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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

Also it should be known to you that:

Total annual interest $ = $ Interest $ \times $ investment.