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Expert-verified Found in: Page 289 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Consider two distinct real numbers, a and b. We define the function${\mathbit{f}}\left(t\right){\mathbf{=}}{\mathbit{d}}{\mathbit{e}}{\mathbit{t}}\left[\begin{array}{ccc}1& 1& 1\\ a& b& t\\ {a}^{2}& {b}^{2}& {t}^{2}\end{array}\right]$a. Show that ${\mathbf{f}}\left(t\right)$ is a quadratic function. What is the coefficient of ${{\mathbit{t}}}^{{\mathbf{2}}}$? b. Explain why ${\mathbit{f}}\left(a\right){\mathbf{=}}{\mathbit{f}}\left(b\right){\mathbf{=}}{\mathbf{0}}$. Conclude that ${\mathbit{f}}\left(t\right){\mathbf{=}}{\mathbit{k}}\left(t-a\right)\left(t-b\right)$ , 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 ${t}^{2}is-a+b$.

b) .$f\left(t\right)=k\left(t-a\right)\left(t-b\right)$

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

See the step by step solution

## Step 1: What is the coefficient of t2

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

$f\left(t\right)=det\left[\begin{array}{ccc}1& 1& 1\\ a& b& t\\ {a}^{2}& {b}^{2}& {t}^{2}\end{array}\right]\phantom{\rule{0ex}{0ex}}f\left(t\right)=b{t}^{2}+{a}^{2}t+a{b}^{2}-{a}^{2}b-{b}^{2}t-a{t}^{2}\phantom{\rule{0ex}{0ex}}f\left(t\right)=\left(-a+b\right){t}^{2}+\left({a}^{2}-{b}^{2}\right)t+\left(a{b}^{2}-{a}^{2}b\right)$

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:

$f\left(a\right)=\left(-a+b\right){a}^{2}+\left({a}^{2}-{b}^{2}\right)a+\left(a{b}^{2}-{a}^{2}b\right)=-{a}^{3}+{a}^{2}b+{a}^{3}-a{b}^{2}+a{b}^{2}-{a}^{2}b=0\phantom{\rule{0ex}{0ex}}f\left(a\right)=\left(-a+b\right){b}^{2}+\left({a}^{2}-{b}^{2}\right)b+\left(a{b}^{2}-{a}^{2}b\right)=-{a}^{3}+{a}^{2}b+{a}^{3}-a{b}^{2}+a{b}^{2}-{a}^{2}b=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Thus,f\left(a\right)=f\left(b\right)=0\phantom{\rule{0ex}{0ex}}Considerk\in \square .Wecompute,\phantom{\rule{0ex}{0ex}}k\left(t-a\right)\left(t-b\right)=k{t}^{2}+k\left(-a-b\right)t+kab\phantom{\rule{0ex}{0ex}}Fork=-a+b,wehaveexactlyf\left(t\right)=k\left(t-a\right)\left(t-b\right)$

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

We solve,

$f\left(t\right)\ne 0\phantom{\rule{0ex}{0ex}}\left(-a+b\right){t}^{2}+\left({a}^{2}-{b}^{2}\right)t+\left(a{b}^{2}-{a}^{2}b\right)\ne 0\phantom{\rule{0ex}{0ex}}t\ne \frac{-{a}^{2}+{b}^{2}±\sqrt{{a}^{4}-2{a}^{2}{b}^{2}+{b}^{4}-4\left(-a+b\right)\left(a{b}^{2}-{a}^{2}-{b}^{2}\right)}}{2\left(-a+b\right)}\phantom{\rule{0ex}{0ex}}t=\frac{\left(b+a\right)\left(b-a\right)±{\left(b-a\right)}^{2}}{2\left(b-a\right)}\phantom{\rule{0ex}{0ex}}t=\frac{b+a±\left(b-a\right)}{2}\phantom{\rule{0ex}{0ex}}t\ne a,b$ ### Want to see more solutions like these? 