Determinants_ Properties and Applications_ Attempt review _ eClass
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Question 1
Correct
Mark 5.00 out of 5.00
Properties of the Determinant
Many of the properties below are discussed in the video Determinants: Properties and Applications 1
. You should know these properties. Many
of these have a simple explanation, but for others the proof is more complicated, so most of these will be skipped. The video provides a partial
proof of property (c) below.
Theorem: Let be an matrix.
(a) For any scalar ,
(b) For any matrix ,
(c) is invertible if and only if 0
. Furthermore, if is invertible, then
(d) For the transpose of , we have
(e) If contains a row or column of zeros, or if contains a row (resp. column) which is a scalar multiple of another row (resp.column) then
0
(f) If and are
matrices such that ,
, and are identical except that the row (resp. column) of is the sum of the
rows
(resp. columns) of and , that is,
then
Proof: We will prove .
Assume: is invertible, so that
.
≠ 0
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by Property
since
.
Note:
The proof of Property (d) uses the Laplace Expansion Theorem.
Note:
We can generalize Property (b) to the product of an arbitrary number of matrices. If are matrices, then
(b)
≠ 0
Correct
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Question 2
Correct
Mark 7.00 out of 7.00
Properties of the Determinant: Examples
Example: Let .
By Property
, . Let's compute directly to get a sense of why this property holds:
2
4
8
Example: Use Property (c) to determine whether the following matrix is invertible:
Since -9
, the matrix
invertible.
(a)
≠ 0
is
Correct
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Question 3
Correct
Mark 8.00 out of 8.00
Example: Let
.
Since column
of is scalar multiple of column 1, we know by Property
that 0
.
Indeed,
0
(by cofactor expansion along column 3)
Recall Property (f):
If , , and are matrices which are identical except that
then
Example: Let
.
Note:
Thus
by Property
0
24
(see example directly above)
24
.
3
(e)
(f)
Correct
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Question 4
Correct
Mark 11.00 out of 11.00
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Example:
Assume , , and are matrices such that , , and .
Evaluate the following determinants: (Enter as a/b, where applicable.)
1. -6
2. 81
3.
4
4.
-128
5.
1/3
6.
3/4
7.
1/12
An example similar to no.8 below is provided in the video
Determinants: Properties and Applications 1 starting at 12:20.
8. by Property
by Property
by Properties
24
(a)
(b)
(b), (c), and (d)
Correct
Marks for this submission: 11.00/11.00.
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Question 5
Correct
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Example:
Is , for all matrices and ?
Correct answer, well done.
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Counterexample:
Take and , where is the identity matrix. Thus
0
0
0
0
Correct answer, well done.
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Then 0
,but 1
1
2
Correct answer, well done.
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Correct answer, well done.
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Correct answer, well done.
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Correct answer, well done.
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Thus
in this case,
Correct answer, well done.
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and hence in general.
Correct answer, well done.
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No
≠
≠
A correct answer is: "No"
A correct answer is .
A correct answer is , which can be typed in as follows: 0
A correct answer is , which can be typed in as follows: 1
A correct answer is , which can be typed in as follows: 1
A correct answer is , which can be typed in as follows: 2
A correct answer is: "≠"
A correct answer is: "≠"
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Question 6
Correct
Mark 2.00 out of 2.00
Applications of the Determinant
We have already discussed how the determinant of a square matrix \(A\) can be used to determine if \(A\) is invertible. Indeed, \(A\) is
invertible if and only if \(\det(A) \)
. This, in turn, gives us an application to linear transformations: a linear transformation \(T:
\mathbb{R}^n \rightarrow \mathbb{R}^n\) is invertible if and only if the standard matrix \([T]\) of \(T\) is invertible if and only if \(\det([T]) \)
.
Determinants are useful in other ways. In this learning activity, we will discuss how determinants can be used to compute the inverse of an
invertible matrix, as well as some applications of determinants to geometry, including the concept of the cross product of two vectors in \
(\mathbb{R}^3\). Later in Block 6, we will use determinants to compute what is known as the characteristic polynomial of a square matrix.
≠ 0
≠ 0
Correct
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Question 7
Correct
Mark 5.00 out of 5.00
The Adjoint Formula for the Inverse of an Invertible Matrix
The concept of the determinant gives us an alternative method for computing the inverse of an invertible matrix. For this method, we require
the following definition:
Definition:
Let \(A\) be an \(n\times n\) matrix with cofactors \(C_{ij} = (-1)^{i+j} \det(A_{ij})\). The adjoint
of \(A\), denoted \
(\operatorname{adj}(A)\), is the
transpose of the matrix of cofactors of \(A\), that is,
\[ \operatorname{adj}(A) = \left[\begin{array}{cccc}C_{11} & C_{12} & \dots & C_{1n} \\ C_{21} & C_{22} & \dots & C_{2n} \\ \vdots & \vdots &
& \vdots \\ C_{n1} & C_{n2} & \dots & C_{nn} \end{array} \right]^{ \ T} = \left[\begin{array}{cccc}C_{11} & C_{21} & \dots & C_{n1} \\ C_{12} &
C_{22} & \dots & C_{n2} \\ \vdots & \vdots & & \vdots \\ C_{1n} & C_{2n} & \dots & C_{nn} \end{array}\right] \]
Theorem:
Let \(A\) be an invertible \(n \times n\) matrix, then
\[A^{-1}=\frac{1}{\det(A)}\; \operatorname{adj}(A)\]
Example: Let \(A = \begin{bmatrix}a&b\\c&d\end{bmatrix}\) be an invertible \(2 \times 2\) matrix, then
\(C_{11} = \)
d
\(C_{12} =\)
-c
Correct answer, well done.
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Correct answer, well done.
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\(C_{21} =\)
-b
\(C_{22} =\)
a
Correct answer, well done.
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Thus,
\(\operatorname{adj}(A) =
\left[\begin{array}{cccc}C_{11} & C_{12} \\ C_{21} & C_{22}
\end{array} \right]^T = \left[\begin{array}{cccc}C_{11} & C_{21} \\ C_{12} & C_{22} \end{array}
\right] = \)
d
-b
-c
a
Correct answer,
well done.
Marks for this
submission: 1.00/1.00.
and hence \[A^{-1}=\frac{1}{\det(A)}\; \operatorname{adj}(A) = \frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\]
So the formula for the inverse of a \(2 \times 2\) matrix that was given in Block 3 is, in fact, the adjoint formula for the inverse of a \(2 \times
2\) matrix.
A correct answer is \( d \), which can be typed in as follows: d
A correct answer is \( -c \), which can be typed in as follows: -c
A correct answer is \( -b \), which can be typed in as follows: -b
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A correct answer is \( a \), which can be typed in as follows: a
A correct answer is \( \left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right] \).
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Question 8
Correct
Mark 6.00 out of 6.00
Example: Use the adjoint formula for the inverse of a matrix to find the inverse of the invertible matrix \[A = {\left[\begin{array}{ccc} 2 & 1 & 4
\\ 1 & 0 & -1 \\ 0 & 1 & 8 \end{array}\right]}\]
Let's start by computing the adjoint of \(A\).
Note: in the second line of this computation, "\(+\)" and "\( - \)" represent the sign of the cofactors \(C_{ij} = (-1)^{i+j}\det(A_{ij})\), which are
determined by the term \((-1)^{i+j}\).
\
(\operatorname{adj}
(A)\)
\
(=\)
\( \left[\begin{array}{cccc}C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} &
C_{33} \end{array} \right]^{ \ T}\)
\
(=\)
\( \left[\begin{array}{cccccc} +\begin{vmatrix}0&-1\\1&8\end{vmatrix}&& -
\begin{vmatrix}1&-1\\0&8\end{vmatrix} && +\begin{vmatrix}1&0\\0&1\end{vmatrix}\; \; \\ -
\begin{vmatrix}1&4\\1&8\end{vmatrix}& &+\begin{vmatrix}2&4\\0&8\end{vmatrix}& & -
\begin{vmatrix}2&1\\0&1\end{vmatrix}\; \; \\ + \begin{vmatrix}1&4\\0&-1\end{vmatrix} &&
- \begin{vmatrix}2&4\\1&-1\end{vmatrix} && + \begin{vmatrix}2&1\\1&0\end{vmatrix} \; \;
\end{array} \right]^{\ T}\)
\
(=\)
\( \left( \begin{array}{c}\hspace{0.1cm}\\ \\ \\ \\\end{array}\right.\) 1
-8
1
-4
16
-2
-1
6
-1
\
(\left. \begin{array}{c}\hspace{0.1cm}\\ \\ \\ \\\end{array}\right)^T\)
Correct
answer, well
done.
Marks for this
submission:
3.00/3.00.
\
(=\)
1
-4
-1
-8
16
6
1
-2
-1
Correct
answer, well
done.
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submission:
1.00/1.00.
Now, we will compute the determinant of \(A\):
\(\det(A) = \)
-2
Correct answer, well done.
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Therefore, the inverse of \(A\) is
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\(A^{-1} = \dfrac{1}{\det(A)}\; \operatorname{adj}(A) = \)
-1/2
2
1/2
4
-8
-3
-1/2
1
1/2
Correct answer, well done.
Marks for this submission: 1.00/1.00.
A correct answer is \( \left[\begin{array}{ccc} 1 & -8 & 1 \\ -4 & 16 & -2 \\ -1 & 6 & -1 \end{array}\right] \).
A correct answer is \( \left[\begin{array}{ccc} 1 & -4 & -1 \\ -8 & 16 & 6 \\ 1 & -2 & -1 \end{array}\right] \).
A correct answer is \( -2 \), which can be typed in as follows: -2
A correct answer is \( \left[\begin{array}{ccc} -\frac{1}{2} & 2 & \frac{1}{2} \\ 4 & -8 & -3 \\ -\frac{1}{2} & 1 & \frac{1}{2} \end{array}\right] \).
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Question 9
Correct
Mark 13.00 out of 13.00
Cross Product
Definition:
If \(\mathbf{u}= [u_1, u_2, u_3]\) and \(\mathbf{v}= [v_1, v_2, v_3]\) are vectors in \(\mathbb{R}^3\), then the cross product
of \
(\mathbf{u}\) and \(\mathbf{v}\) is the vector in \(\mathbb{R}^3\) given by
\[\mathbf{u}\times\mathbf{v}= [u_2v_3−u_3v_2,\, u_3v_1−u_1v_3,\, u_1v_2−u_2v_1]\]
In determinant notation
\[\begin{aligned}\mathbf{u}\times\mathbf{v} &= \left[ \, \begin{vmatrix} u_2 & u_3\\ v_2 & v_3 \end{vmatrix}, -\begin{vmatrix} u_1 & u_3\\ v_1
& v_3 \end{vmatrix}, \begin{vmatrix} u_1 & u_2\\ v_1 & v_2 \end{vmatrix} \, \right] \end{aligned}\]
As explained later, the cross product \( \mathbf{u}\times\mathbf{v}\) is orthogonal to both \( \mathbf{u}\) and \( \mathbf{v}\). It has
applications in mathematics and physics.
Let \(\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\) be the standard basis vectors in \(\mathbb{R}^3\). A convenient way to remember this
formula is to view it as the following determinant
\[\begin{aligned} \mathbf{u} \times \mathbf{v} &=\left| \begin{array}{ccc} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ u_1 & u_2 & u_3 \\
v_1 & v_2 & v_3 \end{array} \right| \\ \\&= \begin{vmatrix} u_2 & u_3\\ v_2 & v_3 \end{vmatrix}\mathbf{e}_1 - \begin{vmatrix} u_1 & u_3\\ v_1
& v_3 \end{vmatrix} \mathbf{e}_2 + \begin{vmatrix} u_1 & u_2\\ v_1 & v_2 \end{vmatrix} \mathbf{e}_3 \\ \\ &= \left[ \, \begin{vmatrix} u_2 &
u_3\\ v_2 & v_3 \end{vmatrix}, -\begin{vmatrix} u_1 & u_3\\ v_1 & v_3 \end{vmatrix}, \begin{vmatrix} u_1 & u_2\\ v_1 & v_2 \end{vmatrix} \,
\right], \end{aligned}\]
where we must
use cofactor expansion along
row 1. Viewing the cross product as this determinant not only provides us with a helpful way to
remember the formula, but it will also enable us to use the properties of the determinant to understand the properties of the cross product,
which will be presented in Question 12.
Example: Let \(\mathbf{u} = [3,2,1]\) and \(\mathbf{v} = [-1,1,0]\). Compute \(\mathbf{u} \times \mathbf{v}\).
\(\mathbf{u} \times
\mathbf{v}\)
\(= \det \) 2
1
1
0
\(\mathbf{e}_1 \; - \; \det \) 3
1
-1
0
\(\mathbf{e}_2 \; + \; \det \)
3
2
-1
1
\(\mathbf{e}_3\)
Correct answer, well done.
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Correct answer, well done.
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Correct answer, well done.
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\(=\) -1
\(\mathbf{e}_1 \; - \; \) 1
\(\mathbf{e}_2 \; + \; \) 5
\(\mathbf{e}_3\)
Correct answer, well done.
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Correct answer, well done.
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Correct answer, well done.
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\(=\)
-1
-1
5
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\(\mathbf{u} \times
\mathbf{v}\)
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Exercise:
Calculate the following cross products.
(a) \([1,2,3]\times[4,3,2]=\)
-5
10
-5
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(b) \([1,1,-1]\times[2,0,2]=\)
2
-4
-2
Correct answer, well done.
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(c) \([2,0,2]\times[1,1,-1]=\)
-2
4
2
Correct answer, well done.
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A correct answer is \( \left[\begin{array}{cc} 2 & 1 \\ 1 & 0 \end{array}\right] \).
A correct answer is \( \left[\begin{array}{cc} 3 & 1 \\ -1 & 0 \end{array}\right] \).
A correct answer is \( \left[\begin{array}{cc} 3 & 2 \\ -1 & 1 \end{array}\right] \).
A correct answer is \( -1 \), which can be typed in as follows: -1
A correct answer is \( 1 \), which can be typed in as follows: 1
A correct answer is \( 5 \), which can be typed in as follows: 5
A correct answer is \( \left[\begin{array}{ccc} -1 & -1 & 5 \end{array}\right] \).
A correct answer is \( \left[\begin{array}{ccc} -5 & 10 & -5 \end{array}\right] \).
A correct answer is \( \left[\begin{array}{ccc} 2 & -4 & -2 \end{array}\right] \).
A correct answer is \( \left[\begin{array}{ccc} -2 & 4 & 2 \end{array}\right] \).
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Question 10
Correct
Mark 6.00 out of 6.00
The Scalar Triple Product
Definition: For \(\mathbf{u} = [u_1, u_2, u_3]\), \(\mathbf{v}= [v_1, v_2, v_3]\), and \(\mathbf{w}= [w_1, w_2, w_3]\), the number given by \
[\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})\] is called a scalar triple product of
\(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\).
Note:
\[\begin{aligned} \mathbf{u}\cdot(\mathbf{v}\times\mathbf{w}) &= [u_1, u_2, u_3] \cdot \left( \left[ \, \begin{vmatrix} v_2 & v_3\\ w_2 &
w_3 \end{vmatrix}, - \begin{vmatrix} v_1 & v_3\\ w_1 & w_3 \end{vmatrix}, \begin{vmatrix} v_1 & v_2\\ w_1 & w_2 \end{vmatrix} \, \right] \right)
\; \; \; \text{(by definition of} \; \; \mathbf{v} \times \mathbf{w} )\\ &= u_1\begin{vmatrix} v_2 & v_3\\ w_2 & w_3 \end{vmatrix}-
u_2\begin{vmatrix} v_1 & v_3\\ w_1 & w_3 \end{vmatrix} +u_3\begin{vmatrix} v_1 & v_2\\ w_1 & w_2 \end{vmatrix}\\ &= \begin{vmatrix} u_1 &
u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix} \; \; \; \hspace{6cm}\text{(by cofactor expansion along row 1)}\end{aligned}\]
Thus, we can compute \(\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})\) using the formula \[\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})
= \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix}\]
Example:
Compute the following scalar triple product using the determinant formula above.
\(\qquad [1,2,1]\cdot([2,0,3]\times[-1,1,0])=\) -7
.
The scalar triple product gives us a geometric description of the determinant of \(3 \times 3\) matrices:
Theorem:
Let \(\mathbf{u}\), \(\mathbf{v}\) and \(\mathbf{w}\) be vectors in \(\mathbb{R}^3\). The volume \(\mathcal{V}\) of the
parallelepiped in \(\mathbb{R}^3\) created by the vectors \(\mathbf{u}\), \(\mathbf{v}\) and \(\mathbf{w}\) is given by the absolute value of \
(\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})\), that is, \[\mathcal{V} = |\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})|\]
Example:
The volume \(\mathcal{V}\) of the parallelepiped formed by \(\mathbf{u}=[1,2,1]\), \(\mathbf{v}=[2,0,3]\), and \(\mathbf{w} = [-1,1,0]\) is \
(\mathcal{V} = |\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})| =\) 7
.
This geometric interpretation of the scalar triple product gives us a method to test whether or not three vectors in \(\mathbb{R}^3\) lie in the
same plane:
Theorem:
Let \(\mathbf{u}\), \(\mathbf{v}\) and \(\mathbf{w}\) be vectors in \(\mathbb{R}^3\).
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Question 11
Correct
Mark 1.00 out of 1.00
Then \(\mathbf{u}\), \(\mathbf{v}\) and \(\mathbf{w}\) lie in the same plane if and only if \(\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w}) =\)
0
.
Example: Do the vectors \(\mathbf{u} = [1,2,1]\), \(\mathbf{v} = [2,0,3]\) and \(\mathbf{w} = [-1,1,0]\) lie in the same plane?
, since \(\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w}) = \)
-7
.
No
≠ 0
Correct
Marks for this submission: 6.00/6.00.
Area of a Parallelogram in \(\mathbb{R}^2\)
The determinant of a \(2 \times 2\) matrix also has a nice geometric interpretation.
Theorem:
Let \(\mathbf{u} = [u_1,u_2]\) and \(\mathbf{v} = [v_1, v_2]\) be vectors in \(\mathbb{R}^2\). Then the area \(\mathcal{A}\) of the
parallelogram in \(\mathbb{R}^2\) determined by \(\mathbf{u}\) and \(\mathbf{v}\) is given by the absolute value of \(\det \begin{bmatrix}u_1
& u_2\\v_1&v_2\end{bmatrix}\), that is,
\[ \mathcal{A} = \left| \, \det \begin{bmatrix}u_1 & u_2\\v_1&v_2\end{bmatrix} \, \right | \]
Example: Find the area \(\mathcal{A}\) of the parallelogram in \(\mathbb{R}^2\) determined by the vectors \(\mathbf{u} = [-1,4]\) and \
(\mathbf{v} = [3,-2]\).
\( \mathcal{A} = \left| \, \det \begin{bmatrix}-1&4\\3&-2\end{bmatrix} \, \right| =\) 10
Correct
Marks for this submission: 1.00/1.00.
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Question 12
Correct
Mark 8.00 out of 8.00
Properties of the Cross Product
Recall: \[\begin{aligned} \mathbf{u} \times \mathbf{v} &=\left| \begin{array}{ccc} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ u_1 & u_2 &
u_3 \\ v_1 & v_2 & v_3 \end{array} \right| \\ \\ &= \left[ \, \begin{vmatrix} u_2 & u_3\\ v_2 & v_3 \end{vmatrix}, -\begin{vmatrix} u_1 & u_3\\ v_1
& v_3 \end{vmatrix}, \begin{vmatrix} u_1 & u_2\\ v_1 & v_2 \end{vmatrix} \, \right], \end{aligned}\] When considering the properties of the
cross product below, keep in mind the properties of the determinant.
For the following statements, let \(\mathbf{u},\,\mathbf{v}\) and \(\mathbf{w}\) be vectors in \(\mathbb{R^3}\) and \(k\in\mathbb{R}\).
(a) \((k\mathbf{u})\times\mathbf{v}=\)
k
\
((\mathbf{u}\times\mathbf{v})=\mathbf{u}\times(k\mathbf{v})\).
\
(\hspace{2cm}\)
(What happens to the determinant above
if one row is multiplied by the scalar \(k\)?)
Correct answer, well done.
Marks for this submission: 1.00/1.00.
(b) \(\mathbf{u}\times(\mathbf{v}+\mathbf{w})=
(\mathbf{u}\times\mathbf{v})+(\mathbf{u}\times\mathbf{w})\).
(c) \((\mathbf{u}+\mathbf{v})\times\mathbf{w}=
(\mathbf{u}\times\mathbf{w})+(\mathbf{v}\times\mathbf{w})\).
(d) \(\mathbf{u}\times\mathbf{0}=\) 0
0
0
\
(=\mathbf{0}\times\mathbf{u}\).
Correct answer, well done.
Marks for this submission: 1.00/1.00.
(e) \(\mathbf{u}\parallel\mathbf{v} \,\, \Leftrightarrow \,\,
\mathbf{u}\times\mathbf{v}=\) 0
0
0
.
(Recall: \(\mathbf{u}\parallel\mathbf{v} \)
means that \(\mathbf{u}\) and \(\mathbf{v}\)
are parallel.)
(What happens to the determinant above if
\(\mathbf{u}\) and \(\mathbf{v}\) are parallel?)
Correct answer, well done.
Marks for this submission: 1.00/1.00.
(f) \(\mathbf{u}\cdot(\mathbf{u}\times\mathbf{v})=\) 0
and \
(\mathbf{v}\cdot(\mathbf{u}\times\mathbf{v})=\) 0
.
Thus, \(\mathbf{u}\times\mathbf{v}\) is to both \
(\mathbf{u}\) and \(\mathbf{v}\).
(What are the determinants of the associated
3x3 matrices for these scalar triple products?)
Correct answer, well done.
Marks for this submission: 1.00/1.00.
Correct answer, well done.
Marks for this submission: 1.00/1.00.
Correct answer, well done.
Marks for this submission: 1.00/1.00.
(g) For \(\mathbf{u}\nparallel\mathbf{v}\) the ordered triple of vectors \((\mathbf{u},\mathbf{v},\mathbf{u}\times\mathbf{v})\) has
right hand
orientation
. i.e. if using your right hand, you point your index finger in the direction of \(\mathbf{u}\) and your middle finger in the direction of
\(\mathbf{v}\) then \(\mathbf{u}\times\mathbf{v}\) will point in the same direction as your thumb if you point it so that it is perpendicular to
the plane made by your index and middle fingers.
orthogonal
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In the diagram below, the observer is viewing the plane spanned by \(\mathbf{u}\) and \(\mathbf{v}\) from slightly above the plane. The
direction of \(\mathbf{u}\times\mathbf{v}\) is in the same direction as the vector. (The red arrow is the one pointing
upwards in the diagram, and the purple down.)
Correct answer, well done.
Marks for this submission: 1.00/1.00.
(h) \(\mathbf{v}\times\mathbf{u}=\) \
((\mathbf{u}\times\mathbf{v})\).
\
(\hspace{3cm}\)
(What happens to the determinant
above
if you interchange row 2 and row 3?)
Correct answer, well done.
Marks for this submission: 1.00/1.00.
(i) The area \(\mathcal{A}\) of the parallelogram in \(\mathbb{R}^3\) determined by the vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by \[
\mathcal{A} = || \mathbf{u} \times \mathbf{v} ||\] Furthermore, if \(\theta\) is the angle between \(\mathbf{u}\) and \(\mathbf{v}\), then \
[\|\mathbf{u}\times\mathbf{v}\|=\|\mathbf{u}\|\,\|\mathbf{v}\|\,\sin\theta\]
red
-
A correct answer is \( k \), which can be typed in as follows: k
A correct answer is \( \left[\begin{array}{ccc} 0 & 0 & 0 \end{array}\right] \).
A correct answer is \( \left[\begin{array}{ccc} 0 & 0 & 0 \end{array}\right] \).
A correct answer is \( 0 \), which can be typed in as follows: 0
A correct answer is \( 0 \), which can be typed in as follows: 0
A correct answer is: "orthogonal"
A correct answer is: "red"
A correct answer is: "-"
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19/22
Question 13
Correct
Mark 9.00 out of 9.00
Example:
Let \(\mathbf{u}\) and \(\mathbf{v}\) be vectors in \(\mathbb{R}^3\) and suppose \(\mathbf{u} \times \mathbf{v} = [2,1,-3]\).
Compute \((4\mathbf{v} + \mathbf{u}) \times (2\mathbf{u})\).
\((4\mathbf{v} + \mathbf{u}) \times (2\mathbf{u})\)
\(=
((4\mathbf{v}) \times (2\mathbf{u})) + (\mathbf{u} \times
(\mathbf{2u}))\)
\(\big(\)by Property \(\big)\)
Correct answer, well
done.
Marks for this submission:
1.00/1.00.
\(=\) 8
\((\mathbf{v} \times \mathbf{u}) + 2 (\mathbf{u} \times
\mathbf{u})\)
\(\big(\)by Property \(\big)\)
Correct answer, well
done.
Marks for this submission:
1.00/1.00.
Correct answer, well
done.
Marks for this submission:
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\(=\)
-8
\((\mathbf{u} \times \mathbf{v}) + 2 (\mathbf{u}\times
\mathbf{u}) \)
\(\big(\)by Property \(\big)\)
Correct answer, well
done.
Marks for this submission:
1.00/1.00.
Correct answer, well
done.
Marks for this submission:
1.00/1.00.
\(=\)
-8
\((\mathbf{u} \times \mathbf{v}) + 2\)
0
0
0
\(\big(\)by Property \(\big)\)
Correct answer, well
done.
Marks for this submission:
1.00/1.00.
Correct answer, well
done.
Marks for this submission:
1.00/1.00.
Correct answer, well
done.
Marks for this submission:
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\(=\) -16
-8
24
Correct answer, well
done.
Marks for this submission:
1.00/1.00.
(c)
(a)
(h)
(e)
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Question 14
Correct
Mark 5.00 out of 5.00
A correct answer is: "(c)"
A correct answer is \( 8 \), which can be typed in as follows: 8
A correct answer is: "(a)"
A correct answer is \( -8 \), which can be typed in as follows: -8
A correct answer is: "(h)"
A correct answer is \( -8 \), which can be typed in as follows: -8
A correct answer is \( \left[\begin{array}{ccc} 0 & 0 & 0 \end{array}\right] \).
A correct answer is: "(e)"
A correct answer is \( \left[\begin{array}{ccc} -16 & -8 & 24 \end{array}\right] \).
Exercise:
Let \(\mathbf{u}=[3,-2,1]\), \(\mathbf{v}=[1,1,1]\) and \(\mathbf{w}=[2,-2,0]\).
The area of the parallelogram formed by \(\mathbf{u}\) and \(\mathbf{v}\) is \(\mathcal{A}=\sqrt{\,}\)
38
.
The volume of the parallelepiped formed by \(\mathbf{u},\,\mathbf{v}\) and \(\mathbf{w}\) is \(\mathcal{V}=\)
2
.
Given that the volume of a parallelepiped is the area of its base times its height, the height of the parallelepiped formed by \
(\mathbf{u},\,\mathbf{v}\) and \(\mathbf{w}\) when its base is viewed as the parallelogram formed by \(\mathbf{u}\) and \(\mathbf{v}\) is \(h=\)
2
\(/\sqrt{\,}\)
38
.
The parallelepiped formed by \(\mathbf{u}\) and \(\mathbf{v}\) and \(k\mathbf{u}\times\mathbf{v}\) will have the same volume as the
parallelepiped formed by \(\mathbf{u},\,\mathbf{v}\) and \(\mathbf{w}\) if \(k=\pm1/\)
19
.
Correct
Marks for this submission: 5.00/5.00.
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Question 15
Correct
Mark 7.00 out of 7.00
Using the Cross Product to Find the Equation of a Plane
We know from Property (f) that the cross product \(\mathbf{u} \times \mathbf{v}\) is orthogonal to both of the vectors \(\mathbf{u}\) and \
(\mathbf{v}\). Therefore, if \(\mathbf{u}\) and \(\mathbf{v}\) are a pair of direction vectors for a plane, then \(\mathbf{u} \times \mathbf{v}\) is a
normal vector for the plane.
An example similar to, yet different from, the one given below can be found in the video
Determinants: Properties and Applications 2 starting
at 7:40.
Example:
Consider the plane \(\mathcal{P}\) in \(\mathbb{R}^3\) that contains the three points \(\mathbf{p_1}=[1,0,1]\), \(\mathbf{p_2}=
[1,0,0]\) and \(\mathbf{p_3}=[0,1,0]\).
Let \(\mathbf{u} = \mathbf{p_2}-\mathbf{p_1} = \) 0
0
-1
and let \(\mathbf{v} = \mathbf{p_3}-\mathbf{p_1} =\)
-1
1
-1
.
Correct answer, well done.
Marks for this submission: 1.00/1.00.
Correct answer, well done.
Marks for this submission: 1.00/1.00.
Since \(\mathbf{u}\) and \(\mathbf{v}\) scalar multiples of each other, they parallel
and hence \(\mathbf{u}\) and \(\mathbf{v}\) serve as direction vectors for \(\mathcal{P}\).
Correct answer, well done.
Marks for this submission: 1.00/1.00.
Correct answer, well done.
Marks for this submission: 1.00/1.00.
Using the cross product of the vectors \(\mathbf{u}\) and \(\mathbf{v}\), we obtain that a normal vector for this plane is
\(\mathbf{n}= \mathbf{u} \times \mathbf{v} =\) 1
1
0
Correct answer, well done.
Marks for this submission: 2.00/2.00.
Now, using this and the fact that the point \(\mathbf{p_1}\) is in the plane we obtain that the general equation for this plane is
x+y-1=0
Correct answer, well done.
Marks for this submission: 1.00/1.00.
(Use \(x, y, z\) as the variables in your computation for this equation.)
There is yet another way to use the determinant to obtain the general equation of a plane in \(\mathbb{R}^3\). It essentially amounts to the
same calculation as with the cross product above, but the argument is different: see the first half of the video Determinants: Properties and
Applications 2
.
are not
are not
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A correct answer is \( \left[\begin{array}{ccc} 0 & 0 & -1 \end{array}\right] \).
A correct answer is \( \left[\begin{array}{ccc} -1 & 1 & -1 \end{array}\right] \).
A correct answer is: "are not"
A correct answer is: "are not"
A correct answer is \( \left[\begin{array}{ccc} 1 & 1 & 0 \end{array}\right] \).
A correct answer is \( y+x=1 \), which can be typed in as follows: y+x = 1
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