Section PDM Properties of Determinants of Matrices

From A First Course in Linear Algebra
Version 1.07
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/

We have seen how to compute the determinant of a matrix, and the incredible fact that we can perform expansion about any row or column to make this computation. In this largely theoretical section, we will state and prove several more intriguing properties about determinants. Our main goal will be the two results in Theorem SMZD and Theorem DRMM, but more specifically, we will see how the value of a determinant will allow us to gain insight into the various properties of a square matrix.

Subsection DRO: Determinants and Row Operations

We start easy with a straightforward theorem whose proof presages the style of subsequent proofs in this subsection.

Theorem DZRC
Determinant with Zero Row or Column
Suppose that $A$ is a square matrix with a row where every entry is zero, or a column where every entry is zero. Then $det (A) = 0$ . $□$

Proof Suppose that $A$ is a square matrix of size $n$ and row $i$ has every entry equal to zero. We compute $det (A)$ via expansion about row $i$ .

\begin{array}{l} det (A) & = \sum_{j = 1}^{n} {(- 1)}^{i + j} {[A]}_{i j} det (A (i | j)) & Theorem DER \\ = \sum_{j = 1}^{n} {(- 1)}^{i + j} 0 det (A (i | j)) & Row i is zeros \\ = \sum_{j = 1}^{n} 0 = 0 \end{array}

The proof for the case of a zero column is entirely similar, or could be derived from an application of Theorem DT employing the transpose of the matrix. $■$

Theorem DRCS
Determinant for Row or Column Swap
Suppose that $A$ is a square matrix. Let $B$ be the square matrix obtained from $A$ by interchanging the location of two rows, or interchanging the location of two columns. Then $det (B) = - det (A)$ . $□$

Proof Begin with the special case where $A$ is a square matrix of size $n$ and we form $B$ by swapping adjacent rows $i$ and $i + 1$ for some $1 \leq i \leq n - 1$ . Notice that the assumption about swapping adjacent rows means that $B (i + 1 | j) = A (i | j)$ for all $1 \leq j \leq n$ , and ${[B]}_{i + 1, j} = {[A]}_{i j}$ for all $1 \leq j \leq n$ . We compute $det (B)$ via expansion about row $i + 1$ .

\begin{array}{l} det (B) & = \sum_{j = 1}^{n} {(- 1)}^{(i + 1) + j} {[B]}_{i + 1, j} det (B (i + 1 | j)) & Theorem DER \\ = \sum_{j = 1}^{n} {(- 1)}^{(i + 1) + j} {[A]}_{i j} det (A (i | j)) & Hypothesis \\ = \sum_{j = 1}^{n} {(- 1)}^{1} {(- 1)}^{i + j} {[A]}_{i j} det (A (i | j)) \\ = (- 1) \sum_{j = 1}^{n} {(- 1)}^{i + j} {[A]}_{i j} det (A (i | j)) \\ = - det (A) & Theorem DER \end{array}

So the result holds for the special case where we swap adjacent rows of the matrix. As any computer scientist knows, we can accomplish any rearrangement of an ordered list by swapping adjacent elements. This principle can be demonstrated by naïve sorting algorithms such as “bubble sort.” In any event, we don’t need to discuss every possible reordering, we just need to consider a swap of two rows, say rows $s$ and $t$ with $1 \leq s < t \leq n$ .

Begin with row $s$ , and repeatedly swap it with each row just below it, including row $t$ and stopping there. This will total $t - s$ swaps. Now swap the former row $t$ , which currently lives in row $t - 1$ , with each row above it, stopping when it becomes row $s$ . This will total another $t - s - 1$ swaps. In this way, we create $B$ through a sequence of $2 (t - s) - 1$ swaps of adjacent rows, each of which adjusts $det (A)$ by a multiplicative factor of $- 1$ . So

det (B) = {(- 1)}^{2 (t - s) - 1} det (A) = {({(- 1)}^{2})}^{t - s} {(- 1)}^{- 1} det (A) = - det (A)

as desired.

The proof for the case of swapping two columns is entirely similar, or could be derived from an application of Theorem DT employing the transpose of the matrix. $■$

So Theorem DRCS tells us the effect of the first row operation (Definition RO) on the determinant of a matrix. Here’s the effect of the second row operation.

Theorem DRCM
Determinant for Row or Column Multiples
Suppose that $A$ is a square matrix. Let $B$ be the square matrix obtained from $A$ by multiplying a single row by the scalar $α$ , or by multiplying a single column by the scalar $α$ . Then $det (B) = α det (A)$ . $□$

Proof Suppose that $A$ is a square matrix of size $n$ and we form the square matrix $B$ by multiplying each entry of row $i$ of $A$ by $α$ . Notice that the other rows of $A$ and $B$ are equal, so $A (i | j) = B (i | j)$ , for all $1 \leq j \leq n$ . We compute $det (B)$ via expansion about row $i$ .

\begin{array}{l} det (B) & = \sum_{j = 1}^{n} {(- 1)}^{i + j} {[B]}_{i j} det (B (i | j)) & Theorem DER \\ = \sum_{j = 1}^{n} {(- 1)}^{i + j} {[B]}_{i j} det (A (i | j)) & Hypothesis \\ = \sum_{j = 1}^{n} {(- 1)}^{i + j} α {[A]}_{i j} det (A (i | j)) & Hypothesis \\ = α \sum_{j = 1}^{n} {(- 1)}^{i + j} {[A]}_{i j} det (A (i | j)) \\ = α det (A) & Theorem DER \end{array}

The proof for the case of a multiple of a column is entirely similar, or could be derived from an application of Theorem DT employing the transpose of the matrix. $■$

Let’s go for understanding the effect of all three row operations. But first we need an intermediate result, but it is an easy one.

Theorem DERC
Determinant with Equal Rows or Columns
Suppose that $A$ is a square matrix with two equal rows, or two equal columns. Then $det (A) = 0$ . $□$

Proof Suppose that $A$ is a square matrix of size $n$ where the two rows $s$ and $t$ are equal. Form the matrix $B$ by swapping rows $r$ and $s$ . Notice that as a consequence of our hypothesis, $A = B$ . Then

\begin{array}{l} det (A) & = \frac{1}{2} (det (A) + det (A)) \\ = \frac{1}{2} (det (A) - det (B)) & Theorem DRCS \\ = \frac{1}{2} (det (A) - det (A)) & Hypothesis, A = B \\ = \frac{1}{2} (0) = 0 \end{array}

The proof for the case of two equal columns is entirely similar, or could be derived from an application of Theorem DT employing the transpose of the matrix. $■$

Now explain the third row operation. Here we go.

Theorem DRCMA
Determinant for Row or Column Multiples and Addition
Suppose that $A$ is a square matrix. Let $B$ be the square matrix obtained from $A$ by multiplying a row by the scalar $α$ and then adding it to another row, or by multiplying a column by the scalar $α$ and then adding it to another column. Then $det (B) = det (A)$ . $□$

Proof Suppose that $A$ is a square matrix of size $n$ . Form the matrix $B$ by multiplying row $s$ by $α$ and adding it to row $t$ . Let $C$ be the auxiliary matrix where we replace row $t$ of $A$ by row $s$ of $A$ . Notice that $A (t | j) = B (t | j) = C (t | j)$ for all $1 \leq j \leq n$ . We compute the determinant of $B$ by expansion about row $t$ .

\begin{array}{l} det (B) & = \sum_{j = 1}^{n} {(- 1)}^{t + j} {[B]}_{t j} det (B (t | j)) & Theorem DER \\ = \sum_{j = 1}^{n} {(- 1)}^{t + j} (α {[A]}_{s j} + {[A]}_{t j}) det (B (s | j)) & Hypothesis \\ = \sum_{j = 1}^{n} {(- 1)}^{t + j} α {[A]}_{s j} det (B (t | j)) \\ + \sum_{j = 1}^{n} {(- 1)}^{t + j} {[A]}_{t j} det (B (t | j)) \\ = α \sum_{j = 1}^{n} {(- 1)}^{t + j} {[A]}_{s j} det (B (t | j)) \\ + \sum_{j = 1}^{n} {(- 1)}^{t + j} {[A]}_{t j} det (B (t | j)) \\ = α \sum_{j = 1}^{n} {(- 1)}^{t + j} {[C]}_{t j} det (C (t | j)) \\ + \sum_{j = 1}^{n} {(- 1)}^{t + j} {[A]}_{t j} det (A (t | j)) \\ = α det (C) + det (A) & Theorem DER \\ = α 0 + det (A) = det (A) & Theorem DERC \end{array}

The proof for the case of adding a multiple of a column is entirely similar, or could be derived from an application of Theorem DT employing the transpose of the matrix. $■$

Is this what you expected? We could argue that the third row operation is the most popular, and yet it has no effect whatsoever on the determinant of a matrix! We can exploit this, along with our understanding of the other two row operations, to provide another approach to computing a determinant. We’ll explain this in the context of an example.

Example DRO
Determinant by row operations
Suppose we desire the determinant of the $4 \times 4$ matrix

A = [\begin{matrix} 2 & 0 & 2 & 3 \\ 1 & 3 & - 1 & 1 \\ - 1 & 1 & - 1 & 2 \\ 3 & 5 & 4 & 0 \end{matrix}]

We will perform a sequence of row operations on this matrix, shooting for an upper triangular matrix, whose determinant will be simply the product of its diagonal entries. For each row operation, we will track the effect on the determinant via Theorem DRCS, Theorem DRCM, Theorem DRCMA.

\begin{array}{l} \to_{}^{R_{1} \leftrightarrow R_{2}} A_{1} & = [\begin{matrix} 1 & 3 & - 1 & 1 \\ 2 & 0 & 2 & 3 \\ - 1 & 1 & - 1 & 2 \\ 3 & 5 & 4 & 0 \end{matrix}] & det (A) & = - det (A_{1}) & Theorem DRCS \\ \to_{}^{- 2 R_{1} + R_{2}} A_{2} & = [\begin{matrix} 1 & 3 & - 1 & 1 \\ 0 & - 6 & 4 & 1 \\ - 1 & 1 & - 1 & 2 \\ 3 & 5 & 4 & 0 \end{matrix}] & = - det (A_{2}) & Theorem DRCMA \\ \to_{}^{1 R_{1} + R_{3}} A_{3} & = [\begin{matrix} 1 & 3 & - 1 & 1 \\ 0 & - 6 & 4 & 1 \\ 0 & 4 & - 2 & 3 \\ 3 & 5 & 4 & 0 \end{matrix}] & = - det (A_{3}) & Theorem DRCMA \\ \to_{}^{- 3 R_{1} + R_{4}} A_{4} & = [\begin{matrix} 1 & 3 & - 1 & 1 \\ 0 & - 6 & 4 & 1 \\ 0 & 4 & - 2 & 3 \\ 0 & - 4 & 7 & - 3 \end{matrix}] & = - det (A_{4}) & Theorem DRCMA \\ \to_{}^{1 R_{3} + R_{2}} A_{5} & = [\begin{matrix} 1 & 3 & - 1 & 1 \\ 0 & - 2 & 2 & 4 \\ 0 & 4 & - 2 & 3 \\ 0 & - 4 & 7 & - 3 \end{matrix}] & = - det (A_{5}) & Theorem DRCMA \\ \to_{}^{- \frac{1}{2} R_{2}} A_{6} & = [\begin{matrix} 1 & 3 & - 1 & 1 \\ 0 & 1 & - 1 & - 2 \\ 0 & 4 & - 2 & 3 \\ 0 & - 4 & 7 & - 3 \end{matrix}] & = 2 det (A_{6}) & Theorem DRCM \\ \to_{}^{- 4 R_{2} + R_{3}} A_{7} & = [\begin{matrix} 1 & 3 & - 1 & 1 \\ 0 & 1 & - 1 & - 2 \\ 0 & 0 & 2 & 11 \\ 0 & - 4 & 7 & - 3 \end{matrix}] & = 2 det (A_{7}) & Theorem DRCMA \\ \to_{}^{4 R_{2} + R_{4}} A_{8} & = [\begin{matrix} 1 & 3 & - 1 & 1 \\ 0 & 1 & - 1 & - 2 \\ 0 & 0 & 2 & 11 \\ 0 & 0 & 3 & - 11 \end{matrix}] & = 2 det (A_{8}) & Theorem DRCMA \\ \to_{}^{- 1 R_{3} + R_{4}} A_{9} & = [\begin{matrix} 1 & 3 & - 1 & 1 \\ 0 & 1 & - 1 & - 2 \\ 0 & 0 & 2 & 11 \\ 0 & 0 & 1 & - 22 \end{matrix}] & = 2 det (A_{9}) & Theorem DRCMA \\ \to_{}^{- 2 R_{4} + R_{3}} A_{10} & = [\begin{matrix} 1 & 3 & - 1 & 1 \\ 0 & 1 & - 1 & - 2 \\ 0 & 0 & 0 & 55 \\ 0 & 0 & 1 & - 22 \end{matrix}] & = 2 det (A_{10}) & Theorem DRCMA \\ \to_{}^{R_{3} \leftrightarrow R_{4}} A_{11} & = [\begin{matrix} 1 & 3 & - 1 & 1 \\ 0 & 1 & - 1 & - 2 \\ 0 & 0 & 1 & - 22 \\ 0 & 0 & 0 & 55 \end{matrix}] & = - 2 det (A_{11}) & Theorem DRCS \\ \to_{}^{\frac{1}{55} R_{4}} A_{12} & = [\begin{matrix} 1 & 3 & - 1 & 1 \\ 0 & 1 & - 1 & - 2 \\ 0 & 0 & 1 & - 22 \\ 0 & 0 & 0 & 1 \end{matrix}] & = - 110 det (A_{12}) & Theorem DRCM \end{array}

The matrix $A_{12}$ is upper triangular, so expansion about the first column (repeatedly) will result in $det (A_{12}) = (1) (1) (1) (1) = 1$ (see Example DUTM) and thus, $det (A) = - 110 (1) = - 110$ .

Notice that our sequence of row operations was somewhat ad hoc, such as the transformation to $A_{5}$ . We could have been even more methodical, and strictly followed the process that converts a matrix to reduced row-echelon form (Theorem REMEF), eventually achieving the same numerical result with a final matrix that equaled the $4 \times 4$ identity matrix. Notice too that we could have stopped with $A_{8}$ , since at this point we could compute $det (A_{8})$ by two expansions about first columns, followed by a simple determinant of a $2 \times 2$ matrix (Theorem DMST).

The beauty of this approach is that computationally we should already have written a procedure to convert matrices to reduced-row echelon form, so all we need to do is track the multiplicative changes to the determinant as the algorithm procedes. Further, for a square matrix of size $n$ this approach requires on the order of $n^{3}$ multiplications, while a recursive application of expansion about a row or column (Theorem DER, Theorem DEC) will require in the vicinity of $(n - 1) (n!)$ multiplications. So even for very small matrices, a computational approach utilizing row operations will have superior run-time. Tracking, and controlling, the effects of round-off errors is another story, best saved for a numerical linear algebra course. $⊠$

Subsection DROEM: Determinants, Row Operations, Elementary Matrices

As a final preparation for our two most important theorems about determinants, we prove a handful of facts about the interplay of row operations and matrix multiplication with elementary matrices with regard to the determinant. But first, a simple, but crucial, fact about the identity matrix.

Theorem DIM
Determinant of the Identity Matrix
For every $n \geq 1$ , $det (I_{n}) = 1$ . $□$

Proof It may be overkill, but this is a good situation to run through a proof by induction on $n$ (Technique I). Is the result true when $n = 1$ ? Yes,

\begin{array}{l} det (I_{1}) & = {[I_{1}]}_{11} & Definition DM \\ = 1 & Definition IM \end{array}

Now assume the theorem is true for the identity matrix of size $n - 1$ and investigate the determinant of the identity matrix of size $n$ with expansion about row 1,

\begin{array}{l} det (I_{n}) & = \sum_{j = 1}^{n} {(- 1)}^{1 + j} {[I_{n}]}_{1 j} det (I_{n} (1 | j)) & Definition DM \\ = {(- 1)}^{1 + 1} {[I_{n}]}_{11} det (I_{n} (1 | 1)) \\ + \sum_{j = 2}^{n} {(- 1)}^{1 + j} {[I_{n}]}_{1 j} det (I_{n} (1 | j)) \\ = 1 det (I_{n - 1}) + \sum_{j = 2}^{n} {(- 1)}^{1 + j} 0 det (I_{n} (1 | j)) & Definition IM \\ = 1 (1) + \sum_{j = 2}^{n} 0 = 1 & Induction Hypothesis \end{array}

■

Theorem DEM
Determinants of Elementary Matrices
For the three possible versions of an elementary matrix (Definition ELEM) we have the determinants,

$det (E_{i, j}) = - 1$
$det (E_{i} (α)) = α$
$det (E_{i, j} (α)) = 1$

□

Proof Swapping rows $i$ and $j$ of the identity matrix will create $E_{i, j}$ (Definition ELEM), so

\begin{array}{l} det (E_{i, j}) & = - det (I_{n}) & Theorem DRCS \\ = - 1 & Theorem DIM \end{array}

Multiplying row $i$ of the identity matrix by $α$ will create $E_{i} (α)$ (Definition ELEM), so

\begin{array}{l} det (E_{i} (α)) & = α det (I_{n}) & Theorem DRCM \\ = α (1) = α & Theorem DIM \end{array}

Multiplying row $i$ of the identity matrix by $α$ and adding to row $j$ will create $E_{i} (α) j$ (Definition ELEM), so

\begin{array}{l} det (E_{i} (α) j) & = det (I_{n}) & Theorem DRCMA \\ = 1 & Theorem DIM \end{array}

■

Theorem DEMMM
Determinants, Elementary Matrices, Matrix Multiplication
Suppose that $A$ is a square matrix of size $n$ and $E$ is any elementary matrix of size $n$ . Then

det (E A) = det (E) det (A)

□

Proof The proof procedes in three parts, one for each type of elementary matrix, with each part very similar to the other two. First, let $B$ be the matrix obtained from $A$ by swapping rows $i$ and $j$ ,

\begin{array}{l} det (E_{i, j} A) & = det (B) & Theorem EMDRO \\ = - det (A) & Theorem DRCS \\ = det (E_{i, j}) det (A) & Theorem DEM \end{array}

Second, let $B$ be the matrix obtained from $A$ by multiplying row $i$ by $α$ ,

\begin{array}{l} det (E_{i} (α) A) & = det (B) & Theorem EMDRO \\ = α det (A) & Theorem DRCM \\ = det (E_{i} (α)) det (A) & Theorem DEM \end{array}

Third, let $B$ be the matrix obtained from $A$ by multiplying row $i$ by $α$ and adding to row $j$ ,

\begin{array}{l} det (E_{i, j} (α) A) & = det (B) & Theorem EMDRO \\ = det (A) & Theorem DRCMA \\ = det (E_{i, j} (α)) det (A) & Theorem DEM \end{array}

Since the desired result holds for each variety of elementary matrix individually, we are done. $■$

Subsection DNMMM: Determinants, Nonsingular Matrices, Matrix Multiplication

If you asked someone with substantial experience working with matrices about the value of the determinant, they’d be likely to quote the following theorem as the first thing to come to mind.

Theorem SMZD
Singular Matrices have Zero Determinants
Let $A$ be a square matrix. Then $A$ is singular if and only if $det (A) = 0$ . $□$

Proof $(\Rightarrow)$ Suppose that $A$ is a singular matrix of size $n$ . Then $A$ is row-equivalent to a square matrix $B$ in reduced row-echelon form (Theorem REMEF). Since $A$ is singular, the matrix $B$ is not the identity matrix (Theorem NMRRI). Therefore, the number of pivot columns is strictly less than $n$ , i.e. $r < n$ , and so $B$ has at least one row of all zeros.

There is a sequence of row operations $R_{1}, R_{2}, R_{3}, \dots, R_{s}$ that will convert $B$ into $A$ . For each of these row operations, there is an elementary matrix $E_{i}$ which effects the row operation by matrix multiplication (Theorem EMDRO). Repeated applications of Theorem EMDRO allow us to write

A = E_{s} E_{s - 1} \dots E_{2} E_{1} B

Then

\begin{array}{l} det (A) & = det (E_{s} E_{s - 1} \dots E_{2} E_{1} B) \\ = det (E_{s}) det (E_{s - 1}) \dots det (E_{2}) det (E_{1}) det (B) & Theorem DEMMM \\ = det (E_{s}) det (E_{s - 1}) \dots det (E_{2}) det (E_{1}) 0 & Theorem DZRC \\ = 0 \end{array}

$(\Leftarrow)$ We will establish the contrapositive of this implication. So begin by assuming that $A$ is nonsingular. Then $A$ is row-equivalent to the identity matrix by Theorem NMRRI. As above, there is a sequence of row operations that will convert $I_{n}$ to $A$ , which can be effected by matrix multiplication by elementary matrices and Theorem DEMMM allows us to “distribute” the determinant through this product. Mimicking the first half of the proof, we would arrive at

det (A) = det (E_{s}) det (E_{s - 1}) \dots det (E_{2}) det (E_{1}) det (I_{n})

We know that $det (I_{n}) = 1 \neq 0$ . From Theorem DEM we can infer that the determinant of an elementary matrix is never zero (note the ban on $α = 0$ for $E_{i} (α)$ in Definition ELEM). So the product on the right is composed of nonzero scalars, and so is also nonzero. This is the result we needed. $■$

For the case of $2 \times 2$ matrices you might compare the application of Theorem SMZD with the combination of the results stated in Theorem DMST and Theorem TTMI.

Example ZNDAB
Zero and nonzero determinant, Archetypes A and B
The coefficient matrix in Archetype A has a zero determinant (check this!) while the coefficient matrix Archetype B has a nonzero determinant (check this, too). These matrices are singular and nonsingular, respectively. This is exactly what Theorem SMZD says, and continues our list of contrasts between these two archetypes. $⊠$

Since Theorem SMZD is an equivalence (Technique E) we can expand on our growing list of equivalences about nonsingular matrices. The addition of the condition $det (A) \neq 0$ is one of the best motivations for learning about determinants.

Theorem NME7
Nonsingular Matrix Equivalences, Round 7
Suppose that $A$ is a square matrix of size $n$ . The following are equivalent.

$A$ is nonsingular.
$A$ row-reduces to the identity matrix.
The null space of $A$ contains only the zero vector, $N (A) = \{0\}$ .
The linear system $ℒ S (A, b)$ has a unique solution for every possible choice of $b$ .
The columns of $A$ are a linearly independent set.
$A$ is invertible.
The column space of $A$ is $ℂ^{n}$ , $C (A) = ℂ^{n}$ .
The columns of $A$ are a basis for $ℂ^{n}$ .
The rank of $A$ is $n$ , $r (A) = n$ .
The nullity of $A$ is zero, $n (A) = 0$ .
The determinant of $A$ is nonzero, $det (A) \neq 0$ .

□

Proof Theorem SMZD says $A$ is singular if and only if $det (A) = 0$ . If we negate each of these statements, we arrive at two contrapositives that we can combine as the equivalence, $A$ is nonsingular if and only if $det (A) \neq 0$ . This allows us to add a new statement to the list found in Theorem NME6. $■$

Computationally, row-reducing a matrix is the most efficient way to determine if a matrix is nonsingular, though the effect of using division in a computer can lead to round-off errors that confuse small quantities with critical zero quantities. Conceptually, the determinant may seem the most efficient way to determine if a matrix is nonsingular. The definition of a determinant uses just addition, subtraction and multiplication, so division is never a problem. And the final test is easy: is the determinant zero or not? However, the number of operations involved in computing a determinant by the definition very quickly becomes so excessive as to be impractical.

Now for the coup de grâce. We will generalize Theorem DEMMM to the case of any two square matrices. You may recall thinking that matrix multiplication was defined in a needlessly complicated manner. For sure, the definition of a determinant seems even stranger. (Though Theorem SMZD might be forcing you to reconsider.) Read the statement of the next theorem and contemplate how nicely matrix multiplication and determinants play with each other.

Theorem DRMM
Determinant Respects Matrix Multiplication
Suppose that $A$ and $B$ are square matrices of the same size. Then $det (A B) = det (A) det (B)$ . $□$

Proof Suppose that $A$ or $B$ is singular. Then either $det (A) = 0$ or $det (B) = 0$ by Theorem SMZD. In either case, $det (A) det (B) = 0$ . By the contrapositive of Theorem NPNT, we know $A B$ is singular as well. So by Theorem SMZD, $det (A B) = 0$ . So in this case, we have the desired equality.

Now assume that $A$ and $B$ are both nonsingular. By Theorem NMPEM there are elementary matrices $E_{1}, E_{2}, E_{3} \dots, E s$ and $E_{s + 1}, E_{s + 2}, E_{s + 3} \dots, E_{s + t}$ such that

\begin{array}{l} A & = E_{1} E_{2} E_{3} \dots E_{s} & B & = E_{s + 1} E_{s + 2} E_{s + 3} \dots E_{s + t} \end{array}

Then

\begin{array}{l} det (A B) & = det (E_{1} E_{2} \dots E_{s} E_{s + 1} E_{s + 2} \dots E_{s + t}) \\ = det (E_{1}) det (E_{2}) \dots det (E_{s}) det (E_{s + 1} E_{s + 2} \dots E_{s + t}) & Theorem DEMMM \\ = det (E_{1} E_{2} \dots E_{s}) det (E_{s + 1} E_{s + 2} \dots E_{s + t}) & Theorem DEMMM \\ = det (A) det (B) \end{array}

■

It’s an amazing thing that matrix multiplication and the determinant interact this way. Might it also be true that $det (A + B) = det (A) + det (B)$ ? (See Exercise PDM.M30.)

Subsection READ: Reading Questions

Condiser the two matrices below, and suppose you already have computed $det (A) = - 120$ . What is $det (B)$ ? Why? $\begin{array}{l} A & = [\begin{matrix} 0 & 8 & 3 & - 4 \\ - 1 & 2 & - 2 & 5 \\ - 2 & 8 & 4 & 3 \\ 0 & - 4 & 2 & - 3 \end{matrix}] & B & = [\begin{matrix} 0 & 8 & 3 & - 4 \\ 0 & - 4 & 2 & - 3 \\ - 2 & 8 & 4 & 3 \\ - 1 & 2 & - 2 & 5 \end{matrix}] \end{array}$
State the theorem that allows us to make yet another extension to our NMEx series of theorems.
What is amazing about the interaction between matrix multiplication and the determinant?

Subsection EXC: Exercises

C30 Each of the archetypes below is a system of equations with a square coefficient matrix, or is a square matrix itself. Compute the determinant of each matrix, noting how Theorem SMZD indicates when the matrix is singular or nonsingular.
Archetype A
Archetype B
Archetype F
Archetype K
Archetype L

Contributed by Robert Beezer

M20 Construct a $3 \times 3$ nonsingular matrix and call it $A$ . Then, for each entry of the matrix, compute the corresponding cofactor, and create a new $3 \times 3$ matrix full of these cofactors by placing the cofactor of an entry in the same location as the entry it was based on. Once complete, call this matrix $C$ . Compute $A C^{t}$ . Any observations? Repeat with a new matrix, or perhaps with a $4 \times 4$ matrix.
Contributed by Robert Beezer Solution [1101]

M30 Construct an example to show that the following statement is not true for all square matrices $A$ and $B$ of the same size: $det (A + B) = det (A) + det (B)$ .
Contributed by Robert Beezer

T10 Theorem NPNT says that if the product of square matrices $A B$ is nonsingular, then the individual matrices $A$ and $B$ are nonsingular also. Construct a new proof of this result making use of theorems about determinants of matrices.
Contributed by Robert Beezer

T15 Use Theorem DRCM to prove Theorem DZRC as a corollary. (See Technique LC.)
Contributed by Robert Beezer

T20 Suppose that $A$ is a square matrix of size $n$ and $α \in ℂ^{}$ is a scalar. Prove that $det (α A) = α^{n} det (A)$ .
Contributed by Robert Beezer

T25 Employ Theorem DT to construct the second half of the proof of Theorem DRCM (the portion about a multiple of a column).
Contributed by Robert Beezer

Subsection SOL: Solutions

M20 Contributed by Robert Beezer Statement [1099]
The result of these computations should be a matrix with the value of $det (A)$ in the diagonal entries and zeros elsewhere. The suggestion of using a nonsingular matrix was partially so that it was obvious that the value of the determinant appears on the diagonal.

This result (which is true in general) provides a method for computing the inverse of a nonsingular matrix. Since $A C^{t} = det (A) I_{n}$ , we can multiply by the reciprocal of the determinant (which is nonzero!) and the inverse of $A$ (it exists!) to arrive at an expression for the matrix inverse:

A^{- 1} = \frac{1}{det (A)} C^{t}

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