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geniusboy [140]
3 years ago
9

PLS HELP.....

Mathematics
1 answer:
Gemiola [76]3 years ago
3 0
The second one because Idk I’m just guessing lol it doesn’t look like it should be there though
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Step-by-step explanation:

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What is the area of a circle with a radius of 1 foot?<br>ооооplz helppp​
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third one

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Select all statements below which are true for all invertible n×n matrices A and B
Mekhanik [1.2K]

Answer:

a. False

b. False

c. True

d. True

e. False

f. False

Step-by-step explanation:

Hi,

We have certain properties for matrices,<em> (where A and B are nxn matrices and I is the identity matrix) </em>:

{(A^{-1})}^{-1} = A

(AB)^{-1} = B^{-1}A^{-1}

(A')^{-1} = (A^{-1})'

(A^{n})^{-1} = (A^{-1})^{n} = A^{-n}

AA^{-1} = A^{-1}A = I

AI = IA = A

Using these properties, we verify the provided statements:

A. False.

None of the properties help verify this statement. We ca use an example for counter:

Let  A =\left[\begin{array}{cc}1&2\\2&0\\\end{array}\right] and  B = \left[\begin{array}{cc}5&1\\3&2\end{array}\right] , we calculate the L.H.S:

A+B = \left[\begin{array}{cc}1+5&2+1\\2+3&0+2\end{array}\right]\\= \left[\begin{array}{cc}6&3\\5&2\end{array}\right]

The square of (A+B):

(A + B)^{2} = \left[\begin{array}{cc}36&9\\25&4\end{array}\right]

Lets calculate the R.H.S:

A^{2} =\left[\begin{array}{cc}1&4\\4&0\\\end{array}\right]\\B^{2} = \left[\begin{array}{cc}25&1\\9&4\end{array}\right]\\2AB = \left[\begin{array}{cc} (1 \times 5) + (2 \times 3)  &(1 \times 1) + (2 \times 2)\\(2 \times 5) + (0 \times 3)& (2 \times 1) + (0 \times 2)\end{array}\right]\\= \left[\begin{array}{cc} 11 &5\\10& 2\end{array}\right]

A^{2} + B^{2} + 2AB = \left[\begin{array}{ccc}1+25+11&4+1+5 \\4+9+10&0+4+2\\\end{array}\right] \\= \left[\begin{array}{ccc}37&10 \\23&6\\\end{array}\right]

This proves that: L.H.S ≠ R.H.S

Hence, A is false.

B. False

This can only hold when the eigenvalues for A are real.

trace (A^{2}) > 0, det (A^{2}) > 0 : \\(A + A^{-1}) = ( I + A^{2} ) A^ {- 1} = ( A ( I + A ^{2} )^ {-1})^ {-1}

C. True

This is a simplification of the distribution property of matrices.

D. True

The property that inverse is possible for any "n" value of the matrix.

E. False

Similar to part A, we can show that this property is invalid for any nxn matrix. Let:

A = \left[\begin{array}{cc}1&2\\0&1\end{array}\right] \\A^{-1} = \left[\begin{array}{cc}1&-2\\0&1\end{array}\right]

L.H.S:

A + A^{-1} = \left[\begin{array}{cc}1+1&2-2\\0+0&1+1\end{array}\right]  = \left[\begin{array}{cc}2&0\\0&2\end{array}\right]

(A + A^{-1})^{9} = \left[\begin{array}{cc}512&0\\0&512\end{array}\right]

R.H.S:

A^{9} = \left[\begin{array}{cc}1&512\\0&1\end{array}\right] \\A^{-9}= \left[\begin{array}{cc}1&-0.001953125\\0&1\end{array}\right]\\\\\\A^{9} + A^{-9} = \left[\begin{array}{cc}2&512\\0&2\end{array}\right] \\

Since, L.H.S ≠ R.H.S, the statement is false.

F. False

This is a basic matrix rule, that commutative property does not apply on matrices.

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3 years ago
consider the quadratic equation ax^2-40x+109=0 determine the value of a if one solution to the quad equation is x=5+3/2i​
IrinaVladis [17]

Answer:

The answer is 4.

Step-by-step explanation:

The value of a in ax^2 - 40x + 109 = 0 when one solution to the quad equation is x = 5 + 3 over 2i is 4.

6 0
3 years ago
Help pls it’s just one simple question
Vadim26 [7]

Answer:

B) FG and KL

Step-by-step explanation: Please brainliest!

4 0
3 years ago
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