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Solve the equation
$open vertical bar table row cell straight x plus straight a end cell cell space straight x end cell cell space straight x end cell row straight x cell space straight x plus straight a end cell cell space straight x end cell row straight x straight x cell space straight x plus straight a end cell end table close vertical bar space equals space 0 comma space space space straight a not equal to space 0.$

72 Views

Long Answer Type

122.

Solve:
$open vertical bar table row cell straight x plus 1 end cell 2 3 row 3 cell straight x plus 2 end cell 1 row 1 2 cell straight x plus 3 end cell end table close vertical bar space equals 0$

69 Views

123.

Prove that:
$open vertical bar table row cell straight a squared plus 1 end cell cell straight a space straight b end cell cell straight a space straight c end cell row cell straight a space straight b end cell cell straight b squared plus 1 end cell cell straight b space straight c end cell row cell straight a space straight c end cell cell straight b space straight c end cell cell straight c squared plus 1 end cell end table close vertical bar space equals space 1 plus straight a squared plus straight b squared plus straight c squared.$

94 Views

Short Answer Type

124.

If x, y, z are different and
$increment space equals space open vertical bar table row straight x cell space space space space straight x squared end cell cell space 1 plus straight x cubed end cell row straight y cell space space space straight y squared end cell cell space 1 plus straight y cubed end cell row straight z cell space space straight z squared end cell cell space 1 plus straight z cubed end cell end table close vertical bar space equals space 0 comma space space then space show space that space 1 plus xyz space equals 0$

70 Views

Long Answer Type

125.

Prove that:
$open vertical bar table row straight x cell space space space straight x squared end cell cell space space 1 plus px cubed end cell row straight y cell space space straight y squared end cell cell space space 1 plus py cubed end cell row straight z cell space space straight z squared end cell cell space space 1 plus pz cubed end cell end table close vertical bar space equals space left parenthesis 1 plus pxyz right parenthesis thin space left parenthesis straight x minus straight y right parenthesis thin space left parenthesis straight y minus straight z right parenthesis thin space left parenthesis straight z minus straight x right parenthesis$
where p is any scalar.

76 Views

Short Answer Type

126.

If $space open vertical bar table row straight a cell space space straight a cubed end cell cell space space straight a to the power of 4 minus 1 end cell row straight b cell space straight b cubed end cell cell space space straight b to the power of 4 minus 1 end cell row straight c cell space straight c cubed end cell cell space space straight c to the power of 4 minus 1 end cell end table close vertical bar space equals space 0 comma$
show that
a + b + c = abc(ab + bc + ca) where a , b , c are all different.

72 Views

127.
Let the three digit numbers A28,3B9 and 62C where A, B, and C are any integers between 0 and 9, be divisible by a fixed integer k. Show that the determinant
$open vertical bar table row straight A cell space space 3 end cell cell space space space 6 end cell row 8 cell space 9 end cell cell space space straight C end cell row 2 cell space straight B end cell cell space space 2 end cell end table close vertical bar$
is divisible by k.

The numbers A28, 3B9, 62C are divisible by k.
Let A28 = ka, 3B9 = k b, 62C = kc where a , b , c are integers.

Let space increment space equals space open vertical bar table row straight A cell space space 3 end cell cell space space 6 end cell row 8 cell space space 9 end cell cell space space straight C end cell row 2 cell space space straight B end cell cell space space 2 end cell end table close vertical bar

equals space open vertical bar table row straight A 3 6 row cell 8 plus 100 straight A plus 10.2 end cell cell space space space space space space space 9 plus 100. space 3 plus 10. space straight B end cell cell space space space space space space space space space straight C plus 100.6 plus 10.2 end cell row 2 straight B 2 end table close vertical bar space space

straight R subscript 2 plus 100 space straight R subscript 1 plus 10 space straight R subscript 3

equals space open vertical bar table row straight A cell space space space space 3 end cell cell space space space space 6 end cell row cell straight A 28 end cell cell space space space 3 straight B 9 end cell cell space space space space 62 straight C end cell row 2 straight B cell space space space 2 end cell end table close vertical bar equals space open vertical bar table row straight A cell space space space 3 end cell cell space space space 6 end cell row cell straight k space straight a end cell cell space space space straight k space straight b end cell cell space space space space straight k space straight c end cell row 2 straight B cell space space 2 end cell end table close vertical bar space equals space straight k space open vertical bar table row straight A cell space space space 3 end cell cell space space space 6 end cell row straight a cell space space straight b end cell cell space space space straight c end cell row 2 cell space space straight B end cell cell space space space 2 end cell end table close vertical bar equals space integral space multiple space of space straight k therefore space space space space space increment space is space divisible space by space straight k.

151 Views

Long Answer Type

128.

Prove that:
$open vertical bar table row cell left parenthesis straight b plus straight c right parenthesis squared end cell cell straight a squared end cell cell straight a squared end cell row cell straight b squared end cell cell left parenthesis straight c plus straight a right parenthesis squared end cell cell straight b squared end cell row cell straight c squared end cell cell straight c squared end cell cell left parenthesis straight a plus straight b right parenthesis squared end cell end table close vertical bar space equals space 2 space abc space left parenthesis straight a plus straight b plus straight c right parenthesis cubed.$

98 Views

129.

Prove that:
$open vertical bar table row cell negative 2 straight a end cell cell space space space space straight a plus straight b end cell cell space space space straight a plus straight c end cell row cell straight b plus straight a end cell cell space space minus 2 straight b end cell cell space space space straight b plus straight c end cell row cell straight c plus straight a end cell cell space space straight c plus straight b end cell cell space space minus 2 straight c end cell end table close vertical bar space equals space 4 left parenthesis straight a plus straight b right parenthesis thin space left parenthesis straight b plus straight c right parenthesis thin space left parenthesis straight c plus straight a right parenthesis.$