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Question

If $a comma blank b comma blank c$ are non-zeros, then the system of equations $open parentheses alpha plus a close parentheses x plus alpha y plus alpha z equals 0 comma blank alpha x plus open parentheses alpha plus b close parentheses y plus alpha z equals 0 comma alpha x plus alpha y plus open parentheses alpha plus c close parentheses z equals 0$ has a non-trivial solution if

$α^{- 1} = - (a^{- 1} + b^{- 1} + c^{- 1})$
$α^{- 1} = a + b + c$
$α + a + b + c = 1$
None of these

The correct answer is: $alpha to the power of negative 1 end exponent equals negative left parenthesis a to the power of negative 1 end exponent plus b to the power of negative 1 end exponent plus c to the power of negative 1 end exponent right parenthesis$

The given system of equations will have a non-trivial solution if
$open vertical bar table row cell alpha plus a end cell alpha alpha row alpha cell alpha plus b end cell alpha row alpha alpha cell alpha plus c end cell end table close vertical bar equals 0$
Operating $R subscript 2 end subscript rightwards arrow R subscript 2 end subscript minus R subscript 1 end subscript$ and $R subscript 3 end subscript rightwards arrow R subscript 3 end subscript minus R subscript 1 end subscript$ , we get
$open vertical bar table row cell alpha plus a end cell alpha alpha row cell negative a end cell b 0 row cell negative a end cell 0 c end table close vertical bar equals 0$
$rightwards double arrow blank alpha a b plus c open parentheses alpha b plus a b plus a alpha close parentheses equals 0 rightwards double arrow alpha open parentheses b c plus c a plus a b close parentheses plus a b c equals 0$
$rightwards double arrow blank fraction numerator 1 over denominator alpha end fraction equals negative open parentheses fraction numerator 1 over denominator a end fraction plus fraction numerator 1 over denominator b end fraction plus fraction numerator 1 over denominator c end fraction close parentheses left parenthesis because a comma blank b comma blank c not equal to 0 right parenthesis$

Related Questions to study

$increment subscript 1 end subscript equals open vertical bar table row cell y to the power of 5 end exponent z to the power of 6 end exponent left parenthesis z to the power of 3 end exponent minus y to the power of 3 end exponent right parenthesis end cell cell x to the power of 4 end exponent z to the power of 6 end exponent left parenthesis x to the power of 3 end exponent minus z to the power of 3 end exponent right parenthesis end cell cell x to the power of 4 end exponent y to the power of 5 end exponent left parenthesis y to the power of 3 end exponent minus x to the power of 3 end exponent right parenthesis end cell row cell y to the power of 2 end exponent z to the power of 3 end exponent left parenthesis y to the power of 6 end exponent minus z to the power of 6 end exponent right parenthesis end cell cell x z to the power of 3 end exponent left parenthesis z to the power of 6 end exponent minus x to the power of 6 end exponent right parenthesis end cell cell x y to the power of 2 end exponent left parenthesis x to the power of 6 end exponent minus y to the power of 6 end exponent right parenthesis end cell row cell y to the power of 2 end exponent z to the power of 3 end exponent left parenthesis z to the power of 3 end exponent minus y to the power of 3 end exponent right parenthesis end cell cell x z to the power of 3 end exponent left parenthesis x to the power of 3 end exponent minus z to the power of 3 end exponent right parenthesis end cell cell x y to the power of 2 end exponent left parenthesis y to the power of 3 end exponent minus x to the power of 3 end exponent right parenthesis end cell end table close vertical bar$ and $increment subscript 2 end subscript equals open vertical bar table row x cell y to the power of 2 end exponent end cell cell z to the power of 3 end exponent end cell row cell x to the power of 4 end exponent end cell cell y to the power of 5 end exponent end cell cell z to the power of 6 end exponent end cell row cell x to the power of 7 end exponent end cell cell y to the power of 8 end exponent end cell cell z to the power of 9 end exponent end cell end table close vertical bar$ . Then $increment subscript 1 end subscript increment subscript 2 end subscript$ is equal to

$increment subscript 1 end subscript equals open vertical bar table row cell y to the power of 5 end exponent z to the power of 6 end exponent left parenthesis z to the power of 3 end exponent minus y to the power of 3 end exponent right parenthesis end cell cell x to the power of 4 end exponent z to the power of 6 end exponent left parenthesis x to the power of 3 end exponent minus z to the power of 3 end exponent right parenthesis end cell cell x to the power of 4 end exponent y to the power of 5 end exponent left parenthesis y to the power of 3 end exponent minus x to the power of 3 end exponent right parenthesis end cell row cell y to the power of 2 end exponent z to the power of 3 end exponent left parenthesis y to the power of 6 end exponent minus z to the power of 6 end exponent right parenthesis end cell cell x z to the power of 3 end exponent left parenthesis z to the power of 6 end exponent minus x to the power of 6 end exponent right parenthesis end cell cell x y to the power of 2 end exponent left parenthesis x to the power of 6 end exponent minus y to the power of 6 end exponent right parenthesis end cell row cell y to the power of 2 end exponent z to the power of 3 end exponent left parenthesis z to the power of 3 end exponent minus y to the power of 3 end exponent right parenthesis end cell cell x z to the power of 3 end exponent left parenthesis x to the power of 3 end exponent minus z to the power of 3 end exponent right parenthesis end cell cell x y to the power of 2 end exponent left parenthesis y to the power of 3 end exponent minus x to the power of 3 end exponent right parenthesis end cell end table close vertical bar$ and $increment subscript 2 end subscript equals open vertical bar table row x cell y to the power of 2 end exponent end cell cell z to the power of 3 end exponent end cell row cell x to the power of 4 end exponent end cell cell y to the power of 5 end exponent end cell cell z to the power of 6 end exponent end cell row cell x to the power of 7 end exponent end cell cell y to the power of 8 end exponent end cell cell z to the power of 9 end exponent end cell end table close vertical bar$ . Then $increment subscript 1 end subscript increment subscript 2 end subscript$ is equal to

The system of equations $a x minus y minus z equals alpha minus 1$ $x minus alpha y minus z equals alpha minus 1$ $x minus y minus alpha z equals alpha minus 1$ Has no solution if $alpha$ is

The system of equations $a x minus y minus z equals alpha minus 1$ $x minus alpha y minus z equals alpha minus 1$ $x minus y minus alpha z equals alpha minus 1$ Has no solution if $alpha$ is

The value of the determinant $open vertical bar table row 1 1 1 row cell blank to the power of m end exponent C subscript 1 end subscript end cell cell blank to the power of m plus 1 end exponent C subscript 1 end subscript end cell cell blank to the power of m plus 2 end exponent C subscript 1 end subscript end cell row cell blank to the power of m end exponent C subscript 2 end subscript end cell cell blank to the power of m plus 1 end exponent C subscript 2 end subscript end cell cell blank to the power of m plus 2 end exponent C subscript 2 end subscript end cell end table close vertical bar$ is equal to

The value of the determinant $open vertical bar table row 1 1 1 row cell blank to the power of m end exponent C subscript 1 end subscript end cell cell blank to the power of m plus 1 end exponent C subscript 1 end subscript end cell cell blank to the power of m plus 2 end exponent C subscript 1 end subscript end cell row cell blank to the power of m end exponent C subscript 2 end subscript end cell cell blank to the power of m plus 1 end exponent C subscript 2 end subscript end cell cell blank to the power of m plus 2 end exponent C subscript 2 end subscript end cell end table close vertical bar$ is equal to

parallel

If $alpha comma blank beta comma blank gamma$ are the angles of a triangle and the system of equations $cos invisible function application open parentheses alpha minus beta close parentheses x plus cos invisible function application open parentheses beta minus gamma close parentheses y plus cos invisible function application open parentheses gamma minus alpha close parentheses z equals 0 cos invisible function application open parentheses alpha plus beta close parentheses x plus cos invisible function application open parentheses beta plus gamma close parentheses y plus cos invisible function application open parentheses gamma plus alpha close parentheses z equals 0 sin invisible function application open parentheses alpha plus beta close parentheses x plus sin invisible function application open parentheses beta plus gamma close parentheses y plus sin invisible function application open parentheses gamma plus alpha close parentheses z equals 0$ Has non-trivial solutions, then triangle is necessarily

If $alpha comma blank beta comma blank gamma$ are the angles of a triangle and the system of equations $cos invisible function application open parentheses alpha minus beta close parentheses x plus cos invisible function application open parentheses beta minus gamma close parentheses y plus cos invisible function application open parentheses gamma minus alpha close parentheses z equals 0 cos invisible function application open parentheses alpha plus beta close parentheses x plus cos invisible function application open parentheses beta plus gamma close parentheses y plus cos invisible function application open parentheses gamma plus alpha close parentheses z equals 0 sin invisible function application open parentheses alpha plus beta close parentheses x plus sin invisible function application open parentheses beta plus gamma close parentheses y plus sin invisible function application open parentheses gamma plus alpha close parentheses z equals 0$ Has non-trivial solutions, then triangle is necessarily

If $x not equal to y not equal to z$ and $open vertical bar table row x cell x to the power of 2 end exponent end cell cell 1 plus x to the power of 3 end exponent end cell row y cell y to the power of 2 end exponent end cell cell 1 plus y to the power of 3 end exponent end cell row z cell z to the power of 2 end exponent end cell cell 1 plus z to the power of 3 end exponent end cell end table close vertical bar equals 0$ , then the value of $x y z$ is

If $x not equal to y not equal to z$ and $open vertical bar table row x cell x to the power of 2 end exponent end cell cell 1 plus x to the power of 3 end exponent end cell row y cell y to the power of 2 end exponent end cell cell 1 plus y to the power of 3 end exponent end cell row z cell z to the power of 2 end exponent end cell cell 1 plus z to the power of 3 end exponent end cell end table close vertical bar equals 0$ , then the value of $x y z$ is

If $f open parentheses x close parentheses equals a plus b x plus c x to the power of 2 end exponent$ and $alpha comma blank beta comma blank gamma$ are the roots of the equation $x to the power of 3 end exponent equals 1$ , then $open vertical bar table row a b c row b c a row c a b end table close vertical bar$ is equal to

If $f open parentheses x close parentheses equals a plus b x plus c x to the power of 2 end exponent$ and $alpha comma blank beta comma blank gamma$ are the roots of the equation $x to the power of 3 end exponent equals 1$ , then $open vertical bar table row a b c row b c a row c a b end table close vertical bar$ is equal to

parallel

If $A equals open square brackets table row alpha 2 row 2 alpha end table close square brackets$ and $open vertical bar A to the power of 3 end exponent close vertical bar equals 125 comma blank$ then the value of $alpha$ is

If $A equals open square brackets table row alpha 2 row 2 alpha end table close square brackets$ and $open vertical bar A to the power of 3 end exponent close vertical bar equals 125 comma blank$ then the value of $alpha$ is

If $omega blank open parentheses not equal to 1 close parentheses$ is a cube root of unity, then value of the determinant $open vertical bar table row 1 cell 1 plus i plus omega to the power of 2 end exponent end cell cell omega to the power of 2 end exponent end cell row cell 1 minus i end cell cell negative 1 end cell cell omega to the power of 2 end exponent minus 1 end cell row cell negative i end cell cell negative i plus omega minus 1 end cell cell negative 1 end cell end table close vertical bar$ is

If $omega blank open parentheses not equal to 1 close parentheses$ is a cube root of unity, then value of the determinant $open vertical bar table row 1 cell 1 plus i plus omega to the power of 2 end exponent end cell cell omega to the power of 2 end exponent end cell row cell 1 minus i end cell cell negative 1 end cell cell omega to the power of 2 end exponent minus 1 end cell row cell negative i end cell cell negative i plus omega minus 1 end cell cell negative 1 end cell end table close vertical bar$ is

If $A subscript 1 end subscript comma blank B subscript 1 end subscript comma blank C subscript 1 end subscript comma blank horizontal ellipsis$ are, respectively, the cofactors of the elements $a subscript 1 end subscript comma blank b subscript 1 end subscript comma blank c subscript 1 end subscript comma blank horizontal ellipsis$ of the determinant $increment equals open vertical bar table row cell a subscript 1 end subscript end cell cell b subscript 1 end subscript end cell cell c subscript 1 end subscript end cell row cell a subscript 2 end subscript end cell cell b subscript 2 end subscript end cell cell c subscript 2 end subscript end cell row cell a subscript 3 end subscript end cell cell b subscript 3 end subscript end cell cell c subscript 3 end subscript end cell end table close vertical bar comma blank increment not equal to 0$ , then the value of $open vertical bar table row cell B subscript 2 end subscript end cell cell C subscript 2 end subscript end cell row cell B subscript 3 end subscript end cell cell C subscript 3 end subscript end cell end table close vertical bar$ is equal to

If $A subscript 1 end subscript comma blank B subscript 1 end subscript comma blank C subscript 1 end subscript comma blank horizontal ellipsis$ are, respectively, the cofactors of the elements $a subscript 1 end subscript comma blank b subscript 1 end subscript comma blank c subscript 1 end subscript comma blank horizontal ellipsis$ of the determinant $increment equals open vertical bar table row cell a subscript 1 end subscript end cell cell b subscript 1 end subscript end cell cell c subscript 1 end subscript end cell row cell a subscript 2 end subscript end cell cell b subscript 2 end subscript end cell cell c subscript 2 end subscript end cell row cell a subscript 3 end subscript end cell cell b subscript 3 end subscript end cell cell c subscript 3 end subscript end cell end table close vertical bar comma blank increment not equal to 0$ , then the value of $open vertical bar table row cell B subscript 2 end subscript end cell cell C subscript 2 end subscript end cell row cell B subscript 3 end subscript end cell cell C subscript 3 end subscript end cell end table close vertical bar$ is equal to

parallel

Let $stack a with rightwards arrow on top$ be a unit vector perpendicular to unit vectors $stack b with rightwards arrow on top a n d stack c with rightwards arrow on top$ inclined at an angle a, then $stack b with rightwards arrow on top cross times stack c with rightwards arrow on top$ is

Let $stack a with rightwards arrow on top$ be a unit vector perpendicular to unit vectors $stack b with rightwards arrow on top a n d stack c with rightwards arrow on top$ inclined at an angle a, then $stack b with rightwards arrow on top cross times stack c with rightwards arrow on top$ is

Let $stack a with rightwards arrow on top comma short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow stack b with rightwards arrow on top short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow a n d short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow stack c with rightwards arrow on top$ be three non-coplanar vectors and $stack d with rightwards arrow on top$ be a non-zero vector which is perpendicular to $open parentheses stack a with rightwards arrow on top plus stack b with rightwards arrow on top plus stack c with rightwards arrow on top close parentheses$ and is represented as $stack d with rightwards arrow on top equals x open parentheses stack a with rightwards arrow on top cross times stack b with rightwards arrow on top close parentheses plus y open parentheses stack b with rightwards arrow on top cross times stack c with rightwards arrow on top close parentheses plus z open parentheses stack c with rightwards arrow on top cross times stack a with rightwards arrow on top close parentheses$ . Then

Let $stack a with rightwards arrow on top comma short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow stack b with rightwards arrow on top short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow a n d short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow stack c with rightwards arrow on top$ be three non-coplanar vectors and $stack d with rightwards arrow on top$ be a non-zero vector which is perpendicular to $open parentheses stack a with rightwards arrow on top plus stack b with rightwards arrow on top plus stack c with rightwards arrow on top close parentheses$ and is represented as $stack d with rightwards arrow on top equals x open parentheses stack a with rightwards arrow on top cross times stack b with rightwards arrow on top close parentheses plus y open parentheses stack b with rightwards arrow on top cross times stack c with rightwards arrow on top close parentheses plus z open parentheses stack c with rightwards arrow on top cross times stack a with rightwards arrow on top close parentheses$ . Then

Let OPQR is a tetrahedron such that O is origin and $stack p with rightwards arrow on top comma short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow stack q with rightwards arrow on top comma short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow stack r with rightwards arrow on top$ are position vectors of P, Q, R respectively and a is the angle which OP makes with face PQR then

Hence required angle is 2.

Let OPQR is a tetrahedron such that O is origin and $stack p with rightwards arrow on top comma short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow stack q with rightwards arrow on top comma short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow stack r with rightwards arrow on top$ are position vectors of P, Q, R respectively and a is the angle which OP makes with face PQR then

Hence required angle is 2.

parallel

On strong heating Mgcl₂ 6H₂ Othe product obtained is

On strong heating Mgcl₂ 6H₂ Othe product obtained is

Chemistry-General

Which one of the following hydroxide is insoluble in water?

Which one of the following hydroxide is insoluble in water?

Chemistry-General

Which is the incorrect statement?

Which is the incorrect statement?

Chemistry-General

parallel

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