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\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
\title{Mathematics Assignment 2 --- Matrices}
\author{Ahmad Saalim Lone, 2019BCSE017}
\date{05 May, 2020}
\maketitle
\section*{Question 1}
If
\(
A =
\begin{bmatrix}
3 & 3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{bmatrix}
\)
find $A^{-1}$ and verify that $A^{-1}A = I = AA^{-1}$.
\subsection*{Solution}
\begin{align*}
A &=
\begin{bmatrix}
3 & 3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{bmatrix} \\
A^{-1} &= \frac{adj(A)}{|A|} \\
|A| &= 3 \times (-3 + 4 ) - 3 \times 2 + 4 \times -2 \\
|A| &= -11 \\
\text{Cofactor matrix of A } &=
\begin{bmatrix}
1 & 2 & -2 \\
7 & 3 & -3 \\
24 & 4 & -15
\end{bmatrix} \\
adj(A) &=
\begin{bmatrix}
1 & 7 & 24 \\
2 & 3 & 4 \\
-2 & -3 & -15
\end{bmatrix} \\
A^{-1} &=
\begin{bmatrix}
\frac{1}{11} & \frac{2}{11} & -\frac{2}{11} \\[6pt]
\frac{7}{11} & \frac{3}{11} & -\frac{3}{11} \\[6pt]
\frac{24}{11} & \frac{4}{11} & -\frac{15}{11}
\end{bmatrix}
\end{align*}
\begin{align*}
A \times A^{-1} &=
\begin{bmatrix}
\frac{1}{11} & \frac{2}{11} & -\frac{2}{11} \\[6pt]
\frac{7}{11} & \frac{3}{11} & -\frac{3}{11} \\[6pt]
\frac{24}{11} & \frac{4}{11} & -\frac{15}{11}
\end{bmatrix}
\begin{bmatrix}
3 & 3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{bmatrix} \\
&=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix} \\
&= I
\end{align*}
Similarly, $A^{-1}\times A = I$
\section*{Question 2}
Find the inverse of
\(
A =\begin{bmatrix}
1 & 2 & 1 \\
3 & 2 & 3 \\
1 & 1 & 2
\end{bmatrix}
\) by applying E-transformation.
\subsection*{Solution}
\[
\begin{bmatrix}
1 & 2 & 1 \\
3 & 2 & 3 \\
1 & 1 & 2
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
A
\]
After applying the following transformations
\begin{align*}
C_3 &\to C_3 - C_1 \\
C_2 &\to \frac{C_1}{2} \\
R_1 &\to R_1 - R_2 \\
R_1 &\to \frac{R_1}{-2} \\
C_1 &\to C_1 - C_3 \\
R_2 &\to R_2 - 3R1 \\
R_2 &\to R_3 - \frac{R_2}{2}
\end{align*}
we get
\begin{align*}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
&=
\begin{bmatrix}
-1 & \frac{1}{4} & \frac{1}{2} \\[6pt]
3 & -\frac{1}{4} & -\frac{1}{2} \\[6pt]
-\frac{5}{2} & \frac{1}{8} & \frac{3}{4}
\end{bmatrix}
\times A \\
\therefore A^{-1} &=
\begin{bmatrix}
-1 & \frac{1}{4} & \frac{1}{2} \\[6pt]
3 & -\frac{1}{4} & -\frac{1}{2} \\[6pt]
-\frac{5}{2} & \frac{1}{8} & \frac{3}{4}
\end{bmatrix}
\end{align*}
\section*{Question 3}
Reduce the matrix
\(
A =
\begin{bmatrix}
1 & -1 & 2 & -3 \\
4 & 1 & 0 & 2 \\
0 & 3 & 0 & 4 \\
0 & 1 & 0 & 2
\end{bmatrix}
\) to the normal form and hence determine its rank.
\subsection*{Solution}
To reduce the matrix we apply the following operations
\begin{align*}
R_1 &\to R_1 + R_3 \\
R_4 &\to R_4 - R_2 \\
R_2 &\to R_2 + R_4 \\
C_2 &\to C_2 - C_3 \\
C_4 &\to C_4 - C_2 \\
R_3 &\to R_3 - R_2 \\
C_2 &\rightleftharpoons C_3 \\
C_2 &\to \frac{C_2}{2} \\
R_3 &\to \frac{R_3}{2} \\
R_4 &\to \frac{R_4}{2} \\
C_4 &\to C_4 - C_2 \\
C_3 &\to C_3 - C_4 \\
C_4 &\rightleftharpoons C_2 \\
R_4 &\to R_4 - R_1 \\
R_4 &\to -R_4 \\
R_1 &\to R_1 - R_4
\end{align*}
At the end we arrive at
\[
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
\]
i.e. $[I_4]$
$rank = 4$
\section*{Question 4}
Reduce the matrix \(A =
\begin{bmatrix}
-2 & -1 & -3 & -1 \\
1 & 2 & 3 & -1 \\
1 & 0 & 1 & 1 \\
0 & 1 & 1 & -1
\end{bmatrix}
\) to Echelon form and find its rank.
\subsection*{Solution}
To convert it into Echelon form, we apply the following transformations.
\begin{align*}
R_1 &\to R_1 + R_2 + R_3 \\
R_2 &\to R_2 - R_3 \\
C_2 &\to C_2 + C_4 \\
C_4 &\to C_4 + C_3 \\
C_4 &\to C_4 - 2 C_1 \\
C_3 &\to C_3 - C_2 \\
C_2 &\to C_2 - C_1 \\
R_2 &\to \frac{R_2}{2} \\
R_1 &\leftrightharpoons R_4 \\
C_4 &\leftrightharpoons C_1 \\
R_4 &\to R_4 - R_2
\end{align*}
We get
\[
\begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}
\]
\[
rank = 3
\]
\section*{Question 5}
Solve
\begin{align*}
x + y + z &= 9 \\
2x + 5y + 7z &= 52 \\
2x + y - z &= 0
\end{align*}
\subsection*{Solution}
\begin{align*}
\Delta &=
\begin{vmatrix}
1 & 1 & 1 \\
2 & 5 & 7 \\
2 & 1 & -1
\end{vmatrix}
= -4
\\
\Delta_1 &=
\begin{vmatrix}
9 & 1 & 1 \\
52 & 5 & 7 \\
0 & 1 & -1
\end{vmatrix}
= -4
\\
\Delta_2 &=
\begin{vmatrix}
1 & 9 & 1 \\
2 & 52 & 7 \\
2 & 0 & -1
\end{vmatrix}
= -12
\\
\Delta_3 &=
\begin{vmatrix}
1 & 1 & 9 \\
2 & 5 & 52 \\
2 & 1 & 0
\end{vmatrix}
= -20
\\
x &= \frac{\Delta_1}{\Delta} = 1 \\
y &= \frac{\Delta_2}{\Delta} = 3 \\
z &= \frac{\Delta_3}{\Delta} = 5
\end{align*}
\section*{Question 6}
Test the consistency of:
\begin{align*}
3x - y + 2z &= 3 \\
2x + y + 3z &= 5 \\
x - 2y - z &= 1
\end{align*}
\subsection*{Solution}
\[
\text{Augemented matrix} =
\begin{bmatrix}
3 & -1 & 2 & : & 3 \\
2 & 1 & 3 & : & 5 \\
1 & -2 & -1 & : & 1
\end{bmatrix}
\]
\[
\Delta =
\begin{vmatrix}
3 & -1 & 2 \\
2 & 1 & 3 \\
1 & -2 & -1
\end{vmatrix}
= 10
\]
\[
\Delta \ne 0 \therefore \text{it is consistent with the unique solution}
\]
\section*{Question 7}
Solve the equations
\begin{align*}
x + 3y - 2z &= 0 \\
2x -y + 4z &= 0 \\
x - 11y + 14z &= 0
\end{align*}
\subsection*{Solution}
\begin{align*}
\Delta =\begin{vmatrix}
1 & 3 & -2 \\
2 & -1 & 4 \\
1 & -11 & 14
\end{vmatrix}
&= 0 \\
\Delta_1 = \Delta_2 = \Delta_3 &= 0 \\
\therefore \text{unique solution is } x = y = z &= 0
\end{align*}
\end{document}