Question 14 Marks
A total amount of ₹7000 is deposited in three different savings bank accounts with annual interest rates of 5%, 8%and $8 \frac{1}{2} \%$ respectively. The total annual interest from these three accounts s is ₹550. Equal amounts have been deposited in the 5% and 8% savings accounts. Find the amount deposited in each of the three accounts, with the help of matrices.
Answer
View full question & answer→Let ₹ x, ₹ y and ₹ z be invested in saving accounts at the rate of 5%, 8% and $8 \frac{1}{2} \%$, respectively.
Then, according to given condition, we have the following system of equations
$
\begin{array}{l}
x+y+z=7000 \ldots(i) \\
\text { and } \frac{5 x}{100}+\frac{8 y}{100}+\frac{17 z}{200}=550 \\
\Rightarrow 10 x+16 x+17 z=110000...(ii)
\end{array}
$
Also, $x - y =0$...(iii)
This system of equations can be written in matrix from as $AX = B$
where, $A =\left[\begin{array}{ccc}1 & 1 & 1 \\ 10 & 16 & 17 \\ 1 & -1 & 0\end{array}\right], X =\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B =\left[\begin{array}{c}7000 \\ 110000 \\ 0\end{array}\right]$
Here, $| A |=\left|\begin{array}{ccc}1 & 1 & 1 \\ 10 & 16 & 17 \\ 1 & -1 & 0\end{array}\right|$
$
\begin{array}{l}
\Rightarrow|A|=1(0+17)-1(0-17)+1(-10-16) \\
=17+17-26=8 \neq 0
\end{array}
$
So, A is non- singular matrix and its inverse exists.
Now, cofactors of elements of $| A |$ are,
$\begin{array}{l}
A_{11}=(-1)^2\left|\begin{array}{cc}
16 & 17 \\
-1 & 0
\end{array}\right|=1(0+17)=17 \\
A_{12}=(-1)^3\left|\begin{array}{cc}
10 & 17 \\
1 & 0
\end{array}\right|=-1(0-17)=17 \\
A_{13}=(-1)^4\left|\begin{array}{cc}
10 & 16 \\
1 & -1
\end{array}\right|=1(-10-16)=-26 \\
A_{21}=(-1)^3\left|\begin{array}{cc}
1 & 1 \\
-1 & 0
\end{array}\right|=-1(0+1)=-1 \\
A_{22}=(-1)^4\left|\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right|=1(0-1)=-1 \\
A_{23}=(-1)^5\left|\begin{array}{cc}
1 & 1 \\
1 & -1
\end{array}\right|=-1(-1-1)=2 \\
A_{31}=(-1)^4\left|\begin{array}{cc}
1 & 1 \\
16 & 17
\end{array}\right|=1(17-16)=1 \\
A_{32}=(-1)^5\left|\begin{array}{cc}
1 & 1 \\
10 & 17
\end{array}\right|=-1(17-10)=-7 \\
A_{33}=(-1)^6\left|\begin{array}{cc}
1 & 1 \\
10 & 16
\end{array}\right|=1(16-10)=6 \\
\therefore \operatorname{adj}(A)=\left[\begin{array}{lll}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
\end{array}\right]^T \\
=\left[\begin{array}{ccc}
17 & 17 & -26 \\
-1 & -1 & 2 \\
1 & -7 & 6
\end{array}\right]^T=\left[\begin{array}{ccc}
17 & -1 & 1 \\
17 & -1 & -7 \\
-26 & 2 & 6
\end{array}\right]
\end{array}
$
Now, $A ^{-1}=\frac{\operatorname{adj}(A)}{|A|}=\frac{1}{8}\left[\begin{array}{ccc}17 & -1 & 1 \\ 17 & -1 & -7 \\ -26 & 2 & 6\end{array}\right]$
and the solution of given system is given by
$
\begin{array}{l}
X=A^{-1} B . \\
\Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{8}\left[\begin{array}{ccc}
17 & -1 & 1 \\
17 & -1 & -7 \\
-26 & 2 & 6
\end{array}\right]\left[\begin{array}{c}
7000 \\
110000 \\
0
\end{array}\right] \\
=\frac{1}{8}\left[\begin{array}{c}
119000-110000+0 \\
119000-110000+0 \\
-182000+220000+0
\end{array}\right]
\end{array}
$
On comparing the corresponding elements, we get x = 1125, y = 1125, z = 4750.
Hence, the amount deposited in each type of account is ₹1125, ₹1125 and ₹4750, respectively.
Then, according to given condition, we have the following system of equations
$
\begin{array}{l}
x+y+z=7000 \ldots(i) \\
\text { and } \frac{5 x}{100}+\frac{8 y}{100}+\frac{17 z}{200}=550 \\
\Rightarrow 10 x+16 x+17 z=110000...(ii)
\end{array}
$
Also, $x - y =0$...(iii)
This system of equations can be written in matrix from as $AX = B$
where, $A =\left[\begin{array}{ccc}1 & 1 & 1 \\ 10 & 16 & 17 \\ 1 & -1 & 0\end{array}\right], X =\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B =\left[\begin{array}{c}7000 \\ 110000 \\ 0\end{array}\right]$
Here, $| A |=\left|\begin{array}{ccc}1 & 1 & 1 \\ 10 & 16 & 17 \\ 1 & -1 & 0\end{array}\right|$
$
\begin{array}{l}
\Rightarrow|A|=1(0+17)-1(0-17)+1(-10-16) \\
=17+17-26=8 \neq 0
\end{array}
$
So, A is non- singular matrix and its inverse exists.
Now, cofactors of elements of $| A |$ are,
$\begin{array}{l}
A_{11}=(-1)^2\left|\begin{array}{cc}
16 & 17 \\
-1 & 0
\end{array}\right|=1(0+17)=17 \\
A_{12}=(-1)^3\left|\begin{array}{cc}
10 & 17 \\
1 & 0
\end{array}\right|=-1(0-17)=17 \\
A_{13}=(-1)^4\left|\begin{array}{cc}
10 & 16 \\
1 & -1
\end{array}\right|=1(-10-16)=-26 \\
A_{21}=(-1)^3\left|\begin{array}{cc}
1 & 1 \\
-1 & 0
\end{array}\right|=-1(0+1)=-1 \\
A_{22}=(-1)^4\left|\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right|=1(0-1)=-1 \\
A_{23}=(-1)^5\left|\begin{array}{cc}
1 & 1 \\
1 & -1
\end{array}\right|=-1(-1-1)=2 \\
A_{31}=(-1)^4\left|\begin{array}{cc}
1 & 1 \\
16 & 17
\end{array}\right|=1(17-16)=1 \\
A_{32}=(-1)^5\left|\begin{array}{cc}
1 & 1 \\
10 & 17
\end{array}\right|=-1(17-10)=-7 \\
A_{33}=(-1)^6\left|\begin{array}{cc}
1 & 1 \\
10 & 16
\end{array}\right|=1(16-10)=6 \\
\therefore \operatorname{adj}(A)=\left[\begin{array}{lll}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
\end{array}\right]^T \\
=\left[\begin{array}{ccc}
17 & 17 & -26 \\
-1 & -1 & 2 \\
1 & -7 & 6
\end{array}\right]^T=\left[\begin{array}{ccc}
17 & -1 & 1 \\
17 & -1 & -7 \\
-26 & 2 & 6
\end{array}\right]
\end{array}
$
Now, $A ^{-1}=\frac{\operatorname{adj}(A)}{|A|}=\frac{1}{8}\left[\begin{array}{ccc}17 & -1 & 1 \\ 17 & -1 & -7 \\ -26 & 2 & 6\end{array}\right]$
and the solution of given system is given by
$
\begin{array}{l}
X=A^{-1} B . \\
\Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{8}\left[\begin{array}{ccc}
17 & -1 & 1 \\
17 & -1 & -7 \\
-26 & 2 & 6
\end{array}\right]\left[\begin{array}{c}
7000 \\
110000 \\
0
\end{array}\right] \\
=\frac{1}{8}\left[\begin{array}{c}
119000-110000+0 \\
119000-110000+0 \\
-182000+220000+0
\end{array}\right]
\end{array}
$
On comparing the corresponding elements, we get x = 1125, y = 1125, z = 4750.
Hence, the amount deposited in each type of account is ₹1125, ₹1125 and ₹4750, respectively.
