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3 years ago

If A^2=A, which matrix is matrix A

Mathematics

2 answers:

Ronch [10]3 years ago

8 0

Consider all options:

$\left[\begin{array}{cc}5&5\\-4&-4\end{array}\right] \cdot \left[\begin{array}{cc}5&5\\-4&-4\end{array}\right] =\left[\begin{array}{cc}5\cdot 5+5\cdot (-4)&5\cdot 5+5\cdot (-4)\\-4\cdot 5+(-4)\cdot (-4)&-4\cdot 5+(-4)\cdot (-4)\end{array}\right]=$

$=\left[\begin{array}{cc}5&5\\-4&-4\end{array}\right].$

This option is true.

$\left[\begin{array}{cc}6&5\\5&6\end{array}\right] \cdot \left[\begin{array}{cc}6&5\\5&6\end{array}\right] =\left[\begin{array}{cc}6\cdot 6+5\cdot 5&6\cdot 5+5\cdot 6\\5\cdot 6+6\cdot 5&5\cdot 5+6\cdot 6\end{array}\right]=$

$=\left[\begin{array}{cc}61&60\\60&61\end{array}\right].$

This option is false.

$\left[\begin{array}{cc}0.5&-0.5\\-0.5&0.5\end{array}\right] \cdot \left[\begin{array}{cc}0.5&-0.5\\-0.5&0.5\end{array}\right] =\left[\begin{array}{cc}0.5\cdot 0.5+(-0.5)\cdot (-0.5)&0.5\cdot (-0.5)+(-0.5)\cdot 0.5\\-0.5\cdot 0.5+0.5\cdot (-0.5)&-0.5\cdot (-0.5)+0.5\cdot 0.5\end{array}\right]=$

$=\left[\begin{array}{cc}0.5&-0.5\\-0.5&0.5\end{array}\right].$

This option is true.

$\left[\begin{array}{cc}0.5&0.5\\-0.5&0.5\end{array}\right] \cdot \left[\begin{array}{cc}0.5&0.5\\-0.5&0.5\end{array}\right] =\left[\begin{array}{cc}0.5\cdot 0.5+0.5\cdot (-0.5)&0.5\cdot 0.5+0.5\cdot 0.5\\-0.5\cdot 0.5+0.5\cdot (-0.5)&-0.5\cdot 0.5+0.5\cdot 0.5\end{array}\right]=$

$=\left[\begin{array}{cc}0&0.5\\-0.5&0\end{array}\right].$

This option is false.

$\left[\begin{array}{cc}-6&-6\\5&5\end{array}\right] \cdot \left[\begin{array}{cc}-6&-6\\5&5\end{array}\right] =\left[\begin{array}{cc}-6\cdot (-6)+(-6)\cdot 5&-6\cdot (-6)+(-6)\cdot 5\\5\cdot (-6)+5\cdot 5&5\cdot (-6)+5\cdot 5\end{array}\right]=$

$=\left[\begin{array}{cc}6&6\\-5&-5\end{array}\right].$

This option is false.

Answer: correct options are A and C.

pentagon [3]3 years ago

7 0

Answer:

Options 1 and 3.

Step-by-step explanation:

By definition the product between two matrices is:

Let's suppose both matrices are 2x2,

$A=\left[\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right]$

$B=\left[\begin{array}{cc}b_{11}&b_{12}\\b_{21}&b_{22}\end{array}\right]$

The product between A and B is:

$AB=\left[\begin{array}{cc}a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22}\\a_{21}b_{11}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22}\end{array}\right]$

IMPORTANT: It's not the same AB then BA the results of both products are differents.

Now we are going to analyze every option:

$A^2=A.A$

Option 1:

$A=\left[\begin{array}{cc}5&5\\-4&-4\end{array}\right]$

$A.A=\left[\begin{array}{cc}5&5\\-4&-4\end{array}\right] .\left[\begin{array}{cc}5&5\\-4&-4\end{array}\right] =\\\\=\left[\begin{array}{cc}5.5+5(-4)&5.5+5(-4)\\(-4).5+(-4).(-4)&(-4).5+(-4)(-4)\end{array}\right] \\\\=\left[\begin{array}{cc}25-20&25-20\\-20+16&-20+16\end{array}\right] \\\\=\left[\begin{array}{cc}5&5\\-4&-4\end{array}\right]=A$

We can see that A.A=A then this is the correct option.

Option 2:

$A=\left[\begin{array}{cc}6&5\\5&6\end{array}\right]\\\\A.A=\left[\begin{array}{cc}6&5\\5&6\end{array}\right].\left[\begin{array}{cc}6&5\\5&6\end{array}\right]\\\\=\left[\begin{array}{cc}6.6+5.5&6.5+5.6\\5.6+6.5&5.5+6.6\end{array}\right]\\\\=\left[\begin{array}{cc}36+25&30+30\\30+30&25+36\end{array}\right]\\\\=\left[\begin{array}{cc}61&60\\60&61\end{array}\right]\neq A$

$A.A\neq A$ Then this option is incorrect.

Option 3:

$A=\left[\begin{array}{cc}0,5&-0,5\\-0,5&0,5\end{array}\right] \\\\A.A=\left[\begin{array}{cc}0,5&-0,5\\-0,5&0,5\end{array}\right].\left[\begin{array}{cc}0,5&-0,5\\-0,5&0,5\end{array}\right]\\\\=\left[\begin{array}{cc}0,5.0,5+(-0,5).(-0,5)&0,5.(-0,5)+(-0,5).(0,5)\\(-0,5).0,5+0,5.(-0,5)&(-0,5).(-0,5)+0,5.0,5\end{array}\right]\\\\=\left[\begin{array}{cc}0,25+0,25&-0,25-0,25\\-0,25-0,25&0,25+0,25\end{array}\right]\\\\=\left[\begin{array}{cc}0,5&-0,5\\-0,5&0,5\end{array}\right]= A$

We can see that this option is also correct.

Option 4:

$A=\left[\begin{array}{cc}0,5&0,5\\-0,5&0,5\end{array}\right] \\\\A.A=\left[\begin{array}{cc}0,5&0,5\\-0,5&0,5\end{array}\right].\left[\begin{array}{cc}0,5&0,5\\-0,5&0,5\end{array}\right]\\\\=\left[\begin{array}{cc}0,5.0,5+0,5.(-0,5)&0,5.0,5+0,5.0,5\\(-0,5).0,5+0,5.(-0,5)&(-0,5).0,5+0,5.0,5\end{array}\right]\\\\=\left[\begin{array}{cc}0,25-0,25&0,25+0,25\\-0,25-0,25&-0,25+0,25\end{array}\right]\\\\=\left[\begin{array}{cc}0&0,5\\-0,5&0\end{array}\right]\neq A$

Then this option is incorrect.

Option 5:

$A=\left[\begin{array}{cc}-6&-6\\5&5\end{array}\right]\\\\A.A=\left[\begin{array}{cc}-6&-6\\5&5\end{array}\right].\left[\begin{array}{cc}-6&-6\\5&5\end{array}\right]\\\\=\left[\begin{array}{cc}(-6).(-6)+(-6).5&(-6).(-6)+(-6).5\\5.(-6)+5.5&5.(-6)+5.5\end{array}\right]\\\\=\left[\begin{array}{cc}36-30&36-30\\-30+25&-30+25\end{array}\right]\\\\=\left[\begin{array}{cc}6&6\\-5&-5\end{array}\right]\neq A$

Then this option is incorrect.

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Find the sum of the positive integers less than 200 which are not multiples of 4 and 7

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Answer:

$12942$ is the sum of positive integers between $1$ (inclusive) and $199$ (inclusive) that are not multiples of $4$ and not multiples $7$ .

Step-by-step explanation:

For an arithmetic series with:

$a_1$ as the first term,
$a_n$ as the last term, and
$d$ as the common difference,

there would be $\displaystyle \left(\frac{a_n - a_1}{d} + 1\right)$ terms, where as the sum would be $\displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n)$ .

Positive integers between $1$ (inclusive) and $199$ (inclusive) include:

$1,\, 2,\, \dots,\, 199$ .

The common difference of this arithmetic series is $1$ . There would be $(199 - 1) + 1 = 199$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}$ .

Similarly, positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $4$ include:

$4,\, 8,\, \dots,\, 196$ .

The common difference of this arithmetic series is $4$ . There would be $(196 - 4) / 4 + 1 = 49$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $7$ include:

$7,\, 14,\, \dots,\, 196$ .

The common difference of this arithmetic series is $7$ . There would be $(196 - 7) / 7 + 1 = 28$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $28$ (integers that are both multiples of $4$ and multiples of $7$ ) include:

$28,\, 56,\, \dots,\, 196$ .

The common difference of this arithmetic series is $28$ . There would be $(196 - 28) / 28 + 1 = 7$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}$ .

The requested sum will be equal to:

the sum of all integers from $1$ to $199$ ,
minus the sum of all integer multiples of $4$ between $1\!$ and $199\!$ , and the sum integer multiples of $7$ between $1$ and $199$ ,
plus the sum of all integer multiples of $28$ between $1$ and $199$ - these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

$19900 - 4900 - 2842 + 784 = 12942$ .

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