Question

Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award x each, y each and z each for the three respective values to 3, 2 and 1 students respectively with a total award money of 1,600. School B wants to spend 2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount for one prize on each value is 900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.

Solution

From the given data, we write the following equations:
$open parentheses straight x space space straight y space space straight z close parentheses open parentheses table row 3 row 2 row 1 end table close parentheses space equals space 1600 open parentheses straight x space space straight y space space straight z close parentheses open parentheses table row 4 row 1 row 3 end table close parentheses space equals space 2300 open parentheses straight x space space straight y space space straight z close parentheses open parentheses table row 1 row 1 row 1 end table close parentheses space equals space 900$
From above system, we get:
3x+2y+z = 1600
4x+y+3z = 2300
x+y+z = 900
Thus we get:
$open parentheses table row 3 cell space 2 end cell cell space 1 end cell row 4 cell space 1 end cell cell space 3 end cell row 1 cell space 1 end cell cell space 1 end cell end table close parentheses open parentheses table row straight x row straight y row straight z end table close parentheses space equals space open parentheses table row 1600 row 2300 row 900 end table close parentheses$
This is of the form
$AX space equals space straight B comma space where space straight A space equals space open parentheses table row 3 cell space space 2 end cell cell space space 1 end cell row 4 cell space 1 end cell cell space space 3 end cell row 1 cell space 1 end cell cell space 1 end cell end table close parentheses semicolon space space space straight X space equals space open parentheses table row straight x row straight y row straight z end table close parentheses space space space and space straight B space equals space open parentheses table row 1600 row 2300 row 900 end table close parentheses open vertical bar straight A close vertical bar space equals space open vertical bar table row 3 cell space 2 end cell cell space 1 end cell row 4 cell space 1 end cell 3 row 1 cell space 1 end cell 1 end table close vertical bar space equals space 3 left parenthesis 1 minus 3 right parenthesis minus 2 left parenthesis 4 minus 3 right parenthesis plus 1 left parenthesis 4 minus 1 right parenthesis space equals space minus 6 minus 2 plus 3 equals space minus 5 not equal to 0 We space need space to space find space straight A to the power of negative 1 end exponent semicolon straight C subscript 11 equals negative 2 semicolon space space space straight C subscript 12 equals negative 1 semicolon space space straight C subscript 13 equals 3 straight C subscript 21 equals negative 1 semicolon space space straight C subscript 22 equals 2 semicolon space space space space space straight C subscript 23 equals negative 1 straight C subscript 31 equals 5 semicolon space space space space space straight C subscript 32 equals negative 5 semicolon space space space straight C subscript 33 equals negative 5$
Therefore, $adj space straight A space equals space open parentheses table row cell negative 2 end cell cell space minus 1 space end cell cell space space 3 end cell row cell negative 1 end cell cell space space 2 end cell cell negative 1 end cell row 5 cell negative 5 end cell cell space minus 5 end cell end table close parentheses to the power of straight T space equals space open parentheses table row cell negative 2 end cell cell space minus 1 end cell cell space space 5 end cell row cell negative 1 end cell cell space space 2 end cell cell negative 5 end cell row 3 cell space minus 1 end cell cell negative 5 end cell end table close parentheses$
$Thus comma space straight A to the power of negative 1 end exponent space equals space fraction numerator adjA over denominator open vertical bar straight A close vertical bar end fraction equals negative 1 fifth open parentheses table row cell negative 2 end cell cell space minus 1 end cell cell space space space 5 end cell row cell negative 1 end cell cell space space 2 end cell cell space minus 5 end cell row 3 cell negative 1 end cell cell space minus 5 end cell end table close parentheses$
Therefore, $straight X equals straight A to the power of negative 1 end exponent straight B$
$open parentheses table row straight x row straight y row straight z end table close parentheses space equals space minus 1 fifth open parentheses table row cell negative 2 end cell cell space space minus 1 end cell cell space space space 5 end cell row cell negative 1 end cell cell space space 2 end cell cell negative 5 end cell row 3 cell space minus 1 end cell cell negative space 5 end cell end table close parentheses space open parentheses table row 1600 row 2300 row 900 end table close parentheses$
$rightwards double arrow open parentheses table row straight x row straight y row straight z end table close parentheses space equals 1 fifth open parentheses table row cell negative 2 cross times 1600 minus 1 cross times 2300 plus 5 cross times 900 end cell row cell negative 1 cross times 1600 plus 2 cross times 2300 minus 5 cross times 900 end cell row cell 3 cross times 1600 minus 1 cross times 2300 minus 5 cross times 900 end cell end table close parentheses rightwards double arrow open parentheses table row straight x row straight y row straight z end table close parentheses equals space minus 1 fifth open parentheses table row cell negative 1000 end cell row cell negative 1500 end cell row cell negative 2000 end cell end table close parentheses rightwards double arrow open parentheses table row straight x row straight y row straight z end table close parentheses space equals space open parentheses table row 200 row 300 row 400 end table close parentheses$
Awards can be given for discipline.

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Matrices

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