Question

An electronics company produces​ transistors, resistors, and computer chips. Each transistor requires 3 units of​ copper, 2 units of​ zinc, and 1 unit of glass. Each resistor requires 3​, 1​, and 2 units of the three​ materials, and each computer chip requires 2​, 1​, and 2 units of these​ materials, respectively. How many of each product can be made with 1600 units of​ copper, 1025 units of​ zinc, and 625 units of​ glass? Solve this exercise by using the inverse of the coefficient matrix to solve a system of equations.

Answer 1

Answer:

475 transistors, 25 resistors and 50 computer chips can be produced.

Step-by-step explanation:

Let us consider, p = Number of transistors.

                           q = Number of resistors.

                            r = Number of computer chips.

The following three linear equations according to question,

$3\times p + 3\times q + 2\times r = 1600\\2\times p + 1\times q + 1\times r = 1025\\1\times p + 2\times q + 2\times r = 625$

The matrix form of any system, Ax = B

Where, A = Coefficient matrix

            B = Constant vector

            x = Variable vector

$A = \left[\begin{array}{ccc}3&3&2\\2&1&1\\1&2&2\end{array}\right], x = \left[\begin{array}{ccc}p\\q\\r\end{array}\right], B = \left[\begin{array}{ccc}1600\\1025\\625\end{array}\right]$

The inverse matrix, $A^{-1}$ can be found by using the following formula,

           $A^{-1} = \frac{1}{det A}\times (C_{A}) ^{T}$

Where, det A = Determinant of matrix A.

                 $C_{A}$ = Matrix of cofactors of A

Now, applying this formula to find $A^{-1}$;

$det A = \left[\begin{array}{ccc}3&3&2\\2&1&1\\1&2&2\end{array}\right] = 3\times(2-2)-3\times(4-1)+2\times(4-1) = -3$

Here, $det A\neq 0$, thus the matrix is invertible.

$C_{A} = \left[\begin{array}{ccc}(2-2)&-(4-1)&(4-1)\\-(6-4)&(6-2)&-(6-3)\\(3-2)&-(3-4)&(3-6)\end{array}\right] = \left[\begin{array}{ccc}0&-3&3\\-2&4&-3\\1&1&-3\end{array}\right] \\(C_{A}) ^{T} = \left[\begin{array}{ccc}0&-3&3\\-2&4&-3\\1&1&-3\end{array}\right] ^{T} = \left[\begin{array}{ccc}0&-2&1\\-3&4&1\\3&-3&-3\end{array}\right]$

$A^{-1} = \frac{1}{-3}\times\left[\begin{array}{ccc}0&-2&1\\-3&4&1\\3&-3&-3\end{array}\right] \\ So, x= \frac{1}{-3} \left[\begin{array}{ccc}0&-2&1\\-3&4&1\\3&-3&-3\end{array}\right]\times\left[\begin{array}{ccc}1600\\1025\\625\end{array}\right]= \frac{1}{-3} \left[\begin{array}{ccc}-1425\\-75\\-150\end{array}\right] = \left[\begin{array}{ccc}475\\25\\50\end{array}\right]$

So, p = 475, q = 25, r = 50.