Resolve this in a way I can understand

We'll expand along the first row:

$$|A| = 4 \begin{vmatrix} 6 & -3 \\ 1 & -5 \end{vmatrix} - 9 \begin{vmatrix} -8 & -3 \\ 3 & -5 \end{vmatrix} + 2 \begin{vmatrix} -8 & 6 \\ 3 & 1 \end{vmatrix}$$

Now, calculate each $2 \times 2$ determinant $(ad-bc)$:

$$|A| = 4((6)(-5) - (-3)(1)) - 9((-8)(-5) - (-3)(3)) + 2((-8)(1) - (6)(3))$$ $$|A| = 4(-30 + 3) - 9(40 + 9) + 2(-8 - 18)$$ $$|A| = 4(-27) - 9(49) + 2(-26)$$ $$|A| = -108 - 441 - 52$$ $$|A| = -601$$

Understanding Transpose Determinant ($|A^T|$):
The transpose of a matrix ($A^T$) is created by flipping the matrix over its diagonal, meaning rows become columns and columns become rows. A key property of determinants is that the determinant of a matrix is always equal to the determinant of its transpose.
So, $|A^T| = |A|$.

Since $|A| = -601$, then $|A^T|$ is also $-601$.

$$|A^T| = -601$$

$$|A| + |A^T| = -601 + (-601)$$ $$|A| + |A^T| = -1202$$

The value of $|A| + |A^T|$ is $\boxed{-1202}$.

Hence or otherwise solve the equations:
$4x+9y+2z = 21$
$-8x+6y-3z = 41$
$3x+y-5z = -73$

Understanding Cramer's Rule:
Cramer's Rule is a method for solving systems of linear equations using determinants. For a system $AX=B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix, the solutions for $x, y, z$ are given by:

$$x = \frac{|A_x|}{|A|}, \quad y = \frac{|A_y|}{|A|}, \quad z = \frac{|A_z|}{|A|}$$

Here, $|A|$ is the determinant of the coefficient matrix. $|A_x|$ is the determinant of a new matrix formed by replacing the $x$-column of $A$ with the constant terms from $B$. Similarly for $|A_y|$ and $|A_z|$.

The coefficient matrix $A$ is:

$$A = \begin{bmatrix} 4 & 9 & 2 \\ -8 & 6 & -3 \\ 3 & 1 & -5 \end{bmatrix}$$

The constant matrix $B$ is:

$$B = \begin{bmatrix} 21 \\ 41 \\ -73 \end{bmatrix}$$

We already calculated $|A| = -601$.

Replace the first column of $A$ with the constants from $B$:

A_x = \begin{bmatrix} 21 & 9 & 2 \\ 41 & 6 & -3 \\ -73 & 1 & -5 \end{vmatrix}

Now, calculate its determinant:

$$|A_x| = 21 \begin{vmatrix} 6 & -3 \\ 1 & -5 \end{vmatrix} - 9 \begin{vmatrix} 41 & -3 \\ -73 & -5 \end{vmatrix} + 2 \begin{vmatrix} 41 & 6 \\ -73 & 1 \end{vmatrix}$$ $$|A_x| = 21((6)(-5) - (-3)(1)) - 9((41)(-5) - (-3)(-73)) + 2((41)(1) - (6)(-73))$$ $$|A_x| = 21(-30 + 3) - 9(-205 - 219) + 2(41 + 438)$$ $$|A_x| = 21(-27) - 9(-424) + 2(479)$$ $$|A_x| = -567 + 3816 + 958$$ $$|A_x| = 4207$$

Now find $x$:

$$x = \frac{|A_x|}{|A|} = \frac{4207}{-601} = -7$$

Replace the second column of $A$ with the constants from $B$:

A_y = \begin{bmatrix} 4 & 21 & 2 \\ -8 & 41 & -3 \\ 3 & -73 & -5 \end{vmatrix}

Now, calculate its determinant:

$$|A_y| = 4 \begin{vmatrix} 41 & -3 \\ -73 & -5 \end{vmatrix} - 21 \begin{vmatrix} -8 & -3 \\ 3 & -5 \end{vmatrix} + 2 \begin{vmatrix} -8 & 41 \\ 3 & -73 \end{vmatrix}$$ $$|A_y| = 4((41)(-5) - (-3)(-73)) - 21((-8)(-5) - (-3)(3)) + 2((-8)(-73) - (41)(3))$$ $$|A_y| = 4(-205 - 219) - 21(40 + 9) + 2(584 - 123)$$ $$|A_y| = 4(-424) - 21(49) + 2(461)$$ $$|A_y| = -1696 - 1029 + 922$$ $$|A_y| = -1803$$

Now find $y$:

$$y = \frac{|A_y|}{|A|} = \frac{-1803}{-601} = 3$$

Replace the third column of $A$ with the constants from $B$:

A_z = \begin{bmatrix} 4 & 9 & 21 \\ -8 & 6 & 41 \\ 3 & 1 & -73 \end{vmatrix}

Now, calculate its determinant:

$$|A_z| = 4 \begin{vmatrix} 6 & 41 \\ 1 & -73 \end{vmatrix} - 9 \begin{vmatrix} -8 & 41 \\ 3 & -73 \end{vmatrix} + 21 \begin{vmatrix} -8 & 6 \\ 3 & 1 \end{vmatrix}$$ $$|A_z| = 4((6)(-73) - (41)(1)) - 9((-8)(-73) - (41)(3)) + 21((-8)(1) - (6)(3))$$ $$|A_z| = 4(-438 - 41) - 9(584 - 123) + 21(-8 - 18)$$ $$|A_z| = 4(-479) - 9(461) + 21(-26)$$ $$|A_z| = -1916 - 4149 - 546$$ $$|A_z| = -6611$$

Now find $z$:

$$z = \frac{|A_z|}{|A|} = \frac{-6611}{-601} = 11$$

The solution to the system of equations is $\boxed{x = -7, y = 3, z = 11}$.