Let p(t)=t^5-6t^4+6t^3-6t^2+5t.a) Factorise p(t) as a product of degree 1 polynomials.

Anish Buchanan 2021-09-10 Answered
Let \(\displaystyle{p}{\left({t}\right)}={t}^{{5}}-{6}{t}^{{4}}+{6}{t}^{{3}}-{6}{t}^{{2}}+{5}{t}\).
a) Factorise p(t) as a product of degree 1 polynomials.
b) Give an example of a matrix with characteristic polynomial p(t). That is, find a matrix A such that \(\displaystyle{P}_{{a}}{\left({t}\right)}={p}{\left({t}\right)}\)
c) Give an example of a degree 3 polynomial q(t) with real coefficients that has two imaginary roots. For the polynomial q(t) you find, give a matrix that has characterestic polynomial q(t).

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Expert Answer

Aniqa O'Neill
Answered 2021-09-11 Author has 22389 answers

The given polynomial is
\(\displaystyle{p}{\left({t}\right)}={t}^{{5}}-{6}{t}^{{4}}+{6}{t}^{{3}}-{6}{t}^{{2}}+{5}{t}\)
We have to find the roots of this polynomial to write it as product of degree 1 polynomials.
a) Finding the roots
\(\displaystyle{p}{\left({t}\right)}={t}^{{5}}-{6}{t}^{{4}}+{6}{t}^{{3}}-{6}{t}^{{2}}+{5}{t}\)
\(\displaystyle{p}{\left({t}\right)}={t}{\left({t}^{{4}}-{6}{t}^{{3}}+{6}{t}^{{2}}-{6}{t}+{5}\right)}\)
As t=1 also satisfies the polynomial so dividing the rest of the polynomial by t-1, we get
\(\displaystyle{p}{\left({t}\right)}={t}{\left({t}-{1}\right)}{\left({t}^{{3}}-{5}{t}^{{2}}+{t}-{5}\right)}\)
Now t=5 satisfies the rest of the polynomial so dividing the cubic polynomial by t-5, we get
\(\displaystyle{p}{\left({t}\right)}={t}{\left({t}-{1}\right)}{\left({t}-{5}\right)}{\left({t}^{{2}}+{1}\right)}\)
Now as we know that
\(\displaystyle{t}^{{2}}+{1}={0}\)
\(\displaystyle\Rightarrow{t}={i},-{i}\)
Hence, the given polynomial can be written as
\(\displaystyle{p}{\left({t}\right)}={t}{\left({t}-{1}\right)}{\left({t}-{5}\right)}{\left({t}+{i}\right)}{\left({t}-{i}\right)}\)
b) We have to find a matrix A such that characteristic of matrix A and the given polynomial p(t) are equal.
That is,
\(\displaystyle{P}_{{A}}{\left({t}\right)}={p}{\left({t}\right)}\)
As we know the roots of characteristic polynomial are eigen values of the matrix .
Hence the required matrix A must have eigen values 0,1,5 ,i and −i.
Also we know that eigen values of diagonal matrix are diagonal entries it self.
Therefore, the required matrix A can be given as
\(A=\begin{bmatrix}0&0&0&0&0\\0&1&0&0&0\\0&0&5&0&0\\0&0&0&i&0\\0&0&0&0&-i\end{bmatrix}\)
c) We have to give an example of 3 degree polynomial with real coefficients and two imaginary roots.
The polynomial q(t) can be taken as
\(\displaystyle{q}{\left({t}\right)}={\left({t}^{{3}}-{5}{t}^{{2}}+{t}-{5}\right)}\)
It has three roots such that
\(\displaystyle{q}{\left({t}\right)}={\left({t}-{5}\right)}{\left({t}+{1}\right)}{\left({t}-{i}\right)}\)
where two of the roots are imaginary.
Now the required matrix whose characteristic polynomial is equal to the polynomial q(t) can be given as
\(B=\begin{bmatrix}5&0&0\\0&i&0\\0&0&-i\end{bmatrix}\)
Clearly, \(\displaystyle{P}_{{B}}{\left({t}\right)}={q}{\left({t}\right)}\)

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