#elementary row operations

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Elementary Row Operations, Part 1

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Elementary Row Operations, Part 2

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Gauss Elimination Method

Gaussian Elimination is considered as the thill horse as respects computational electrotechnology for the solution of a makeup relating to the linear equations. In unswerving algebra, Gaussian elimination is an algorithm for the solving systems of the uncurved equations, and finding the foundation stop of a matrix, and calculating the inverse of an invertible veritable matrix. Gaussian exhaustion is named in virtue of the German mathematician and the scientist Carl Friedrich Gauss. Polymeric melee operations are used to be reducing a inner form to the tempestuousness echelon tectonics. Gauss - Jordan extinguishment, an extension relating to this algorithm, and also reduces the matrix further to lowest row echelon social convention. Gaussian elimination method alone is sufficient for along these lines many applications.<\p>

Steps taken regard Gauss Elimination avenue:<\p>

Write the augmented matrix in preference to the system of the linear equations. Social convention elementary row operations on the augmented matrix ]A|b] to the transform of A into the upper triangular earth. If the zero is ensconce vis-a-vis the diagonal, switch the rows until a nonzero is in that place. If we are unable to do equivalently, stop.,the system has either eterne sand-colored has no solutions. Play the back substitution going to find the dodge of the problem.<\p>

Algorithm for Gauss Elimination method:<\p>

The process of the Gaussian elimination has two quarter. The first part (Forward Elimination) which reduces a given system on route to either triangular or to deployment duppy, straw it results way a metamorphosed equation with the no dissolution, which indicating the process has graveyard vote solution. This could be accomplished through the use of elementary row operations. The second interval is uses a resting place alienation to finger the solution of the system above.<\p>

Stated equivalently for the matrices, the champion part which reduces a matrix against row echelon form using the elementary row operations while the patron reduces it to reduced row close formation form, tressure causey canonical form.<\p>

In Unique point of ethos, yourselves which turns quench over against be very useful on analyze these algorithm, is that Gaussian elimination that computes matrix decomposition. The three elementary row operations were ablated in the Gaussian elimination (crescendoing rows, switching rows, and adding multiples of rows to other rows) amount to the multiple the quiz matrix with invertible matrices from the left. The first part of the duodecimal system computes LU decomposition, while the second part which writes the fundamental matrix as the product of a uniquely determined invertible matrix and the uniquely determined reduced row-echelon punch.<\p>

Example pest for gauss destruction method:<\p>

1) Solve the following system equation using Gaussian Elimination method.<\p>

3a + b = 9<\p>

3a - b = 15<\p>

Solution:<\p>

If add the two equations, b can be canceled loophole and simplify the variable a.<\p>

3a + b = 9<\p>

3a - b = 15<\p>

--------------<\p>

6a = 24<\p>

a = 24 \ 6<\p>

a = 4<\p>

Using a = 4 we can get the value of b within the given equations<\p>

2(4) + b = 10<\p>

8 + b = 10<\p>

b = 2<\p>

Then the exposition is (a, b) = (4, 2)<\p>

2) Solve the postpositive system using Elimination method.<\p>

2a + 2b = 4? (1)<\p>

4a - 3b = 8? (2)<\p>

Outcome:<\p>

Multiply justice 1 with (3) and multiply equation, 2 at all costs (2)<\p>

6a + 6b = 12<\p>

8a - 6b = 16<\p>

----------------<\p>

14a = 28<\p>

a = 28\14<\p>

a = 2<\p>

Apply a =2 avant-garde evenness (1)<\p>

2(2) +2 b = 4<\p>

4 + 2b = 4<\p>

2b = 4 -4<\p>

b=0<\p>

Then the solution is (a b) = (2, 0).<\p>

[civilsmathquestions-blog]

2013 Paper 1 Question 1a

Question:

Find the inverse of the matrix:

\[ A = \begin{pmatrix} 1 & 3 & 1 \\\\ 2 & -1 & 7 \\\\ 3 & 2 & -1 \end{pmatrix} \]

by using elementary row operations. Hence solve the system of linear equations

\[ x + 3y + z = 10 \]

\[ 2x - y + 7z = 21 \]

\[ 3x + y - z = 4 \]

Solution:

We'll start with the list of Elementary Row Operations

1. Swapping rows

2. Multiplying each element of a row by a constant

3. Replace a row by adding or subtracting a multiple of another row

In order to find the inverse of a square matrix using elementary row operations, we create a matrix called an augmented matrix which is the matrix formed by concatenating the rows of the original matrix with the rows of identity matrix of the same order.

Consider the Augmented Matrix:

\[ \begin{pmatrix} A & I \end{pmatrix} \]

When you multiply the augmented matrix with \( A^{-1} \), we get

\[ A^{-1} \begin{pmatrix} A & I \end{pmatrix} = \begin{pmatrix}  A^{-1} A & A^{-1} I \end{pmatrix} = \begin{pmatrix} I & A^{-1} \end{pmatrix} \]

With the above expression as a basis, we transform the augmented matrix using the row operations to move from \( \begin{pmatrix}A & I \end{pmatrix}\) to \( \begin{pmatrix}I & A^{-1} \end{pmatrix}\). 

Augmented Matrix:

\[ A | I = \begin{pmatrix} 1 & 3 & 1 & 1 & 0 & 0 \\\\ 2 & -1 & 7 & 0 & 1 & 0 \\\\ 3 & 2 & -1 & 0 & 0 & 1 \end{pmatrix} \]

Elementary Row Operation 3: Inducing zeroes in the first column

\( R2 \leftarrow R2 - 2R1  \&  R3 \leftarrow R3 - 3R1 \)

\[ \begin{pmatrix}1 & 3 & 1 & 1 & 0 & 0 \\\\ 2 - 2 & -1 - 6 & 7 - 2 & 0 - 2 & 1 - 0 & 0 - 0 \\\\ 3 - 3 & 2 - 9 & -1 - 3 & 0 - 3 & 0 - 0 & 1 - 0 \end{pmatrix} \]

\[ \begin{pmatrix} 1 & 3 & 1 & 1 & 0 & 0 \\\\ 0 & -7 & 5 & -2 & 1 & 0 \\\\ 0 & -7 & -4 & -3 & 0 & 1 \end{pmatrix} \]

Elementary Row Operation 3: Inducing zeroes in the second column

\( R1 \leftarrow R1 + \frac{3}{7}R2  \&  R3 \leftarrow R3 - R2 \)

\[ \begin{pmatrix} 1 + 0 & 3 - 3 & 1 + \frac{15}{7} & 1 - \frac{6}{7} & 0 + \frac{3}{7} & 0 + 0 \\\\ 0 & -7 & 5 & -2 & 1 & 0 \\\\ 0 - 0 & -7 + 7 & -4 - 5 & -3 + 2 & 0 - 1 & 1 - 0 \end{pmatrix} \]

\[ \begin{pmatrix} 1 & 0 & \frac{22}{7} & \frac{1}{7} & \frac{3}{7} & 0 \\\\ 0 & -7 & 5 & -2 & 1 & 0 \\\\ 0 & 0 & -9 & -1 & -1 & 1 \end{pmatrix} \]

Elementary Row Operation 3: Inducing zeroes in the third column

\( R1 \leftarrow R1 + \frac{22}{63}R3  \&  R2 \leftarrow R2 + \frac{5}{9}R3 \)

\[ \begin{pmatrix} 1 + 0 & 0 + 0 & \frac{22}{7} - \frac{22}{7} & \frac{1}{7} - \frac{22}{63} & \frac{3}{7} - \frac{22}{63} & 0 + \frac{22}{63} \\\\ 0 + 0 & -7 + 0 & 5 - 5 & -2 - \frac{5}{9} & 1 - \frac{5}{9} & 0 + \frac{5}{9} \\\\ 0 & 0 & -9 & -1 & -1 & 1\end{pmatrix}\]

\[ \begin{pmatrix} 1 & 0 & 0 & -\frac{13}{63} & \frac{5}{63} & \frac{22}{63} \\\\ 0 & -7 & 0 & -\frac{23}{9} & \frac{4}{9} & \frac{5}{9} \\\\ 0 & 0 & -9 & -1 & -1 & 1 \end{pmatrix}\]

Elementary Row Operation 2: Getting the \( \begin{pmatrix} I & A^{-1} \end{pmatrix} \) format

\( R2 \leftarrow -\frac{1}{7}R2  \&  R3 \leftarrow -\frac{1}{9}R3 \)

\[ \begin{pmatrix} 1 & 0 & 0 & -\frac{13}{63} & \frac{5}{63} & \frac{22}{63} \\\\ 0 & 1 & 0 & \frac{23}{63} & -\frac{4}{63} & -\frac{5}{63} \\\\ 0 & 0 & 1 & \frac{1}{9} & \frac{1}{9} & -\frac{1}{9} \end{pmatrix}\]

Now that we got the augmented matrix in the format we need, we have

\[ A^{-1} = \begin{pmatrix} -\frac{13}{63} & \frac{5}{63} & \frac{22}{63} \\\\ \frac{23}{63} & -\frac{4}{63} & -\frac{5}{63} \\\\ \frac{1}{9} & \frac{1}{9} & -\frac{1}{9} \end{pmatrix}\]

The set of linear equations can be represented as

\[ Ax = b \]

where

\[ A = \begin{pmatrix} 1 & 3 & 1 \\\\ 2 & -1 & 7 \\\\ 3 & 2 & -1 \end{pmatrix} \]

\[ x = \begin{pmatrix} x \\\\ y \\\\ z \end{pmatrix} \]

\[ b = \begin{pmatrix} 10 \\\\ 21 \\\\ 4 \end{pmatrix} \]

\[ Ax = b \implies A^{-1}Ax = A^{-1}b \implies Ib = A^{-1}b \implies x = A^{-1}b \]

\[ x = A^{-1}b \]

\[ = \begin{pmatrix} -\frac{13}{63} & \frac{5}{63} & \frac{22}{63} \\\\ \frac{23}{63} & -\frac{4}{63} & -\frac{5}{63} \\\\ \frac{1}{9} & \frac{1}{9} & -\frac{1}{9} \end{pmatrix} \begin{pmatrix} 10 \\\\ 21 \\\\ 4 \end{pmatrix} \]

\[ = \begin{pmatrix} -\frac{130}{63}+\frac{105}{63}+\frac{88}{63} \\\\ \frac{230}{63}-\frac{84}{63}-\frac{20}{63} \\\\ \frac{10}{9}+\frac{21}{9}-\frac{4}{9} \end{pmatrix}\]

\[ = \begin{pmatrix} 1 \\\\ 2 \\\\ 3 \end{pmatrix}\]