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Maths-

General

Easy

Question

Matrix A = $open square brackets table row x 3 2 row 1 y 4 row 2 2 z end table close square brackets$ , If xyz = 60 and 8x + 4y + 3z = 20, then A(adj A) is equal to:

$[\begin{matrix} 64 & 0 & 0 \\ 0 & 64 & 0 \\ 0 & 0 & 64 \end{matrix}]$
$[\begin{matrix} 88 & 0 & 0 \\ 0 & 88 & 0 \\ 0 & 0 & 88 \end{matrix}]$
$[\begin{matrix} 68 & 0 & 0 \\ 0 & 68 & 0 \\ 0 & 0 & 68 \end{matrix}]$
$[\begin{matrix} 34 & 0 & 0 \\ 0 & 34 & 0 \\ 0 & 0 & 34 \end{matrix}]$

The correct answer is: $open square brackets table row 68 0 0 row 0 68 0 row 0 0 68 end table close square brackets$

$because$ |A| = 68
$therefore$ A (adj A) = |A|. I

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Let A = $open square brackets table row 1 0 0 row 2 1 0 row 3 2 1 end table close square brackets$ . If U₁, U₂, U₃ are column matrices satisfying AU₁ = $open square brackets table row 1 row 0 row 0 end table close square brackets$ , AU₂ = $open square brackets table row 2 row 3 row 0 end table close square brackets$ , AU₃ = $open square brackets table row 2 row 3 row 1 end table close square brackets$ and U is 3 × 3 matrix whose columns are U₁, U₂ and U₃. Then |U| =

Let A = $open square brackets table row 1 0 0 row 2 1 0 row 3 2 1 end table close square brackets$ . If U₁, U₂, U₃ are column matrices satisfying AU₁ = $open square brackets table row 1 row 0 row 0 end table close square brackets$ , AU₂ = $open square brackets table row 2 row 3 row 0 end table close square brackets$ , AU₃ = $open square brackets table row 2 row 3 row 1 end table close square brackets$ and U is 3 × 3 matrix whose columns are U₁, U₂ and U₃. Then |U| =

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If A = $open square brackets table row 4 cell x plus 2 end cell row cell 2 x – 3 end cell cell x plus 1 end cell end table close square brackets$ is symmetric, then x =

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chemistry-General

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A = $open square brackets table row 1 cell tan invisible function application x end cell row cell – tan invisible function application x end cell 1 end table close square brackets$ then let us define a function f(x) = dt.(A^TA^–1) then which of the following can not be the value of $fraction numerator f open parentheses f open parentheses f open parentheses f............ f left parenthesis x right parenthesis close parentheses close parentheses close parentheses over denominator n t i m e s end fraction$ is (n ≥ 2)

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