Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Consider two distinct real numbers, a and b. We define the function
$f (t) = d e t [\begin{matrix} 1 & 1 & 1 \\ a & b & t \\ a^{2} & b^{2} & t^{2} \end{matrix}]$
a. Show that $f (t)$ is a quadratic function. What is the coefficient of $t^{2}$ ? b. Explain why $f (a) = f (b) = 0$ . Conclude that $f (t) = k (t - a) (t - b)$ , for some constant k . Find k , using your work in part (a). c. For which values of t is the matrix invertible?

Therefore, the determinant of the given matrix is given by,

a). The coefficient of is $t^{2} i s - a + b$ .

b) . $f (t) = k (t - a) (t - b)$

C) The matrix invertible value of t is $t \neq a, b$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: What is the coefficient of t2

a) Show that $f (t)$ is a quadratic function.

$f (t) = d e t [\begin{matrix} 1 & 1 & 1 \\ a & b & t \\ a^{2} & b^{2} & t^{2} \end{matrix}] f (t) = b t^{2} + a^{2} t + a b^{2} - a^{2} b - b^{2} t - a t^{2} f (t) = (- a + b) t^{2} + (a^{2} - b^{2}) t + (a b^{2} - a^{2} b)$

Thus, f is a quadratic function, the coefficient of $t^{2}$ being localid="1659512383139" $- a + b$ .

Step 2: Explain why f(a)=f(b)=0

b) We compute:

$f (a) = (- a + b) a^{2} + (a^{2} - b^{2}) a + (a b^{2} - a^{2} b) = - a^{3} + a^{2} b + a^{3} - a b^{2} + a b^{2} - a^{2} b = 0 f (a) = (- a + b) b^{2} + (a^{2} - b^{2}) b + (a b^{2} - a^{2} b) = - a^{3} + a^{2} b + a^{3} - a b^{2} + a b^{2} - a^{2} b = 0 T h u s, f (a) = f (b) = 0 C o n s i d e r k \in □ . W e c o m p u t e, k (t - a) (t - b) = k t^{2} + k (- a - b) t + k a b F o r k = - a + b, w e h a v e e x a c t l y f (t) = k (t - a) (t - b)$

Step 3: For which values of t is the matrix invertible.

We solve,

$f (t) \neq 0 (- a + b) t^{2} + (a^{2} - b^{2}) t + (a b^{2} - a^{2} b) \neq 0 t \neq \frac{- a^{2} + b^{2} \pm \sqrt{a^{4} - 2 a^{2} b^{2} + b^{4} - 4 (- a + b) (a b^{2} - a^{2} - b^{2})}}{2 (- a + b)} t = \frac{(b + a) (b - a) \pm {(b - a)}^{2}}{2 (b - a)} t = \frac{b + a \pm (b - a)}{2} t \neq a, b$

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Linear Algebra With Applications

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Short Answer

Step by Step Solution

Step 1: What is the coefficient of t2

Step 2: Explain why f(a)=f(b)=0

Step 3: For which values of t is the matrix invertible.

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Linear Algebra With Applications

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Short Answer

Step by Step Solution

Step 1: What is the coefficient of t2

Step 2: Explain why f(a)=f(b)=0

Step 3: For which values of t is the matrix invertible.

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