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Solve by matrix method:
$3 over straight x plus 4 over straight y plus 7 over straight z space equals space 14 2 over straight x minus 1 over straight y plus 3 over straight z space equals space 4 1 over straight x plus 2 over straight y minus 3 over straight z space equals space 0$

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302.
Solve the system of the following equations:
$2 over straight x plus 3 over straight y plus 10 over straight z space equals space 4 4 over straight x minus 6 over straight y plus 5 over straight z space equals space 1 6 over straight x plus 9 over straight y minus 20 over straight z space equals space 2$

The given equations are
$2 over straight x plus 3 over straight y plus 10 over straight z space equals space 4 4 over straight x minus 6 over straight y plus 5 over straight z space equals space 1 6 over straight x plus 9 over straight y minus 20 over straight z space equals space 2$

Put $1 over straight x space equals space straight a comma space space space 1 over straight y space equals space straight b comma space space space space 1 over straight z equals straight c$
∴ given equations become
2a + 3b + 10c = 4
4a – 6b + 5c = 1
6a + 9b – 20c = 2
These equations can be written as
   $open square brackets table row 2 cell space space space space space 3 end cell cell space space space space space 10 end cell row 4 cell space space minus 6 end cell cell space space space space space 5 end cell row 6 cell space space space space 9 end cell cell space space minus 20 end cell end table close square brackets space open square brackets table row straight a row straight b row straight c end table close square brackets space equals open square brackets table row 4 row 1 row 2 end table close square brackets$

or    $or space space space space AX space equals space straight B space where space straight A space equals space open square brackets table row 2 cell space space space space space space 3 end cell cell space space space space 10 end cell row 4 cell space space minus 6 end cell cell space space space space space 5 end cell row 6 cell space space space space space space 9 end cell cell space space minus 20 end cell end table close square brackets comma space space space straight X space equals space open square brackets table row straight a row straight b row straight c end table close square brackets comma space space straight B space equals space open square brackets table row 4 row 1 row 2 end table close square brackets$
$open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space space space space 3 end cell cell space space space space space 10 end cell row 4 cell space minus 6 end cell cell space space space space space 5 end cell row 6 cell space space space space space 9 end cell cell space space minus 20 end cell end table close vertical bar space equals space 2 open vertical bar table row cell negative 6 end cell cell space space space space space space space space 5 end cell row 9 cell space space minus 20 end cell end table close vertical bar space minus space space 3 space open vertical bar table row 4 cell space space space space space 5 end cell row 6 cell space minus 20 end cell end table close vertical bar space plus space 10 space open vertical bar table row 4 cell space space space minus 6 end cell row 6 cell space space space space space space space 9 end cell end table close vertical bar space space space space space space space space equals 2 space left parenthesis 120 minus 45 right parenthesis space minus space 3 left parenthesis negative 80 minus 30 right parenthesis space plus space 10 space left parenthesis 36 plus 36 right parenthesis space space space space space space space space space equals 2 space left parenthesis 75 right parenthesis space minus space 3 space left parenthesis negative 110 right parenthesis space plus space 10 space left parenthesis 72 right parenthesis space equals space 150 plus 330 plus 720 space equals space 1200 space not equal to space 0 therefore space space space space straight A to the power of negative 1 end exponent space exists.$

Co-factors of the elements of first row of | A | are
$open vertical bar table row cell negative 6 end cell cell space space space space space space 5 end cell row 9 cell space minus 20 end cell end table close vertical bar comma space space space space minus open vertical bar table row 4 cell space space space space space space 5 end cell row 6 cell space space minus 20 end cell end table close vertical bar comma space space space space space open vertical bar table row 4 cell space space space space minus 6 end cell row 6 cell space space space space space space 9 end cell end table close vertical bar$
i.e.   120 – 45, – (– 80 – 30), 36 + 36 i.e. 75, 110, 72 respectively.
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row 3 cell space space space space space 10 end cell row 9 cell space space minus 20 end cell end table close vertical bar comma space space space space open vertical bar table row 2 cell space space space space 10 end cell row 6 cell space space minus 20 end cell end table close vertical bar comma space space space minus open vertical bar table row 2 cell space space space space 3 end cell row 6 cell space space space space 9 end cell end table close vertical bar$
i.e. – (– 60 – 90), – 40 – 60, – (18 – 18) i.e. 150, 100, 0 respectively.
Co-factors of the elements of third row of | A | are
$open vertical bar table row cell space space 3 end cell cell space space space 10 end cell row cell negative 6 end cell cell space space space space 5 end cell end table close vertical bar comma space space space space space minus open vertical bar table row 2 cell space space space space 10 end cell row 4 cell space space space space 5 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space space space space space space 3 end cell row 4 cell space space space minus 6 end cell end table close vertical bar$
i.e. 15 + 60, – (10 – 40), – 12 – 12 i.e. 75, 30, –24 respectively.
$therefore space space space space space adj space straight A space equals space open square brackets table row 75 cell space space space 110 end cell cell space space space 72 end cell row 150 cell space minus 100 end cell cell space space 0 end cell row 75 cell space space space 30 end cell cell space minus 24 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 75 cell space space space space space 150 end cell cell space space space 75 end cell row 110 cell space space minus 100 end cell cell space space space 30 end cell row 72 cell space space space space 0 end cell cell negative 24 end cell end table close square brackets space space space space space straight A to the power of negative 1 end exponent space equals space fraction numerator adj space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space 1 over 1200 open square brackets table row 75 cell space space space space space 150 end cell cell space space space 75 end cell row 110 cell space space minus 100 end cell cell space space 30 end cell row 72 cell space space 0 end cell cell space space minus 24 end cell end table close square brackets Now comma space space space AX space equals space space straight B space space space space space rightwards double arrow space space space space space straight X space equals space straight A to the power of negative 1 end exponent straight B therefore space space space space space open square brackets table row straight a row straight b row straight c end table close square brackets space equals space 1 over 1200 open square brackets table row 75 cell space space space space 150 end cell cell space space space space 75 end cell row 110 cell space minus 100 end cell cell space space space 30 end cell row 72 cell space space 0 end cell cell negative 24 end cell end table close square brackets space open square brackets table row 4 row 1 row 2 end table close square brackets rightwards double arrow space space space space space space open square brackets table row straight a row straight b row straight c end table close square brackets space equals space 1 over 1200 open square brackets table row cell 300 plus 150 plus 150 end cell row cell 440 minus 100 plus 60 end cell row cell 288 plus 0 minus 48 end cell end table close square brackets rightwards double arrow space space space space space space space space open square brackets table row straight a row straight b row straight c end table close square brackets space equals space 1 over 1200 open square brackets table row 600 row 400 row 240 end table close square brackets space space space space space rightwards double arrow space space space space space open square brackets table row straight a row straight b row straight c end table close square brackets space equals space open square brackets table row cell 1 half end cell row cell 1 third end cell row cell 1 fifth end cell end table close square brackets therefore space space space space space straight a space equals space 1 half comma space space space space straight b space equals space 1 third comma space space space straight c space equals space 1 fifth space space space rightwards double arrow space space space straight x space equals space 2 comma space space space space straight y space equals space 3 comma space space space straight z space equals space 5.$

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303.

If $straight A space equals space open square brackets table row 2 cell space space space minus 3 end cell cell space space space space 5 end cell row 3 cell space space space space space 2 end cell cell space minus 4 end cell row 1 cell space space space 1 end cell cell space minus 2 end cell end table close square brackets comma$ find A^-^1,Using A^-¹, solve the following system of linear equations.
2x – 3y + 5z = 16
3x + 2y – 4z = – 4
x + y – 2z = – 3

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304.

If $straight A space equals space open square brackets table row 2 cell space space space minus 3 end cell cell space space space space 5 end cell row 3 cell space space space space 2 end cell cell negative 4 end cell row 1 cell space space space 1 end cell cell negative 2 end cell end table close square brackets comma$ find A ^–1 .
Using A^–1, solve the following system of linear equations.
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3

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305.

Compute A^–1 for the following matrix
$straight A space equals space open square brackets table row cell negative 1 end cell cell space space space space space 2 end cell cell space space space space 5 end cell row cell space space 2 end cell cell space space minus 3 end cell cell space space space 1 end cell row cell negative 1 end cell cell space space 1 end cell cell space space space space 1 end cell end table close square brackets.$
Hence solve the system of equations.
– x + 2y+ 5 z = 2
2x – 3y + Z = 15
– x + y + z = – 3

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306.

Compute A^–1 for the following matrix
$straight A space equals space open square brackets table row 1 cell space space space space space 2 end cell cell space space space 5 end cell row 1 cell space minus 1 end cell cell space minus 1 end cell row 2 cell space space space 3 end cell cell space minus 2 end cell end table close square brackets.$
Hence, solve the system of equations:
x + 2y + 5z = 10
x – y – z = – 2
2x + 3y – 2z = – 1

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307.

If $straight A equals space open square brackets table row 3 cell space space space minus 2 end cell cell space space space space space 1 end cell row 2 cell space space space space 1 end cell cell space minus 3 end cell row cell negative 1 end cell cell space space space 2 end cell cell space space space 1 end cell end table close square brackets comma$ Find A^–1.

Using A solve the following systems of linear equations:
3x – 2y + z = 2
2y + y – 3z = – 5
– x + 2y + z = 6.

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Short Answer Type

308.

Investigate for what values of a and b the simultaneous equations:
x + y + z = 6
x + 2y + 3z = 10
x + 2y + az = b have a unique solution.

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Long Answer Type

309.

The sum of three numbers is 6. If we multiply third number by 3 and add second number to it, we get 11. By adding first and third numbers, we get double of the second number. Represent it algebraically and find the numbers using matrix method.

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310. The cost of 4 kg. onion, 3 kg. wheat and 2 kg. rice is Rs. 60. The cost of 2 kg. onion, 4 kg. wheat and 6 kg. rice is Rs. 90. The cost of 6 kg. onion. 2 kg. wheat and 3 kg. rice is Rs. 70. Find cost of each item per kg. by matrix method.

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302. Solve the system of the following equations:

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