Find the average value of the sum of the squares of three numbers x, y, z, where each number is between 0 and 2.

Mathematics

1 answer:

mr_godi [17]3 years ago

8 0

answer:

Step-by-step explanation:

#'s between zero and two = one

the only number between 0 and two is 1

x = 1 y = 1 z = 1

the average value of the sum of the squares of x, y and z = 1

1, 1, 1

the average is 1

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A=(2 3)<br> (-1 4)<br><br> Calculate A^2-6A+11I

disa [49]

Answer:

A² - 6A + 11 I = $\left[\begin{array}{ccc}0&0\\0&0\end{array}\right]$

Step-by-step explanation:

Given the matrix

$A=\left[\begin{array}{ccc}2&3\\-1&4\end{array}\right]$

Calculate A² - 6A + 11 I

$A^2 = A*A= \left[\begin{array}{ccc}2&3\\-1&4\end{array}\right] *\left[\begin{array}{ccc}2&3\\-1&4\end{array}\right] = \left[\begin{array}{ccc}2*2-3*1&2*3+3*4\\-1*2-4*1&-1*3+4*4\end{array}\right] =\left[\begin{array}{ccc}1&18\\-6&13\end{array}\right]$

$6A=6*\left[\begin{array}{ccc}2&3\\-1&4\end{array}\right] =\left[\begin{array}{ccc}12&18\\-6&24\end{array}\right]$

$11 I = 11 * \left[\begin{array}{ccc}1&0\\0&1\end{array}\right] =\left[\begin{array}{ccc}11&0\\0&11\end{array}\right]$

∴ A² - 6A + 11 I = $\left[\begin{array}{ccc}1&18\\-6&13\end{array}\right] -\left[\begin{array}{ccc}12&18\\-6&24\end{array}\right] +\left[\begin{array}{ccc}11&0\\0&11\end{array}\right] =\left[\begin{array}{ccc}0&0\\0&0\end{array}\right]$