\[A=\begin{bmatrix}0&0&-3\\2&3&-1\\0&0&10\end{bmatrix}\]
\(A\) is singular \(A\) is nonsingular
Properties of the Determinant
We begin by summarizing the properties of determinants we introduced in previous sections.
- The determinant of the identity matrix is equal to 1. (Lemma lemma:detofid)
- The determinant of a triangular matrix is equal to the product of the main diagonal entries. (Lemma lemma:triangulardet)
- The determinant of the transpose is equal to the determinant of the matrix. (Theorem th:detoftrans)
- If a matrix contains a row of zeros, then its determinant is equal to 0. (Lemma lemma:det0lemma)
- If two rows of a matrix are the same, then the determinant of the matrix is equal to 0. (Lemma lemma:det0lemma)
- If one row of a matrix is a scalar multiple of another row, then the determinant of the matrix is equal to 0. (Lemma lemma:det0lemma)
-
If \(B\) is obtained from \(A\) by interchanging two different rows, then
\[\det {B}=-\det {A}\] -
If \(B\) is obtained from \(A\) by multiplying one of the rows of \(A\) by a non-zero constant \(k\). Then
\[\det {B}=k\det {A}\] -
If \(B\) is obtained from \(A\) by adding a multiple of one row of \(A\) to another row, then
\[\det {B}=\det {A}\]
(The last three properties comprise Theorem th:elemrowopsanddet)
In this section we will prove the following important results:
- (a)
- A square matrix is singular if and only if its determinant is equal to 0.
- (b)
- The determinant of a product is the product of the determinants.
To get us started, we need the following lemma.
Let \(A\) be a square matrix, and let \(E\) be an elementary matrix, then
\[\det {EA}=\det {E}\det {A}\]
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. Determine whether \(A\) is an invertible matrix without using elementary row
operations.
\[A=\begin{bmatrix}3&4&-8\\1&8&-10\\1&-2&1\end{bmatrix}\]
Compute the determinant of \(A\). You will find that \(\det {A}=0\). By Theorem 4 we conclude that \(A\)
is not invertible.
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.
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, \(\det {(AB)^{-1}}=\answer {3/2}\)