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Question

For suitable matrices A, B; the false statement is-

(AB)^T =A^TB^T
(A^T)^T= A
(A – B)^T =A^T– B^T
(A^T)–1 = (A^–1 )^T

hint Hint:

The correct answer is: (AB)^T =A^TB^T

For suitable matrices A, B we are asked to find the false statement
we know that by property

Therefore choice 1 is correct

If A = $open square brackets table row a b row b a end table close square brackets$ , then |A +A^T| equals -

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If A= $open square brackets table row cell cos invisible function application theta end cell cell negative sin invisible function application theta end cell row cell sin invisible function application theta end cell cell cos invisible function application theta end cell end table close square brackets$ , then AA^T equals-

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If A = $open square brackets table row 3 2 row 1 4 end table close square brackets$ and B = $open square brackets table row cell negative 1 end cell 2 row cell negative 1 end cell 1 end table close square brackets$ , then correct statement is -

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If A, B, C, are three matrices, then A^T + B^T + C^T is -

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If A and B are matrices of order m × n and n × m respectively, then the order of matrix B^T (A^T)^T is -

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If A = $open square brackets table row 1 2 row cell negative 1 end cell 2 end table close square brackets$ and B = $open square brackets table row 3 4 row 2 cell negative 2 end cell end table close square brackets$ , then (AB)^T is-

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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 -

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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 =

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

Maths-General

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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

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

Maths-General

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Can’t find what you’re looking for?

For suitable matrices A, B; the false statement is-

(AB)^T =A^TB^T
(A^T)^T= A
(A – B)^T =A^T– B^T
(A^T)–1 = (A^–1 )^T

The correct answer is: (AB)^T =A^TB^T

For suitable matrices A, B we are asked to find the false statement
we know that by property

Related Questions to study

If A = $open square brackets table row a b row b a end table close square brackets$ , then |A +A^T| equals -

If A = $open square brackets table row a b row b a end table close square brackets$ , then |A +A^T| equals -

If A = $open square brackets table row 3 2 row 1 4 end table close square brackets$ and B = $open square brackets table row cell negative 1 end cell 2 row cell negative 1 end cell 1 end table close square brackets$ , then correct statement is -

If A = $open square brackets table row 3 2 row 1 4 end table close square brackets$ and B = $open square brackets table row cell negative 1 end cell 2 row cell negative 1 end cell 1 end table close square brackets$ , then correct statement is -

If A, B, C, are three matrices, then A^T + B^T + C^T is -

If A, B, C, are three matrices, then A^T + B^T + C^T is -

If A and B are matrices of order m × n and n × m respectively, then the order of matrix B^T (A^T)^T is -

If A and B are matrices of order m × n and n × m respectively, then the order of matrix B^T (A^T)^T is -

If A = $open square brackets table row 1 2 row cell negative 1 end cell 2 end table close square brackets$ and B = $open square brackets table row 3 4 row 2 cell negative 2 end cell end table close square brackets$ , then (AB)^T is-

If A = $open square brackets table row 1 2 row cell negative 1 end cell 2 end table close square brackets$ and B = $open square brackets table row 3 4 row 2 cell negative 2 end cell end table close square brackets$ , then (AB)^T is-

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 -

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 -

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

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For suitable matrices A, B; the false statement is-

(AB)T =ATBT (AT)T= A (A – B)T =AT – BT (AT)–1 = (A–1 )T

The correct answer is: (AB)T =ATBT

For suitable matrices A, B we are asked to find the false statementwe know that by property

Related Questions to study

If A =, then |A +AT| equals -

If A =, then |A +AT| equals -

If A= , then AAT equals-

If A= , then AAT equals-

If A = and B =, then correct statement is -

If A = and B =, then correct statement is -

If A, B, C, are three matrices, then AT + BT + CT is -

If A, B, C, are three matrices, then AT + BT + CT is -

If A and B are matrices of order m × n and n × m respectively, then the order of matrix BT (AT)T is -

If A and B are matrices of order m × n and n × m respectively, then the order of matrix BT (AT)T is -

If A = and B = , then (AB)T is-

If A = and B = , then (AB)T is-

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

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

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-

equals-

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

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(AB)^T =A^TB^T
(A^T)^T= A
(A – B)^T =A^T– B^T
(A^T)–1 = (A^–1 )^T

The correct answer is: (AB)^T =A^TB^T

For suitable matrices A, B we are asked to find the false statement
we know that by property

If A = $open square brackets table row a b row b a end table close square brackets$ , then |A +A^T| equals -

If A = $open square brackets table row a b row b a end table close square brackets$ , then |A +A^T| equals -

If A = $open square brackets table row 3 2 row 1 4 end table close square brackets$ and B = $open square brackets table row cell negative 1 end cell 2 row cell negative 1 end cell 1 end table close square brackets$ , then correct statement is -

If A = $open square brackets table row 3 2 row 1 4 end table close square brackets$ and B = $open square brackets table row cell negative 1 end cell 2 row cell negative 1 end cell 1 end table close square brackets$ , then correct statement is -

If A, B, C, are three matrices, then A^T + B^T + C^T is -

If A, B, C, are three matrices, then A^T + B^T + C^T is -

If A and B are matrices of order m × n and n × m respectively, then the order of matrix B^T (A^T)^T is -

If A and B are matrices of order m × n and n × m respectively, then the order of matrix B^T (A^T)^T is -

If A = $open square brackets table row 1 2 row cell negative 1 end cell 2 end table close square brackets$ and B = $open square brackets table row 3 4 row 2 cell negative 2 end cell end table close square brackets$ , then (AB)^T is-

If A = $open square brackets table row 1 2 row cell negative 1 end cell 2 end table close square brackets$ and B = $open square brackets table row 3 4 row 2 cell negative 2 end cell end table close square brackets$ , then (AB)^T is-

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 -

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 -

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