Find PA, MA for A = What about over Q? Justify your answers. 01 2 1 2 0 over R. Is A diagonalisable? 2 0 1 Select one: O Over R we obtain P(x) = (x − 2)(x² + x + 3) which does not factor further, and A ‡ 213 so mд = PA is not the product of distinct linear factors, so A is not diagonalisable. The same holds over Q. O Over R we obtain P(x) = (x + 2)(x − √√3)² but (A +213)(A - √√313) = 0 so mÃ₁ = (x + 2)(x − √√3) and A is diagonalisable. But over Q we have PA(x) = (x − 3)(x² - 2x + 3) which does not factorise further and A 213 so m₁ = PA is not the product of distinct linear factors, so not diagonalisable. O Over R we obtain P(x) = (x − √2)(x² + x - 3) which factors further to give 3 distinct real roots. Hence m₁ = PA and A is diagonalisable. Over there are no roots at all so mA = PA but A is not diagonalisable. O Over R we obtain pÃ(x) = (x − 3)(x − √√3)(x + √√3) which has 3 distinct roots. So m = PA and A is diagonalisable. But over Q we just have PA(x) = (x − 3)(x² - 3) which does not factorise further and A‡ 313 so mA = PA is not the product of distinct linear factors, so A is not diagonalisable. None of the others apply O
Find PA, MA for A = What about over Q? Justify your answers. 01 2 1 2 0 over R. Is A diagonalisable? 2 0 1 Select one: O Over R we obtain P(x) = (x − 2)(x² + x + 3) which does not factor further, and A ‡ 213 so mд = PA is not the product of distinct linear factors, so A is not diagonalisable. The same holds over Q. O Over R we obtain P(x) = (x + 2)(x − √√3)² but (A +213)(A - √√313) = 0 so mÃ₁ = (x + 2)(x − √√3) and A is diagonalisable. But over Q we have PA(x) = (x − 3)(x² - 2x + 3) which does not factorise further and A 213 so m₁ = PA is not the product of distinct linear factors, so not diagonalisable. O Over R we obtain P(x) = (x − √2)(x² + x - 3) which factors further to give 3 distinct real roots. Hence m₁ = PA and A is diagonalisable. Over there are no roots at all so mA = PA but A is not diagonalisable. O Over R we obtain pÃ(x) = (x − 3)(x − √√3)(x + √√3) which has 3 distinct roots. So m = PA and A is diagonalisable. But over Q we just have PA(x) = (x − 3)(x² - 3) which does not factorise further and A‡ 313 so mA = PA is not the product of distinct linear factors, so A is not diagonalisable. None of the others apply O
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.3: Lines
Problem 74E
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