the management committe of a residential colony decided to award some of its members ( say x ) for honesty, - Maths - Determinants

Question

the management committe of a residential colony decided to award
some of its members ( say x ) for honesty, some ( say y ) for
helping others and some others ( say z ) for supervising the
workers to keep the colony neat and clean The sum of all the
awardees is 12 Three times the sum of awardees for cooperation and
supervision added to two times the no of awardees for honesty is 33
If the sum of the no of awardees for honesty and supervision is
twice the no of awardees for helping others, using matrix
method
Find the no of awardees of each category
apart from these values, suggest one more value which the
management of the colony must include for awaards

Ayush Jha · Accepted Answer

Let awards for honesty be x , for helping others be y and for supervision be z.

So, we have the following three relations-

$x + y + z = 12 2 x + 3 y + 3 z = 33 x - 2 y + z = 0$

So, let,

$A = | \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & - 2 & 1 \end{matrix} | X = | \begin{matrix} x \\ \underset{z}{y} \end{matrix} | B = | \begin{matrix} 12 \\ \underset{0}{33} \end{matrix} | X = A^{- 1} B$

Now,

$A^{- 1} = \frac{a d j (A)}{| A |} W e h a v e, | A | = 3 a d j (A) = | \begin{matrix} 9 & - 3 & 0 \\ 1 & 0 & - 1 \\ - 7 & 3 & 1 \end{matrix} | A^{- 1} = | \begin{matrix} 3 & - 1 & 0 \\ \frac{1}{3} & 0 & - \frac{1}{3} \\ - \frac{7}{3} & 1 & \frac{1}{3} \end{matrix} |$

Therefore,

$| \begin{matrix} x \\ \underset{z}{y} \end{matrix} | = | \begin{matrix} 3 & - 1 & 0 \\ \frac{1}{3} & 0 & - \frac{1}{3} \\ - \frac{7}{3} & 1 & \frac{1}{3} \end{matrix} | | \begin{matrix} 12 \\ \underset{0}{33} \end{matrix} | | \begin{matrix} x \\ \underset{z}{y} \end{matrix} | = | \begin{matrix} 3 \\ \underset{5}{4} \end{matrix} |$

Hence x= 3, y=4, z=5