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15. If [tex]\(|a| = 4\)[/tex], [tex]\(|b| = \sqrt{3}\)[/tex], and [tex]\(a, b = -3\)[/tex], the magnitude of [tex]\(a - b\)[/tex] is equal to:
A. 7
B. 5
C. [tex]\(\sqrt{13}\)[/tex]
D. [tex]\(\sqrt{22}\)[/tex]

16. If [tex]\(A\)[/tex] is perpendicular to [tex]\(B\)[/tex], then the cosine of the angle between [tex]\(A\)[/tex] and [tex]\(A - B\)[/tex] is:
A. [tex]\(\frac{|A - B|}{|A|}\)[/tex]
B. [tex]\(\frac{|A - B|}{|B|}\)[/tex]
C. [tex]\(\frac{|A|}{|A - B|}\)[/tex]
D. [tex]\(\frac{|B|}{|A - B|}\)[/tex]

17. If [tex]\(\theta\)[/tex] is the angle between the vector [tex]\(A = 2i - j\)[/tex] and [tex]\(B = i + j\)[/tex], then [tex]\(\cos \theta\)[/tex] is equal to:
A. [tex]\(\frac{1}{2}\)[/tex]
B. [tex]\(\frac{\sqrt{3}}{2}\)[/tex]
C. [tex]\(\frac{2}{\sqrt{5}}\)[/tex]
D. [tex]\(\frac{1}{\sqrt{10}}\)[/tex]

18. Which of the following is a vector parallel and in the opposite direction to [tex]\(A = 3\)[/tex]?
A. [tex]\(\frac{3}{2}i - 2j\)[/tex]
B. [tex]\(-\frac{3}{2}i + 2j\)[/tex]
C. [tex]\(-6i - 8j\)[/tex]
D. [tex]\(6i - 8j\)[/tex]

19. Let [tex]\(|a|\)[/tex] denote the length of vector [tex]\(a\)[/tex]. Suppose [tex]\(|a| = 3\)[/tex], [tex]\(b\)[/tex] is not a vector, and the angle between [tex]\(a\)[/tex] and [tex]\(b\)[/tex] is [tex]\(\theta = \frac{\pi}{3}\)[/tex]. What is the magnitude of [tex]\(|a - 2b|\)[/tex]?
A. [tex]\(\sqrt{5}\)[/tex]
B. [tex]\(\sqrt{7}\)[/tex]
C. [tex]\(\sqrt{8}\)[/tex]
D. [tex]\(\sqrt{19}\)[/tex]

20. If [tex]\(v = (2, -2\sqrt{3})\)[/tex] is a positive vector, then what are the magnitude and direction, respectively, of the vector?
A. [tex]\(-4, 60^{\circ}\)[/tex]
B. [tex]\(4, 300^{\circ}\)[/tex]
C. [tex]\(12, -60^{\circ}\)[/tex]
D. [tex]\(12, 300^{\circ}\)[/tex]

21. If [tex]\(u = i + 3j + 4k\)[/tex] and [tex]\(v = 2i + 7j - 5k\)[/tex], then the cross product [tex]\(u \times v\)[/tex] is equal to:
A. [tex]\(43i + 13j + k\)[/tex]
B. [tex]\(43i - 13j + k\)[/tex]
C. [tex]\(-43i + 13j + k\)[/tex]
D. [tex]\(-43i + 13j - k\)[/tex]

22. If [tex]\(u = 3i + 4j + 5k\)[/tex] and [tex]\(v = 7i + 8j + 9k\)[/tex], then the magnitude of [tex]\(u \times v\)[/tex] is equal to:
A. [tex]\(-4\sqrt{6}\)[/tex]
B. [tex]\(4\sqrt{6}\)[/tex]
C. [tex]\(-2\sqrt{6}\)[/tex]
D. [tex]\(2\sqrt{6}\)[/tex]

23. Let [tex]\(u\)[/tex] and [tex]\(v\)[/tex] be two vectors with [tex]\(|u| = 3\)[/tex] and [tex]\(|v| = 6\)[/tex], and [tex]\(\theta = \frac{\pi}{3}\)[/tex]. Find [tex]\(|3u - 2v|\)[/tex].

24. Let [tex]\(u\)[/tex] and [tex]\(v\)[/tex] be two vectors with [tex]\(|u| = 7\)[/tex] and [tex]\(|v| = 3\)[/tex], and [tex]\(\theta = \frac{\pi}{3}\)[/tex]. Find [tex]\(|u + v|\)[/tex].

25. If [tex]\(u = (3, 6)\)[/tex], [tex]\(v = (8, 4)\)[/tex], and [tex]\(w = (2, 7)\)[/tex], then [tex]\(u - (v - w) =\)[/tex].

26. Let [tex]\(A = \begin{pmatrix} 3 & x - 1 \\ 1 & x \end{pmatrix}\)[/tex]. If [tex]\(\det(A) = 0\)[/tex], determine the values of [tex]\(x\)[/tex].

27. Using Cramer's rule (if possible), solve the following system of linear equations:
[tex]\[
\begin{cases}
2x + 3y - 5z = 1 \\
x + y - z = 2 \\
2y + z = 8
\end{cases}
\][/tex]

Sagot :

Let's tackle question 26:

Given matrix [tex]\( A \)[/tex] is:
[tex]\[ A = \begin{pmatrix} 3 & x-1 \\ 1 & x \end{pmatrix} \][/tex]

We need to find the values of [tex]\( x \)[/tex] such that [tex]\(\operatorname{det}(A) = 0\)[/tex].

Step-by-Step Solution:

1. Write the Determinant Formula:

The determinant of a [tex]\( 2 \times 2 \)[/tex] matrix
[tex]\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \][/tex]
is calculated using the formula:
[tex]\[ \text{det} = ad - bc \][/tex]

2. Apply the Formula to Our Matrix:

Given our matrix [tex]\( A = \begin{pmatrix} 3 & x-1 \\ 1 & x \end{pmatrix} \)[/tex], we identify:
[tex]\[ a = 3, \quad b = x-1, \quad c = 1, \quad d = x \][/tex]

The determinant of [tex]\( A \)[/tex] is therefore:
[tex]\[ \text{det}(A) = (3)(x) - (1)(x-1) \][/tex]

3. Simplify the Determinant Expression:

[tex]\[ \text{det}(A) = 3x - (x - 1) = 3x - x + 1 = 2x + 1 \][/tex]

4. Set the Determinant to Zero:

[tex]\[ 2x + 1 = 0 \][/tex]

5. Solve for [tex]\( x \)[/tex]:

[tex]\[ 2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2} \][/tex]

Thus, the value of [tex]\( x \)[/tex] that makes the determinant of matrix [tex]\( A \)[/tex] equal to zero is:
[tex]\[ x = -\frac{1}{2} \][/tex]