#### Provide solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 10

\begin{aligned} &x=2000, y=3000, z=5000 \end{aligned}

Given:

Total investment=10000

Let x, y, z be the investments,

\begin{aligned} &x+y+z=10000 \\ &0.1 x+0.12 y+0.15 z=1310 \\ &-0.1 x-0.12 y-0.15 z=190 \end{aligned}

Hint:

X=A-1B is used for the problem

Solution:

Let x, y, z be the investments at the rates of interest of 10%,12%,13% per annum respectively

Total investment=10000

\begin{aligned} &x+y+z=10000 \\ \end{aligned}

Income from the 1st investment of Rs x

$=\operatorname{Rs} \frac{10 x}{100}=0.1 x$

Income from 2nd investment of Rs y

$=\operatorname{Rs} \frac{12y}{100}=0.12 y$

Income from 3rd investment of Rs z

$=\operatorname{Rs} \frac{15z}{100}=0.15 z$

\begin{aligned} &\therefore \text { Total annual income }=\operatorname{Rs}(0.1 x+0.12 y+0.15 z)\\ &0.1 x+0.12 y+0.15 z=1310 \end{aligned}

Its given that the combined income from the first two income is Rs190 short of the income from the third.

\begin{aligned} &0.1 x+0.12 y+0.15 z=190 \\ &-0.1 x-0.12 y+0.15 z=190 \\ &\text { Thus, } x+y+z=10000 \\ &0.1 x+0.12 y+0.15 z=1310 \\ &-0.1 x-0.12 y+0.15 z=190 \end{aligned}

\begin{aligned} &{\left[\begin{array}{ccc} 1 & 1 & 1 \\ 0.1 & 0.12 & 0.15 \\ -0.1 & -0.12 & 0.15 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 10000 \\ 1310 \\ 190 \end{array}\right]} \\ &A X=B \\ &A=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 0.1 & 0.12 & 0.15 \\ -0.1 & -0.12 & 0.15 \end{array}\right] \end{aligned}

\begin{aligned} &|A|=1(0.15 \times 0.12+0.15 \times 0.12)-1(0.15 \times 0.1+0.15 \times 0.1)+1(-0.1 \times 0.12+0.12 \times 0.1) \\ &|A|=0.036-0.0370 \\ &|A|=0.006 \end{aligned}

Let i,j be the cofactors of the elements

\begin{aligned} &a_{i j} \text { in } A=\left[a_{i j}\right] \\ &c_{11}=(-1)^{1+1}\left|\begin{array}{cc} 0.12 & 0.15 \\ -0.12 & 0.15 \end{array}\right|=0.036, c_{12}=(-1)^{1+2}\left|\begin{array}{cc} 0.1 & 0.15 \\ -0.1 & 0.15 \end{array}\right|=-0.036 \\ &c_{21}=(-1)^{2+1}\left|\begin{array}{cc} 1 & 1 \\ -0.12 & 0.15 \end{array}\right|=-0.27, c_{22}=(-1)^{2+2}\left|\begin{array}{cc} 1 & 1 \\ -0.1 & 0.15 \end{array}\right|=0.25 \end{aligned}

\begin{aligned} &c_{31}=(-1)^{3+1}\left|\begin{array}{cc} 1 & 1 \\ 0.12 & 0.15 \end{array}\right|=0.3, c_{32}=(-1)^{3+2}\left|\begin{array}{cc} 1 & 1 \\ 0.1 & 0.15 \end{array}\right|=-0.05 \\ &c_{13}=(-1)^{1+3}\left|\begin{array}{cc} 1 & 0.12 \\ -0.1 & -0.12 \end{array}\right|=0, c_{23}=(-1)^{2+3}\left|\begin{array}{cc} 1 & 1 \\ -0.1 & -0.12 \end{array}\right|=0.02 \\ &c_{33}=(-1)^{3+3}\left|\begin{array}{cc} 1 & 1 \\ 0.1 & 0.12 \end{array}\right|=0.02 \end{aligned}

\begin{aligned} &a d j A=\left[\begin{array}{ccc} 0.036 & -0.03 & 0 \\ -0.27 & 0.25 & 0.02 \\ 0.03 & -0.05 & 0.02 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 0.036 & -0.27 & 0.03 \\ -0.03 & 0.25 & -0.05 \\ 0 & 0.02 & 0.02 \end{array}\right] \end{aligned}

\begin{aligned} &A^{-1}=\frac{1}{|A|} a d j A \\ &=\frac{1}{0.006}\left[\begin{array}{ccc} 0.036 & -0.27 & 0.03 \\ -0.03 & 0.25 & -0.05 \\ 0 & 0.02 & 0.02 \end{array}\right] \\ &X=A^{-1} B \end{aligned}

\begin{aligned} &X=\left[\begin{array}{ccc} 0.036 & -0.27 & 0.03 \\ -0.03 & 0.25 & -0.05 \\ 0 & 0.02 & 0.02 \end{array}\right]\left[\begin{array}{c} 10000 \\ 1310 \\ 190 \end{array}\right] \\ &=\frac{1}{0.006}\left[\begin{array}{c} 360-353.7+5.7 \\ -300+327.5-9.5 \\ 0+26.2+3.8 \end{array}\right] \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1000}{6}\left[\begin{array}{l} 12 \\ 18 \\ 30 \end{array}\right]=\left[\begin{array}{c} 2000 \\ 3000 \\ 5000 \end{array}\right]} \\ &x=2000, y={{3000}, z=5000} \end{aligned}

Thus the three investments are of Rs 2000, Rs.3000, Rs 5000.