Question

A trust fund has Rs$30,000$ that must be invested in two different types of bounds. The first bond pays $5\% $ interest per year and the second pays $7\% $ interest per year. Using matrix multiplication, determine how to divide Rs$30,000$ among the two types of bounds if the trust fund must obtain an annual total interest of Rs$1800$.

Answer 1

Hint:Assume investment on one of the bonds as a variable. Then you automatically get the investment on the second bond. After that form two matrices one for investments and one for interests per year. At last, get the annual interest and equate it to the value given in the question.

Complete step-by-step answer:
Let the investment on first bond be Rs $x$
Total investment$ = $Rs$30,000$
Investment on second bond$ = $Rs$\left( {30,000 - x} \right)$
Now let us represent investment per bond by the matrix $A$
$A = \left[ {\begin{array}{*{20}{c}}
  x \\
  {30000 - x}
\end{array}} \right]$
$A$ is a $2 \times 1$ matrix which means it has $2$ rows and $1$ columns.
Now we will represent interest per year by a matrix.
Interest paid by the first bond$ = 5\% $
Interest paid by the second bond$ = 7\% $.
$B = \left[ {\begin{array}{*{20}{c}}
  {5\% }&{7\% }
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{5}{{100}}}&{\dfrac{7}{{100}}}
\end{array}} \right]$
$B$ is the $1 \times 2$ matrix which means it has $1$ row and 2 columns.
Now the question arises why we took $B$ as the $1 \times 2$ matrix and not $2 \times 1$ like $A$
This answer is to the fact that matrix multiplication takes place only when the number of columns of the first matrix is equal to the number of rows of the second matrix.
As it is said that we have to solve it by matrix multiplication, therefore it is necessary to take the order of matrices $A$ and $B$ in such a way that matrix multiplication is possible.
Now total annual interest$ = $ interest per bond $ \times $ investment per bond
$1800 = {\left[ {\begin{array}{*{20}{c}}
  {\dfrac{5}{{100}}}&{\dfrac{7}{{100}}}
\end{array}} \right]_{1 \times 2}} \times {\left[ {\begin{array}{*{20}{c}}
  x \\
  {30000 - x}
\end{array}} \right]_{2 \times 1}}$
             $ = {\left[ {\dfrac{{5x}}{{100}} + \dfrac{7}{{100}}\left( {30000 - x} \right)} \right]_{1 \times 1}}$
             $ = \dfrac{{5x + 21000 - 7x}}{{100}}$
$180000 = 210000 - 2x$
$2x = 30000$
$x = 15000$
Hence amount invested at $5\% = $ Rs$15000$
Amount invested at $7\% =$Rs$\left( {30,000 - x} \right)$= $ Rs$15000

Note:The most important part about this question is to choose the order of the matrix for matrix multiplication. Students choose the wrong order of the matrix and are not able to multiply the matrices or multiply them in the wrong way.
Also it should be known to you that:
Total annual interest $ = $ Interest $ \times $ investment.