Exam-Style Problem

9231 P13 - Jun 2017 - Q11O - 14 marks

6254

The matrix \(A\), given by \(A=\begin{pmatrix}1&-1&0&2\\3&-1&4&0\\5&-8&-6&19\\-2&3&2&-7\end{pmatrix}\), represents a transformation from \(\mathbb{R}^4\) to \(\mathbb{R}^4\).

(i) Find the rank of \(A\) and show that \(\left\{\begin{pmatrix}2\\2\\-1\\0\end{pmatrix},\begin{pmatrix}1\\3\\0\\1\end{pmatrix}\right\}\) is a basis for the null space of the transformation.

(ii) Show that if \(A\mathbf{x}=p\begin{pmatrix}1\\3\\5\\-2\end{pmatrix}+q\begin{pmatrix}-1\\-1\\-8\\3\end{pmatrix}\), where \(p\) and \(q\) are given real numbers, then \(\mathbf{x}=\begin{pmatrix}p+2\lambda+\mu\\q+2\lambda+3\mu\\-\lambda\\\mu\end{pmatrix}\), where \(\lambda\) and \(\mu\) are real numbers.

(iii) Find the values of \(p\) and \(q\) such that \(p\begin{pmatrix}1\\3\\5\\-2\end{pmatrix}+q\begin{pmatrix}-1\\-1\\-8\\3\end{pmatrix}=\begin{pmatrix}3\\7\\18\\-7\end{pmatrix}\).

(iv) Find the solution of \(A\mathbf{x}=\begin{pmatrix}3\\7\\18\\-7\end{pmatrix}\) of the form \(\mathbf{x}=\begin{pmatrix}4\\9\\m\\n\end{pmatrix}\), where \(m\) and \(n\) are positive integers to be found.

Solution Checked by expert

Answer: (i) \(\operatorname{rank}(A)=2\), and a basis for the null space is \(\left\{\begin{pmatrix}2\\2\\-1\\0\end{pmatrix},\begin{pmatrix}1\\3\\0\\1\end{pmatrix}\right\}\).

(ii) \(\mathbf{x}=\begin{pmatrix}p+2\lambda+\mu\\q+2\lambda+3\mu\\-\lambda\\\mu\end{pmatrix}\).

(iii) \(p=2\), \(q=-1\).

(iv) \(\mathbf{x}=\begin{pmatrix}4\\9\\1\\4\end{pmatrix}\).

Row-reducing \(A\) gives an echelon form with two non-zero rows, for example

\(\begin{pmatrix}1&-1&0&2\\0&1&2&-3\\0&0&0&0\\0&0&0&0\end{pmatrix}\).

Therefore \(\operatorname{rank}(A)=2\).

To find the null space, solve \(A\mathbf{x}=\mathbf{0}\). From the echelon form, with \(\mathbf{x}=(x,y,z,t)^T\),

\(x-y+2t=0\), and \(y+2z-3t=0\).

Let \(z=\lambda\) and \(t=\mu\). Then

\(y=-2\lambda+3\mu\), and \(x=y-2\mu=-2\lambda+\mu\).

Thus

\(\mathbf{x}=\lambda\begin{pmatrix}-2\\-2\\1\\0\end{pmatrix}+\mu\begin{pmatrix}1\\3\\0\\1\end{pmatrix}\).

Replacing the first vector by its negative gives the equivalent basis

\(\left\{\begin{pmatrix}2\\2\\-1\\0\end{pmatrix},\begin{pmatrix}1\\3\\0\\1\end{pmatrix}\right\}\).

The first two columns of \(A\) are \(\begin{pmatrix}1\\3\\5\\-2\end{pmatrix}\) and \(\begin{pmatrix}-1\\-1\\-8\\3\end{pmatrix}\). Hence

\(A\begin{pmatrix}p\\q\\0\\0\end{pmatrix}=p\begin{pmatrix}1\\3\\5\\-2\end{pmatrix}+q\begin{pmatrix}-1\\-1\\-8\\3\end{pmatrix}\).

Therefore the general solution is a particular solution plus the null space:

\(\mathbf{x}=\begin{pmatrix}p\\q\\0\\0\end{pmatrix}+\lambda\begin{pmatrix}2\\2\\-1\\0\end{pmatrix}+\mu\begin{pmatrix}1\\3\\0\\1\end{pmatrix}\).

\(\mathbf{x}=\begin{pmatrix}p+2\lambda+\mu\\q+2\lambda+3\mu\\-\lambda\\\mu\end{pmatrix}\).

Now solve

\(p\begin{pmatrix}1\\3\\5\\-2\end{pmatrix}+q\begin{pmatrix}-1\\-1\\-8\\3\end{pmatrix}=\begin{pmatrix}3\\7\\18\\-7\end{pmatrix}\).

From the first two components,

\(p-q=3\), and \(3p-q=7\).

Subtracting gives \(2p=4\), so \(p=2\). Then \(2-q=3\), so \(q=-1\).

For \(A\mathbf{x}=\begin{pmatrix}3\\7\\18\\-7\end{pmatrix}\), use \(p=2\) and \(q=-1\). Then

\(\mathbf{x}=\begin{pmatrix}2+2\lambda+\mu\\-1+2\lambda+3\mu\\-\lambda\\\mu\end{pmatrix}\).

We need the first two components to be \(4\) and \(9\), so

\(2\lambda+\mu=2\), and \(2\lambda+3\mu=10\).

Subtracting gives \(2\mu=8\), so \(\mu=4\). Then \(2\lambda+4=2\), so \(\lambda=-1\).

Therefore

\(\mathbf{x}=\begin{pmatrix}4\\9\\1\\4\end{pmatrix}\).