Determinants of Square Matrices

Section 2.8 Determinants of Square Matrices

Objectives

Define the determinant as a function with a set of properties.
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Find the determinant of a $2 \times 2$ matrix
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Use the Laplace expansion method to find the determinant of a matrix.
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Use row operations to find the determinant of a matrix.
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Understand the geometry of the determinant.
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The determinant is a useful function for specifying when matrix is invertible or singular. For example, in section Section 2.3, we saw that for

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\begin{equation} A = \begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix}\tag{2.8.1} \end{equation}

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that the inverse matrix exists if and only if $ad-bc \neq 0\text{.}$ The determinant of this matrix will be the function $ad-bc\text{.}$ We will see two different methods for finding the determinant of an $n \times n$ matrix.

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Subsection 2.8.1 Properties of the Determinant

Definition 2.8.1.

Consider an $n \times n$ matrix $A$ and denote the rows by $\vec{\rho}_1, \vec{\rho}_2, \ldots, \vec{\rho}_n\text{.}$ An $n \times n$ determinant is a function $\det: {\cal M}_{n \times n} \rightarrow \mathbb{R}$ such that

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$\det(\vec{\rho}_1,\ldots, k \vec{\rho}_i + \vec{\rho}_j, \ldots, \vec{\rho}_n) = \det(\vec{\rho}_1, \ldots, \vec{\rho}_j , \ldots, \vec{\rho}_n)$ for $i \neq j$ ,
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$\det(\vec{\rho}_1,\ldots, \vec{\rho}_i, \ldots, \vec{\rho}_j, \ldots, \vec{\rho}_n) = -\det(\vec{\rho}_1, \ldots, \vec{\rho}_j, \ldots, \vec{\rho}_i, \ldots, \vec{\rho}_n)$ for $i \neq j\text{,}$
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$\det(\vec{\rho}_1,\ldots, k \vec{\rho}_i, \ldots, \vec{\rho}_n) = k \det(\vec{\rho}_1, \ldots, \vec{\rho}_i , \ldots, \vec{\rho}_n)\text{,}$
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$\det(I) =1\text{,}$ where $I$ is the identity matrix.
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The $\vec{\rho}_i$’s are the rows of the matrix and often the notation $|T| = \det(T)$ is used.

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In more colloquial terms, these properties are:

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if one multiplies a row by a constant and adds to a second row that the determinant is unchanged.
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If one swaps rows in a matrix, the determinant of the new matrix is the opposite sign of the original matrix.
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If one multiplies any row of a matrix by $k$ then the determinant of the new matrix is $k$ times the determinant of the original matrix.
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The determinant of the identity matrix is 1.
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Notice that the above explain the properties of a matrix, but how does one compute it? If

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\begin{equation*} A = \begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix} \end{equation*}

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then define $f(A) = ad-bc\text{.}$ We will show that $f$ satisfies the properties of the determinant, therefore is the determinant formula for a $2 \times 2$ matrix.

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\begin{equation*} kR_1 +R_2 \rightarrow R_2 \qquad A' = \begin{bmatrix} a \amp b \\ ka +c \amp kb + d \end{bmatrix} \end{equation*}

and

\begin{align*} \det(A') \amp = a(kb+d) - b(ka+c)\\ \amp = kab+ad-kab-bc \\ \amp = ad-bc \\ \amp = \det(A) \end{align*}

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\begin{equation*} R_1 \leftrightarrow R_2 \qquad A' = \begin{bmatrix} c \amp d \\ a \amp b \end{bmatrix} \end{equation*}

and

\begin{equation*} \det(A') = cb-ad = -\det(A) \end{equation*}

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\begin{equation*} kR_1 \rightarrow R_1 \qquad A' = \begin{bmatrix} ka \amp kb \\ c \amp d \end{bmatrix} \end{equation*}

and

\begin{align*} \det(A') \amp = kad-kbc \\ \amp =k(ad-bc) \\ \amp = k\det(A) \end{align*}

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And finally if

\begin{equation*} A' = \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix} \end{equation*}

then

\begin{equation*} \det(A') = 1(1)-0(0) = 1 \end{equation*}

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Remark 2.8.2.

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\begin{equation*} A = \begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix} \end{equation*}

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then

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\begin{equation*} \det(A) = ad - bc \end{equation*}

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Lemma 2.8.3.

The following are additional properties of a determinant.

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A matrix with two identical rows has a zero determinant.
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A matrix with a row of zeros has a zero determinant.
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A matrix is nonsingular if and only if its determinant is nonzero.
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The determinant of a matrix in echelon form is the product of its diagonal elements.
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Proof.

Consider a matrix with two identical rows. Swapping the two rows changes the sign of the determinant, however leaves the matrix unchanged. This implies that the determinant must be zero.
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Call the row $i$ with the row of zeros. Let $j$ be any other row. Apply the row operation $R_j + R_i \to R_i$ and now rows $i$ and $j$ are identical and the determinant has not changed. From the previous statement, the determinant is 0.
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Recall that for a matrix to be nonsingular, that a linear system with coefficients from the matrix has a unique system or the row reduced echelon form of the matrix has no row of zeros. Let $T$ be the original matrix and $\hat{T}$ is the matrix in echelon form. If the $\hat{T}$ is nonsingular, then it is the identity matrix, which has the nonzero determinant 1.
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A singular matrix $T$ reduces to a matrix $\hat{T}$ with a row of zeros. By #2, this determinant is 0.
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Consider a matrix in echelon form:
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\begin{align*} |\hat{T}| \amp = \begin{vmatrix} t_{1,1} \amp t_{1,2} \amp \cdots \amp t_{1,n} \\ 0 \amp t_{2,2} \amp \amp t_{2,n} \\ \vdots \amp \amp \ddots \amp \\ 0 \amp 0 \amp \vdots \amp t_{n,n} \\ \end{vmatrix} \end{align*}

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Using property (3) of the determinant defintion,
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\begin{align*} \amp = t_{1,1} t_{2,2} \cdots t_{n,n} \begin{bmatrix} 1 \amp t_{1,2}/t_{1,1} \amp \cdots \amp t_{1,n}/t_{1,1} \\ 0 \amp 1 \amp \amp t_{2,n}/t_{2,2} \\ \vdots \amp \amp \ddots \amp \\ 0 \amp 0 \amp \vdots \amp 1\\ \end{bmatrix} \end{align*}

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The matrix can be reduced to the identity matrix, with determinant 1
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\begin{align*} \amp = t_{1,1} t_{2,2} \cdots t_{n,n} \end{align*}

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The 1st, 2nd and 4th statements of the lemma are very helpful for calculating the determinant. The 4th statement gives a very general way to calculate a determinant. Use the elementary row operations to put the matrix in echelon form (recall to keep track of multiplying any row by a constant and row swaps). The determinant will then be the product of the diagonal elements.
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The third statement of the lemma above gives a function (the determinant) that determines whether or not a matrix is singular.
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Subsection 2.8.2 Calculating Determinants

We now calculate the determinant of two square matrices. The first example is a $2 \times 2$ matrix and we use both the formula as well as Gauss’ method.

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Example 2.8.4.

Find the determinant of the following matrix using a) the formula for $2 \times 2$ determinants and b) using Gauss’ method.

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\begin{equation*} T = \begin{bmatrix} 3 \amp 2 \\ 1 \amp -2 \end{bmatrix} \end{equation*}

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Solution.

Using the formula $|T| = ad-bc=-6-2=-8\text{.}$

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Using Gauss’s method,

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\begin{align*} \qquad |T| \amp= \begin{vmatrix} 3 \amp 2 \\ 1 \amp -2 \end{vmatrix} \\ R_1 \leftrightarrow R_2 \qquad -|T|\amp= \begin{vmatrix} 1 \amp -2 \\ 3 \amp 2 \end{vmatrix} \\ -3R_1 + R_2 \rightarrow R_2 \qquad -|T|\amp= \begin{vmatrix} 1 \amp -2 \\ 0 \amp 8 \end{vmatrix} = 8 \end{align*}

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So $|T|=-8\text{.}$

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This shows that although Gauss’ method succeeds in finding the determinant, it takes more operations than the simple formula.

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Example 2.8.5.

Use Gauss’s method to find the determinants of the following matrices:

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\begin{align*} T \amp = \begin{bmatrix} 3 \amp 0 \amp 2 \\ 1 \amp 4 \amp 0 \\ 0 \amp 2 \amp 5 \end{bmatrix} \amp S \amp = \begin{bmatrix} 0 \amp 1 \amp 3 \amp -4 \\ 2 \amp 0 \amp 2 \amp 7 \\ 0 \amp 0 \amp 6 \amp 8 \\ 1 \amp 0 \amp 10 \amp 6 \end{bmatrix} \end{align*}

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Solution.

For both examples, we use row operations and keep track of any row swaps (introducing a $-1$) or multiples.

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\begin{align*} |T| = \amp \begin{vmatrix} 3 \amp 0 \amp 2 \\ 1 \amp 4 \amp 0 \\ 0 \amp 2 \amp 5 \end{vmatrix}\\ R_1 \leftrightarrow R_2 \qquad -|T|= \amp \begin{vmatrix} 1 \amp 4 \amp 0 \\ 3 \amp 0 \amp 2 \\ 0 \amp 2 \amp 5 \end{vmatrix} \\ -3 R_1 + R_2 \rightarrow R_2 \qquad -|T| = \amp \begin{vmatrix} 1 \amp 4 \amp 0 \\ 0 \amp -12 \amp 2 \\ 0 \amp 2 \amp 5 \end{vmatrix} \\ R_2 \leftrightarrow R_3 \qquad |T|= \amp \begin{vmatrix} 1 \amp 4 \amp 0 \\ 0 \amp 2 \amp 5 \\ 0 \amp -12 \amp 2 \\ \end{vmatrix} \\ 6 R_2 + R_3 \rightarrow R_3 \qquad |T| = \amp \begin{vmatrix} 1 \amp 4 \amp 0 \\ 0 \amp 2 \amp 5 \\ 0 \amp 0 \amp 32 \\ \end{vmatrix} = 64 \end{align*}

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\begin{align*} |S| \amp = \begin{vmatrix} 0 \amp 1 \amp 3 \amp -4 \\ 2 \amp 0 \amp 2 \amp 7 \\ 0 \amp 0 \amp 6 \amp 8 \\ 1 \amp 0 \amp 10 \amp 6 \end{vmatrix} \\ R_1 \leftrightarrow R_4 \qquad -|S| \amp = \begin{vmatrix} 1 \amp 0 \amp 10 \amp 6 \\ 2 \amp 0 \amp 2 \amp 7 \\ 0 \amp 0 \amp 6 \amp 8 \\ 0 \amp 1 \amp 3 \amp -4 \\ \end{vmatrix} \\ -2R_1 + R_2 \rightarrow R_2 \qquad -|S| \amp = \begin{vmatrix} 1 \amp 0 \amp 10 \amp 6 \\ 0 \amp 0 \amp -18 \amp -5 \\ 0 \amp 0 \amp 6 \amp 8 \\ 0 \amp 1 \amp 3 \amp -4 \\ \end{vmatrix} \\ R_2 \leftrightarrow R_4 \qquad |S| \amp= \begin{vmatrix} 1 \amp 0 \amp 10 \amp 6 \\ 0 \amp 1 \amp 3 \amp -4 \\ 0 \amp 0 \amp 6 \amp 8 \\ 0 \amp 0 \amp -18 \amp -5 \\ \end{vmatrix} \\ 3 R_3 + R_4 \rightarrow R_4 \qquad |S| \amp = \begin{vmatrix} 1 \amp 0 \amp 10 \amp 6 \\ 0 \amp 1 \amp 3 \amp -4 \\ 0 \amp 0 \amp 6 \amp 8 \\ 0 \amp 0 \amp 0 \amp 19 \\ \end{vmatrix} = 6 (19) = 114 \end{align*}

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Subsection 2.8.3 Expansion Method for finding the Determinant

Although Gauss’ method is a very robust and in general efficient method for finding determinants, a method called the Laplace Expansion method can be quite helpful at times as well. Before defining this, we need to know a matrix minor and cofactor first.

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Definition 2.8.6.

For any $n\times n$ matrix $T$ , the $(n - 1)\times(n - 1)$ matrix formed by deleting row $i$ and column $j$ of $T$ is the $i,j$ minor of $T$. The $i,j$ cofactor $T_{i,j}$ of $T$ is $(-1)^{i+j}$ times the determinant of the $i, j$ minor of $T$ and denoted $(-1)^{i+j} |T_{i,j}|\text{.}$

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Example 2.8.7.

Find the $T_{1,1}$ and $T_{2,3}$ minors and cofactors of the matrix

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\begin{equation*} T = \begin{bmatrix} 1 \amp 2 \amp 3 \\ 4 \amp 5 \amp 6 \\ 7 \amp 8 \amp 9 \end{bmatrix} \end{equation*}

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Solution.

Recall that $T_{i,j}$ minor is found by removing the $i$th row and $j$th column or

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\begin{align*} T_{1,1} \amp = \begin{bmatrix} 5 \amp 6 \\ 8 \amp 9 \end{bmatrix} \amp T_{2,3} \amp = \begin{bmatrix} 1 \amp 2 \\ 7 \amp 8 \end{bmatrix} \end{align*}

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and the cofactors are the determinants of each of these matrices times $(-1)^{i+j}$ or

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\begin{align*} (-1)^{1+1} |T_{1,1}| \amp= (1) (45-24) = 21 \\ (-1)^{2+3} |T_{2,3}| \amp = (-1) (8-14) = 6 \end{align*}

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Now that we have the prerequisites, the following is the Laplace Expansion method for finding a determinant.

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Theorem 2.8.8. Laplace Expansion of Determinants.

The determinant of an $n \times n$ matrix $T$ can be found by expanding by cofactors on row $i$ or column $j$ of $T\text{.}$ That is

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\begin{equation*} |T| = t_{i,1} T_{i,1} + t_{i,2} T_{i,2} + \cdots + t_{i,n} T_{i,n} \end{equation*}

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for any $i$ satisfying $1 \leq i \leq n$ or

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\begin{equation*} |T| = t_{1,j} T_{1,j} + t_{2,j} T_{2,j} + \cdots + t_{n,j} T_{n,j} \end{equation*}

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for any $j$ satisfying $1 \leq j \leq n\text{.}$

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Example 2.8.9.

Use the expansion formula to find the determinants of the matrices in Example 2.8.5, namely

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Solution.

In the case of $T\text{,}$ we will expand across the first row and use the formula for the $2\times 2$ determinant.

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\begin{align*} |T| \amp = (-1)^{1+1} (3) \begin{vmatrix}4 \amp 0 \\ 2 \amp 5 \end{vmatrix} + (-1)^{1+2} (0) \begin{vmatrix} 1 \amp 0 \\ 0 \amp 5 \end{vmatrix} + (-1)^{1+3} (2) \begin{vmatrix} 1 \amp 4 \\ 0 \amp 2 \end{vmatrix} \\ \amp = 3 (20) + (2) (2-0) = 64 \end{align*}

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and for $S\text{,}$ we’ll expand down the 2nd column because all but one is zero. And because of this, I won’t show the cofactors of $T_{1,2}, T_{1,3}$ and $T_{1,4}\text{.}$

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\begin{align*} |S| \amp = (-1)^{1+2} (1) \begin{vmatrix} 2 \amp 2 \amp 7 \\ 0 \amp 6 \amp 8 \\ 1 \amp 10 \amp 6 \end{vmatrix} + 0 + 0 + 0 \end{align*}

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and now to find this $3\times 3$ determinant, expand about the 2nd row

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\begin{align*} |S| \amp = (-1) \bigl( (-1)^{2+2} (6) \begin{vmatrix} 2 \amp 7 \\ 1 \amp 6 \end{vmatrix} + (-1)^{2+3} (8) \begin{vmatrix} 2 \amp 2 \\ 1 \amp 10 \end{vmatrix} \bigr) \end{align*}

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and now use the formula for $2 \times 2$ determinants.

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\begin{equation*} |S| = -(6 (12-7) - 8 (20-2)) = -(30- 144) =114 \end{equation*}

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Subsection 2.8.4 Geometry of Determinants

In the previous section, the determinant was introduced as a function that determines whether or not a matrix was singular due to whether or not the function was 0. In this section, we will look at a geometric approach to the determinant and show that it can be used to determine areas (and volumes) of regions bounded by vectors. We will show that this geometric approach is identical (in the two-dimensional case) as the properties in Definition 2.8.1.

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Consider the parallelogram formed by two vectors. In the argument below, it is important that the vector $\langle x_1, y_1 \rangle$ is below and to the right of the vector $\langle x_2, y_2 \rangle\text{.}$

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Figure 2.8.10. Plot of two vectors in $\mathbb{R}^2$ forming a parallelogram.
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The area of the parallelogram can be determined by taking the area of the enclosing rectangle and subtracting out the rectangles $A$ and $F$ and triangles $B, C, D$ and $E$ as shown below:

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Figure 2.8.11. Finding the area of the parallelogram
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\begin{align*} \text{area of parallelogram} \amp = \text{area of enclosing rect} \\ \amp \qquad - \text{area of rectangle $A$} - \text{area of triangle $B$} \\ \amp \qquad \cdots - \text{area of rectangle $F$} \\ \amp = (x_1+x_2)(y_1+y_2) - x_2 y_1 - \frac{1}{2} x_1 y_1 \\ \amp \qquad - \frac{1}{2} x_2 y_2 - \frac{1}{2} x_2 y_2 - \frac{1}{2} x_1 y_1 - x_2 y_1 \\ \amp = x_1 y_2 - x_2 y_1 \end{align*}

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and note that

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\begin{equation*} \begin{vmatrix} x_1 \amp x_2 \\ y_1 \amp y_2 \end{vmatrix} = x_1 y_2 - x_2 y_1 \end{equation*}

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And this result is identical to the determinant seen above. Again, as noted, the vectors were set up to have a positive area, however in general, one can define the area as the absolute value of the determinant.

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Subsubsection 2.8.4.1 Transformation of the Vectors and the size of the Parallelogram

From above, the area of the parallelogram is the determinant of the vectors that are along the sides.

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Consider two vectors in $\mathbb{R}^2$ and rotate them so one is on the $x$-axis. Also take $\vec{u}$ and multiply it by a factor of $k$

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Figure 2.8.12. Scaling a parallelogram
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From this geometric argument, the area of the parallelogram formed by the vectors $\vec{v}$ and $k\vec{u}$ appears to $k$ times larger. This is property 3 of Definition Definition 2.8.1.

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Next, let’s look at transformation $\vec{u} + k\vec{v}\text{.}$ The picture on the left is the original two vectors and that on the right is the transformed vectors (with $k$ about 0.2 in this picture). The original area and the transformed area are identical in this case since neither the height of the parallelogram nor its width has changed.

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Figure 2.8.13. Skewing a parallelogram
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This property shows that replacing a row with a constant times another row plus the current row results in an unchanged area is consistent with property 1 of Definition 2.8.1.

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The other transformation related to the determinant is property 2 of Definition 2.8.1 or in other words, if one switched the order of the vectors (row swaps), that the determinant changes sign. The area does not change because the area is the absolute value of the determinant.

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Definition 2.8.14.

In $\mathbb{R}^n\text{,}$ the parallelpiped formed by $\langle \vec{v}_1, \vec{v}_2, \ldots, \vec{v}_n \rangle$ includes the set

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\begin{equation*} \{ t_1 \vec{v}_1 + t_2 \vec{v}_2 + \cdots + t_n \vec{v}_n\; | \; t_1, t_2, \ldots, t_n \in [0,1] \} \end{equation*}

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The volume of the parallelepiped is the absolute value of the determinant of the matrix whose columns are $\vec{v}_1, \vec{v}_2, \ldots, \vec{v}_n\text{.}$

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Example 2.8.15.

Find the volume of the parallelpiped formed by the vectors:

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\begin{equation*} \begin{bmatrix} 3 \\ 0 \\ 2 \end{bmatrix}, \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} \end{equation*}

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Solution.

The volume is the absolute value of the determinant of the matrix with these three columns. We’ll use Gauss’ method to find the determinant.

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\begin{align*} |A| \amp= \begin{vmatrix} 3 \amp -1 \amp 2 \\ 0 \amp 2 \amp 3 \\ 2 \amp 0 \amp 1 \end{vmatrix} \\ 3R_3 \rightarrow R_3 \qquad 3|A| \amp= \begin{vmatrix} 3 \amp -1 \amp 2 \\ 0 \amp 2 \amp 3 \\ 6 \amp 0 \amp 3 \end{vmatrix} \\ -2 R_1 + R_3 \rightarrow R_3 \qquad 3|A| \amp= \begin{vmatrix} 3 \amp -1 \amp 2 \\ 0 \amp 2 \amp 3 \\ 0 \amp 2 \amp -1 \end{vmatrix} \\ -R_2 + R_3 \rightarrow R_3 \qquad 3|A| \amp= \begin{vmatrix} 3 \amp -1 \amp 2 \\ 0 \amp 2 \amp 3 \\ 0 \amp 0 \amp -4 \end{vmatrix} \end{align*}

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and multiplying down the diagonal, $3|A| = -24\text{,}$ so $|A|=-8\text{.}$ This means that the volume is 8 units.

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Subsection 2.8.5 Properties of Determinants

Need something here.

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