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3 years ago

Please help solve! I tried but I’m messing this up, would really appreciate it!

Mathematics

1 answer:

STALIN [3.7K]3 years ago

7 0

Answer:

K = $\frac{4r}{n} -a$ n $\neq 0$

Step-by-step explanation:

We have our equation $r = \frac{n}{4} (a+K)$ and are looking for K

First we have to switch sides :

$\frac{n}{4} (a+K) = r$

Since we have a denominator of 4, we multiply both sides by 4 :

$n (a+K) = 4r$

We then divide both sides by n :

$a+K = \frac{4r}{n} ; n\neq 0$

Subtract a from both sides :

$K = \frac{4r}{n} -a; n\neq 0$

^^^

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Answer:

Entries of I^k are are also identity elements.

Step-by-step explanation:

a) For the 2×2 identity matrix I, show that I² =I

$I^{2}=\left[\begin{array}{cc}1&0\\0&1\end{array}\right] \times \left[\begin{array}{cc}1&0\\0&1\end{array}\right] \\\\=\left[\begin{array}{cc}1\times 1+0\times 0&1\times 0+0\times 1\\0\times 1+1\times 0&0\times 0+1\times1\end{array}\right] \\\\=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

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b) For the n×n identity matrix I, show that I² =I

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$\delta I_{ij}=1\quad if\quad i=j\\\delta I_{ij}=0\quad if\quad i\ne j\\$

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$\delta I_{ij}=1\quad if\quad i=j\\\delta I_{ij}=0\quad if\quad i\ne j\\$

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