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If A = $open square brackets table row 2 0 0 row 0 2 0 row 0 0 2 end table close square brackets$ , then value of A^–1 is-

$[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$
$[\begin{matrix} 1 / 2 & 0 & 0 \\ 0 & 1 / 2 & 0 \\ 0 & 0 & 1 / 2 \end{matrix}]$
$[\begin{matrix} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & - 2 \end{matrix}]$
None of these

The correct answer is: $open square brackets table row cell 1 divided by 2 end cell 0 0 row 0 cell 1 divided by 2 end cell 0 row 0 0 cell 1 divided by 2 end cell end table close square brackets$

We are given the matrix A and we have to find $A to the power of negative 1 end exponent$
$S o comma a u g m e n t space t h e m a t r i x space w i t h space t h e space i d e n t i t y space m a t r i x colon open square brackets table row 2 0 0 1 0 0 row 0 2 0 0 1 0 row 0 0 2 0 0 1 end table close square brackets D i v i d e space r o w space 1 space b y space 2 colon R subscript 1 equals R subscript 1 over 2 equals open square brackets table row 1 0 0 cell 1 half end cell 0 0 row 0 2 0 0 1 0 row 0 0 2 0 0 1 end table close square brackets D i v i d e r o w space 2 space b y space space 2 colon R subscript 2 equals R subscript 2 over 2 open square brackets table row 1 0 0 cell 1 half end cell 0 0 row 0 1 0 0 cell 1 half end cell 0 row 0 0 2 0 0 1 end table close square brackets D i v i d e space r o w space 3 space b y space 2 colon R subscript 3 equals R subscript 3 over 2 open square brackets table row 1 0 0 cell 1 half end cell 0 0 row 0 1 0 0 cell 1 half end cell 0 row 0 0 1 0 0 cell 1 half end cell end table close square brackets$

Therefore the correct option is choice 2

If A = $open square brackets table row 1 2 row 3 cell negative 5 end cell end table close square brackets$ , B = $open square brackets table row 1 0 row 0 2 end table close square brackets$ and X is a matrix such that A = BX, then X equals -

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Matrix $open square brackets table row lambda cell negative 1 end cell 4 row cell negative 3 end cell 0 1 row cell negative 1 end cell 1 2 end table close square brackets$ is not invertible if-

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chemistry-

If A = $open square brackets table row 1 cell negative 2 end cell 3 row 4 0 cell negative 1 end cell row cell negative 3 end cell 1 5 end table close square brackets$ , then (adj A)₂₃ =

Maths-General

General

maths-

If A = $open square brackets table row 1 2 3 row 2 3 4 row 0 0 2 end table close square brackets$ , then the value of adj (adj A) is-

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The number of values of $theta$ in [0, 2 $straight pi$ ] for which at least two roots are equal

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Maths-

Let A be a square matrix. Then which of the following is not a symmetric matrix -

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Maths-

If A is a square matrix, then A– $straight A to the power of straight prime$ is -

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maths-

If A is symmetric matrix and B is a skew- symmetric matrix, then for n $element of$ N, false statement is -

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General

Maths-

If A – $straight A to the power of straight prime$ = 0, then $straight A to the power of straight prime$ is -

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Maths-

If A is a matrix of order 3 × 4, then both AB^T and B^TA are defined if order of B is -

Maths-General

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 $A A to the power of straight prime$ equals -

maths-General

General

Maths-

Consider the cubic equation $x cubed minus left parenthesis 1 plus cos space theta plus sin space theta right parenthesis x squared plus left parenthesis cos space theta sin space theta plus cos space theta plus sin space theta right parenthesis x minus sin space theta cos space theta equals 0$ whose roots are x₁, x₂, and x₃
The value of $x subscript 1 superscript 2 plus x subscript 2 superscript 2 plus x subscript 3 superscript 2$ equals

In this question, we have to given the cubic equation and its root x1,x2,x3 and we have to find the
$x 1 squared plus x 2 squared plus x 3 squared$ .Use, $left parenthesis x 1 plus x 2 plus x 3 right parenthesis squared equals x 1 squared plus x 2 squared plus x 3 squared plus 2 left parenthesis x 1 x 2 plus x 2 x 3 plus x 3 x 1 right parenthesis$ and sum of roots is $fraction numerator negative c over denominator a end fraction$ and product of roots is b/a.

Consider the cubic equation $x cubed minus left parenthesis 1 plus cos space theta plus sin space theta right parenthesis x squared plus left parenthesis cos space theta sin space theta plus cos space theta plus sin space theta right parenthesis x minus sin space theta cos space theta equals 0$ whose roots are x₁, x₂, and x₃
The value of $x subscript 1 superscript 2 plus x subscript 2 superscript 2 plus x subscript 3 superscript 2$ equals

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If A =, then value of A–1 is-

None of these

The correct answer is:

We are given the matrix A and we have to find

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If A = $open square brackets table row 2 0 0 row 0 2 0 row 0 0 2 end table close square brackets$ , then value of A^–1 is-

$[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$
$[\begin{matrix} 1 / 2 & 0 & 0 \\ 0 & 1 / 2 & 0 \\ 0 & 0 & 1 / 2 \end{matrix}]$
$[\begin{matrix} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & - 2 \end{matrix}]$
None of these

The correct answer is: $open square brackets table row cell 1 divided by 2 end cell 0 0 row 0 cell 1 divided by 2 end cell 0 row 0 0 cell 1 divided by 2 end cell end table close square brackets$

If A = $open square brackets table row 1 2 row 3 cell negative 5 end cell end table close square brackets$ , B = $open square brackets table row 1 0 row 0 2 end table close square brackets$ and X is a matrix such that A = BX, then X equals -

If A = $open square brackets table row 1 2 row 3 cell negative 5 end cell end table close square brackets$ , B = $open square brackets table row 1 0 row 0 2 end table close square brackets$ and X is a matrix such that A = BX, then X equals -

Matrix $open square brackets table row lambda cell negative 1 end cell 4 row cell negative 3 end cell 0 1 row cell negative 1 end cell 1 2 end table close square brackets$ is not invertible if-

Matrix $open square brackets table row lambda cell negative 1 end cell 4 row cell negative 3 end cell 0 1 row cell negative 1 end cell 1 2 end table close square brackets$ is not invertible if-

(adj A^T) – (adj A)^T equals-

(adj A^T) – (adj A)^T equals-

If A = $open square brackets table row 1 cell negative 2 end cell 3 row 4 0 cell negative 1 end cell row cell negative 3 end cell 1 5 end table close square brackets$ , then (adj A)₂₃ =

If A = $open square brackets table row 1 cell negative 2 end cell 3 row 4 0 cell negative 1 end cell row cell negative 3 end cell 1 5 end table close square brackets$ , then (adj A)₂₃ =

If A = $open square brackets table row 1 2 3 row 2 3 4 row 0 0 2 end table close square brackets$ , then the value of adj (adj A) is-

If A = $open square brackets table row 1 2 3 row 2 3 4 row 0 0 2 end table close square brackets$ , then the value of adj (adj A) is-

The number of values of $theta$ in [0, 2 $straight pi$ ] for which at least two roots are equal

The number of values of $theta$ in [0, 2 $straight pi$ ] for which at least two roots are equal

If A is a square matrix, then A– $straight A to the power of straight prime$ is -

If A is a square matrix, then A– $straight A to the power of straight prime$ is -

If A is symmetric matrix and B is a skew- symmetric matrix, then for n $element of$ N, false statement is -

If A is symmetric matrix and B is a skew- symmetric matrix, then for n $element of$ N, false statement is -

If A – $straight A to the power of straight prime$ = 0, then $straight A to the power of straight prime$ is -

If A – $straight A to the power of straight prime$ = 0, then $straight A to the power of straight prime$ is -

If A is a matrix of order 3 × 4, then both AB^T and B^TA are defined if order of B is -

If A is a matrix of order 3 × 4, then both AB^T and B^TA are defined if order of B is -