Properties of the Determinant

We begin by summarizing the properties of determinants we introduced in previous sections.

In this section we will prove the following important results:

1.
A square matrix is singular if and only if its determinant is equal to 0.
2.
The determinant of a product is the product of the determinants.

To get us started, we need the following lemma.

Recall that if \(E\) is obtained from \(I\) using an elementary row operation, then the same elementary row operation carries \(A\) to \(EA\). There are three types of elementary row operations and three types of elementary matrices, so we will have to consider three cases.

Case 1. Suppose \(E\) is obtained from \(I\) by interchanging two rows, then

\[\det {E}=-1\quad \text {and}\quad \det {EA}=-\det {A}\]
so
\[\det {EA}=\det {E}\det {A}\]

Case 2. Suppose \(E\) is obtained from \(I\) by multiplying one of the rows of \(I\) by a non-zero constant \(k\), then

\[\det {E}=k\quad \text {and}\quad \det {EA}=k\det {A}\]
so
\[\det {EA}=\det {E}\det {A}\]

Case 3. Suppose \(E\) is obtained from \(I\) by adding a scalar multiple of one row to another row, then

\[\det {E}=1\quad \text {and}\quad \det {EA}=\det {A}\]
so
\[\det {EA}=\det {E}\det {A}\]

Invertibility and the Determinant

Recall that we first introduced determinants in the context of invertibility of \(2\times 2\) matrices. Specifically, we found that \(A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\) is invertible if and only if \(\det {A}\neq 0\). (A logically equivalent statement is: \(A\) is singular if and only if \(\det {A}=0\).) We are now in the position to prove this result for all square matrices.

Let \(A\) be a square matrix. To determine whether \(A\) is singular we need to find \(\mbox {rref}(A)\). In Elementary Matrices we found that there exist elementary matrices \(E_1,\ldots ,E_k\) such that

\[E_k\ldots E_2E_1A=\mbox {rref}(A)\]
so
\[\det {(E_k\ldots E_2E_1A)}=\det {\big (\mbox {rref}(A)\big )}\]
By repeated application of Lemma 2, we find that
\[\det {E_k}\ldots \det {E_2}\det {E_1}\det {A}=\det {\big (\mbox {rref}(A)\big )}\]
Suppose that \(A\) is singular, then \(\mbox {rref}(A)\neq I\). But then \(\mbox {rref}(A)\) contains a row of zeros, and \(\det {\big (\mbox {rref}(A)\big )}=0\). (Lemma ??) Since determinants of elementary matrices are non-zero, we conclude that \(\det {A}=0\).

Conversely, suppose \(\det {A}=0\), then

\[\det {\big (\mbox {rref}(A)\big )}=\det {E_k}\ldots \det {E_2}\det {E_1}\det {A}=0\]
But then \(\mbox {rref}(A)\neq I\), so \(A\) is singular.

Determinant of a Product

Suppose \(A\) is invertible, then \(A\) can be written as a product of elementary matrices. (Theorem ??)

\[A=E_1E_2\ldots E_k\]
Then, by repeated application of Lemma 2, we get
\begin{align*}\det {AB}&=\det {(E_1E_2\ldots E_kB)}\\ &=\det {E_1}\det {E_2}\ldots \det {E_k}\det {B}\\ &=\det {(E_1E_2\ldots E_k)}\det {B}\\ &=\det {A}\det {B} \end{align*}

Now suppose that \(A\) is not invertible. Then \(AB\) is also not invertible. So, \(\det {A}=0\) and \(\det {AB}=0\). Thus \(\det {AB}=0=\det {A}\det {B}\).

The following theorem is a nice consequence of Theorem 8. We leave the proof to the reader. (Practice Problem )

Practice Problems

Without doing written computations, determine whether matrix \(A\) is singular.
\[A=\begin{bmatrix}0&0&-3\\2&3&-1\\0&0&10\end{bmatrix}\]
\(A\) is singular \(A\) is nonsingular
Without doing written computations, determine whether matrix \(A\) is singular.
\[A=\begin{bmatrix}2&-3&-1\\0&4&5\\0&0&-3\\\end{bmatrix}\]
\(A\) is singular \(A\) is nonsingular
Show that all matrices of the form
\[\begin{bmatrix}x&x+1&x+2\\x+3&x+4&x+5\\x+6&x+7&x+8\end{bmatrix}\]
are singular.
Find values of \(x\) for which the given matrix is singular.
\[\begin{bmatrix}2-x&1\\5&6-x\end{bmatrix}\]
List values of \(x\) in an increasing order.

Answer:

\[x=\answer {1}\quad x=\answer {7}\]
Suppose \(A\) and \(B\) are \(2\times 2\) matrices such that \(\det {A}=2\) and \(\det {B}=\frac {1}{3}\). Then, \(\det {AB^{-1}}=\answer {6}\)
Suppose \(A\) and \(B\) are \(2\times 2\) matrices such that \(\det {A}=2\) and \(\det {B}=\frac {1}{3}\). Then, \(\det {(AB)^{-1}}=\answer {3/2}\)
Suppose \(A\) and \(B\) are \(2\times 2\) matrices such that \(\det {A}=2\) and \(\det {B}=\frac {1}{3}\). Then, \(\det {2AB}=\answer {8/3}\)
Prove or give a counterexample.
\[\det {(A+B)}=\det {(A^T+B^T)}\]
Prove that for a nonsingular matrix \(A\)
\[\det {A^{-1}}=\frac {1}{\det {A}}\]
(Theorem 10)
Suppose \(A\) is an invertible matrix such that
\[A^{-1}=A^T\]
What are the possible values for \(\det {A}\)? List your answers in increasing order.

Answer: \(\det {A}=\answer {-1},\quad \answer {1}\)