Eigenvalues, Eigenvectors, and Eigenspaces_ Part 2_ Attempt review _ eClass

3/27/24, 1:24 AM Eigenvalues, Eigenvectors, and Eigenspaces: Part 2: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=15222924&cmid=7601911 1/14 Started on Monday, 25 March 2024, 9:11 PM State Finished Completed on Monday, 25 March 2024, 10:06 PM Time taken 55 mins 38 secs Grade 66.00 out of 66.00 ( 100 %)

3/27/24, 1:24 AM Eigenvalues, Eigenvectors, and Eigenspaces: Part 2: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=15222924&cmid=7601911 2/14 Question 1 Correct Mark 5.00 out of 5.00 The Relationship between the Algebraic and Geometric Multiplicities of a Matrix Recall the following definitions for an matrix . When considering the characteristic polynomial of , we view as a variable. Definition: The algebraic multiplicity of an eigenvalue of , denoted , is the number of times appears as a root of the characteristic polynomial of . Equivalently, it is the number of times appears as a factor of . Definition: Let be an eigenvalue of (so here is viewed as a fixed number). The geometric multiplicity of , denoted , is the dimension of the eigenspace corresponding to . If is an eigenvalue of a square matrix , then we have the following inequalities involving the algebraic multiplicity and the geometric multiplicity : Theorem: When is an eigenvalue, there must be at least one non-zero eigenvector, hence the eigenspace is non-zero and its dimension is at least one: this explains the first inequality. The explanation why the second inequality is true is more complicated and will not be provided in this course. It is possible for to be strictly less than , as the next example shows. Example Let The characteristic polynomial of is , so is an eigenvalue with algebraic multiplicity 2  , and is an eigenvalue with algebraic multiplicity 1  . Without further calculations, we may determine from the inequalities that the geometric multiplicity of is 1  . For the other eigenvalue, , all that the inequalities tell us is that 2  . However, after some row operations, we find that the matrix has rank , so the dimension of its null space is 1  , showing that the geometric multiplicity of is 1  . By computing reduced row-echelon forms, we find that the eigenspace associated to is spanned by and the eigenspace associated to is spanned by . This example illustrates the following important case of the theorem above. Important special case: If the algebraic multiplicity of an eigenvalue is 1, so , then the inequality in the theorem above reduces to This can only be true when . Correct Marks for this submission: 5.00/5.00.

3/27/24, 1:24 AM Eigenvalues, Eigenvectors, and Eigenspaces: Part 2: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=15222924&cmid=7601911 4/14 Question 3 Correct Mark 10.00 out of 10.00 Example: Let \(A = {\left[\begin{array}{ccc} 3 & 0 & 0 \\ 5 & 8 & 0 \\ 6 & 7 & -12 \end{array}\right]}\). The eigenvalues of \(A\) are 3 , 8 , and -12 . Correct answer, well done. Marks for this submission: 1.00/1.00. Since \(\lambda = 0\) an eigenvalue of \(A\), the matrix \(A\) invertible. Correct answer, well done. Marks for this submission: 1.00/1.00. Correct answer, well done. Marks for this submission: 1.00/1.00. Let's verify this conclusion by computing the determinant of \(A\). Since \(\det(A) = \) -288 , the matrix \(A\) invertible. 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. Example: Let \(B = {\left[\begin{array}{ccc} -1 & 0 & 1 \\ 2 & 6 & -14 \\ 1 & 0 & -1 \end{array}\right]}\). The eigenvalues of \(B\) are -2 , 6 , and 0 . Correct answer, well done. Marks for this submission: 2.00/2.00. Since \(\lambda = 0\) an eigenvalue of \(B\), the matrix \(B\) invertible. Correct answer, well done. Marks for this submission: 1.00/1.00. Correct answer, well done. Marks for this submission: 1.00/1.00. is not is ≠ 0 is is is not A correct answer is \( 3 \), which can be typed in as follows: 3 A correct answer is \( 8 \), which can be typed in as follows: 8 A correct answer is \( -12 \), which can be typed in as follows: -12 A correct answer is: "is not" A correct answer is: "is" A correct answer is \( -288 \), which can be typed in as follows: -288 A correct answer is: "≠ 0" A correct answer is: "is"

3/27/24, 1:24 AM Eigenvalues, Eigenvectors, and Eigenspaces: Part 2: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=15222924&cmid=7601911 5/14 Question 4 Correct Mark 2.00 out of 2.00 A correct answer is \( -2 \), which can be typed in as follows: -2 A correct answer is \( 0 \), which can be typed in as follows: 0 A correct answer is \( 6 \), which can be typed in as follows: 6 A correct answer is: "is" A correct answer is: "is not" The Invertible Matrix Theorem revisited Based upon the material covered in Block 5, we may expand the Invertible Matrix Theorem to the following: The Invertible Matrix Theorem The following statements are equivalent for an \(n \times n\) matrix \(A\). This means that if one is true, then all the others are true; if one is false, then all the others are false. \(\bullet\) \(A\) is invertible. \(\bullet\) The equation \(A\mathbf{x} = \mathbf{0}\) has only the trivial solution. \(\bullet\) The equation \(A\mathbf{x} = \mathbf{b}\) has a solution for every \(\mathbf{b} \in \mathbb{R}^n\). \(\bullet\) The equation \(A\mathbf{x} = \mathbf{b}\) has a unique solution for every \(\mathbf{b} \in \mathbb{R}^n\). \(\bullet\) The reduced row-echelon form of \(A\) is \(I_n\). \(\bullet\) A row echelon form of \( A\) has \( n \) non-zero rows. \(\bullet\) The nullity of \(A\) is \(0\). \(\bullet\) The rank of \(A\) is \(n\). \(\bullet\) The columns of \(A\) are linearly independent. \(\bullet\) The columns of \(A\) span \(\mathbb{R}^n\). \(\bullet\) The columns of \(A\) form a basis for \(\mathbb{R}^n\). \(\bullet\) The rows of \(A\) are linearly independent. \(\bullet\) The rows of \(A\) span \(\mathbb{R}^n\). \(\bullet\) The rows of \(A\) form a basis for \(\mathbb{R}^n\). \(\bullet\) \(\text{det}(A)\)  . \(\bullet\) \(0\)  an eigenvalue of \(A\) ≠ 0 is not Correct Marks for this submission: 2.00/2.00.

3/27/24, 1:24 AM Eigenvalues, Eigenvectors, and Eigenspaces: Part 2: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=15222924&cmid=7601911 7/14 Recall from Question 10 of the learning activity "Eigenvalues, Eigenvectors, and Eigenspaces: Part 1" that the eigenvalues of \(A\) are \(\lambda = 0\), \(\lambda = 1\), and \(\lambda = 2\), and the corresponding eigenspaces are \[ E_0 = \operatorname{Span}\left( \mathbf{u} = \begin{bmatrix}0\\0\\1\end{bmatrix} \right), \; \; E_1 = \operatorname{Span}\left( \mathbf{v} = \begin{bmatrix}-1\\0\\1\end{bmatrix} \right), \] \[ \text{and} \; \; E_2 = \operatorname{Span}\left( \mathbf{w} = \begin{bmatrix}-2\\-1\\1\end{bmatrix} \right) \] Therefore, since \(\mathbf{u} \in E_0\), \(\mathbf{v} \in E_1\), and \(\mathbf{w} \in E_2\) are eigenvectors of \(A\) corresponding to different eigenvalues, the set \[ \left\{ \begin{bmatrix}0\\0\\1\end{bmatrix}, \begin{bmatrix}-1\\0\\1\end{bmatrix}, \begin{bmatrix}-2\\-1\\1\end{bmatrix}\right\}\] is  . linearly independent Correct Marks for this submission: 6.00/6.00.

3/27/24, 1:24 AM Eigenvalues, Eigenvectors, and Eigenspaces: Part 2: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=15222924&cmid=7601911 8/14 Question 6 Correct Mark 7.00 out of 7.00 Matrices with no Repeated Eigenvalues Suppose \(A\) is an \(n \times n\) matrix. Since its characteristic polynomial \( C_A(\lambda) \) has degree \( n\), a basic fact from algebra tells us that \( C_A(\lambda) \) has at most \(n \) roots. At one extreme, it may not have any root in the set of real numbers \( \mathbb{R}\). This is the case, for instance, of the matrix \( \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \) whose characteristic polynomial is \( \lambda^2 +1 \). It can also have just one root of multiplicity \( n\): for instance, if \( A \) is a triangular matrix with all its diagonal entries equal to \( \lambda_1 \), then \( C_A(\lambda) = (-1)^n (\lambda-\lambda_1)^n.\) Another extreme possibility is when \( A\) has \(n\) distinct eigenvalues \(\lambda_1,\ldots,\lambda_n\). Because \(A\) has size \(n \times n\), the characteristic polynomial \(C_A(\lambda)\) has degree n  . On the other hand, \(\quad C_A(\lambda) = (-1)^n(\lambda - \lambda_1)^{m_1}(\lambda - \lambda_2)^{m_2} \cdots (\lambda - \lambda_n)^{m_n} ,\) where \(m_i \geq 1\) is the algebraic multiplicity of \(\lambda_i\) (that is, \(m_i = \operatorname{alg}(\lambda_i)\)). Therefore, since \ (C_A(\lambda)\) has degree n  , and the polynomial \( (-1)^n(\lambda - \lambda_1)(\lambda - \lambda_2)\cdots (\lambda - \lambda_n) \) has the same degree, it follows that \( m_i = \,\) 1  for all \(i \in \{1,\ldots,n\}\). Hence, by the inequalities \(1 \leq \operatorname{geo}(\lambda_i) \leq m_i\), we have \(\operatorname{geo}(\lambda_i)= \,\) 1  . This means that every eigenspace of \(A\) has dimension 1  . Suppose that \(\mathbf{v}_i\) is an eigenvector of \(A\) with eigenvalue \(\lambda_i\), for \(i \in \{1,\ldots,n\}\). Because these eigenvectors all correspond to different eigenvalues, they are linearly independent by Question 5. Hence, they form \(n\) linearly independent vectors in \ (\mathbb{R}^n\), so in fact they form a basis for \(\mathbb{R}^n\). This proves the following theorem. Theorem If \(A\) is an \(n \times n\) matrix with \(n\) distinct eigenvalues, then \(\bullet\) \(A\) has \(n\) eigenspaces, each of dimension \(1\), \(\bullet\) \(\mathbb{R}^n\) has a basis consisting of eigenvectors, one basis vector from each eigenspace. The \( 3\times 3 \) matrix in Question 10 in Part 1 had 3 distinct eigenvalues. Here is another example of such a matrix. Example The \(3 \times 3\) matrix \(\quad A = \begin{bmatrix}6 & 13 & -7 \\ -1 & -2 & 1 \\ 1 & 1 & -2\end{bmatrix} \) has characteristic polynomial \( C_A(\lambda) = -(\lambda + 1)(\lambda - 1)(\lambda - 2)\) and therefore has 3  distinct eigenvalues, namely, -1,1,2  (enter your answer as a list of numbers in increasing numerical order, separated by commas but no spaces). By row reducing \(A - \lambda_i I\), where \(\lambda_1,\lambda_2,\lambda_3\) are the eigenvalues of \(A\) in increasing numerical order, we find that eigenvectors for the three eigenvalues are, respectively, \(\quad \mathbf{v}_1 = \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix} ,\quad \mathbf{v}_2 = \begin{bmatrix}4 \\ -1 \\ 1\end{bmatrix} ,\quad \mathbf{v}_3 = \begin{bmatrix}5 \\ -1 \\ 1\end{bmatrix} .\) These three vectors therefore form a basis of \(\mathbb{R}^3\) consisting of eigenvectors of \(A\). Correct Marks for this submission: 7.00/7.00.

3/27/24, 1:24 AM Eigenvalues, Eigenvectors, and Eigenspaces: Part 2: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=15222924&cmid=7601911 10/14 Question 8 Correct Mark 7.00 out of 7.00 Example: Let \(A = {\left[\begin{array}{ccc} -1 & -1 & 1 \\ 0 & -2 & 0 \\ 2 & -2 & 0 \end{array}\right]}\). Recall from Question 9 of the learning activity "Eigenvalues, Eigenvectors, and Eigenspaces: Part 1" that \(\lambda = -2\) is an eigenvalue of \ (A\) and the corresponding eigenspace is \[E_{-2} = \operatorname{Span}\left( {\left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]}\; ,\; {\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]}\right)\] Note: The eigenvalues of \(A\) are \(\lambda = 1\) and \(\lambda = -2\). Since \(\lambda =\) 0 is not an eigenvalue of \(A\), we know that \(A\) is invertible. Correct answer, well done. Marks for this submission: 1.00/1.00. (a) Compute \(A^7{\left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]}\). \(A^7{\left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]} = (-2)^7{\left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]} =\) -128 -128 0 Correct answer, well done. Marks for this submission: 1.00/1.00. (b) Compute \(A^{-4}{\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]}\). \(A^{-4} {\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]} = \dfrac{1}{(-2)^{4}} {\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]} =\) -1/16 0 1/16 Correct answer, well done. Marks for this submission: 1.00/1.00. (c) Compute \(A^3{\left[\begin{array}{c} -3 \\ 2 \\ 5 \end{array}\right]}\). Since \({\left[\begin{array}{c} -3 \\ 2 \\ 5 \end{array}\right]} = \) 2 \({\left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]} + \) 5 \ ({\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]} \in E_{-2}\), the vector \({\left[\begin{array}{c} -3 \\ 2 \\ 5 \end{array}\right]}\) is an eigenvector of \(A\) with eigenvalue -2 . 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. Thus, \(A^3{\left[\begin{array}{c} -3 \\ 2 \\ 5 \end{array}\right]} =\) 24 -16 -40 Correct answer, well done. Marks for this submission: 1.00/1.00.

3/27/24, 1:24 AM Eigenvalues, Eigenvectors, and Eigenspaces: Part 2: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=15222924&cmid=7601911 11/14 A correct answer is \( 0 \), which can be typed in as follows: 0 A correct answer is \( \left[\begin{array}{c} -128 \\ -128 \\ 0 \end{array}\right] \). A correct answer is \( \left[\begin{array}{c} -\frac{1}{16} \\ 0 \\ \frac{1}{16} \end{array}\right] \). A correct answer is \( 2 \), which can be typed in as follows: 2 A correct answer is \( 5 \), which can be typed in as follows: 5 A correct answer is \( -2 \), which can be typed in as follows: -2 A correct answer is \( \left[\begin{array}{c} 24 \\ -16 \\ -40 \end{array}\right] \).

3/27/24, 1:24 AM Eigenvalues, Eigenvectors, and Eigenspaces: Part 2: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=15222924&cmid=7601911 13/14 Question 10 Correct Mark 12.00 out of 12.00 Linear Combinations of Eigenvectors Working with eigenvectors can greatly simplify computations, since matrix multiplication with an eigenvector is just scalar multiplication. As we will see in the following theorem, there are also computational advantages to working with linear combinations of eigenvectors. Theorem: Let \(A\) be an \(n \times n\) matrix and let \(\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m\) be eigenvectors of \(A\) corresponding to the eigenvalues \(\lambda_1, \lambda_2, \ldots \lambda_m\), respectively. If \(\mathbf{x}\) is a linear combination of these eigenvectors, say \[\mathbf{x} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_m \mathbf{v}_m,\] then for any non-negative integer \(k\) (or any integer if \(A\) is invertible), \[A^k \mathbf{x} = c_1\lambda_1^k\mathbf{v}_1 + c_2\lambda_2^k\mathbf{v}_2 + \cdots + c_m \lambda_m^k\mathbf{v}_m\] Let's illustrate how to use this theorem through an example. Example: Let \(A = {\left[\begin{array}{ccc} -1 & -1 & 1 \\ 0 & -2 & 0 \\ 2 & -2 & 0 \end{array}\right]}\). Recall from Question 9 of the learning activity "Eigenvalues, Eigenvectors, and Eigenspaces: Part 1" that the eigenvalues of \(A\) are \(\lambda = 1\) and \(\lambda = -2\) and the corresponding eigenspaces are \[ E_1 = \operatorname{Span}\left( \mathbf{u} = {\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right]}\right) \; \; \text{and} \; \; \; \; E_{-2} = \operatorname{Span}\left(\mathbf{v}= {\left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]}\; ,\;\mathbf{w} = {\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]}\right)\] Note: For any integer \(k\), since \(\mathbf{u} \in E_1\), we have \(A^k\mathbf{u} = (1)^k\mathbf{u} = \mathbf{u}\) and since \(\mathbf{v}, \mathbf{w} \in E_{-2}\), we have \(A^k\mathbf{v} = (-2)^k\mathbf{v}\) and \(A^k\mathbf{w} = (-2)^k\mathbf{w}\). (a) Compute \(A^3{\left[\begin{array}{c} -2 \\ -4 \\ 13 \end{array}\right]}\). We first note that \({\left[\begin{array}{c} -2 \\ -4 \\ 13 \end{array}\right]}\) can be written as a linear combination of the eigenvectors \ (\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\). Indeed, \[{\left[\begin{array}{c} -2 \\ -4 \\ 13 \end{array}\right]} = 5 {\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right]} - 4{\left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]} + 3{\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]}\] Thus, \(A^3{\left[\begin{array}{c} -2 \\ -4 \\ 13 \end{array}\right]}\) \ (=\) \( A^3\left(5 {\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right]} - 4{\left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]} + 3{\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]}\right)\) \ (=\) \( A^3\left(5 {\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right]}\right) - A^3 \left(4{\left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]}\right) + A^3\left(3{\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]}\right)\) \ (=\) \( 5A^3{\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right]} - 4A^3{\left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]} + 3A^3{\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]}\) \ (=\) \( 5(\) 1 \()^3{\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right]} - 4(\) -2 \()^3{\left[\begin{array} {c} 1 \\ 1 \\ 0 \end{array}\right]} + 3(\) -2 \()^3{\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]}\), since \(\mathbf{u} \in E_1\) and \(\mathbf{v}, \mathbf{w} \in E_{-2}\) 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. \ (=\) 5 \({\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right]}\) \( + \) 32 \({\left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]}\) \( - \) 24 \({\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]}\)

3/27/24, 1:24 AM Eigenvalues, Eigenvectors, and Eigenspaces: Part 2: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=15222924&cmid=7601911 14/14 Thus, \(A^3{\left[\begin{array}{c} -2 \\ -4 \\ 13 \end{array}\right]}\) \ (=\) \( A^3\left(5 {\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right]} - 4{\left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]} + 3{\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]}\right)\) 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. \ (=\) 61 32 -14 Correct answer, well done. Marks for this submission: 1.00/1.00. (b) Compute \(A^5{\left[\begin{array}{c} 1 \\ -2 \\ 6 \end{array}\right]}\). We first note that \({\left[\begin{array}{c} 1 \\ -2 \\ 6 \end{array}\right]}\) can be written as a linear combination of the eigenvectors \ (\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\). Indeed, \({\left[\begin{array}{c} 1 \\ -2 \\ 6 \end{array}\right]} =\) 3 \({\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right]} - \) 2 \ ({\left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]} + \) 0 \({\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]}\) 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. Thus, \( A^5{\left[\begin{array}{c} 1 \\ -2 \\ 6 \end{array}\right]} = \) 67 64 6 Correct answer, well done. Marks for this submission: 2.00/2.00. A correct answer is \( 1 \), which can be typed in as follows: 1 A correct answer is \( -2 \), which can be typed in as follows: -2 A correct answer is \( -2 \), which can be typed in as follows: -2 A correct answer is \( 5 \), which can be typed in as follows: 5 A correct answer is \( 32 \), which can be typed in as follows: 32 A correct answer is \( 24 \), which can be typed in as follows: 24 A correct answer is \( \left[\begin{array}{c} 61 \\ 32 \\ -14 \end{array}\right] \). A correct answer is \( 3 \), which can be typed in as follows: 3 A correct answer is \( 2 \), which can be typed in as follows: 2 A correct answer is \( 0 \), which can be typed in as follows: 0 A correct answer is \( \left[\begin{array}{c} 67 \\ 64 \\ 6 \end{array}\right] \).