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Determinants

Short Answer Type

311.

For what values of k, the system of linear equations

x + y + z = 2
2x + y - z = 3
3x + 2y + kz = 4

has a unique solution?

1308 Views

Long Answer Type

312.

Using properties of determinants, prove that

$open vertical bar table row cell left parenthesis straight x plus straight y right parenthesis squared end cell zx zy row zx cell left parenthesis straight z plus straight y right parenthesis squared end cell xy row zy xy cell left parenthesis straight z plus straight x right parenthesis squared end cell end table close vertical bar space space equals space 2 xyz left parenthesis straight x plus straight y plus straight z right parenthesis cubed$

1198 Views

Short Answer Type

313.

Using the properties of determinants, solve the following for x:

$open vertical bar table row cell straight x plus 2 end cell cell space space straight x plus 6 end cell cell space straight x minus 1 end cell row cell straight x plus 6 end cell cell straight x minus 1 end cell cell straight x plus 2 end cell row cell straight x minus 1 end cell cell straight x plus 2 end cell cell straight x plus 6 end cell end table close vertical bar space equals space 0$

751 Views

314.

If $open vertical bar table row cell 3 straight x end cell cell space 7 end cell row cell negative 2 end cell cell space 4 end cell end table close vertical bar space equals open vertical bar table row 8 cell space 7 end cell row 6 cell space 4 end cell end table close vertical bar comma$ find the value of x.

196 Views

315.

Use Properties of determinants, prove that:
$open vertical bar table row cell 1 plus straight a end cell cell space 1 end cell cell space 1 end cell row 1 cell 1 plus straight b end cell 1 row 1 1 cell 1 plus straight c end cell end table close vertical bar space equals space abc plus bc plus ca plus ab$

389 Views

316.

If $open vertical bar table row cell straight x plus 1 end cell cell space space straight x minus 1 end cell row cell straight x minus 3 end cell cell space straight x plus 2 end cell end table close vertical bar space equals space open vertical bar table row 4 cell space space minus 1 end cell row 1 cell space space space space 3 end cell end table close vertical bar comma$ then write the value of x.

193 Views

317.

Using properties of determinants prove the following:
$open vertical bar table row 1 cell space space straight x end cell cell space space straight x squared end cell row cell straight x squared end cell cell space 1 end cell straight x row straight x cell space straight x end cell 1 end table close vertical bar space equals space left parenthesis 1 minus straight x cubed right parenthesis squared$

247 Views

318.

Using properties of determinants, prove that

594 Views

Long Answer Type

319.
Use product $open square brackets table row 1 cell negative 1 end cell 2 row 0 2 cell negative 3 end cell row 3 cell negative 2 end cell 4 end table close square brackets space space open square brackets table row cell negative 2 end cell 0 1 row 9 2 cell negative 3 end cell row 6 1 cell negative 2 end cell end table close square brackets$ to solve the system of equations x + 3z = 9,
–x + 2y – 2z = 4, 2x – 3y + 4z = –3

straight A space equals open square brackets table row 1 cell negative 1 end cell 2 row 0 2 cell negative 3 end cell row 3 cell negative 2 end cell 4 end table close square brackets space straight B space equals space open square brackets table row cell negative 2 end cell 0 1 row 9 2 cell negative 3 end cell row 6 1 cell negative 2 end cell end table close square brackets AxB space equals space open square brackets table row 1 cell negative 1 end cell 2 row 0 2 cell negative 3 end cell row 3 cell negative 2 end cell 4 end table close square brackets space open square brackets table row cell negative 2 end cell 0 1 row 9 2 cell negative 3 end cell row 6 1 cell negative 2 end cell end table close square brackets space equals open square brackets table row cell negative 2 minus 9 plus 12 space end cell cell space 0 minus 2 plus 2 space end cell cell space 1 plus 3 minus 4 space end cell row cell 0 plus 18 minus 18 space end cell cell space 0 plus 4 minus 3 space end cell cell space 0 minus 6 plus 6 end cell row cell negative 6 minus 18 plus 24 space end cell cell space 0 minus 4 plus 4 space end cell cell space 3 plus 6 minus 8 end cell end table close square brackets equals open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets

Since A X B = I,
therefore, B = A-1..... (i)
Now, the given system of equations is
x+3z =9
-x+2y-2z =4
2x-3y+4z = -3
This can also be represented as,
$open square brackets table row 1 0 3 row cell negative 1 end cell 2 cell negative 2 end cell row 2 cell negative 3 end cell 4 end table close square brackets 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 9 row 4 row cell negative 3 end cell end table close square brackets Here comma space we space can space observe space that space open square brackets table row 1 0 3 row cell negative 1 end cell 2 cell negative 2 end cell row 2 cell negative 3 end cell 4 end table close square brackets space equals space straight A to the power of straight T So comma space straight A to the power of straight T 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 9 row 4 row cell negative 3 end cell end table close square brackets Multiply space the space above space expression space by space left parenthesis straight A to the power of straight T right parenthesis to the power of negative 1 end exponent open square brackets table row straight x row straight y row straight z end table close square brackets space equals space left parenthesis straight A to the power of straight T right parenthesis to the power of negative 1 end exponent space open square brackets table row 9 row 4 row cell negative 3 end cell end table close square brackets open square brackets table row straight x row straight y row straight z end table close square brackets space equals space left parenthesis straight A to the power of straight T right parenthesis to the power of negative 1 end exponent space open square brackets table row 9 row 4 row cell negative 3 end cell end table close square brackets open square brackets table row straight x row straight y row straight z end table close square brackets space equals space left parenthesis straight B right parenthesis to the power of straight T space space space space space open square brackets table row 9 row 4 row cell negative 3 end cell end table close square brackets equals open square brackets table row cell negative 2 end cell 0 1 row 9 2 cell negative 3 end cell row 6 1 cell negative 2 end cell end table close square brackets to the power of straight T open square brackets table row 9 row 4 row cell negative 3 end cell end table close square brackets equals open square brackets table row cell negative 18 plus 36 minus 18 end cell row cell 0 plus 8 minus 3 end cell row cell 9 minus 12 plus 6 end cell end table close square brackets equals open square brackets table row 0 row 5 row 3 end table close square brackets$
Hence x=0 y=5 and =3