Using Matrices, solve the following: A practical Humber student invested 20,000 in two stocks. The two stocks yielded 7% and 11% simple interest in the first year. The total interest received was $1,880. How much does the student invested in each stock?

Question
Matrices
asked 2020-11-27
Using Matrices, solve the following: A practical Humber student invested 20,000 in two stocks. The two stocks yielded 7% and 11% simple interest in the first year. The total interest received was $1,880. How much does the student invested in each stock?

Answers (1)

2020-11-28
Step 1
Let x be the amount invested at 7%
y be the amount invested at 11%
Total amount invested = 20000
Thus, x+y=20000 ...(1)
Interest received at 7% =0.07x
Interest received at 11%=0.11y
Total interest received =1880
Thus, 0.07x+0.11y=1880 ...(2)
step 2
Hence we have two system of equations
x+y=20000
0.07x+0.11y=1880
we solve it by matrices
It can be expressed as
\(\begin{bmatrix}1 & 1 \\0.07 & 0.11 \end{bmatrix}\begin{bmatrix}x \\y \end{bmatrix}=\begin{bmatrix}20000 \\1880 \end{bmatrix}\)
Augmented Matrix for Ax=b is
\(\begin{bmatrix}1 & 1 &20000 \\0.07 & 0.11 &1880 \end{bmatrix}\)
\(R_2 \rightarrow R_2-0.07R_1 \begin{bmatrix}1 & 1&20000 \\0.07 & 0.11&1880 \end{bmatrix}\)
Which can be expressed as
x+y=20000
0.04y=480
\(\Rightarrow y=12000\) And, \(x=20000-12000=8000\) Hence \(\begin{bmatrix}x \\y \end{bmatrix}=\begin{bmatrix}8000 \\12000 \end{bmatrix}\)
Step 3
ANSWER:The student invested 8,000 at 7% interest and 12,000 at 11% interest
0

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