Answer to Question #191560 in Linear Algebra for EUGINE HAWEZA

Question

Write notes on how to add, subtract and multiply matrices, and show how they may apply to solving a three system of equation , site examples on the applications of matrices to solving real world business problems

Answer 1

How to add and subtract matix:-

Rule1: A matrix only be 'added to' or 'subtracted from' another matrix if the both matrices have the same dimensions.

Rule 2: To add or subtract two matrices, just add or subtract the corresponding entries, and place this sum or difference in the corresponding position in the matrix which results.

Example

Add the matrices.

$\begin{bmatrix} 1 & 5\\ -4 & 3 \end{bmatrix} + \begin{bmatrix} 2 & -1\\ 4 & -1 \end{bmatrix}$

First note that both metrices are 2×2

matrices, so we add corresponding entries

$\begin{bmatrix} 3 & 4\\ 0 & 2 \end{bmatrix}$

Subtraction the matrices

$$\begin{bmatrix} 4 & 5 &6\\ 2 & 3 &4 \end{bmatrix} - \begin{bmatrix} 2 & 4 & 6\\ 1 & 2 & 3 \end{bmatrix}$$

First note that both metrices are 2×3

matrices, so we subtract corresponding entries

$\begin{bmatrix} 2&1&0 \\ 1&1&1 \end{bmatrix}$

Multiplying a Matrix by Another Matrix

multiply a matrix by another matrix we need to do the "dot product" of rows and columns ...

see with an example:

To work out the answer for the 1st row and 1st column:

$\begin{bmatrix} 1&2&3\\ 4 &5&6 \end{bmatrix} × \begin{bmatrix} 7&8\\ 9&10\\ 11&12 \end{bmatrix}$

The "Dot Product" is where we multiply matching members, then sum up:

(1, 2, 3) • (7, 9, 11) = 1×7 + 2×9 + 3×11

  = 58

We match the 1st members (1 and 7), multiply them, likewise for the 2nd members (2 and 9) and the 3rd members (3 and 11), and finally sum them up.

Similarly

(1, 2, 3) • (8, 10, 12) = 1×8 + 2×10 + 3×12

  = 64

Similarly

(4, 5, 6) • (7, 9, 11) = 4×7 + 5×9 + 6×11

  = 139

And

(4, 5, 6) • (8, 10, 12) = 4×8 + 5×10 + 6×12

  = 154

Finally we get result matrix

$\begin{bmatrix} 58&64\\ 139&154 \end{bmatrix}$

Solving system of three equation , examples on the applications of matrices to solving real world business problems

In the spirit of Christmas and New Years' resolutions, suppose that what we were on a diet and needed to eat precisely 245

245 calories, 6

6 grams of protein, and 7

7 grams of fat for breakfast. Unfortunately, I open my cupboard to see that all I have is three boxes of cereal: Cheerios, Cinnamon Toast Crunch, and Rice Krispies. There nutritional information per serving is as follows:

Cereal                     Calories     Protein     Fat
Cheerios                   120          4           2
Cinnamon Toast Crunch      130          3           5
Rice Krispies              105          1           2

Now, normally, I would dive in and gorge myself on Cinnamon Toast Crunch

because they're delicious - but, I need to stick to my new years resolution.

First, I denote c=

 servings of cheerios, 

t= servings of Cinnamon Toast Crunch, and 

r= servings of Rice Krispies. Then, I form the following system of 3

3 equations in 3

3 unknowns:

120c+130t+105r=245 calories

4c+3t+r=6 grams of protein

2c+5t+2r=7 grams of fat

Now, we find it solution..

Step 1: Find the determinant of the coefficient matrix.

$\begin{vmatrix} 120 & 130 & 105\\ 4 & 3 &1\\ 2&5&2 \end{vmatrix}$

Now find determinant as

=120(3×2-1×5)+130(1×2-4×2)+105(4×5-3×2)=810

Step2: Find the determinant of the x - matrix ( D_x ). X - matrix is formed by replacing the x-column values with the answer-column values

$\begin{vmatrix} 245 & 130 & 105\\ 6 & 3 &1\\ 7&5&2 \end{vmatrix}$

Now find determinant = 540

Step3: Find the determinant of the y - matrix ( Dy ). Y - matrix is formed by replacing the y-column values with the answer-column values

$\begin{vmatrix} 120 & 245 & 105\\ 4 & 6 &1\\ 2&7&2 \end{vmatrix}$

Now find determinant = 810

Step4: Find the determinant of the z - matrix ( D_z) Z- matrix is formed by replacing the z-column values with the answer-column values

$\begin{vmatrix} 120 & 130 & 245\\ 4 & 3 &6\\ 2&5&7 \end{vmatrix}$

Now find determinant = 270

Cramers Rule says that the solutions are

x=540/810=2/3

y=810/810=1

z=270/810=1/3