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Determinants

Long Answer Type

231.

Examine the consistency of the system of equations:
3x – y – 2z = 2
2y – z = – 1
3x – 5y = 3

80 Views

Short Answer Type

232.

Examine the consistency of the system of equations:
5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = – 1

90 Views

Long Answer Type

233.
Examine the consistencies of the system of equations:
3x – y + 2z = 3
2x + y + 3z =5
x - 2y - z = 1

The given equations are
3x – y + 2z = 3
2x + y + 3z = 5
x – 2 – z = 1
These equations can be written as
$open square brackets table row 3 cell space space minus 1 end cell cell space space space space space space 2 end cell row 2 cell space space space space space 1 end cell cell space space space space space space 3 end cell row 1 cell space space minus 2 end cell cell space space minus 1 end cell end table close square brackets space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 3 row 5 row 1 end table close square brackets$

or $AX space equals space straight B space where space straight A space equals space open square brackets table row 3 cell space space minus 1 end cell cell space space 2 end cell row 2 cell space space 1 end cell cell space 3 end cell row 1 cell space minus 2 end cell cell space 1 end cell end table close square brackets comma space space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space straight B space equals space open square brackets table row 3 row 5 row 1 end table close square brackets$
$open vertical bar straight A close vertical bar space equals space open vertical bar table row 3 cell space minus 1 end cell cell space space space space 2 end cell row 2 cell space space space 1 end cell cell space space space space space 3 space end cell row 1 cell negative 2 end cell cell space minus 1 end cell end table close vertical bar space equals space 3 left parenthesis negative 1 plus 6 right parenthesis minus left parenthesis negative 1 right parenthesis minus left parenthesis negative 2 minus 3 right parenthesis plus 2 left parenthesis negative 4 minus 1 right parenthesis space space space space space equals space 15 minus 5 minus 10 space equals space 0$
Co-factors of the elements of the first row of | A | are
$open vertical bar table row cell space space 1 end cell cell space space space space space space 3 end cell row cell negative 2 end cell cell space space minus 1 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space space space space space 3 end cell row 1 cell space space space minus 1 end cell end table close vertical bar comma space space open vertical bar table row 2 cell space space space space space space 1 end cell row 1 cell space space space space minus 2 end cell end table close vertical bar$

or – 1 + 6, – (– 2, – 3), – 4 – 1 or 5, 5 – 5 respectively.
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row cell negative 1 end cell cell space space space space space space space 2 end cell row cell negative 2 end cell cell space space space minus 1 end cell end table close vertical bar comma space space space space open vertical bar table row 3 cell space space space space space space 2 end cell row 1 cell space space minus 1 end cell end table close vertical bar comma space space space minus open vertical bar table row 3 cell space space space minus 1 end cell row 1 cell space space space minus 2 end cell end table close vertical bar or space space space minus 5 comma space space minus 5 comma space space 5 space respectively.$
Co-factors of the elements of the third row of | A | are
$open vertical bar table row cell negative 1 end cell cell space space space 2 end cell row 1 cell space space 3 end cell end table close vertical bar comma space space minus open vertical bar table row 3 cell space space space space space 2 end cell row 2 cell space space space space space 3 end cell end table close vertical bar comma space space open vertical bar table row 3 cell space space space space minus 1 end cell row 2 cell space space space space space space 1 end cell end table close vertical bar$
or – 5, – 5, 5 respectively.
$therefore space space adj. space straight A space equals space open square brackets table row cell space space space 5 end cell cell space space space space space space space 5 end cell cell space space space minus 5 end cell row cell negative 5 end cell cell space space space minus 5 end cell cell space space space space space space 5 end cell row cell negative 5 end cell cell space space space minus 5 end cell cell space space space space space 5 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 5 cell space space space minus 5 end cell cell space space space space minus 5 end cell row 5 cell space space minus 5 end cell cell space space minus 5 end cell row cell negative 5 end cell cell space space space space space 5 end cell cell space space space space space 5 end cell end table close square brackets space space space space space space space left parenthesis adj. space straight A right parenthesis thin space straight B space equals space open square brackets table row cell space space 5 end cell cell space space space minus 5 end cell cell space space space space minus 5 end cell row cell space 5 end cell cell space space minus 5 end cell cell space space space minus 5 end cell row cell negative 5 space end cell cell space space space space space 5 end cell cell space space space space space 5 end cell end table close square brackets space open square brackets table row 3 row 5 row 1 end table close square brackets space space space space space space space space space space space space space space space space space space space space space space space equals space open square brackets table row cell 15 minus 25 minus 5 end cell row cell 15 minus 25 minus 5 end cell row cell negative 15 plus 25 plus 5 end cell end table close square brackets space equals space open square brackets table row cell negative 15 end cell row cell negative 15 end cell row 15 end table close square brackets not equal to space straight O$

given equations have no solution.

91 Views

234.

Examine the consistencies of the system of equations:
x - y+ z = 3
2x - y – z = 2
– x – 2y + 2z = 1

74 Views

Short Answer Type

235. Solve the following system of equations using matrix method:
2x + 5y = 1
1x + 2y = 7

72 Views

Long Answer Type

236. Solve system of linear equations, using matrix method:
5x + 2y = 4
7x + 3y = 5

74 Views

Short Answer Type

237. Solve system of linear equations, using matrix method:
2x - y = -2
3x + 4y = 3

93 Views

238. Solve system of linear equations, using matrix method:
4x – 3y = 3
3x – 5y = 7

75 Views

Long Answer Type

239. Solve system of linear equations, using matrix method:
5x + 2y = 3
3x + 2y = 5

74 Views

Short Answer Type

240.

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

79 Views

Previous Year Papers

Test Series

Test Yourself

Long Answer Type

Short Answer Type

Long Answer Type

233. Examine the consistencies of the system of equations:3x – y + 2z = 32x + y + 3z =5x - 2y - z = 1

Short Answer Type

Long Answer Type

Short Answer Type

Long Answer Type

Short Answer Type

233.
Examine the consistencies of the system of equations:
3x – y + 2z = 3
2x + y + 3z =5
x - 2y - z = 1