Properties of determinants

We create a $4\times4$ matrix while leaving the 4th column empty $$\left(\begin{matrix} 1 & 1 & 2 & \fbox{ } \\ -1 & 0 & 3 & \fbox{ } \\ -2 & 1 & -1 & \fbox{ } \\ 0 & 0 & 1 & \fbox{ } \end{matrix} \right)$$

We demand that the 4th column is a linear combination of the columns C1 and C2.

$C4=C1+C2$ (There are infinite possibilities)

Then the $4\times4$ matrix is $$\left(\begin{matrix} 1 & 1 & 2 & 2 \\ -1 & 0 & 3 & -1 \\ -2 & 1 & -1 & -1 \\ 0 & 0 & 1 & 0 \end{matrix} \right)$$

The determinant can be calculated now. Is it necessary to do it? By construction the property 2.c) is satisfied, so the determinant is empty.

First of all, we create the $3\times3$ matrix $$A=\left(\begin{matrix} 1 & 0 & 1 \\ 2 & 2 & 1 \\ -2 & 1 & -4 \end{matrix} \right)$$ The construction of the transpose matrix is done by exchanging lines and columns, that is to say $$A^t=\left(\begin{matrix} 1 & 2 & -2 \\ 0 & 2 & 1 \\ 1 & 1 & -4 \end{matrix} \right)$$ The determinant can be now computed: $$det(A^t)=\left|\begin{matrix} 1 & 2 & -2 \\ 0 & 2 & 1 \\ 1 & 1 & -4 \end{matrix} \right|=-8+0+2-(-4)-1-0=-3$$ Calculate also the det (A). The result should be the same, so calculating the det (A) can be a way to verify that we have not made any mistakes in the calculations. $$det(A)=\left|\begin{matrix} 1 & 0 & 1 \\ 2 & 2 & 1 \\ -2 & 1 & -4 \end{matrix} \right|=-8+2+0-(-4)-1-0=-3$$ The results are the same.

The first step is to construct the matrix, in this case $3\times3$, in the most general way possible but with a repeated column. That one is $$A=\left(\begin{matrix} a & d & a \\ b & e & b \\ c & f & c \end{matrix} \right)$$ let's now calculate the determinant (we can do it by means of the general method or using the rule of Sarrus) $$det(A)=\left|\begin{matrix} a & d & a \\ b & e & b \\ c & f & c \end{matrix} \right|=\cancel{a\cdot e\cdot c}+\bcancel{b\cdot f \cdot a} + \xcancel{c \cdot d \cdot b}$$ $$ - \cancel{a \cdot e \cdot c} - \bcancel{b\cdot f \cdot a} - \xcancel{c \cdot d \cdot b} = 0$$