Question

The sum of two numbers is 64. Twice the larger number plus five times the smaller number is 20. Find the two numbers.

Answer 1

Let x and y be the two numbers and x > y.


From your story:

$$x+y = 64;$$

$$2x +5y = 20$$

which gives $$ \{\{x\to 100,y\to -36\}\}$$

Answer 2

You can use simple row reduction to work out these problems.

a + b = 64
2a + 5b = 20

[1 1][64]
[2 5][20]

[1 1][64]
[0 3][-108] R2 - 2R1

You already have your answer here (-108/3 = -36, 64--36 = 100) but I'll finish anyway:

[1 1][64]
[0 1][-36] R2/3

[1 0][100] R1 - R2
[0 1][-36]

a = 100, b = -36

Answer 3

^^ That is not the solution of someone who have few screw missing

Answer 4

btw is this a kind of short cut for the using finding the inverse and then multiply ?

Answer 5

Basically, yes. You can calculate the inverse of the product matrix and multiply it by the resultant matrix and have the same result. Or if you have a graphics calculator (or a calculator that supports matrices) it can do it for you. It's the first way I learnt to solve systems of linear equations.

The inverse way takes a little longer. Since we have a + b = 64 and 2a + 5b = 20 we can assign it to a matrix:

\[\Large \left[ {\begin{array}{*{20}{c}}
a&b\\
{2a}&{5b}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{64}\\
{20}
\end{array}} \right]\]
To work it out via an inverse matrix, just find the determinant of the square matrix:
\[\Large \begin{array}{l}
\left[ {\begin{array}{*{20}{c}}
a&b\\
c&d
\end{array}} \right] = ad - cb\\
\left[ {\begin{array}{*{20}{c}}
1&1\\
2&5
\end{array}} \right] = 1 \cdot 5 - 2 \cdot 1 = 3
\end{array}\]
Then flip the product matrix around and multiply it be the inverse of 3 (i.e. 1/3):
\[\Large \left[ {\begin{array}{*{20}{c}}
5&{ - 1}\\
{ - 2}&1
\end{array}} \right] \cdot \frac{1}{3} = \left[ {\begin{array}{*{20}{c}}
{5/3}&{ - 1/3}\\
{ - 2/3}&{1/3}
\end{array}} \right]\]
Then multiply that by the resultant matrix:
\[\Large \left[ {\begin{array}{*{20}{c}}
{5/3}&{ - 1/3}\\
{ - 2/3}&{1/3}
\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}
{64}\\
{20}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{(5/3) \cdot 64 + ( - 1/3) \cdot 20}\\
{( - 2/3) \cdot 64 + (1/3) \cdot 20}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{100}\\
{ - 36}
\end{array}} \right]\]

Answer 6

Well by flipping the product matrix around (s)he meant swap the position of the diagonal elements and change the sign of the off-diagonal.

Answer 7

He ;) And yes, that's what I meant. I suppose I should have been clearer, 'flipping' sounds like simply turning upside down:

\[\Large \left[ {\begin{array}{*{20}{c}}
a&b\\
c&d
\end{array}} \right] \Rightarrow \left[ {\begin{array}{*{20}{c}}
d&{ - b}\\
{ - c}&a
\end{array}} \right]\]

I haven't played with matrices since... a long time ago... so I can't really remember all the rules - but point of the story is, row reduction is handy for simultaneous equation solving ;)