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Determinants

Long Answer Type

271.

Use matrix method to solve the following system of equations:
2x – 3y + 3z = 1
2x+ 2y+ 3z = 2
3x – 2y + 2z = 3

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272.

Use matrix method to solve the following system of equations:
2x – 3y + 3z = 1
2x+ 2y+ 3z = 2
3x – 2y + 2z = 3

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273.
Use matrix method to solve the following system of equations:
2x – 3y + 3z = 1
2x+ 2y+ 3z = 2
3x – 2y + 2z = 3

The given equations are
2x – 3y + 3z = 1
2x+ 2y+ 3z = 2
3x – 2y + 2z = 3
These equations can be written as
$open square brackets table row 2 cell space space minus 3 end cell cell space space space 3 end cell row 2 cell space space space space 2 end cell cell space space 3 end cell row 3 cell space minus 2 end cell cell space space 2 end cell end table close square brackets space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 1 row 2 row 3 end table close square brackets$

or AX = B where $straight A space equals space open square brackets table row 2 cell space space minus 3 end cell cell space space 3 end cell row 2 cell space space space space space 2 end cell cell space space 3 end cell row 3 cell space space minus 2 end cell cell space space 2 end cell end table close square brackets comma space space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space space space straight B space equals space open square brackets table row 1 row 2 row 3 end table close square brackets$

$open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space space minus 3 end cell cell space space 3 end cell row 2 cell space space space space 2 end cell cell space space 3 end cell row 3 cell space minus 2 end cell cell space space 2 end cell end table close vertical bar space equals space 2 open vertical bar table row cell space space 2 end cell cell space space space 3 end cell row cell negative 2 end cell cell space space space space 2 space end cell end table close vertical bar minus left parenthesis negative 3 right parenthesis space open vertical bar table row 2 cell space space 3 end cell row 3 cell space space 2 end cell end table close vertical bar plus 3 space open vertical bar table row 2 cell space space space space space 2 end cell row 3 cell space space minus 2 end cell end table close vertical bar space space space space space space space equals 2 space left parenthesis 4 plus 6 right parenthesis space plus space 3 space left parenthesis 4 minus 9 right parenthesis space plus space 3 space left parenthesis negative 4 minus 6 right parenthesis space space space space space space space space equals 20 minus 15 minus 30 space equals space minus 25 space not equal to space 0 space therefore space space space space straight A to the power of negative 1 end exponent space exists.$
Co-factors of the elements of first row of | A | are
$open vertical bar table row 2 cell space space space space 3 end cell row cell negative 2 end cell cell space space space space 2 end cell end table close vertical bar comma space space space space space minus open vertical bar table row 2 cell space space space space 3 end cell row 3 cell space space space space 2 end cell end table close vertical bar comma space space open vertical bar table row 2 cell space space space space space 2 end cell row 3 cell space minus 2 end cell end table close vertical bar$

i.e., 10, 5, – 10 respectively
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row cell negative 3 end cell cell space space space 3 end cell row cell negative 2 end cell cell space space space 2 end cell end table close vertical bar comma space space space space open vertical bar table row 2 cell space space space 3 end cell row 3 cell space space 2 end cell end table close vertical bar comma space space space minus open vertical bar table row 2 cell space space space minus 3 end cell row 3 cell space space minus 2 end cell end table close vertical bar$
i.e., 0, – 5, – 5 respectively
Co-factors of the elements of third row of | A | are
$open vertical bar table row cell negative 3 end cell cell space space space 3 end cell row 2 cell space space space 3 end cell end table close vertical bar comma space space space space minus open vertical bar table row 2 cell space space space 3 end cell row 2 cell space space space 3 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space minus 3 end cell row 2 cell space space space space 2 end cell end table close vertical bar$

i.e., 15, 0, 10 respectively
$therefore space space space adj space straight A space equals space open square brackets table row 10 cell space space space space space 5 end cell cell space space space space space minus 10 end cell row 0 cell space minus 5 end cell cell space space space space minus 5 end cell row cell negative 15 end cell cell space space space space 0 end cell cell space space space space space space space 10 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 10 cell space space space space 0 end cell cell space space space minus 15 end cell row 5 cell space space minus 5 end cell cell space space space space space 0 end cell row cell negative 10 end cell cell space space minus 5 end cell cell space space space space space 10 end cell end table close square brackets 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 minus 1 over 25 open square brackets table row 10 cell space space space space 0 end cell cell space space space minus 15 end cell row cell space space 5 end cell cell space space minus 5 end cell cell space space space space space space 0 end cell row cell negative 10 end cell cell space minus 5 end cell cell space space space space space space 10 end cell end table close square brackets Now comma space space AX space equals space straight B space space space space space space rightwards double arrow space space space space straight X space equals space space straight A to the power of negative 1 end exponent straight B 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 equals space minus 1 over 25 open square brackets table row 10 cell space space space 0 end cell cell space space minus 15 end cell row cell space 5 end cell cell negative 5 end cell cell space space space space 0 end cell row cell negative 10 end cell cell negative 5 end cell cell space space 10 end cell end table close square brackets space open square brackets table row 1 row 2 row 3 end table close square brackets 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 minus 1 over 25 open square brackets table row cell 10 plus 0 minus 45 end cell row cell 5 minus 10 plus 0 end cell row cell negative 10 minus 10 plus 30 end cell end table close square brackets space space space 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 minus 1 over 25 open square brackets table row cell negative 35 end cell row cell negative 5 end cell row 10 end table close square brackets rightwards double arrow 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 space equals space open square brackets table row cell 7 over 5 end cell row cell 1 fifth end cell row cell fraction numerator negative 2 over denominator 5 end fraction end cell end table close square brackets space space space space space rightwards double arrow space space space straight x space equals space 7 over 5 comma space space space straight y space equals space 1 fifth comma space space space straight z space equals space minus 2 over 5 space$

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274. Use matrix method to solve the following system of equations:
y + 2z = 4
2z + x = 5
x + 2y = 7

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275. Use matrix method to solve the following system of equations:
3x – 4y + 2z = – 1
2x + 3y + 5z = 7
x + z = 2

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276. Use matrix method to solve the following system of equations:
x – y = 5
y + z = 3
z + x = 4

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277. Use matrix method to solve the following system of equations:
x + y + z = 3
x + 2y + 3z = 4
x + 4y + 9z = 6

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278. Use matrix method to solve the following system of equations:
3x + y + z = 3
2x – y –z = 2
–x –y + z = 1

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279. Use matrix method to solve the following system of equations:
x + 2y – 3z = 6
3x + 2y – 2z = 3
2x – y + z = 2

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280. Use matrix method to solve the following system of equations:
x + y + z = 5
2 x + y – z = 2
2 x – y + z = 2

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Previous Year Papers

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Test Yourself

Long Answer Type

273. Use matrix method to solve the following system of equations:2x – 3y + 3z = 12x+ 2y+ 3z = 23x – 2y + 2z = 3

273.
Use matrix method to solve the following system of equations:
2x – 3y + 3z = 1
2x+ 2y+ 3z = 2
3x – 2y + 2z = 3