23) Given f(x) = x² + x-2, defined over [-3, -1]. Starting from xo= -3, the absolute error Appr. - Exact after the first generated root by Newton's method is: (A) 0.75 (C) 1 (D) None (B) 0.5

Linear Algebra: A Modern Introduction
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Chapter2: Systems Of Linear Equations
Section2.2: Direct Methods For Solving Linear Systems
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Q 23 please
19) Using Inverse method, Row 2 (R2) just after zeroing element A21 in (A/I) is:
(A) {0 1 -25/3 2/3}
(B) (0 18/3 1/3 2/3}
(C) (0 25/3-2/3 1}
(D) None
20) Using Inverse method, column 4 (C4) just after getting a unity diagonal is:
(A) {1/6 -2/25)T
(B) (0 3/25)
(C) (1/6 2/25)
(D) None
21) The firt column (C2) in the inverse matrix of A is:
(A) {3/50 -2/25)
(B) (4/25 -3/25) (C) (3/25 1)T (D) None
22) The exact solution for x2 is:
(A) 0.54
(B) 0.16
C) 0.57
(D) None
Problem 5: Using root finding methods answer question (23, 24 and 25).
23) Given f(x) = x² + x - 2, defined over [-3, -1]. Starting from xo= -3, the absolute
error Appr. - Exact after the first generated root by Newton's method is:
(A) 0.75
(B) 0.5
(C) 1
(D) None
24) Given f(x) = 7 -0.6x, By using Newton's method, the angel produced by the
tangent at any starting point close to the root is approximately:
(A)-40.18°
(B)-30.96
(C)-16.74
(D) None
25) Given f(x) = x² + x - 2. the fixed point using the fixed point method
(x = g(x)), where g(x) is a positive root squared function is:
(A)-2
(B) 1
(C)-2, 1
(D) None
Transcribed Image Text:19) Using Inverse method, Row 2 (R2) just after zeroing element A21 in (A/I) is: (A) {0 1 -25/3 2/3} (B) (0 18/3 1/3 2/3} (C) (0 25/3-2/3 1} (D) None 20) Using Inverse method, column 4 (C4) just after getting a unity diagonal is: (A) {1/6 -2/25)T (B) (0 3/25) (C) (1/6 2/25) (D) None 21) The firt column (C2) in the inverse matrix of A is: (A) {3/50 -2/25) (B) (4/25 -3/25) (C) (3/25 1)T (D) None 22) The exact solution for x2 is: (A) 0.54 (B) 0.16 C) 0.57 (D) None Problem 5: Using root finding methods answer question (23, 24 and 25). 23) Given f(x) = x² + x - 2, defined over [-3, -1]. Starting from xo= -3, the absolute error Appr. - Exact after the first generated root by Newton's method is: (A) 0.75 (B) 0.5 (C) 1 (D) None 24) Given f(x) = 7 -0.6x, By using Newton's method, the angel produced by the tangent at any starting point close to the root is approximately: (A)-40.18° (B)-30.96 (C)-16.74 (D) None 25) Given f(x) = x² + x - 2. the fixed point using the fixed point method (x = g(x)), where g(x) is a positive root squared function is: (A)-2 (B) 1 (C)-2, 1 (D) None
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