Question

An amount of ₹ 5000 is put into three investments at 6%, 7% and 8% per annum respectively. The total annual income from these investments is ₹358. If the total annual income from first two investments is ₹70 more than the income from the third, find the amount of each investment by the matrix method. 

Answer 1

Let these investments be ₹x, ₹y and ₹z, respectively. 

Then, x + y + z = 5000

6x/100 + 7y/100 + 8z/100 = 358

6x + 7y + 8z = 35800

And, 6x/100 + 7y/100 = 8z/100 + 70

6x + 7y - 8z = 7000. 

Representing in the matrix form, 

AX = B

\(\begin{bmatrix}1&1&1\\0&1&2\\0&0&-16\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5000\\5800\\-28800\end{bmatrix}\)

\(R_3\rightarrow\frac{R_3}{-16}\)

\(\begin{bmatrix}1&1&1\\0&1&2\\0&0&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5000\\5800\\1800\end{bmatrix}\)

\(R_1\rightarrow R_1-R_2\)

\(\begin{bmatrix}1&0&-1\\0&1&2\\0&0&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-800\\5800\\1800\end{bmatrix}\)

\(R_1\rightarrow R_1+R_3\)

\(R_2\rightarrow R_2-2R_3\)

\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1000\\2200\\1800\end{bmatrix}\)

\(\Rightarrow\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1000\\2200\\1800\end{bmatrix}\)

\(\Rightarrow\) x = 1000,

y = 2200

and z = 1800.