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Fed [463]
3 years ago
6

Suppose C is a 3 x 3 matrix such that det (C) = 4. Show that det (C+C) is equal to 32

Mathematics
1 answer:
Ber [7]3 years ago
7 0

Step-by-step explanation:

Let's consider C is a matrix given by

\left[\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right]

them determinant of matrix C can be written as

\begin{vmatrix}a & b & c\\ d & e & f\\  g & h & i \end{vmatrix}\ =\ 4.....(1)

Now,

det (C+C)\ =\ \begin{vmatrix}a & b & c\\ d & e & f\\  g & h & i \end{vmatrix}\ +\ \begin{vmatrix}a & b & c\\ d & e & f\\  g & h & i \end{vmatrix}

                  =\ \begin{vmatrix}2a & 2b & 2c\\ 2d & 2e & 2f\\  2g & 2h & 2i \end{vmatrix}

                   =\ 2\times 2\times 2\times \begin{vmatrix}a & b & c\\ d & e & f\\  g & h & i \end{vmatrix}

                   =\ 8\times 4\ \ \ \ \ \ \ \         from\ eq.(1)

                    = 32      

Hence, det (C+C) = 32

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Answer with Step-by-step explanation:

The given differential euation is

\frac{dy}{dx}=(y-5)(y+5)\\\\\frac{dy}{(y-5)(y+5)}=dx\\\\(\frac{A}{y-5}+\frac{B}{y+5})dy=dx\\\\\frac{1}{100}\cdot (\frac{10}{y-5}-\frac{10}{y+5})dy=dx\\\\\frac{1}{100}\cdot \int (\frac{10}{y-5}-\frac{10}{y+5})dy=\int dx\\\\10[ln(y-5)-ln(y+5)]=100x+10c\\\\ln(\frac{y-5}{y+5})=10x+c\\\\\frac{y-5}{y+5}=ke^{10x}

where

'k' is constant of integration whose value is obtained by the given condition that y(2)=0\\

\frac{0-5}{0+5}=ke^{20}\\\\k=\frac{-1}{e^{20}}\\\\\therefore k=-e^{-20}

Thus the solution of the differential becomes

 \frac{y-5}{y+5}=e^{10x-20}

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3 years ago
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Y=5x-4
anyanavicka [17]

Answer:

\displaystyle -5x + y = -92\:or\:y = 5x - 92

Step-by-step explanation:

−12 = 5[16] + b

80

\displaystyle -92 = b \\ \\ y = 5x - 92

If you want it in <em>Standard</em><em> </em><em>Form</em>:

y = 5x - 92

- 5x - 5x

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\displaystyle -5x + y = -92

I am joyous to assist you anytime.

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Masja [62]

(√3 + √11)² + (√3 - √11)²

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  • ( a - b )² = a² + b² - 2ab

<em>Now </em><em>,</em>

(√3 + √11)² + (√3 - √11)²

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Hence , The value of (√3 + √11)² + (√3 - √11)² is 28 .

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