30. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are '80, '60 and'40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Choose the correct answer in Exercises 31 and 32.

Let   $A=\left[\begin{array}{ccc}\hfill x_1\hfill & \hfill y_1\hfill & \hfill z_1\hfill \\ \hfill x_2\hfill & \hfill y_2\hfill & \hfill z_2\hfill \\ \hfill x_3\hfill & \hfill y_3\hfill & \hfill z_3\hfill \end{array}\right]$

Given  $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 0\hfill & \hfill 2\hfill \end{array}\right]$   ...(1)

$\left[\begin{array}{ccc}\hfill x_1+z_1\hfill & \hfill x_2+z_2\hfill & \hfill x_3+z_3\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 0\hfill & \hfill 2\hfill \end{array}\right]$

∴    x₁ + z₁ = 2                … (2)

x₂ + z₂ = 0               … (3)

x₃ + z₃ = 0                … (4)

Given   $A=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill -4\hfill & \hfill 0\hfill & \hfill 4\hfill \end{array}\right]$

$\left[\begin{array}{ccc}\hfill -x_1+z_1\hfill & \hfill -x_2+z_2\hfill & \hfill -x_3+z_3\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 0\hfill & \hfill 4\hfill \end{array}\right]$

⇒   – x₁ + z₁ = −4             … (5)

–x₂ + z₂ = 0                … (6)

–x₃ + z₃ = 4                … (7)

Given  $A=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \end{array}\right]$

$\left[\begin{array}{ccc}\hfill y_1\hfill & \hfill y_2\hfill & \hfill y_3\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \end{array}\right]$

∴     y₁ = 0, y₂ = 2, y₃ = 0

∴     from (2), (3), (4), (5), (6) and (7)

x₁ = 3, x₂ = 0, x₃ = –1

y₁ = 0, y₂ = 2, y₃ = 0

z₁ = –1, z₂ = 0, z₃ = 3

$A=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right]$  

Now (A – 3I)  $\left[\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z\hfill \end{array}\right]$  = $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \end{array}\right]$  

$\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \end{array}\right]$  

$\left[\begin{array}{ccc}\hfill -z\hfill & \hfill -y\hfill & \hfill -x\hfill \end{array}\right]$  =   $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \end{array}\right]$

[z = 1], [y = –2], [x = –3]