# 15.3 Properties of Determinants

The first property, which we deduce from the definition of determinant and what we already know about areas and volumes, is the value of the determinant of an array with all its non-zero entries on the main diagonal. Such an array describes a figure which is a rectangle or rectangular parallelepiped, with sides that are parallel to the $$x$$ and $$y$$ and $$z$$ and whatever axes. We already know that the magnitude of this determinant must be the product of its diagonal entries. The sign we define to be that of this product.

$\begin{pmatrix} 5 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

Thus the determinants of the three arrays above are $$5$$, $$-1$$ and $$2$$, respectively.

This is wonderful?

No, not yet. We really want to be able to evaluated more general determinants. The sign of a determinant with given rows depends on the order in which you choose to list the rows which represent the edges of the figure, as we shall see.

We are interested in the area of parallelograms that are tilted, so that sides are not perpendicular to one another, or that are rotated, so that the sides are not parallel to axes.

And here is the wonderful fact: If you fix the base of a parallelogram, (one side of it,) then its area is the height of the top of parallelogram above that base multiplied by the length of the base. It does not matter how much the parallelogram tilts, it is only the distance perpendicular between the top and bottom that counts.

A similar property holds in any dimension: The size of the n dimensional figure is the size of its $$n-1$$ dimensional base, times the height of the top of the figure perpendicular to its base.

This tells us: we can add any multiple of one row of the array to any other row, without changing its determinant. This is so because in any dimension we can choose any face containing the origin to be a base, and all but one of the lines from the origin to its neighbors, which define the rows of the array, lie in that base. Changing the line not in the face by any vector in the face will not change the height of the figure; it can only change the way the figure tilts.

(By the way this suggests the usual way to compute determinants. We add multiples of rows to other rows to get rid of all the tilting so that the determinant is then the product of its diagonal elements. This is called row reduction.)

Another wonderful fact that follows from the first two is: the determinant is linear in any of the rows (or columns) of its array. This means that if you multiply some row by $$7$$ the value of the determinant goes up by a factor of $$7$$. It also means that if you take two arrays that differ only in some one row, like the following two, which differ only in their first rows:

$\begin{pmatrix} 1 & 2 \\ 5 & 7 \end{pmatrix}, \begin{pmatrix} 2 & 2 \\ 5 & 7 \end{pmatrix}$

then the determinant of the array gotten by summing in the row that differs and keeping the others the same, (getting here $$3 \enspace 4$$ for the first row and $$5 \enspace 7$$ for the second) is the sum of the determinants of the two arrays you started with.

This statement represents the fact that the height of the summed figure above the base is the sum of the heights of the two summand figures.

Exercise 17.5 Show, by adding rows to one another appropriately, that interchanging two rows of an array changes the sign of its determinant. (Hint add a row to another, subtract the other way and add back the first way; or something like that)

And what good is all this?