Question

Question 1

Stonewall receives ¢250 per year in simple interest from an amount of money he invested in ADB, Barclays and GCB. Suppose ADB pays an interest of 2%, Barclays pays an interest of 4% and GCB pays an interest of 5% per annum and an amount of ¢350 more was invested in Barclays than the amount invested in ADB and GCB combined. Also, the amount invested in Barclays is 2 times the amount invested in GCB.

a) Write down the three linear equations and represent them in the matrix form ?? = ?.

b) Find the amount of money Stonewall invested in ADB, Barclays and GCB using Matrix inversion.

Answer 1

Lets x - invested in ADB,

y - invested inBarclays ,

z - invested in GCB.

$0.02x+0.04y+0.05z=250$

$y=350+(x+z)$

$y=2z$

We have a system

$0.02x+0.04y+0.05z=250\\ -x+y-z=350\\ y-2z=0$

$A=\begin{pmatrix} 0.02 & 0.04&0.05 \\ -1& 1&-1\\ 0&1&-2 \end{pmatrix}, X=\begin{pmatrix} x \\ y\\ z \end{pmatrix}, \\B=\begin{pmatrix} 250\\ 350\\ 0 \end{pmatrix}$

$AX=B\\ X=A^{-1}\cdot B\\ A^{-1}:$

$\begin{pmatrix} 0.02 & 0.04&0.05&|1&0&0 \\ -1& 1&-1&|0&1&0\\ 0&1&-2&|0&0&1 \end{pmatrix}\leftrightarrow$

Ir$\leftrightarrow$IIr

$\begin{pmatrix} -1&1&-1&|0&1&0 \\ 0.02&0.04&0.05&|1&0&0\\ 0&1&-2&|0&0&1 \end{pmatrix}\leftrightarrow$

IIr+Ir(0.02)

Ir(-1)

$\begin{pmatrix} 1&-1&1&|0&-1&0 \\ 0&0.06&0.03&|1&0.02&0\\ 0&1&-2&|0&0&1 \end{pmatrix}\leftrightarrow$

IIr$\leftrightarrow$ IIIr

$\begin{pmatrix} 1&-1&1&|0&-1&0 \\ 0&1&-2&|0&0&1\\ 0&0.06&0.03&|1&0.02&0\\ \end{pmatrix}\leftrightarrow$

IIIr+IIr(-0.06)

$\begin{pmatrix} 1&-1&1&|0&-1&0 \\ 0&1&-2&|0&0&1\\ 0&0&0.15&|1&0.02&-0.06\\ \end{pmatrix}\leftrightarrow$

IIIr($\frac{20}{3}$)

$\begin{pmatrix} 1&-1&1&|0&-1&0 \\ 0&1&-2&|0&0&1\\ 0&0&1&|\frac{20}{3}&\frac{2}{15}&-\frac{2}{5}\\ \end{pmatrix}\leftrightarrow$

IIr+IIIr(2)

Ir+IIIr(-1)

$\begin{pmatrix} 1&-1&0&|-\frac{20}{3}&-\frac{17}{15}&\frac{2}{5} \\ 0&1&0&|\frac{40}{3}&\frac{4}{15}&\frac{1}{5}\\ 0&0&1&|\frac{20}{3}&\frac{2}{15}&-\frac{2}{5}\\ \end{pmatrix}\leftrightarrow$

Ir+IIr

$\begin{pmatrix} 1&0&0&|\frac{20}{3}&-\frac{13}{15}&\frac{3}{5} \\ 0&1&0&|\frac{40}{3}&\frac{4}{15}&\frac{1}{5}\\ 0&0&1&|\frac{20}{3}&\frac{2}{15}&-\frac{2}{5}\\ \end{pmatrix}$

$A^{-1}=\begin{pmatrix} \frac{20}{3}&-\frac{13}{15}&\frac{3}{5} \\ \frac{40}{3}&\frac{4}{15}&\frac{1}{5}\\ \frac{20}{3}&\frac{2}{15}&-\frac{2}{5}\\ \end{pmatrix}$

$X=\begin{pmatrix} \frac{20}{3}&-\frac{13}{15}&\frac{3}{5} \\ \frac{40}{3}&\frac{4}{15}&\frac{1}{5}\\ \frac{20}{3}&\frac{2}{15}&-\frac{2}{5}\\ \end{pmatrix}\cdot\begin{pmatrix} 250\\ 350\\ 0 \end{pmatrix}=\\ =\begin{pmatrix} \frac{4090}{3}\\ \frac{10280}{3}\\ \frac{5140}{3} \end{pmatrix}$

x=1363$\frac{1}{3}$

y=3426$\frac{2}{3}$

z=1713$\frac{1}{3}$