diff --git a/cs2104/labs/assignment.pl b/cs2104/labs/assignment.pl
index a36b648..18c5181 100644
--- a/cs2104/labs/assignment.pl
+++ b/cs2104/labs/assignment.pl
@@ -215,9 +215,7 @@ nth(Idx, [_|Tail], Elem) :-
   Idx1 is Idx - 1,
   nth(Idx1, Tail, Elem).
 
-% Taufiq
-nth(0, [V | _], V) :- !.
-nth(I, [_|T], V) :- nth(I1, T, V), I is I1 + 1, I > 0.    
+    
 
 %%%%%%%%%%%%%%
 
@@ -258,9 +256,6 @@ pos_helper(Target, [_|Tail], Acc, Index) :-
   Acc1 is Acc + 1,
   pos_helper(Target, Tail, Acc1, Index).
 
-% Taufiq
-pos(X, [X | _], 0).
-pos(X, [_ | T], I) :- pos(X, T, I1), I is I1 + 1.
 
 %%%%%%%%%%%%%%
 
@@ -294,9 +289,6 @@ all_pos_helper(Target, [_|Tail], Acc, Rest) :-
   Acc1 is Acc + 1,
   all_pos_helper(Target, Tail, Acc1, Rest).
 
-% Taufiq
-all_pos(X, L, []) :- not(member(X, L)).
-all_pos(X, L, LI) :- bagof(I, pos(X, L, I), LI).
 
 %%%%%%%%%%%%%%
 
diff --git a/cs2109s/labs/ps0/ps0.py b/cs2109s/labs/ps0/ps0.py
index 9572cd1..b327756 100644
--- a/cs2109s/labs/ps0/ps0.py
+++ b/cs2109s/labs/ps0/ps0.py
@@ -538,9 +538,8 @@ def average_increase_in_cases(n_cases_increase, n_adj_entries_avg=7):
     '''
 
     sd_win = np.lib.stride_tricks.sliding_window_view(n_cases_increase, 2 * n_adj_entries_avg + 1, axis=1)
-    avg = np.mean(sd_win, axis=2)
+    avg = np.floor(np.mean(sd_win, axis=2))
     res = np.pad(avg, ((0, 0), (n_adj_entries_avg, n_adj_entries_avg)), 'constant', constant_values=(np.nan, np.nan))
-    print(res)
     return res
 
 
@@ -552,7 +551,7 @@ def test_26():
     assert(np.array_equal(actual, expected, equal_nan=True))
 
 
-test_26()
+# test_26()
 
 # Task 2.7
 def is_peak(n_cases_increase_avg, n_adj_entries_peak=7):
@@ -605,9 +604,23 @@ def is_peak(n_cases_increase_avg, n_adj_entries_peak=7):
     Hint: to determine `n_adj_entries_avg` from `n_cases_increase_avg`,
     `np.count_nonzero` and `np.isnan` may be helpful.
     '''
+    # get the number of paddings to add (count nans and / 2)
+    paddings = int(np.count_nonzero(np.isnan(n_cases_increase_avg[0])) / 2)
+    nanless = n_cases_increase_avg[:, paddings: -paddings]
+    # find peaks (get the sliding window)
+    sd_win = np.lib.stride_tricks.sliding_window_view(nanless, 2 * n_adj_entries_peak + 1, axis=1)
+    mids = sd_win[:, :, n_adj_entries_peak]
+    left_vals = np.max(sd_win[:, :, :n_adj_entries_peak], axis=2)
+    right_vals = np.max(sd_win[:, :, n_adj_entries_peak+1:], axis=2)
+    peaks = (mids > left_vals) & (mids >= right_vals)
 
-    # TODO: add your solution here and remove `raise NotImplementedError`
-    raise NotImplementedError
+    means = np.nanmean(nanless, axis=1)[:, np.newaxis] * 0.1
+
+    significant = mids > means
+
+    result = np.logical_and(peaks, significant)
+    res = np.pad(result, ((0, 0), (paddings+n_adj_entries_peak, paddings+ n_adj_entries_peak)), 'constant', constant_values=(False, False))
+    return res
 
 
 def test_27():
@@ -627,7 +640,7 @@ def test_27():
                         [False, False, False, False, False, False, False, False, False]])
     assert np.all(actual2 == expected2)
 
-#test_27()
+test_27()
 
 def visualise_increase(n_cases_increase, n_cases_increase_avg=None):
     '''
diff --git a/ma1522/lec_01.tex b/ma1522/lec_01.tex
index 7eda3d9..bbc8715 100644
--- a/ma1522/lec_01.tex
+++ b/ma1522/lec_01.tex
@@ -166,9 +166,12 @@ To solve a linear system we perform operations:
   \item add a constant multiple of an equation to another
 \end{itemize}
 
+Likewise, for a augmented matrix, the operations are on the \textbf{rows} of the augmented matrix
 
 \begin{itemize}
   \item Multiply row by nonzero constant
   \item Interchange 2 rows
   \item add a constant multiple of a row to another row
 \end{itemize}
+
+To note: all these operations are revertible
diff --git a/ma1522/lec_02.tex b/ma1522/lec_02.tex
index e69de29..a2e8acc 100644
--- a/ma1522/lec_02.tex
+++ b/ma1522/lec_02.tex
@@ -0,0 +1,352 @@
+\subsection{Recap}
+Given the linear equation $a_1x_1 +  a_2x_2 + ... + a_nx_n = b$
+
+\begin{enumerate}
+  \item $a_1 = a_2 = ... = a_n = b = 0$ zero equation
+
+    Solution: $x_1 = t_1, x_2 = t_2, ... = x_n = t_n$
+
+  \item $a_1 = a_2 = ... = a_n = 0 \neq b$ inconsistent
+
+    No Solution
+
+  \item Not all $a_1 ... a_n$ are zero. 
+
+    Set $n-1$ of $x_i$ as params, solve for last variable
+\end{enumerate}
+
+\subsection{Elementary Row Operations Example}
+
+\begin{center}
+
+  \begin{minipage}{.3\linewidth}
+    \systeme{x+y+3z = 0, 2x-2y+2z=4, 3x+9y=3}
+  \end{minipage}%
+  \begin{minipage}{.3\linewidth}
+    \begin{equation*}
+      \begin{amatrix}{3}
+        1 & 1 & 3 & 0 \\
+        2 & 2 & 2 & 4 \\
+        3 & 9 & 0 & 3
+      \end{amatrix}
+    \end{equation*}
+  \end{minipage}
+\end{center}
+
+\subsection{Row Equivalent Matrices}
+2 Augmented Matrices are row equivalent if one can be obtained from the other by a series of elementary row operations
+
+Given a augmented matrix $A$, how to find a row equivalent augmented  matrix B of which is of a \textbf{simple} form?
+
+\subsection{Row Echelon Form}
+
+\begin{defn}[Row Echelon Form (Simple)]
+  Augmented Matrix is in row-echelon form if
+  \begin{itemize}
+    \item Zero rows are grouped together at the bottom
+    \item For any 2 successive nonzero rows, The first nonzero number in the lower row appears to the right of the first nonzero number on the higher row
+      $\begin{amatrix}{4}
+        0 & 0 & 1 & 2 & 3 \\
+        0 & 0 & 0 & 1 & 2 \\
+      \end{amatrix}$
+    \item Leading entry if a nonzero row is a \textbf{pivot point}
+    \item Column of augmented matrix is called 
+      \begin{itemize}
+        \item \textbf{Pivot Column} if it contains a pivot point
+        \item \textbf{Non Pivot Column} if it contains no pivot point
+      \end{itemize} 
+    \item Pivot Column contains exactly 1 pivot point
+
+      \# of pivots = \# of leading entries = \# of nonzero rows
+  \end{itemize}
+\end{defn}
+
+Examples of row echlon form: 
+
+\begin{equation*}
+  \begin{amatrix}{2}
+    3 & 2 & 1 \\
+  \end{amatrix}
+  \begin{amatrix}{2}
+    1 & -1 & 0 \\
+    0 & 1 & 0 \\
+  \end{amatrix}
+  \begin{amatrix}{2}
+    2 & 1 & 0 \\
+    0 & 1 & 0 \\
+    0 & 0 & 1 \\
+  \end{amatrix}
+  \begin{amatrix}{3}
+    1 & 2 & 3 & 4 \\
+    0 & 1 & 1 & 2 \\
+    0 & 0 & 2 & 3 \\
+  \end{amatrix}
+  \begin{amatrix}{4}
+    0 & 1 & 2 & 8 & 1 \\
+    0 & 0 & 0 & 0 & 3 \\
+    0 & 0 & 0 & 0 & 0 \\
+    0 & 0 & 0 & 0 & 0 \\
+  \end{amatrix}
+\end{equation*}
+
+Examples of NON row echlon form: 
+
+\begin{equation*}
+  \begin{amatrix}{2}
+    0 & \textbf{1} & 0 \\
+    \textbf{1} & 0 & 0 \\
+  \end{amatrix}
+  \begin{amatrix}{2}
+    0 & 0 & \textbf{1} \\
+    \textbf{1} & -1 & 0 \\
+    0 & 0 & 1 \\
+  \end{amatrix}
+  \begin{amatrix}{3}
+    \textbf{1} & 0 & 2 & 1 \\
+    0 & \textbf{1} & 0 & 2 \\
+    0 & \textbf{1} & 1 & 3 \\
+  \end{amatrix}
+  \begin{amatrix}{4}
+    \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} \\
+    1 & 0 & 2 & 0 & 1 \\
+    0 & 0 & 0 & 1 & 3 \\
+    0 & 0 & 0 & 0 & 0 \\
+  \end{amatrix}
+\end{equation*}
+\subsection{Reduced Row-Echelon Form}
+
+\begin{defn}[Reduced Row-Echelon Form]
+  Suppose an augmented matrix is in row-echelon form. It is in \textbf{reduced row-echelon form} if
+  \begin{itemize}
+    \item Leading entry of every nonzero row is 1
+      \subitem Every pivot point is one
+    \item In each pivot column, except the pivot point, all other entries are 0.
+  \end{itemize}
+\end{defn}
+
+Examples of reduced row-echelon form: 
+
+\begin{equation*}
+  \begin{amatrix}{2}
+    1 & 2 & 3 \\
+  \end{amatrix}
+  \begin{amatrix}{2}
+    0 & 0 & 0 \\
+    0 & 0 & 0 \\
+  \end{amatrix}
+  \begin{amatrix}{2}
+    1 & 0 & 0 \\
+    0 & 1 & 0 \\
+    0 & 0 & 1 \\
+  \end{amatrix}
+  \begin{amatrix}{3}
+    1 & 0 & 0 & 1 \\
+    0 & 1 & 0 & 2 \\
+    0 & 0 & 1 & 3 \\
+  \end{amatrix}
+  \begin{amatrix}{4}
+    0 & 1 & 2 & 0 & 1 \\
+    0 & 0 & 0 & 1 & 3 \\
+    0 & 0 & 0 & 0 & 0 \\
+    0 & 0 & 0 & 0 & 0 \\
+  \end{amatrix}
+\end{equation*}
+
+Examples of row-echelon form but not reduced: (pivot point is not 1 / all other elements \textbf{in pivot column} must be zero)
+
+\begin{equation*}
+  \begin{amatrix}{2}
+    \textbf{3} & 2 & 1 \\
+  \end{amatrix}
+  \begin{amatrix}{2}
+    1 & \textbf{-1} & 0 \\
+    0 & 1 & 0 \\
+  \end{amatrix}
+  \begin{amatrix}{2}
+    \textbf{2} & 0 & 0 \\
+    0 & 1 & 0 \\
+    0 & 0 & 1 \\
+  \end{amatrix}
+  \begin{amatrix}{3}
+    \textbf{-1} & 2 & 3 & 4 \\
+    0 & 1 & 1 & 2 \\
+    0 & 0 & 2 & 3 \\
+  \end{amatrix}
+  \begin{amatrix}{4}
+    0 & 1 & 2 & \textbf{8} & 1 \\
+    0 & 0 & 0 & \textbf{4} & 3 \\
+    0 & 0 & 0 & 0 & 0 \\
+    0 & 0 & 0 & 0 & 0 \\
+  \end{amatrix}
+\end{equation*}
+
+To note: 2nd matrix has -1 in the pivot column, but 5th matrix has 2 in a non-pivot column so its fine
+
+\subsection{Solving Linear System}
+
+If Augmented Matrix is in reduced row-echelon form, then solving it is easy
+
+\begin{equation*}
+  \begin{amatrix}{3}
+    1 & 0 & 0 & 1 \\
+    0 & 1 & 0 & 2 \\
+    0 & 0 & 1 & 3 \\
+  \end{amatrix}
+  \text{then } x_1 = 1, x_2 = 2, x_3 = 3
+\end{equation*}
+
+\begin{note}
+  \begin{itemize}
+    \item If any equations in the system is inconsistent, the whole system is inconsistent
+  \end{itemize}
+\end{note}
+
+\subsubsection{Examples}
+
+Augmented Matrix: $\begin{amatrix}{4}
+  1 & -1 & 0 & 3 & -2 \\
+  0 & 0 & 1 & 2 & 5 \\
+  0 & 0 & 0 & 0 & 0 \\
+\end{amatrix}$
+
+\begin{itemize}
+  \item The zero row can be ignored. 
+    \systeme{x_1 - x_2 + 3x_4 = -2, x_3 + 2x_4 = 5}
+  \item Degree of freedom(\# cols): 4, number of restrictions (\# pivot cols): 2, arbitrary vars(\# non pivot cols): 4-2 = 2. Set this to the non-pivot cols
+\end{itemize}
+\begin{enumerate}
+  \item Let $x_4 = t$ and sub into 2nd eqn
+    \subitem $x_3 + 2t = 5 \Rightarrow x_3 = 5-2t$
+  \item sub $x_4 = t$ into 1st eqn
+    \subitem $x_1 - x_2 + 3t = -2$
+    \subitem Let $x_2 = s$. Then $x_1 = -2 + s - 3t$
+  \item Infinitely many sols with ($s$ and $t$ as arbitrary params)
+    \subitem $x_1 = -2 + s - 3t, x_2 = s, x_3 = 5-2t, x_4 = t$
+
+\end{enumerate}
+
+Augmented Matrix: $\begin{amatrix}{5}
+  0 & 2 & 2 & 1 & -2 & 2 \\
+  0 & 0 & 1 & 1 & 1 & 3 \\
+  0 & 0 & 0 & 0 & 2 & 4 \\
+\end{amatrix}$
+\begin{itemize}
+  \item \systeme{0x_1 + 2x_2 + 2x_3 + 1x_4 -2x_5 = 2,x_3 + x_4 +x_5 = 3,2x_5 = 4}
+  \item Degree of freedom: 5, number of restrictions: 3, arbitrary vars: 5-3 = 2
+\end{itemize}
+
+\begin{enumerate}
+  \item by 3rd eqn, $2x_5 = 4 \Rightarrow x_5 = 2$
+  \item sub $x_5 = 2$ into 2nd eqn
+    \subitem $x_3 + x_4 + 2 = 3 \Rightarrow x_3 + x_4 = 1$
+    \subitem let $x_4 = t$. Then $x_3 = 1-t$
+  \item sub $x_5 = 2, x_3 = 1-t, x_4 = t$ into 1st eqn
+    \subitem $2x_2 + 2(1-t) + t - 2(2) = 2 \Rightarrow 2x_2 -t = 4 \Rightarrow x_2 = \frac{t}{2} + 2$
+  \item system has inf many solns: $x_1 = s, x_2 = \frac{t}{2} + 2, x_3 = 1-t, x_4 = t, x_5 = 2$ where $s$ and $t$ are arbitrary
+\end{enumerate}
+
+\subsubsection{Algorithm}
+Given the augmented matrix is in row-echelon form.
+\begin{enumerate}
+  \item Set variables corresponding to non-pivot columns to be arbitrary parameters
+  \item Solve variables corresponding to pivot columns by back substitution (from last eqn to first)
+\end{enumerate}
+
+
+\subsection{Gaussian Eliminiation}
+
+\begin{defn}[Gaussian Elimination]\ \\
+  \begin{enumerate}
+    \item Find the left most column which is not entirely zero
+    \item If top entry of such column is 0, replace with nonzero number by swapping rows
+    \item For each row below top row, add multiple of top row so that leading entry becomes 0
+    \item Cover top row and repeat to remaining matrix
+  \end{enumerate}
+\end{defn}
+
+\begin{note}[Algorithm with Example]\ \\
+  $\begin{amatrix}{6}
+    0 & 0 & 0 & 2 & 4 & 2 & 8 \\
+    0 & 1 & 2 & 4 & 5 & 3 &-9 \\
+    0 &-2 &-4 &-5 &-4 & 3 & 6 \\
+  \end{amatrix}$
+  \begin{enumerate}
+    \item Find the left most column which is not all zero (2nd column)
+    \item Check top entry of the selection. If its zero, replace it by a nonzero number by interchanging the top row with another row below
+      \subitem $\begin{amatrix}{6}
+        0 & 1 & 2 & 4 & 5 & 3 &-9 \\
+        0 & 0 & 0 & 2 & 4 & 2 & 8 \\
+        0 &-2 &-4 &-5 &-4 & 3 & 6 \\
+      \end{amatrix}$
+    \item For each row below the top row, adda suitable multiple of top row so that leading entry becomes 0.
+      \subitem $2R_1 + R_3$ will ensure that the -2 turns to 0
+      \subitem $\begin{amatrix}{6}
+        0 & 1 & 2 & 4 & 5 & 3 &-9 \\
+        0 & 0 & 0 & 2 & 4 & 2 & 8 \\
+        0 & 0 & 0 & 3 & 6 & 9 &-12\\
+      \end{amatrix}$
+    \item Cover top row and repeat procedure to the remaining matrix
+      \subitem $\begin{amatrix}{6}
+        0 & 1 & 2 & 4 & 5 & 3 &-9 \\
+        \hline
+        0 & 0 & 0 & 2 & 4 & 2 & 8 \\
+        0 & 0 & 0 & 3 & 6 & 9 &-12\\
+      \end{amatrix}$
+      \subitem Look at $C_4$. $R_3 \times -1.5R_2$ will set $R_3C_4$ to zero.
+      \subitem $\begin{amatrix}{6}
+        0 & 1 & 2 & 4 & 5 & 3 &-9 \\
+        \hline
+        0 & 0 & 0 & 2 & 4 & 2 & 8 \\
+        0 & 0 & 0 & 0 & 0 & 6 & -24\\
+      \end{amatrix}$
+      \subitem This is now in row echelon form.
+  \end{enumerate}
+      Only use $R_i \Leftrightarrow R_j or R_i + CR_j$ in this method.
+\end{note}
+
+\subsection{Gauss-Jordan Elimination}
+\begin{defn}[Gauss Joran Elimination]\ \\
+  \begin{enumerate}
+    \item[1-4.] Use Gaussian Eliminiation to get row-echelon form
+      \setcounter{enumi}{4}
+    \item For each nonzero row, multiply a suitable constant so pivot point becomes 1
+    \item Begin with last nonzero row and work backwords
+      \subitem Add suitable multiple of each row to the rows above to introduce 0 above pivot point
+  \end{enumerate}
+  \begin{itemize}
+    \item Every matrix has a unique reduced row-echelon form.
+    \item Every nonzero matrix has infinitely many row-echelon ofrm
+  \end{itemize}
+\end{defn}
+\begin{note}[Gauss Jordan Elimination Example] Suppose an augmented matrix is in row-echelon form. 
+  $\begin{amatrix}{5}
+    1 & 2 & 4 & 5 & 3 & -9 \\
+    0 & 0 & 2 & 4 & 2 & 8 \\
+    0 & 0 & 0 & 0 & 6 & -24 \\
+  \end{amatrix}$
+  \begin{enumerate}
+    \item All pivot points must be 1
+      \subitem multiply $R_2$ by $\frac{1}{2}$ and $R_3$ by $\frac{1}{6}$
+      \subitem $\begin{amatrix}{5}
+        1 & 2 & 4 & 5 & 3 & -9 \\
+        0 & 0 & 1 & 2 & 1 & 4 \\
+        0 & 0 & 0 & 0 & 1 & -4 \\
+      \end{amatrix}$
+    \item In each pivot col, all entries other than pivot point must be 0. Work backwards
+      \subitem $R_1 + -3R_1$, $R_2 + -R_1$
+      \subitem $\begin{amatrix}{5}
+        1 & 2 & 4 & 5 & 0 & 3 \\
+        0 & 0 & 1 & 2 & 0 & 8 \\
+        0 & 0 & 0 & 0 & 1 & -4 \\
+      \end{amatrix}$
+      \subitem $R_1 + -4R_2$
+      \subitem $\begin{amatrix}{5}
+        1 & 2 & 0 & -3 & 0 & -29 \\
+        0 & 0 & 1 & 2 & 0 & 8 \\
+        0 & 0 & 0 & 0 & 1 & -4 \\
+      \end{amatrix}$
+
+  \end{enumerate}
+\end{note}
+
+
diff --git a/ma1522/master.pdf b/ma1522/master.pdf
index 21a01db..0996854 100644
Binary files a/ma1522/master.pdf and b/ma1522/master.pdf differ
diff --git a/ma1522/master.tex b/ma1522/master.tex
index c114c8c..12f37c0 100644
--- a/ma1522/master.tex
+++ b/ma1522/master.tex
@@ -69,6 +69,10 @@
 This is a note
 \end{note}
 
+\begin{defn}[Some Term]
+This is a definition
+\end{defn}
+
 % Maybe I need to add one more part: Examples.
 % Set style and colour later.
 
diff --git a/ma1522/preamble.tex b/ma1522/preamble.tex
index 4a068b9..146e28e 100644
--- a/ma1522/preamble.tex
+++ b/ma1522/preamble.tex
@@ -9,7 +9,10 @@
 \usepackage[english]{babel}
 \usepackage{framed}
 \usepackage[dvipsnames]{xcolor}
+\usepackage{soul}
 \usepackage{tcolorbox}
+\usepackage{systeme}
+\usepackage{multicol}
 
 \colorlet{LightGray}{White!90!Periwinkle}
 \colorlet{LightOrange}{Orange!15}
@@ -34,6 +37,10 @@
 \declaretheorem[style=notesty,numbered=no]{note}
 \tcolorboxenvironment{note}{colback=LightRed}
 
+\declaretheoremstyle[name=Definition,]{defnsty}
+\declaretheorem[style=defnsty,numbered=no]{defn}
+\tcolorboxenvironment{defn}{colback=LightGray}
+
 \setstretch{1.2}
 \geometry{
     textheight=9in,