Question

An outbreak of chickenpox hit the local public schools. Approximately \[15\%\] of the male and female juniors and \[25\%\] of the male and female seniors are the currently healthy, \[35\%\] of the male and the female juniors and \[30\%\] of the male and female seniors are currently sick, and \[50\%\] of the male and female juniors and \[45\%\] of the male and female seniors are carriers of the chicken pox. There are \[100\] male juniors, \[80\] male seniors ,\[120\] female juniors and \[100\] female seniors. Using two matrices and one matrix equation, find out how many males and how many females are healthy, sick and carriers.

Answer 1

Hint: In order to find the number of females and males that are healthy, sick and carriers, we must be considering two matrices in which one of the matrix would be representing the number of females and males in two different rows and the other matrix would be representing the number of seniors and juniors who are healthy, sick and carriers. Upon solving these two matrices we will be obtaining the required values.

Complete step-by-step solution:
Now let us learn about matrices. A matrix is generally an array of numbers. It can have a desirable number of rows and columns. We can perform operations such as addition, subtraction, multiplying a matrix to another matrix or multiplying a matrix to a constant, division and also transposition of a matrix. There are different types of matrices. They are: null matrix, identity matrix, upper triangular matrix, lower triangular matrix , square matrix, rectangular matrix etc.
Now let us start solving our given problem.
Firstly let us represent the matrix for the number of males and females of both senior and junior.
The matrix would be,
\[\left[ \begin{matrix}
   100 & 80 \\
   120 & 100 \\
\end{matrix} \right]\]
Here the first row represents males and the second row represents females.
Now let us represent the second matrix of seniors and juniors who are healthy, sick and carriers.
The matrix we get is,
\[\left[ \begin{matrix}
  &\text{0.15}&\text{0.35}&\text{0.50} \\
 & \text{0.25}&\text{0.30}&\text{0.45} \\
\end{matrix} \right]\]
Here the first row represents the juniors and the second row represents the seniors regarding the healthy, sick and carriers.
Now in order to find the total number let us multiply the matrices. We get,
\[\left[ \begin{matrix}
   100 & 80 \\
   120 & 100 \\
\end{matrix} \right]\times \left[ \begin{matrix}
  & \text{0.15}&\text{0.35}&\text{0.50} \\
 & \text{0.25}&\text{0.30}&\text{0.45} \\
\end{matrix} \right]\]
Upon solving this,
\[\left[ \begin{matrix}
  & 35 & 59 & 86 \\
 & 43 & 72 & 105 \\
\end{matrix} \right]\]
So we get \[35\] healthy males, \[\text{59}\] sick males and \[\text{86}\] carrier males, \[\text{43}\] healthy females, \[\text{72}\] sick females and \[\text{105}\] female carriers.

Note: While multiplying the matrices, we must always consider the first row of the first matrix and the first column of the second matrix. We must be aware regarding the representation of elements in the matrix as well as the criteria. We can apply the concept of matrices in computer graphics, electromagnetism, optics etc.