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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:
$2 over straight x plus 3 over straight y plus 10 over straight z space equals space 4 4 over straight x minus 6 over straight y plus 5 over straight z space equals space 1 6 over straight x plus 9 over straight y minus 20 over straight z space equals space 2$

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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.

Let first, second and third numbers be denoted by x, y and z respectively.
From given conditions,
x + y + z = 6
y + 3z = 11 or 0x + y + 3z = 11
x + z = 2y or x – 2y + z = 0
These equations can be written as
$open square brackets table row 1 cell space space space space space space 1 end cell cell space space space 1 end cell row 0 cell space space space space space space 1 end cell cell space space space 3 end cell row 1 cell space space space minus 2 end cell cell space space 1 end cell end table close square brackets space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 6 row 11 row 0 end table close square brackets comma$
or $AX space equals space straight B space space where space straight A space equals space open square brackets table row 1 cell space space space space 1 end cell cell space space space 1 end cell row 0 cell space space space space 1 end cell cell space space space 3 end cell row 1 cell space minus 2 end cell cell space space space 1 end cell end table close square brackets comma space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space space straight B space equals space open square brackets table row 6 row 11 row 0 end table close square brackets space$
$open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space 1 end cell cell space space space 1 end cell row 0 cell space space 1 end cell cell space space space 3 end cell row 1 cell negative 2 end cell cell space space 1 end cell end table close vertical bar space equals space 1 space open vertical bar table row 1 cell space space space 3 end cell row cell negative 2 end cell cell space space space 1 end cell end table close vertical bar minus space 1 open vertical bar table row 0 cell space space space 3 end cell row 1 cell space space space 1 end cell end table close vertical bar plus 1 space open vertical bar table row 0 cell space space space space space space space 1 end cell row 1 cell space space space minus 2 end cell end table close vertical bar space space space space space space space space equals 1 left parenthesis 1 plus 6 right parenthesis space minus space 1 left parenthesis 0 minus 3 right parenthesis space plus space 1 left parenthesis 0 minus 1 right parenthesis space equals space 7 plus 3 minus 1 space equals space 9 space not equal to space 0 therefore space space space space space space space space straight A to the power of negative 1 end exponent space exists.$
Co-factors of the elements of first row of | A | are
$open vertical bar table row cell space space 1 end cell cell space space space 3 end cell row cell negative 2 end cell cell space space space 1 end cell end table close vertical bar comma space space space space minus open vertical bar table row cell space 0 end cell cell space space space 3 end cell row cell space 1 end cell cell space space space 1 end cell end table close vertical bar comma space space space open vertical bar table row 0 cell space space space space space space space 1 end cell row 1 cell space space space minus 2 end cell end table close vertical bar$
i.e., 1 + 6, – (0 - 3), 0 – 1 i.e. 7, 3, – 1 respectively.
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row cell space space 1 end cell cell space space space space 1 end cell row cell negative 2 end cell cell space space space space 1 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space space 1 end cell row 1 cell space space space 1 end cell end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space space space space space 1 end cell row 1 cell space space minus 2 end cell end table close vertical bar$
i.e. – (1 + 2), 1 – 1, – (– 2 – 1) i.e. - 3, 0, 3 respectively
Co-factors of the elements of third row of | A | are
$open vertical bar table row 1 cell space space space space 1 end cell row 1 cell space space space space 3 end cell end table close vertical bar comma space space space space space minus open vertical bar table row 1 cell space space space space 1 end cell row 0 cell space space space space 3 end cell end table close vertical bar comma space space space space open vertical bar table row 1 cell space space space space 1 end cell row 0 cell space space space space 1 end cell end table close vertical bar$
i.e. 3 – 1(3 – 0), 1 – 0 i.e. 2, – 3, 1 respectively.
$therefore space space space space space space space adj space straight A space equals space open square brackets table row cell space space space 7 end cell cell space space space 3 end cell cell space space space minus 1 end cell row cell negative 3 end cell cell space space space space 0 end cell cell space space space space space 3 end cell row cell space space 2 end cell cell space minus 3 end cell cell space space space space 1 end cell end table close square brackets to the power of apostrophe space equals open square brackets table row cell space 7 end cell cell space space minus 3 end cell cell space space space space space space space 2 end cell row cell space 3 end cell cell space space space space space 0 end cell cell space space space minus 3 end cell row cell negative 1 end cell cell space space space space space 3 end cell cell space space space space space space space 1 end cell end table close square brackets space therefore space space space space space space space space space straight A to the power of negative 1 end exponent space equals space fraction numerator adj space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space 1 over 9 open square brackets table row cell space space space 7 end cell cell space space minus 3 end cell cell space space space space space space 2 end cell row cell space space 3 end cell cell space space space space 0 end cell cell space space minus 3 end cell row cell negative 1 end cell cell space space space space 3 end cell cell space space space space space space 1 end cell end table close square brackets$

Now, $Now comma space space space AX space equals space straight B space space space space space space space space space rightwards double arrow space space space space straight X space equals space straight A to the power of negative 1 end exponent straight B$
$therefore space space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 over 9 open square brackets table row cell space space 7 end cell cell space space minus 3 end cell cell space space space space space 2 end cell row cell space space 3 end cell cell space space space 0 end cell cell space minus 3 end cell row cell negative 1 end cell cell space space space 3 end cell cell space space space space 1 end cell end table close square brackets space open square brackets table row 6 row 11 row 0 end table close square brackets to the power of apostrophe space space space rightwards double arrow space space space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 over 9 open square brackets table row cell 42 minus 33 plus 0 end cell row cell 18 plus 0 minus 0 end cell row cell negative 6 plus 33 plus 0 end cell end table close square brackets rightwards double arrow space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 over 9 open square brackets table row 9 row 18 row 27 end table close square brackets space space space space space rightwards double arrow space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 1 row 2 row 3 end table close square brackets therefore space space space space space straight x space equals space 1 comma space space space space space straight y space equals space 2 comma space space space straight z space equals space 3. therefore space space space space space space required space numbers space are space 1 comma space 2 comma space space 3.$

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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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