A trust fund has ₹ 30,000 that must be invested in two different types of bonds. The first bond pays 5% intere

Exercise 3.2 · Q19

Q. 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

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Concept understanding — Matrix Equation Solving

Solving a System of Equations by the Matrix Method

A system of linear equations can be written as a single matrix equation and solved in one clean step using the inverse of a matrix. This is the Class-12 "matrix method" for simultaneous equations.

Writing the system as $AX=B$

Take the system

a_1x+b_1y+c_1z=d_1,\quad a_2x+b_2y+c_2z=d_2,\quad a_3x+b_3y+c_3z=d_3.

Collect the coefficients, the unknowns, and the constants into matrices:

A=\begin{pmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{pmatrix},\quad X=\begin{pmatrix}x\\y\\z\end{pmatrix},\quad B=\begin{pmatrix}d_1\\d_2\\d_3\end{pmatrix}.

Then the whole system is just

AX=B.

Solving when $A$ is invertible

If $\det(A)\neq0$ , then $A^{-1}$ exists, and multiplying both sides on the left by $A^{-1}$ gives

X=A^{-1}B,\qquad\text{where }\;A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A).

So you compute $\det(A)$ , then $\operatorname{adj}(A)$ , form $A^{-1}$ , and multiply by $B$ . The single column $X=A^{-1}B$ hands you $x$ , $y$ , $z$ at once, and because $A^{-1}$ is unique, the solution is unique.

Watch out

Multiply in the correct order: $X=A^{-1}B$ , not $BA^{-1}$ . Matrix multiplication is not commutative, and $BA^{-1}$ is not even defined here.

When $\det(A)=0$

If $\det(A)=0$ , $A^{-1}$ does not exist and the inverse method fails. The system is then either inconsistent (no solution) or has infinitely many solutions. Decide which by computing $(\operatorname{adj}A)\,B$ :

$(\operatorname{adj}A)\,B\neq O$ → no solution (inconsistent).
$(\operatorname{adj}A)\,B=O$ → infinitely many solutions (consistent, dependent). …

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Write the two conditions (total investment and total interest) as a matrix equation $A\mathbf{x}=\mathbf{b}$ and solve using $\mathbf{x}=A^{-1}\mathbf{b}$ . (a) For ₹1800 interest: ₹15,000 in the 5% bond and ₹15,000 in the 7% bond. (b) For ₹2000 interest: ₹5,000 in the 5% bond and ₹25,000 in the 7% bond.

Setting it up

Let $x$ = amount in the first bond (5%) and $y$ = amount in the second bond (7%). The two conditions are:

x + y = 30000, \qquad 0.05x + 0.07y = I,

where $I$ is the target interest. In matrix form:

\begin{bmatrix} 1 & 1 \\ 0.05 & 0.07 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 30000 \\ I \end{bmatrix}.

Inverting the coefficient matrix

For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , $A^{-1} = \dfrac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ .

Here $\det(A) = (1)(0.07) - (1)(0.05) = 0.02 \neq 0$ , so

A^{-1} = \frac{1}{0.02}\begin{bmatrix} 0.07 & -1 \\ -0.05 & 1 \end{bmatrix} = \begin{bmatrix} 3.5 & -50 \\ -2.5 & 50 \end{bmatrix}.

Thus …

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Concept understanding — Matrix Equation Solving

Solving a System of Equations by the Matrix Method

Writing the system as $AX=B$

Solving when $A$ is invertible

When $\det(A)=0$

Setting it up

Inverting the coefficient matrix

Method: Modelling a word problem as a matrix equation $AX=B$

Steps

Common Mistakes

Setting it up

Inverting the coefficient matrix

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