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If $open square brackets table row alpha beta row gamma cell negative alpha end cell end table close square brackets$ is square root of I2, then $alpha comma beta$ and $gamma$ will satisfy the relation -

1+ $α^{2} + β γ$ = 0
1– $α^{2} + β γ$ = 0
1+ $α^{2} - β γ$ = 0
–1+ $α^{2} + β γ$ = 0

The correct answer is: –1+ $alpha squared plus beta gamma$ = 0

To find the relation between $alpha comma beta$ and $gamma$ .
$open square brackets table row alpha beta row gamma cell negative alpha end cell end table close square brackets$
$open square brackets table row alpha beta row gamma cell negative alpha end cell end table close square brackets$

$open square brackets table row cell alpha squared plus beta gamma end cell 0 row 0 cell alpha squared plus beta gamma end cell end table close square brackets$

$Comparing both sides,$

–1+ $alpha squared plus beta gamma$ = 0

Therefore, relation between $alpha comma beta$ and $gamma$ is –1+ $alpha squared plus beta gamma$ = 0.

If A = $open square brackets table row 2 0 1 row 2 1 3 row 1 cell negative 1 end cell 0 end table close square brackets$ then A² – 5A + 6I =

Maths-General

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Maths-

If A = $open square brackets table row 0 cell negative i end cell row i 0 end table close square brackets$ , B = $open square brackets table row 0 1 row 1 0 end table close square brackets$ , C = $open square brackets table row 1 0 row 0 cell negative 1 end cell end table close square brackets$ , then which of the following statement is true-

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Maths-

If A = $open square brackets table row 0 1 row 1 0 end table close square brackets$ and B = $open square brackets table row 0 cell negative i end cell row i 0 end table close square brackets$ , then -

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Maths-

$open square brackets table row 1 cell negative tan invisible function application alpha divided by 2 end cell row cell tan invisible function application alpha divided by 2 end cell 1 end table close square brackets$ $open square brackets table row cell tan invisible function application alpha divided by 2 end cell 1 row 1 0 end table close square brackets$ equals-

Maths-General

General

Maths-

Statement - I For any real value of $theta not equal to left parenthesis 2 n plus 1 right parenthesis pi$ or $left parenthesis 2 straight n plus 1 right parenthesis pi divided by 2 comma straight n element of straight I$ the value of the expression $straight y equals fraction numerator cos squared begin display style space end style theta minus 1 over denominator cos squared begin display style space end style theta plus cos space theta end fraction text is end text straight y less or equal than$ 0 or y≥2 (either less than or equal to zero or greater than or equal to two)
Statement - II $sec space theta element of left parenthesis negative straight infinity comma negative 1 right square bracket union left square bracket 1 comma straight infinity right parenthesis$ for all real values of $theta$ .

In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not, is same as assertion and reason

Statement - I For any real value of $theta not equal to left parenthesis 2 n plus 1 right parenthesis pi$ or $left parenthesis 2 straight n plus 1 right parenthesis pi divided by 2 comma straight n element of straight I$ the value of the expression $straight y equals fraction numerator cos squared begin display style space end style theta minus 1 over denominator cos squared begin display style space end style theta plus cos space theta end fraction text is end text straight y less or equal than$ 0 or y≥2 (either less than or equal to zero or greater than or equal to two)
Statement - II $sec space theta element of left parenthesis negative straight infinity comma negative 1 right square bracket union left square bracket 1 comma straight infinity right parenthesis$ for all real values of $theta$ .

Maths-General

In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not, is same as assertion and reason

General

Maths-

If A = $open square brackets table row cell cos invisible function application alpha end cell cell sin invisible function application alpha end cell row cell negative sin invisible function application alpha end cell cell cos invisible function application alpha end cell end table close square brackets$ , then A² equals-

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Maths-

The root of the equation [x 1 2] $open square brackets table attributes columnalign left left left columnspacing 1em end attributes row cell 0 1 1 end cell row cell 1 0 1 end cell row cell 1 1 0 end cell end table close square brackets$ $open square brackets table row x row cell negative 1 end cell row 1 end table close square brackets$ = 0 is-

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Maths-

If A = $open square brackets table row 3 1 row 7 5 end table close square brackets$ and A² + kI = 8A, then k equals

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Maths-

If A and B are matrices of order m × n and n × n respectively, then which of the following are defined-

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chemistry-

The multiplicative inverse of A = $open square brackets table row cell cos invisible function application theta end cell cell negative sin invisible function application theta end cell row cell sin invisible function application theta end cell cell cos invisible function application theta end cell end table close square brackets$ is

maths-General

General

chemistry-

(A) and (B) are:

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A = $open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets$ , then A³ – 4A² – 6A is equal to -

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If $open square brackets table row alpha beta row gamma cell negative alpha end cell end table close square brackets$ is square root of I2, then $alpha comma beta$ and $gamma$ will satisfy the relation -

1+ $α^{2} + β γ$ = 0
1– $α^{2} + β γ$ = 0
1+ $α^{2} - β γ$ = 0
–1+ $α^{2} + β γ$ = 0

The correct answer is: –1+ $alpha squared plus beta gamma$ = 0

Related Questions to study

If A = $open square brackets table row 2 0 1 row 2 1 3 row 1 cell negative 1 end cell 0 end table close square brackets$ then A² – 5A + 6I =

If A = $open square brackets table row 2 0 1 row 2 1 3 row 1 cell negative 1 end cell 0 end table close square brackets$ then A² – 5A + 6I =

If A = $open square brackets table row 0 1 row 1 0 end table close square brackets$ and B = $open square brackets table row 0 cell negative i end cell row i 0 end table close square brackets$ , then -

If A = $open square brackets table row 0 1 row 1 0 end table close square brackets$ and B = $open square brackets table row 0 cell negative i end cell row i 0 end table close square brackets$ , then -

If A = $open square brackets table row 3 1 row 7 5 end table close square brackets$ and A² + kI = 8A, then k equals

If A = $open square brackets table row 3 1 row 7 5 end table close square brackets$ and A² + kI = 8A, then k equals

If A and B are matrices of order m × n and n × n respectively, then which of the following are defined-

If A and B are matrices of order m × n and n × n respectively, then which of the following are defined-

the products (A) and (B) are:

the products (A) and (B) are:

A matrix A = (aij) m x n is said to be a square matrix if-

A matrix A = (aij) m x n is said to be a square matrix if-

For any square matrix A = [aij], aij = 0, when i $not equal to$ j, then A is-

For any square matrix A = [aij], aij = 0, when i $not equal to$ j, then A is-

(A) and (B) are:

(A) and (B) are:

A = $open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets$ , then A³ – 4A² – 6A is equal to -

A = $open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets$ , then A³ – 4A² – 6A is equal to -

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If is square root of I2, then and will satisfy the relation -

1+ = 0 1– = 0 1+ = 0 –1+ = 0

The correct answer is: –1+ = 0

To find the relation between and .Given, ​ = ​ x = = Comparing both sides, –1+ = 0

Related Questions to study

If A = then A2 – 5A + 6I =

If A = then A2 – 5A + 6I =

If A =, B =, C =, then which of the following statement is true-

If A =, B =, C =, then which of the following statement is true-

If A = and B = , then -

If A = and B = , then -

equals-

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Statement - I For any real value of or the value of the expression 0 or y≥2 (either less than or equal to zero or greater than or equal to two)Statement - II for all real values of .

Statement - I For any real value of or the value of the expression 0 or y≥2 (either less than or equal to zero or greater than or equal to two)Statement - II for all real values of .

If A =, then A2 equals-

If A =, then A2 equals-

The root of the equation [x 1 2] = 0 is-

The root of the equation [x 1 2] = 0 is-

If A = and A2 + kI = 8A, then k equals

If A = and A2 + kI = 8A, then k equals

If A and B are matrices of order m × n and n × n respectively, then which of the following are defined-

If A and B are matrices of order m × n and n × n respectively, then which of the following are defined-

the products (A) and (B) are:

the products (A) and (B) are:

A matrix A = (aij) m x n is said to be a square matrix if-

A matrix A = (aij) m x n is said to be a square matrix if-

For any square matrix A = [aij], aij = 0, when i j, then A is-

For any square matrix A = [aij], aij = 0, when i j, then A is-

The multiplicative inverse of A =is

The multiplicative inverse of A =is

(A) and (B) are:

(A) and (B) are:

A = , then A3 – 4A2 – 6A is equal to -

A = , then A3 – 4A2 – 6A is equal to -

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If $open square brackets table row alpha beta row gamma cell negative alpha end cell end table close square brackets$ is square root of I2, then $alpha comma beta$ and $gamma$ will satisfy the relation -

1+ $α^{2} + β γ$ = 0
1– $α^{2} + β γ$ = 0
1+ $α^{2} - β γ$ = 0
–1+ $α^{2} + β γ$ = 0

The correct answer is: –1+ $alpha squared plus beta gamma$ = 0

If A = $open square brackets table row 2 0 1 row 2 1 3 row 1 cell negative 1 end cell 0 end table close square brackets$ then A² – 5A + 6I =

If A = $open square brackets table row 2 0 1 row 2 1 3 row 1 cell negative 1 end cell 0 end table close square brackets$ then A² – 5A + 6I =

If A = $open square brackets table row 0 1 row 1 0 end table close square brackets$ and B = $open square brackets table row 0 cell negative i end cell row i 0 end table close square brackets$ , then -

If A = $open square brackets table row 0 1 row 1 0 end table close square brackets$ and B = $open square brackets table row 0 cell negative i end cell row i 0 end table close square brackets$ , then -

If A = $open square brackets table row 3 1 row 7 5 end table close square brackets$ and A² + kI = 8A, then k equals

If A = $open square brackets table row 3 1 row 7 5 end table close square brackets$ and A² + kI = 8A, then k equals

For any square matrix A = [aij], aij = 0, when i $not equal to$ j, then A is-

For any square matrix A = [aij], aij = 0, when i $not equal to$ j, then A is-

A = $open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets$ , then A³ – 4A² – 6A is equal to -

A = $open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets$ , then A³ – 4A² – 6A is equal to -