) Check the orders and explain the product order. (3) Multiply to get the interest as a $1\times1$ matrix. (4) Set it equal to the target and solve, for (a) and (b). (5) Check the interest earned and interpret.

Basics you need

1A table becomes a matrix

Put one kind of quantity along a row and the matching rates down a column; labelling rows and columns tells you what each entry means. Basics: rows, columns and addresses

2Row times column

A $1\times n$ row times an $n\times1$ column is the $1\times1$ matrix $[a_1b_1 + \dots + a_nb_n]$ ⁠. Basics: pair up, multiply, add

3Percent as a fraction

5% means $\frac{5}{100}$ ⁠, so the yearly interest on ₹ $P$ at 5% is ₹ $\frac{5P}{100}$ ⁠.

4Solving a linear equation

Clear fractions by multiplying both sides by the same non-zero number, collect $x$ ⁠, divide by its coefficient. Basics: solving equations

Solution

Name the unknown and build the matrices

Let ₹ $x$ be invested in the first bond. Since the whole ₹30,000 is invested, the second bond gets ₹ $(30000 - x)$ ⁠. For the split to make sense, $0 \le x \le 30000$ ⁠.

Bond 1 Bond 2

Amount invested (₹) $x$ $30000 - x$

Interest rate per year $\frac{5}{100}$ $\frac{7}{100}$

The amounts form a row matrix $X$ (one row, columns labelled Bond 1, Bond 2), and the rates form a column matrix $R$ (rows labelled Bond 1, Bond 2):
$X = \left[\;x \quad 30000 - x\;\right], \quad R = \begin{bmatrix} \frac{5}{100} \\[1mm] \frac{7}{100} \end{bmatrix}$
Using one unknown $x$ and writing the other amount as $30000 - x$ builds the condition "total is ₹30,000" straight into the matrix, so only the interest condition is left to impose.

A row matrix is $1\times n$ ⁠; a column matrix is $n\times1$ ⁠. Lesson: types of matrices
Check the orders and choose the product order

$X$ is $1\times2$ and $R$ is $2\times1$ ⁠. The columns of $X$ (2) equal the rows of $R$ (2), so $XR$ is defined, of order $1\times1$ ⁠: a single number in brackets, the total interest in rupees.

We want each amount multiplied by its own rate and the results added. In $XR$ ⁠, the Bond 1 column of $X$ meets the Bond 1 row of $R$ ⁠, and Bond 2 meets Bond 2, which is exactly that pairing. The other order, $RX$ ⁠, is also defined but is $2\times2$ ⁠, and its off-diagonal entries such as $\frac{5}{100}(30000 - x)$ multiply a bond 2 amount by the bond 1 rate, which means nothing here.

$(1\times2)(2\times1) = 1\times1$ ⁠. Lesson: matrix multiplication
Multiply to get the total interest
$XR = \left[\;x \quad 30000 - x\;\right]\begin{bmatrix} \frac{5}{100} \\[1mm] \frac{7}{100} \end{bmatrix}$
The only entry is row 1 of $X$ times column 1 of $R$ ⁠:
$\begin{aligned} c_{11} &= x\cdot\frac{5}{100} + (30000 - x)\cdot\frac{7}{100}\\ &= \frac{5x + 210000 - 7x}{100}\\ &= \frac{210000 - 2x}{100} \end{aligned}$
So the yearly interest is ₹ $\frac{210000 - 2x}{100}$ ⁠, that is,
$XR = \left[\frac{210000 - 2x}{100}\right].$
$7 \times 30000 = 210000$ ⁠, and $5x - 7x = -2x$ ⁠. Keeping the rates as fractions over 100 avoids decimal slips.

	Bond 1	Bond 2
Amount invested (₹)	$x$	$30000 - x$
Interest rate per year	$\frac{5}{100}$	$\frac{7}{100}$

Part (a): total interest ₹1800

Set the product equal to $[1800]$ and solve
$\left[\frac{210000 - 2x}{100}\right] = \begin{bmatrix} 1800 \end{bmatrix}$
Two $1\times1$ matrices are equal when their single entries are equal:
$\begin{aligned} \frac{210000 - 2x}{100} &= 1800\\ 210000 - 2x &= 180000\\ -2x &= -30000\\ x &= 15000 \end{aligned}$
So bond 1 gets ₹15,000 and bond 2 gets ₹ $30000 - 15000$ = ₹15,000.

Multiply both sides by 100 to clear the fraction, subtract 210000 from both sides, then divide by $-2$ ⁠. The value $x = 15000$ lies in $0 \le x \le 30000$ ⁠, so both amounts are genuine (non-negative) investments.

Equality of matrices: corresponding entries equal (Definition 2). Basics: one unknown at a time
Check and interpret

Interest: $15000\cdot\frac{5}{100} = 750$ and $15000\cdot\frac{7}{100} = 1050$ ⁠; total $750 + 1050 = 1800$ ✓.

In words: to earn ₹1800 a year, the trust should put ₹15,000 in the 5% bond and ₹15,000 in the 7% bond.

Part (b): total interest ₹2000

Set the product equal to $[2000]$ and solve
$\left[\frac{210000 - 2x}{100}\right] = \begin{bmatrix} 2000 \end{bmatrix}$ $\begin{aligned} \frac{210000 - 2x}{100} &= 2000\\ 210000 - 2x &= 200000\\ -2x &= -10000\\ x &= 5000 \end{aligned}$
So bond 1 gets ₹5,000 and bond 2 gets ₹ $30000 - 5000$ = ₹25,000.

The same steps as in Part (a), Step 1. A higher target interest needs more money in the 7% bond, so $x$ (the 5% amount) goes down. Again $0 \le 5000 \le 30000$ ⁠, so the split is possible.
Check and interpret

Interest: $5000\cdot\frac{5}{100} = 250$ and $25000\cdot\frac{7}{100} = 1750$ ⁠; total $250 + 1750 = 2000$ ✓.

In words: to earn ₹2000 a year, the trust should put ₹5,000 in the 5% bond and ₹25,000 in the 7% bond.

Why both targets are reachable: the interest $\frac{210000 - 2x}{100}$ falls steadily as $x$ goes from 0 to 30000, from ₹2100 (all in the 7% bond) down to ₹1500 (all in the 5% bond). Any target between ₹1500 and ₹2100 has exactly one split; ₹1800 and ₹2000 are both in that range.

Result

(a) For ₹1800 a year: ₹15,000 in the first (5%) bond and ₹15,000 in the second (7%) bond.
(b) For ₹2000 a year: ₹5,000 in the first (5%) bond and ₹25,000 in the second (7%) bond.
(Textbook answer: (a) ₹15000, ₹15000; (b) ₹5000, ₹25000.)

Check it with numbers

Use the matrix product directly with the answers, writing the rates as decimals 0.05 and 0.07.

\begin{aligned} \text{(a)}&: 15000(0.05) + 15000(0.07)\\ &= 750 + 1050 = 1800\\ \text{(b)}&: 5000(0.05) + 25000(0.07)\\ &= 250 + 1750 = 2000 \end{aligned}

Both match the targets ✓, and in both cases the two amounts add to 30,000 ✓.

Common mistakes

Using 5 and 7 instead of $\frac{5}{100}$ and $\frac{7}{100}$ ⁠. The interest would come out 100 times too large.
Two unknowns but one equation. If you call the amounts $x$ and $y$ ⁠, you also need $x + y = 30000$ ⁠; writing $30000 - x$ uses it automatically.
Computing $RX$ ⁠. It is $2\times2$ and mixes each amount with the wrong rate; the total interest is the $1\times1$ product $XR$ ⁠.
Forgetting to answer both amounts. The question asks how to divide the money, so give both shares in each part.

What to take away

"Quantity times rate, summed" is a row matrix times a column matrix; label the rows and columns so the matching pairs line up.
Build any fixed total into the matrix (here $30000 - x$ ⁠) so you are left with one unknown.
After solving, check the answer makes sense (non-negative amounts) and recompute the interest from it.

Every fact used, and where it comes from

Fact	Used in	Source
Row and column matrices	Building the matrices	Section 3.3 (i), (ii) (Lesson: types of matrices)
$(1\times2)(2\times1)$ product is $1\times1$ ⁠; entry = row times column	Orders; the product	Section 3.4.5 (Lesson: multiplication)
Equal $1\times1$ matrices have equal entries	Part (a), Step 1; Part (b), Step 1	Definition 2, Section 3.3.1
Simple interest for one year at $r$ % on ₹ $P$ is ₹ $\frac{Pr}{100}$	Building the matrices; the product	Meaning of percent
A word problem written as a matrix product	Whole page	Example 19 (election campaign costs)
All amounts and interests	Every step, Check	Recomputed in the site's verification script

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Exercise 3.2 · Q19: Trust fund bonds

Basics you need

Solution

Name the unknown and build the matrices

Check the orders and choose the product order

Multiply to get the total interest

Part (a): total interest ₹1800

Set the product equal to [1800][1800][1800] and solve

Check and interpret

Part (b): total interest ₹2000

Set the product equal to [2000][2000][2000] and solve

Check and interpret

Every fact used, and where it comes from

Set the product equal to $[1800]$ and solve

Set the product equal to $[2000]$ and solve