How to Find the determinant of the following matrices using only determinant properties:
A = $$
\begin{bmatrix}
4 & -3 & 5 \\
5 & 2 & 0 \\
2 & 0 & 4 \\
\end{bmatrix}
$$
B = $$
\begin{bmatrix}
2 & 0 & 1 & 4 \\
3 & 2 & -4 & -2 \\
2 & 3 & -1 & 0 \\
11 & 8 & -4 & 6 \\
\end{bmatrix}
$$
C = $$
\begin{bmatrix}
4 & 2 & 3 & -4 \\
3 & -2 & 1 & 5\\
-2 & 0 & 1 & -3 \\
8 & -2 & 6 & 4 \\
\end{bmatrix}
$$
I have tried every property of determinants, yet I fail to find the solution the way I want, I definitely know the solution using the rules, but I want to use only the properties like just what happen to the determinant when you swipe rows.
Best Answer
If you are really allowed nothing, only row operations, this is basically doing Gaussian elimination until you get matrix which has zeroes under the diagonal. But you must in each step also check how the determinant is changed.
For the third determinant WolframAlpha returns $-30$.
You could calculate it using row operations:
$$\det\begin{pmatrix} 4 & 2 & 3 & -4 \\ 3 & -2 & 1 & 5\\ -2 & 0 & 1 & -3 \\ 8 & -2 & 6 & 4 \\ \end{pmatrix}= 2\det\begin{pmatrix} 4 & 2 & 3 &-4 \\ 3 &-2 & 1 & 5 \\ -2 & 0 & 1 &-3 \\ 4 &-1 & 3 & 2 \\ \end{pmatrix}= 2\det\begin{pmatrix} 0 & 2 & 5 &-10\\ 1 &-2 & 2 & 2 \\ -2 & 0 & 1 &-3 \\ 0 &-1 & 5 &-4 \\ \end{pmatrix}= 2\det\begin{pmatrix} 0 & 2 & 5 &-10\\ 1 &-2 & 2 & 2 \\ 0 &-4 & 5 & 1 \\ 0 &-1 & 5 &-4 \\ \end{pmatrix}= -2\det\begin{pmatrix} 1 &-2 & 2 & 2 \\ 0 & 2 & 5 &-10\\ 0 &-4 & 5 & 1 \\ 0 &-1 & 5 &-4 \\ \end{pmatrix}= -2\det\begin{pmatrix} 1 &-2 & 2 & 2 \\ 0 & 0 &15 &-18\\ 0 & 0 &-15& 17 \\ 0 &-1 & 5 &-4 \\ \end{pmatrix}= -2\det\begin{pmatrix} 1 &-2 & 2 & 2 \\ 0 &-1 & 5 &-4 \\ 0 & 0 &15 &-18\\ 0 & 0 &-15& 17 \\ \end{pmatrix}= -2\det\begin{pmatrix} 1 &-2 & 2 & 2 \\ 0 &-1 & 5 &-4 \\ 0 & 0 &15 &-18\\ 0 & 0 & 0 & -1 \\ \end{pmatrix}= (-2)\cdot15=-30$$
But the computations are simpler if you are also allowed to use Laplace expansion, column operations and other properties of determinants: $$2\det\begin{pmatrix} 0 & 2 & 5 &-10\\ 1 &-2 & 2 & 2 \\ 0 &-4 & 5 & 1 \\ 0 &-1 & 5 &-4 \\ \end{pmatrix}= -2\det\begin{pmatrix} 2 & 5 &-10\\ -4 & 5 & 1 \\ -1 & 5 &-4 \\ \end{pmatrix}= -10\det\begin{pmatrix} 2 & 1 &-10\\ -4 & 1 & 1 \\ -1 & 1 &-4 \\ \end{pmatrix}= -10\det\begin{pmatrix} 0 & 3 &-18\\ 0 &-3 & 17 \\ -1 & 1 &-4 \\ \end{pmatrix}= 10\det\begin{pmatrix} 3 &-18\\ -3 & 17 \\ \end{pmatrix}= 10\det\begin{pmatrix} 3 &-18\\ -3 & 17 \\ \end{pmatrix}=10(3\cdot17-3\cdot18)=-30$$
I'll leave the first two determinants for you.