#augmented matrix

[math-journal3]

(2)<-- 1.2 - 3 --> (4)

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[er-cryptid]

Augmented Matrix

-- one use of matrices is representing a system of equations

-- this is called an augmented matrix

-- to write a system of equations as an augmented matrix, use brackets on the outside

-- let each term be represented by its coefficient with the coefficient’s sign      -- if it is -2, it will stay -2      -- if it is +1, it will stay +1

-- make a dotted vertical line to separate the equation from what it equals

[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

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

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

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

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

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

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

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

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