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4.4 Summary—Important Facts about Determinants

In this section, we summarize the important properties of the determinant needed for the remainder of the text. The results contained in this section have been derived in Sections 4.2 and 4.3; consequently, the facts presented here are stated without proofs.

The determinant of an $n \times n$ $n \times n$ matrix A having entries from a field F is a scalar in F, denoted by det(A) or $| A |$ $| A |$ , and can be computed as follows.

If A is $1 \times 1$ $1 \times 1$ , then $det (A) = A_{11},$ $det (A) = A_{11},$ the single entry of A.
If A is $2 \times 2,$ $2 \times 2,$ then $det (A) = A_{11} A_{22} - A_{12} A_{21}$ $det (A) = A_{11} A_{22} - A_{12} A_{21}$ . For example,

$det (\begin{array}{r} - 1 & 2 \\ 5 & 3 \end{array}) = (- 1) (3) - (2) (5) = - 13.$ $det (\begin{array}{r} - 1 & 2 \\ 5 & 3 \end{array}) = (- 1) (3) - (2) (5) = - 13.$
If A is $n \times n$ $n \times n$ for $n > 2,$ $n > 2,$ then, for each i, we can evaluate the determinant by cofactor expansion along row i as

$det (A) = \sum_{j = 1}^{n} {(- 1)}^{i + j} A_{i j} \cdot det ({\tilde{A}}_{i j}),$ $det (A) = \sum_{j = 1}^{n} {(- 1)}^{i + j} A_{i j} \cdot det ({\tilde{A}}_{i j}),$

or, for each j, we can evaluate the determinant by cofactor expansion along column j as

$det (A) = \sum_{i = 1}^{n} {(- 1)}^{i + j} A_{i j} \cdot det ({\tilde{A}}_{i j}),$ $det (A) = \sum_{i = 1}^{n} {(- 1)}^{i + j} A_{i j} \cdot det ({\tilde{A}}_{i j}),$

where ${\tilde{A}}_{i j}$ ${\tilde{A}}_{i j}$ is the $(n - 1) \times (n - 1)$ $(n - 1) \times (n - 1)$ matrix obtained by deleting row i and column j from A.

In the formulas above, the scalar ${(- 1)}^{i + j} det ({\tilde{A}}_{i j})$ ${(- 1)}^{i + j} det ({\tilde{A}}_{i j})$ is called the cofactor of the row i column j entry of A. In this language, the determinant of A is evaluated as the sum of terms obtained by multiplying each entry of some row or column of A by the cofactor of that entry. Thus det(A) is expressed in terms of n determinants of $(n - 1) \times (n - 1)$ $(n - 1) \times (n - 1)$ matrices. These determinants are then evaluated in terms of determinants of $(n - 2) \times (n - 2)$ $(n - 2) \times (n - 2)$ matrices, and so forth, until $2 \times 2$ $2 \times 2$ matrices are obtained. The determinants of the $2 \times 2$ $2 \times 2$ matrices are then evaluated as in item 2.

Let us consider two examples of this technique in evaluating the determinant of the $4 \times 4$ $4 \times 4$ matrix

A = (\begin{array}{r} 2 & 1 & 1 & 5 \\ 1 & 1 & - 4 & - 1 \\ 2 & 0 & - 3 & 1 \\ 3 & 6 & 1 & 2 \end{array}) .

$A = (\begin{array}{r} 2 & 1 & 1 & 5 \\ 1 & 1 & - 4 & - 1 \\ 2 & 0 & - 3 & 1 \\ 3 & 6 & 1 & 2 \end{array}) .$

To evaluate the determinant of A by expanding along the fourth row, we must know the cofactors of each entry of that row. The cofactor of $A_{41} = 3$ $A_{41} = 3$ is ${(- 1)}^{4 + 1} det (B)$ ${(- 1)}^{4 + 1} det (B)$ , where

B = (\begin{array}{r} 1 & 1 & 5 \\ 1 & - 4 & - 1 \\ 0 & - 3 & 1 \end{array}) .

$B = (\begin{array}{r} 1 & 1 & 5 \\ 1 & - 4 & - 1 \\ 0 & - 3 & 1 \end{array}) .$

Let us evaluate this determinant by expanding along the first column. We have

\begin{array}{l} det (B) & = & {(- 1)}^{1 + 1} (1) det (\begin{array}{r} - 4 & - 1 \\ - 3 & 1 \end{array}) + {(- 1)}^{2 + 1} (1) det (\begin{array}{r} 1 & 5 \\ - 3 & 1 \end{array}) \\ + & {(- 1)}^{3 + 1} (0) det (\begin{array}{r} 1 & 5 \\ - 4 & - 1 \end{array}) \\ = & 1 (1) [(- 4) (1) - (- 1) (- 3)] + (- 1) (1) [(1) (1) - (5) (- 3)] + 0 \\ = & - 7 - 16 + 0 = - 23 . \end{array}

$\begin{array}{l} det (B) & = & {(- 1)}^{1 + 1} (1) det (\begin{array}{r} - 4 & - 1 \\ - 3 & 1 \end{array}) + {(- 1)}^{2 + 1} (1) det (\begin{array}{r} 1 & 5 \\ - 3 & 1 \end{array}) \\ + & {(- 1)}^{3 + 1} (0) det (\begin{array}{r} 1 & 5 \\ - 4 & - 1 \end{array}) \\ = & 1 (1) [(- 4) (1) - (- 1) (- 3)] + (- 1) (1) [(1) (1) - (5) (- 3)] + 0 \\ = & - 7 - 16 + 0 = - 23 . \end{array}$

Thus the cofactor of $A_{41}$ $A_{41}$ is ${(- 1)}^{5} (- 23) = 23$ ${(- 1)}^{5} (- 23) = 23$ . Similarly, the cofactors of $A_{42}, A_{43},$ $A_{42}, A_{43},$ and $A_{44}$ $A_{44}$ are 8, 11, and $- 13$ $- 13$ respectively. We can now evaluate the determinant of A by multiplying each entry of the fourth row by its cofactor; this gives

\det (A) = 3 (23) + 6 (8) + 1 (11) + 2 (- 13) = 102 .

$\det (A) = 3 (23) + 6 (8) + 1 (11) + 2 (- 13) = 102 .$

For the sake of comparison, let us also compute the determinant of A by expansion along the second column. The reader should verify that the cofactors of $A_{12}, A_{22},$ $A_{12}, A_{22},$ and $A_{42}$ $A_{42}$ are $- 14$ $- 14$ 40, and 8, respectively. Thus

\begin{array}{rcl} det (A) & = & {(- 1)}^{1 + 2} (1) det (\begin{array}{r} 1 & - 4 & - 1 \\ 2 & - 3 & 1 \\ 3 & 1 & 2 \end{array}) + {(- 1)}^{2 + 2} (1) det (\begin{array}{r} 2 & 1 & 5 \\ 2 & - 3 & 1 \\ 3 & 1 & 2 \end{array}) \\ + {(- 1)}^{3 + 2} (0) det (\begin{array}{r} 2 & 1 & 5 \\ 1 & - 4 & - 1 \\ 3 & 1 & 2 \end{array}) + {(- 1)}^{4 + 2} (6) det (\begin{array}{r} 2 & 1 & 5 \\ 1 & - 4 & - 1 \\ 2 & - 3 & 1 \end{array}) \\ = & 14 + 40 + 0 + 48 = 102 . \end{array}

$\begin{array}{rcl} det (A) & = & {(- 1)}^{1 + 2} (1) det (\begin{array}{r} 1 & - 4 & - 1 \\ 2 & - 3 & 1 \\ 3 & 1 & 2 \end{array}) + {(- 1)}^{2 + 2} (1) det (\begin{array}{r} 2 & 1 & 5 \\ 2 & - 3 & 1 \\ 3 & 1 & 2 \end{array}) \\ + {(- 1)}^{3 + 2} (0) det (\begin{array}{r} 2 & 1 & 5 \\ 1 & - 4 & - 1 \\ 3 & 1 & 2 \end{array}) + {(- 1)}^{4 + 2} (6) det (\begin{array}{r} 2 & 1 & 5 \\ 1 & - 4 & - 1 \\ 2 & - 3 & 1 \end{array}) \\ = & 14 + 40 + 0 + 48 = 102 . \end{array}$

Of course, the fact that the value 102 is obtained again is no surprise since the value of the determinant of A is independent of the choice of row or column used in the expansion.

Observe that the computation of det(A) is easier when expanded along the second column than when expanded along the fourth row. The difference is the presence of a zero in the second column, which makes it unnecessary to evaluate one of the cofactors (the cofactor of $A_{32}$ $A_{32}$ ). For this reason, it is beneficial to evaluate the determinant of a matrix by expanding along a row or column of the matrix that contains the largest number of zero entries. In fact, it is often helpful to introduce zeros into the matrix by means of elementary row operations before computing the determinant. This technique utilizes the first three properties of the determinant.

Properties of the Determinant

If B is a matrix obtained by interchanging any two rows or interchanging any two columns of an $n \times n$ $n \times n$ matrix A, then $det (B) = - det (A)$ $det (B) = - det (A)$ .
If B is a matrix obtained by multiplying each entry of some row or column of an $n \times n$ $n \times n$ matrix A by a scalar k, then $det (B) = k \cdot det (A)$ $det (B) = k \cdot det (A)$ .
If B is a matrix obtained from an $n \times n$ $n \times n$ matrix A by adding a multiple of row i to row j or a multiple of column i to column j for $i \neq j$ $i \neq j$ then $det (B) = det (A)$ $det (B) = det (A)$ .

As an example of the use of these three properties in evaluating determinants, let us compute the determinant of the $4 \times 4$ $4 \times 4$ matrix A considered previously. Our procedure is to introduce zeros into the second column of A by employing property 3, and then to expand along that column. (The elementary row operations used here consist of adding multiples of row 1 to rows 2 and 4.) This procedure yields

\begin{array}{rcl} det (A) & = & det (\begin{array}{r} 2 & 1 & 1 & 5 \\ 1 & 1 & - 4 & - 1 \\ 2 & 0 & - 3 & 1 \\ 3 & 6 & 1 & 2 \end{array}) = det (\begin{array}{r} 2 & 1 & 1 & 5 \\ - 1 & 0 & - 5 & - 6 \\ 2 & 2 & - 3 & 1 \\ - 9 & 0 & - 5 & - 28 \end{array}) \\ = & 1 {(- 1)}^{1 + 2} det (\begin{array}{r} - 1 & - 5 & - 6 \\ 2 & - 3 & 1 \\ - 9 & - 5 & - 28 \end{array}) . \end{array}

$\begin{array}{rcl} det (A) & = & det (\begin{array}{r} 2 & 1 & 1 & 5 \\ 1 & 1 & - 4 & - 1 \\ 2 & 0 & - 3 & 1 \\ 3 & 6 & 1 & 2 \end{array}) = det (\begin{array}{r} 2 & 1 & 1 & 5 \\ - 1 & 0 & - 5 & - 6 \\ 2 & 2 & - 3 & 1 \\ - 9 & 0 & - 5 & - 28 \end{array}) \\ = & 1 {(- 1)}^{1 + 2} det (\begin{array}{r} - 1 & - 5 & - 6 \\ 2 & - 3 & 1 \\ - 9 & - 5 & - 28 \end{array}) . \end{array}$

The resulting determinant of a $3 \times 3$ $3 \times 3$ matrix can be evaluated in the same manner: Use type 3 elementary row operations to introduce two zeros into the first column, and then expand along that column. This results in the value $- 102$ $- 102$ Therefore

det (A) = 1 {(- 1)}^{1 + 2} (- 102) = 102.

$det (A) = 1 {(- 1)}^{1 + 2} (- 102) = 102.$

The reader should compare this calculation of det(A) with the preceding ones to see how much less work is required when properties 1, 2, and 3 are employed.

In the chapters that follow, we often have to evaluate the determinant of matrices having special forms. The next two properties of the determinant are useful in this regard:

The determinant of an upper triangular matrix is the product of its diagonal entries. In particular, $det (I) = 1$ $det (I) = 1$ .
If two rows (or columns) of a matrix are identical, then the determinant of the matrix is zero.

As an illustration of property 4, notice that

det (\begin{array}{r} - 3 & 1 & 2 \\ 0 & 4 & 5 \\ 0 & 0 & - 6 \end{array}) = (- 3) (4) (- 6) = 72.

$det (\begin{array}{r} - 3 & 1 & 2 \\ 0 & 4 & 5 \\ 0 & 0 & - 6 \end{array}) = (- 3) (4) (- 6) = 72.$

Property 4 provides an efficient method for evaluating the determinant of a matrix:

Use Gaussian elimination and properties 1, 2, and 3 above to reduce the matrix to an upper triangular matrix.
Compute the product of the diagonal entries.

For instance,

\begin{array}{l} det (\begin{array}{r} 1 & - 1 & 2 & 1 \\ 2 & - 1 & - 1 & 4 \\ - 4 & 5 & - 10 & - 6 \\ 3 & - 2 & 10 & - 1 \end{array}) = det (\begin{array}{r} 1 & - 1 & 2 & 1 \\ 0 & 1 & - 5 & 2 \\ 0 & 1 & - 2 & - 2 \\ 0 & 1 & 4 & - 4 \end{array}) \\ \begin{array}{rcl} = & det (\begin{array}{r} 1 & - 1 & 2 & 1 \\ 0 & 1 & - 5 & 2 \\ 0 & 0 & 3 & - 4 \\ 0 & 0 & 9 & - 6 \end{array}) = det (\begin{array}{r} 1 & - 1 & 2 & 1 \\ 0 & 1 & - 5 & 2 \\ 0 & 0 & 3 & - 4 \\ 0 & 0 & 0 & 6 \end{array}) \\ = & 1 \cdot 1 \cdot 3 \cdot 6 = 18 . \end{array} \end{array}

$\begin{array}{l} det (\begin{array}{r} 1 & - 1 & 2 & 1 \\ 2 & - 1 & - 1 & 4 \\ - 4 & 5 & - 10 & - 6 \\ 3 & - 2 & 10 & - 1 \end{array}) = det (\begin{array}{r} 1 & - 1 & 2 & 1 \\ 0 & 1 & - 5 & 2 \\ 0 & 1 & - 2 & - 2 \\ 0 & 1 & 4 & - 4 \end{array}) \\ \begin{array}{rcl} = & det (\begin{array}{r} 1 & - 1 & 2 & 1 \\ 0 & 1 & - 5 & 2 \\ 0 & 0 & 3 & - 4 \\ 0 & 0 & 9 & - 6 \end{array}) = det (\begin{array}{r} 1 & - 1 & 2 & 1 \\ 0 & 1 & - 5 & 2 \\ 0 & 0 & 3 & - 4 \\ 0 & 0 & 0 & 6 \end{array}) \\ = & 1 \cdot 1 \cdot 3 \cdot 6 = 18 . \end{array} \end{array}$

The next three properties of the determinant are used frequently in later chapters. Indeed, perhaps the most significant property of the determinant is that it provides a simple characterization of invertible matrices. (See property 7.)

For any $n \times n$ $n \times n$ matrices A and B, $det (A B) = det (A) \cdot det (B)$ $det (A B) = det (A) \cdot det (B)$ .
An $n \times n$ $n \times n$ matrix A is invertible if and only if $det (A) \neq 0$ $det (A) \neq 0$ . Furthermore, if A is invertible, then $det (A^{- 1}) = \frac{1}{det (A)}$ $det (A^{- 1}) = \frac{1}{det (A)}$ .
For any $n \times n$ $n \times n$ matrix A, the determinants of A and $A^{t}$ $A^{t}$ are equal.

For example, property 7 guarantees that the matrix A on page 233 is invertible because $det (A) = 102$ $det (A) = 102$ .

The final property, stated as Exercise 15 of Section 4.3, is used in Chapter 5. It is a simple consequence of properties 6 and 7.

If A and B are similar matrices, then $det (A) = det (B)$ $det (A) = det (B)$ .

Exercises

Label the following statements as true or false.
1. (a) The determinant of a square matrix may be computed by expanding the matrix along any row or column.
2. (b) In evaluating the determinant of a matrix, it is wise to expand along a row or column containing the largest number of zero entries.
3. (c) If two rows or columns of A are identical, then $det(A)=0$ $det(A)=0$ .
4. (d) If B is a matrix obtained by interchanging two rows or two columns of A, then $det (B) = det (A)$ $det (B) = det (A)$ .
5. (e) If B is a matrix obtained by multiplying each entry of some row or column of A by a scalar, then $det (B) = det (A)$ $det (B) = det (A)$ .
6. (f) If B is a matrix obtained from A by adding a multiple of some row to a different row, then $det (B) = det (A)$ $det (B) = det (A)$ .
7. (g) The determinant of an upper triangular $n \times n$ $n \times n$ matrix is the product of its diagonal entries.
8. (h) For every $A \in M_{n \times n} (F), det (A^{t}) = - det (A)$ $A \in M_{n \times n} (F), det (A^{t}) = - det (A)$ .
9. (i) If $A, B \in M_{n \times n} (F),$ $A, B \in M_{n \times n} (F),$ then $det (A B) = det (A) \cdot det (B)$ $det (A B) = det (A) \cdot det (B)$ .
10. (j) If Q is an invertible matrix, then $det (Q^{- 1}) = {[det (Q)]}^{- 1}$ $det (Q^{- 1}) = {[det (Q)]}^{- 1}$ .
11. (k) A matrix Q is invertible if and only if $det (Q) \neq 0$ $det (Q) \neq 0$ .
Evaluate the determinant of the following $2 \times 2$ $2 \times 2$ matrices.
1. (a) $(\begin{array}{r} 4 & - 5 \\ 2 & 3 \end{array})$ $(\begin{array}{r} 4 & - 5 \\ 2 & 3 \end{array})$
2. (b) $(\begin{array}{r} - 1 & 7 \\ 3 & 8 \end{array})$ $(\begin{array}{r} - 1 & 7 \\ 3 & 8 \end{array})$
3. (c) $(\begin{array}{r} 2 + i & - 1 + 3 i \\ 1 - 2 i & 3 - i \end{array})$ $(\begin{array}{r} 2 + i & - 1 + 3 i \\ 1 - 2 i & 3 - i \end{array})$
4. (d) $(\begin{array}{c} 3 & 4 i \\ - 6 i & 2 i \end{array})$ $(\begin{array}{c} 3 & 4 i \\ - 6 i & 2 i \end{array})$
Evaluate the determinant of the following matrices in the manner indicated.
1. (a) $(\begin{array}{r} 0 & 1 & 2 \\ - 1 & 2 & - 3 \\ 2 & 3 & 0 \end{array})$ $(\begin{array}{r} 0 & 1 & 2 \\ - 1 & 2 & - 3 \\ 2 & 3 & 0 \end{array})$
  
  along the first row
2. (b) $(\begin{array}{r} 1 & 0 & 2 \\ 0 & 1 & 5 \\ - 1 & 3 & 0 \end{array})$ $(\begin{array}{r} 1 & 0 & 2 \\ 0 & 1 & 5 \\ - 1 & 3 & 0 \end{array})$
  
  along the first column
3. (c) $(\begin{array}{r} 0 & 1 & 2 \\ - 1 & 0 & - 3 \\ 2 & 3 & 0 \end{array})$ $(\begin{array}{r} 0 & 1 & 2 \\ - 1 & 0 & - 3 \\ 2 & 3 & 0 \end{array})$
  
  along the second column
4. (d) $(\begin{array}{r} 1 & 0 & 2 \\ 0 & 1 & 5 \\ - 1 & 3 & 0 \end{array})$ $(\begin{array}{r} 1 & 0 & 2 \\ 0 & 1 & 5 \\ - 1 & 3 & 0 \end{array})$
  
  along the third row
5. (e) $(\begin{array}{c} 0 & 1 + i & 2 \\ - 2 i & 0 & 1 - i \\ 3 & 4 i & 0 \end{array})$ $(\begin{array}{c} 0 & 1 + i & 2 \\ - 2 i & 0 & 1 - i \\ 3 & 4 i & 0 \end{array})$
  
  along the third row
6. (f) $(\begin{array}{c} i & 2 + i & 0 \\ - 1 & 3 & 2 i \\ 0 & - 1 & 1 - i \end{array})$ $(\begin{array}{c} i & 2 + i & 0 \\ - 1 & 3 & 2 i \\ 0 & - 1 & 1 - i \end{array})$
  
  along the third column
7. (g) $(\begin{array}{r} 0 & 2 & 1 & 3 \\ 1 & 0 & - 2 & 2 \\ 3 & - 1 & 0 & 1 \\ - 1 & 1 & 2 & 0 \end{array})$ $(\begin{array}{r} 0 & 2 & 1 & 3 \\ 1 & 0 & - 2 & 2 \\ 3 & - 1 & 0 & 1 \\ - 1 & 1 & 2 & 0 \end{array})$
  
  along the fourth column
8. (h) $(\begin{array}{r} 1 & - 1 & 2 & - 1 \\ - 3 & 4 & 1 & - 1 \\ 2 & - 5 & - 3 & 8 \\ - 2 & 6 & - 4 & 1 \end{array})$ $(\begin{array}{r} 1 & - 1 & 2 & - 1 \\ - 3 & 4 & 1 & - 1 \\ 2 & - 5 & - 3 & 8 \\ - 2 & 6 & - 4 & 1 \end{array})$
  
  along the fourth row
Evaluate the determinant of the following matrices by any legitimate method.
1. (a) $(\begin{array}{r} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array})$ $(\begin{array}{r} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array})$
2. (b) $(\begin{array}{r} - 1 & 3 & 2 \\ 4 & - 8 & 1 \\ 2 & 2 & 5 \end{array})$ $(\begin{array}{r} - 1 & 3 & 2 \\ 4 & - 8 & 1 \\ 2 & 2 & 5 \end{array})$
3. (c) $(\begin{array}{r} 0 & 1 & 1 \\ 1 & 2 & - 5 \\ 6 & - 4 & 3 \end{array})$ $(\begin{array}{r} 0 & 1 & 1 \\ 1 & 2 & - 5 \\ 6 & - 4 & 3 \end{array})$
4. (d) $(\begin{array}{r} 1 & - 2 & 3 \\ - 1 & 2 & - 5 \\ 3 & - 1 & 2 \end{array})$ $(\begin{array}{r} 1 & - 2 & 3 \\ - 1 & 2 & - 5 \\ 3 & - 1 & 2 \end{array})$
5. (e) $(\begin{array}{c} i & 2 & - 1 \\ 3 & 1 + i & 2 \\ - 2 i & 1 & 4 - i \end{array})$ $(\begin{array}{c} i & 2 & - 1 \\ 3 & 1 + i & 2 \\ - 2 i & 1 & 4 - i \end{array})$
6. (f) $(\begin{array}{c} - 1 & 2 + i & 3 \\ 1 - i & i & 1 \\ 3 i & 2 & - 1 + i \end{array})$ $(\begin{array}{c} - 1 & 2 + i & 3 \\ 1 - i & i & 1 \\ 3 i & 2 & - 1 + i \end{array})$
7. (g) $(\begin{array}{r} 1 & 0 & - 2 & 3 \\ - 3 & 1 & 1 & 2 \\ 0 & 4 & - 1 & 1 \\ 2 & 3 & 0 & 1 \end{array})$ $(\begin{array}{r} 1 & 0 & - 2 & 3 \\ - 3 & 1 & 1 & 2 \\ 0 & 4 & - 1 & 1 \\ 2 & 3 & 0 & 1 \end{array})$
8. (h) $(\begin{array}{r} 1 & - 2 & 3 & - 12 \\ - 5 & 12 & - 14 & 19 \\ - 9 & 22 & - 20 & 31 \\ - 4 & 9 & - 14 & 15 \end{array})$ $(\begin{array}{r} 1 & - 2 & 3 & - 12 \\ - 5 & 12 & - 14 & 19 \\ - 9 & 22 & - 20 & 31 \\ - 4 & 9 & - 14 & 15 \end{array})$
Suppose that $M \in M_{n \times n} (F)$ $M \in M_{n \times n} (F)$ can be written in the form

$M = (\begin{array}{r} A & B \\ O & I \end{array}),$ $M = (\begin{array}{r} A & B \\ O & I \end{array}),$

where A is a square matrix. Prove that $det (M) = det (A)$ $det (M) = det (A)$ .
^† Prove that if $M \in M_{n \times n} (F)$ $M \in M_{n \times n} (F)$ can be written in the form

$M = (\begin{array}{r} A & B \\ O & C \end{array}),$ $M = (\begin{array}{r} A & B \\ O & C \end{array}),$

where A and C are square matrices, then $det (M) = det (A) \cdot det (C)$ $det (M) = det (A) \cdot det (C)$ . Visit goo.gl/pGMdpX for a solution.

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Table of Contents for 4.4 Summary&#8212;Important Facts about Determinants Summary—Important Facts about Determinants

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4.4 Summary—Important Facts about Determinants Summary—Important Facts about Determinants