01Zsolvingeqns


 * Solving Simultaneous Equations Using Matrices **

You have previously learnt algebraic methods of solving simultaneous linear equations.

Matrices can be used to do the same thing.

Forming a Matrix Equation

For two equations in the form: ax + by = e. . . **(1)** cx + dy = f. . . **(2)**


 * 1) Place the coefficients into a square matrix, (equation 1 in first row, equation 2 in second row etc)
 * 2) multiplied by a column matrix with the variables (x and y),
 * 3) then, "="
 * 4) then a column matrix with the values from the right side of the equations

like this: math \left[ \begin{matrix} a&b \\ c&d \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] = \left[ \begin{matrix} e \\ f \\ \end{matrix} \right] math


 * Example 1 **

Put these simultaneous equations into matrix form: 3x + 5y = 14 4x – 2y = 10

__**Solution**__

math \left[ \begin{matrix} 3&5 \\ 4&-2 \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] = \left[ \begin{matrix} 14 \\ 10 \\ \end{matrix} \right] math

Notice Multiplying out the left side of this equation gives: math \left[ \begin{matrix} 3x+5y \\ 4x-2y \\ \end{matrix} \right] = \left[ \begin{matrix} 14 \\ 10 \\ \end{matrix} \right] math which is the two equations listed above (in matrix notation)


 * Example 2 **

Put these simultaneous equations into matrix form: x + 2y – z = 6 2x – y – z = 1 x + 2z = –1

__**Solution**__

math \left[ \begin{matrix} 1&2&-1 \\ 2&-1&-1 \\ 1&0&2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right] = \left[ \begin{matrix} 6 \\ 1 \\ -1 \\ \end{matrix} \right] math

Notice In equation 3, y had no value so its place in the matrix is occupied with a 0

Solving a Matrix equation

In algebra, to solve the equation 3x = 6, we would divide both sides by 3. This could be written as multiplying both sides by 3 –1.

3x = 6

3 –1 × 3x = 3 –1 × 6

1x = 2

In matrices, we do the same thing. To solve Ax = b, we multiply both sides by A –1 where A –1 is the Inverse of A

Ax = b

A –1 × Ax = A –1 × b

Recall that A –1 × A = I {where I is the Identity Matrix}

Ix = A –1 × b

x = A –1 × b

This means that for the matrix equation: math \left[ \begin{matrix} a&b \\ c&d \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] = \left[ \begin{matrix} e \\ f \\ \end{matrix} \right] math

the solution is: math \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] = \left[ \begin{matrix} a&b \\ c&d \\ \end{matrix} \right] ^ {-1} \left[ \begin{matrix} e \\ f \\ \end{matrix} \right] math

Recall that math \left[ \begin{matrix} a&b \\ c&d \\ \end{matrix} \right] ^ {-1} = \dfrac{1}{ad-bc}\left[ \begin{matrix} d&-b \\ -c&a \\ \end{matrix} \right] math


 * Example 3 **

Solve these simultaneous equations using matrices: 3x + 5y = 14 4x – 2y = 10

__**Solution**__

math \left[ \begin{matrix} 3&5 \\ 4&-2 \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] = \left[ \begin{matrix} 14 \\ 10 \\ \end{matrix} \right] math

therefore math \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] = \left[ \begin{matrix} 3&5 \\ 4&-2 \\ \end{matrix} \right]^{-1} \left[ \begin{matrix} 14 \\ 10 \\ \end{matrix} \right] math

math \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] = \dfrac{1}{-26} \left[ \begin{matrix} -2&-5 \\ -4&3 \\ \end{matrix} \right] \left[ \begin{matrix} 14 \\ 10 \\ \end{matrix} \right] math

math \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] = - \dfrac{1}{26} \left[ \begin{matrix} -78 \\ -26 \\ \end{matrix} \right] math

math \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] = \left[ \begin{matrix} 3 \\ 1 \\ \end{matrix} \right] math

hence the solution is (x ``=`` 3, y ``=`` 1)

The Classpad can be used to achieve the same result.


 * Example 4 **

Solve these simultaneous equations using matrices on a CAS calculator: x + 2y – z = 6 2x – y – z = 1 x + 2z = –1

__**Solution**__

math \left[ \begin{matrix} 1&2&-1 \\ 2&-1&-1 \\ 1&0&2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right] = \left[ \begin{matrix} 6 \\ 1 \\ -1 \\ \end{matrix} \right] math

therefore

math \left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right] = \left[ \begin{matrix} 1&2&-1 \\ 2&-1&-1 \\ 1&0&2 \end{matrix} \right]^{-1} \left[ \begin{matrix} 6 \\ 1 \\ -1 \\ \end{matrix} \right] math

Using the Classpad, gives: math \left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right] = \left[ \begin{matrix} 1 \\ 2 \\ -1 \\ \end{matrix} \right] math

hence the solution is (x ``=`` 1, y ``=`` 2, z ``=`` –1)

For another site that explains the same idea, go here: MathsIsFun .