# Engineering analysis discussion questions

*label*Science

*timer*Asked: Feb 19th, 2019

*account_balance_wallet*$5

### Question Description

When you give solutions, make sure you give step-by-step procedures and explanations to show your reasoning and justifications.

**Attachment preview**

1. (Points 10) Matrix A is singular: True or False? justify.

2. (Points 20) Find A – B, B + A + +,

3. (Points 20) Is it true: AB = BA? justify.

4. (Points 30) -.X = d, find X using Gauss-Jordan elimination method.

• Step 1: Write down the system of linear equations

• Step 2: Construct the augmented coefficient matrix.

• Step 3: Apply Gauss-Jordan elimination.

• Step 4: Reduced to row-echelon form.

5. (Points 20) Find the inverse of matrix B using Gauss-Jordan elimination method.

## Tutor Answer

Hey there. I've managed to solve the problems. Kindly go through them and let me know if any of the solutions isn't clear.

Concepts:

1.

2.

3.

4.

5.

6.

Identity matrices

Gauss Jordan elimination method

Matrix transposition

Inverse of a matrix

Matrix addition

Matrix multiplication

Question 1

Solution

A matrix is singular if and only if the determinant is zero.

𝑎11 𝑎12

For a matrix A = [𝑎21 𝑎22

𝑎31 𝑎32

𝑎13

𝑎23]

𝑎33

The determinant can be found by rewriting the matrix in the below form where column 1 and 2

are repeated;

𝑎11 𝑎12

[𝑎21 𝑎22

𝑎31 𝑎32

𝑎13 𝑎11

𝑎23 𝑎21

𝑎33 𝑎31

𝑎12

𝑎22]

𝑎32

The determinant is then obtained by subtracting the lagging product sum from the leading

product sum as shown below;

Determinant = (a11*a22*a33 + a12*a23*a31 + a13*a21*a32) –

(a31*a22*a13 + a32*a23*a11 + a33*a21*a12)

1 0 2

For the matrix A = [1 2 0]

0 1 2

The matrix above can be rewritten as;

1 0 2 1 0

[1 2 0 1 2 ]

0 1 2 0 1

The determinant is therefore given by;

Det = (1*2*2 + 0*0*0 + 2*1*1) – (0*2*2 + 1*0*1 + 2*1*0)

Det = (4 + 0 + 2 ) – (0 + 0 + 0)

Det = 6

Since the determinant of matrix A is not zero. The matrix is therefore not singular i.e the

statement is false as justified above

Question Two

Finding A – B

I.

For two matrices A and B their difference can be expressed as below;

𝑎11 𝑎12

[𝑎21 𝑎22

𝑎31 𝑎32

𝑎13

𝑏11

𝑎23] - [𝑏21

𝑎33

𝑏31

𝑏12

𝑏22

𝑏32

1

For A = [1

0

0 2

2 0

2 0] and B = [2 1

1 2

0 2

1

A – B = [1

0

0 2

2

2 0] - [2

1 2

0

II.

𝑏13

𝑎11 − 𝑏11

𝑏23] = [𝑎21 − 𝑏21

𝑏33

𝑎31 − 𝑏31

𝑎12 − 𝑏12

𝑎22 − 𝑏22

𝑎32 − 𝑏32

𝑎13 − 𝑏13

𝑎23 − 𝑏23]

𝑎33 − 𝑏33

1

0]

1

0 1

−1 0 1

1 0] = [−1 1 1]

2 1

0 −1 1

Finding B + A + I3

Note that I3 is a 3x3 identity matrix which can be expressed as;

1

I3 = [0

0

0 0

1 0]

0 1

Therefore B + A + I3 will be given by;

2 0 1

1 0 2

1 0

B + A + I3 = [2 1 0] + [1 2 0] + [0 1

0 2 1

0 1 2

0 0

4 0 3

B + A + I3 = [3 4 0]

0 3 4

2+1+1 0+0+0

0

0] = [2 + 1 + 0 1 + 2 + 1

0+0+0 2+1+0

1

1+2+0

0 + 0 + 0]

1+2+1

Question Three

Consider two matrices;

𝑎11 𝑎12 𝑎13

𝑏11

A = [𝑎21 𝑎22 𝑎23] 𝑩 = [𝑏21

𝑎31 𝑎32 𝑎33

𝑏31

AB =

𝑎11 ∗ 𝑏11 + 𝑎12 ∗ 𝑏21 + 𝑎13 ∗ 𝑏31

[𝑎21 ∗ 𝑏11 + 𝑎22 ∗ 𝑏21 + 𝑎23 ∗ 𝑏31

𝑎31 ∗ 𝑏11 + 𝑎32 ∗ 𝑏21 + 𝑎33 ∗ 𝑏3...

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