Consider the planes given by the equations x + 2y + 3z = 4 3x + 4y + 5z = 2 5x + 4y + (k - 1)z = k – 8. Writing these as an augmented matrix and applying Gauss-elimination, we obtain the following augment matrix in row echelon form: 1 2 4 1 2 5 (k – 2)(k + 2) |k+2 This augmented matrix has infinitely many solutions when: A. k = 2. B. k + 2 and k + -2. C. k = -2 The planes intersect in a line when the augmented matrix has infinitely many solutions. Substituting the value of k for which the augment matrix has infinitely many solutions, the solution of the augmented matrix is: D. (x, y, z) = (-6 +t, 5 – 2t, t). E. (x, y, z) = (t, 5 – 2t, –6+ t).

Linear Algebra: A Modern Introduction
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ISBN:9781285463247
Author:David Poole
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Chapter6: Vector Spaces
Section6.2: Linear Independence, Basis, And Dimension
Problem 3AEXP
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Consider the planes given by the equations 

Consider the planes given by the equations
r + 2y + 3z = 4
3x + 4y + 5z = 2
5x + 4y + (k –- 1)z = k – 8.
Writing these as an augmented matrix and applying Gauss-elimination, we obtain the following augment matrix in row echelon form:
1
2
4
1
2
0 (k – 2)(k + 2) k+ 2
This augmented matrix has infinitely many solutions when:
A. k = 2.
B. k + 2 and k + -2.
C. k = -2
The planes intersect in a line when the augmented matrix has infinitely many solutions. Substituting the value of k for which the
augment matrix has infinitely many solutions, the solution of the augmented matrix is:
D. (x, y, z) = (-6+t, 5 – 2t, t).
E. (x, y, z) = (t, 5 – 2t, –6+ t).
Transcribed Image Text:Consider the planes given by the equations r + 2y + 3z = 4 3x + 4y + 5z = 2 5x + 4y + (k –- 1)z = k – 8. Writing these as an augmented matrix and applying Gauss-elimination, we obtain the following augment matrix in row echelon form: 1 2 4 1 2 0 (k – 2)(k + 2) k+ 2 This augmented matrix has infinitely many solutions when: A. k = 2. B. k + 2 and k + -2. C. k = -2 The planes intersect in a line when the augmented matrix has infinitely many solutions. Substituting the value of k for which the augment matrix has infinitely many solutions, the solution of the augmented matrix is: D. (x, y, z) = (-6+t, 5 – 2t, t). E. (x, y, z) = (t, 5 – 2t, –6+ t).
Writing these as an augmented matrix and applying Gauss-elimination, we obtain the following augment matrix in row echelon form:
1
2
3
4
0 1
0 0 (k – 2)(k+ 2) | k + 2
This augmented matrix has infinitely many solutions when:
A. k = 2.
B. k + 2 and k # -2.
C. k = -2
The planes intersect in a line when the augmented matrix has infinitely many solutions. Substituting the value of k for which the
augment matrix has infinitely many solutions, the solution of the augmented matrix is:
D. (x, y, z) = (-6+t, 5 – 2t, t).
E. (г, у, 2) — (t, 5 — 2t, —6 + t).
F. (х, у, 2) 3D (5 — 2t, t, —6 + t).
Then the vector equation of the line where the planes intersect is:
G. r = (5, 0, –6)+ t(-2,1, 1).
H. r = (-6, 5, 0) +t(1, –2, 1).
1. т 3 (0,5, —6) + t(1,—2, 1).
Transcribed Image Text:Writing these as an augmented matrix and applying Gauss-elimination, we obtain the following augment matrix in row echelon form: 1 2 3 4 0 1 0 0 (k – 2)(k+ 2) | k + 2 This augmented matrix has infinitely many solutions when: A. k = 2. B. k + 2 and k # -2. C. k = -2 The planes intersect in a line when the augmented matrix has infinitely many solutions. Substituting the value of k for which the augment matrix has infinitely many solutions, the solution of the augmented matrix is: D. (x, y, z) = (-6+t, 5 – 2t, t). E. (г, у, 2) — (t, 5 — 2t, —6 + t). F. (х, у, 2) 3D (5 — 2t, t, —6 + t). Then the vector equation of the line where the planes intersect is: G. r = (5, 0, –6)+ t(-2,1, 1). H. r = (-6, 5, 0) +t(1, –2, 1). 1. т 3 (0,5, —6) + t(1,—2, 1).
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