For those who don't know part 3 (Determinants)

After solving the problem, I noticed that the problem solution to this problem is given by this video clip from Khan Academy. Abdulrahman Khaled should give credit the originator of the problem and mentioned that the problem is not original.

Anyway, as mentioned in the video, the easy way to solve this to make the matrix into either an upper or lower triangular matrix with row operations and then the determinant is given by the product of the diagonal elements. Two row operation rules are involved.

1. Replacing $row_i$ with $row_i - C\dot{}row_j$ , where $C$ is a constant, will not change the value of the determinant.
2. Whenever two rows are swapped the determinant of the new matrix is multiplied by $-1$ .

I am showing another way of applying the same method here.

$[X]=\begin{bmatrix} 1 & 2 & 2 & 1 \\ 1 & 2 & 4 & 2 \\ 2&7&5&2 \\ -1&4&-6&3 \end{bmatrix}$

$\begin{matrix} row_1 \rightarrow row_1 \\ row_2 - 2row_1 \rightarrow row_2 \\ row_3 - 2row_1 \rightarrow row_3 \\ row_4 - 3row_1 \rightarrow row_4 \end {matrix} \Rightarrow \begin{bmatrix} 1&2&2&1\\ -1&-2&0&0\\ 0&3&1&0\\ -4&-2&-12&0 \end{bmatrix}$

$\begin{matrix} row_1 \rightarrow row_1 \\ row_2 \rightarrow row_2 \\ row_3 \rightarrow row_3 \\ row_4 +12row_3 \rightarrow row_4 \end {matrix} \Rightarrow \begin{bmatrix} 1&2&2&1\\ -1&-2&0&0\\ 0&3&1&0\\ -4&34&0&0 \end{bmatrix}$

$\begin{matrix} row_1 \rightarrow row_1 \\ row_2 \rightarrow row_2 \\ row_3 \rightarrow row_3 \\ row_4 +17row_2 \rightarrow row_4 \end {matrix} \Rightarrow \begin{bmatrix} 1&2&2&1\\ -1&-2&0&0\\ 0&3&1&0\\ -21&0&0&0 \end{bmatrix}$

$\begin{matrix} row_4 \rightarrow row_1 \\ row_2 \rightarrow row_2 \\ row_3 \rightarrow row_3 \\ row_1 \rightarrow row_4 \end {matrix} \Rightarrow - \begin{bmatrix} -21&0&0&0\\ -1&-2&0&0\\ 0&3&1&0\\ 1&2&2&1 \end{bmatrix}$

Therefore, $\det {[X]} = X = -(-21)(-2)(1)(1) = \boxed {-42}$

• So, I tried solving this problem using Chió's method, to reduce the 4x4 matrix in a 3x3 one.
• I have this picture, for those who don't even heard about this method/tool. It is simple, and it works in any 3x3 or bigger matrix.
• After using Chió's, the matrix was reduced to:
• | $\boxed{0} \boxed{2} \boxed{1}$ |
• | $\boxed{3} \boxed{1} \boxed{0}$ |
• | $\boxed{6} \boxed{-4} \boxed{4}$ | ...(i'm sorry i'm not so skilled in writing solutions).
• Solving this matrix, we'll find out $Det = (0 + 0 - 12) - (6 + 24 + 0) = 42$ .

My solution is by Expansion of Minors. I considered the first column. $det~X = \begin{vmatrix} 1 & 2 & 2 & 1 \\ 1 & 2 & 4 & 2 \\ 2 & 7 & 5 & 2 \\ -1 & 4 & -6 & 3 \end{vmatrix} =1 \begin{vmatrix} 2 & 4 & 2 & 2 & 4\\ 7 & 5 & 2 & 7 & 5 \\ 4 & -6 & 3 & 4 & -6 \end{vmatrix}-1\begin{vmatrix} 2 & 2 & 1 & 2 & 2\\ 7 & 5 & 2 & 7 & 5 \\ 4 & -6 & 3 & 4 & -6 \end{vmatrix}+2\begin{vmatrix} 2 & 2 & 1 & 2 & 2\\ 2 & 4 & 2 & 2 & 4 \\ 4 & -6 & 3 & 4 & -6 \end{vmatrix}+1\begin{vmatrix} 2 & 2 & 1 & 2 & 2\\ 2 & 4 & 2 & 2 & 4 \\ 7 & 5 & 2 & 7 & 5 \end{vmatrix}$

$=1[30+32-84-(40-24+84)]-1[30+16-42-(20-24+42)]+2[24+16-12-(16-24+12)]+1[16+28+10-(28+20+8)]$

$=-122+34+48-2$

$=-42$