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sergiy2304 [10]
3 years ago
15

Write a script (Program 2) to perform t he following matrix operations. Use output commands to clearly output each problem with

a problem description and result.
(a) Create the following three matrices: A = [ 9 3 5 /2 4 2/ 6 10 5 ] B = [ 1 4 4/ 0 9 4/ 2 6 1 ] C = [ 8 15 1/ 10 8 2/ 8 5 10 ]

(b) Calculate A + B and B + A to show that addition of matrices is commutative.

(c) Calculate A +( B + C ) and ( A + B )+ C to show that addition of matrices is associative.

(d) Calculate 8( A + C ) and 8A + 8C to show that multiplication by a scalar is distributive.

(e) Calculate A( B + C ) and AB + AC to show that matrix multiplication is distributive.

(f) Does AB = BA ?

(g) Find AT (transpose of A ).

(h) Calculate B-1 (inverse of B ).

(i) Calculate the product BB-1 .

(j) Create a 2x2 matrix consisting of the lower left elements of B .

(k) Create a row vector that is the vector sum of the first row of A and the third row of C .

(l) Calculate the dot product of the second row of B and the third column of C

Engineering
1 answer:
Simora [160]3 years ago
7 0

Find the attachments for complete solution

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amid [387]

Answer:

A) See Attachment B) See Attachment C) 186.153

Explanation:

C) Per phase Voltage= 220/∛3

                                  = 6.037

S=V²/Z

  = (6.037)²/(0.01+i0.05)

  = 62051.282-i310256.41

Active power losses per phase= 62.051kW

total Active power losses= 62.051×3

                                         =186.153kW

6 0
4 years ago
The pressure and temperature at the beginning of compression of a cold air-standard Diesel cycle are 100 kPa and 300K, respectiv
hram777 [196]

Answer:

Cut off ratio=2.38

Explanation:

Given that

T_1=300K

P_1=100KPa

P_2=P_3=7200KPa

T_3=2250K

Lets take T_1 is the temperature at the end of compression process

For air γ=1.4

\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{\dfrac{\gamma -1}{\gamma}}

\dfrac{T_2}{300}=\left(\dfrac{7200}{100}\right)^{\dfrac{1.4-1}{1.4}}

T_2=1070K

At constant pressure

\dfrac{T_3}{T_2}=\dfrac{V_3}{V_2}

\dfrac{2550}{1070}=\dfrac{V_3}{V_2}

\dfrac{V_3}{V_2}=2.83

So cut off ratio

cut\ off\ ratio =\dfrac{V_3}{V_2}

Cut off ratio=2.38

6 0
3 years ago
Sedimentary rock forms at or near the
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5 0
3 years ago
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Problem Statement: Air flows at a rate of 0.1 kg/s through a device as shown below. The pressure and temperature of the air at l
Tema [17]

Answer:

The answer is "+9.05 kw"

Explanation:

In the given question some information is missing which can be given in the following attachment.

The solution to this question can be defined as follows:

let assume that flow is from 1 to 2 then

Q= 1kw

m=0.1 kg/s

From the steady flow energy equation is:

m\{n_1+ \frac{v^2_1}{z}+ gz_1 \}+Q= m \{h_2+ \frac{v^2_2}{2}+ gz_2\}+w\\\\\ change \ energy\\\\0.1[1.005 \times 800]-1= 0.01[1.005\times 700]+w\\\\w= +9.05 \ kw\\\\

If the sign of the work performed is positive, it means the work is done on the surrounding so, that the expected direction of the flow is right.

8 0
3 years ago
An automobile engine provides 632 Joules of work to push the pistons and generates 2203 Joules of heat that must be carried away
Ronch [10]

Answer:

The change in internal energy of the engine is -1,571 Joules.

Explanation:

Change in internal energy (∆U) = energy output (U2) - energy generated (U1)

U1 is the energy generated = 2203 Joules

U2 is the energy output = 632 Joules

Change in internal energy (∆U) = 632 - 2203 = -1,571 Joules

3 0
3 years ago
Read 2 more answers
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