Exam-Style Problem

9231 P11 - Nov 2013 - Q6 - 31 marks

6413

The linear transformation \(\mathrm{T}: \mathbb{R}^{4} \rightarrow \mathbb{R}^{4}\) is represented by the matrix \(\mathbf{M}\), where
\(\mathbf{M}=\left(\begin{array}{rrrr} 1 & -3 & -1 & 2 \\ 4 & -10 & 0 & 2 \\ 1 & -1 & 3 & -4 \\ 5 & -12 & 1 & 1 \end{array}\right) .\)

Find, in either order, the rank of \(\mathbf{M}\) and a basis for the null space \(K\) of T .

Evaluate
\(\mathbf{M}\left(\begin{array}{r} 1 \\ -2 \\ -3 \\ -4 \end{array}\right),\)
and hence show that every solution of
\(\mathbf{M x}=\left(\begin{array}{r} 2 \\ 16 \\ 10 \\ \end{array}\right)\)
has the form
\(\mathbf{x}=\left(\begin{array}{r} 1 \\ -2 \\ -3 \\ -4 \end{array}\right)+\lambda \mathbf{e}_{1}+\mu \mathbf{e}_{2},\)
where \(\lambda\) and \(\mu\) are real numbers and \(\left\{\mathbf{e}_{1}, \mathbf{e}_{2}\right\}\) is a basis for \(K\).

Solution

Answer: The rank of \(\mathbf{M}\) is \(2\), and a basis for the null space \(K\) is \(\{(-5,-2,1,0)^T,(7,3,0,1)^T\}\).

Also, \(\mathbf{M}\begin{pmatrix}1\\-2\\-3\\-4\end{pmatrix}=\begin{pmatrix}2\\16\\10\\22\end{pmatrix}\), so every solution of \(\mathbf{Mx}=\begin{pmatrix}2\\16\\10\\22\end{pmatrix}\) has the form \(\mathbf{x}=\begin{pmatrix}1\\-2\\-3\\-4\end{pmatrix}+\lambda \mathbf{e}_1+\mu \mathbf{e}_2\), where \(\{\mathbf{e}_1,\mathbf{e}_2\}\) is any basis for \(K\).

We row-reduce \(\mathbf{M}\):

\(\begin{pmatrix}1&-3&-1&2\\4&-10&0&2\\1&-1&3&-4\\5&-12&1&1\end{pmatrix}\)

Use the first row as a pivot row.

\(R_2\leftarrow R_2-4R_1\), \(R_3\leftarrow R_3-R_1\), \(R_4\leftarrow R_4-5R_1\):

\(\begin{pmatrix}1&-3&-1&2\\0&2&4&-6\\0&2&4&-6\\0&3&6&-9\end{pmatrix}\)

Now divide the second row by \(2\):

\(\begin{pmatrix}1&-3&-1&2\\0&1&2&-3\\0&2&4&-6\\0&3&6&-9\end{pmatrix}\)

Then eliminate below the pivot in column 2:

\(R_3\leftarrow R_3-2R_2\), \(R_4\leftarrow R_4-3R_2\), giving

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

There are two non-zero rows, so \(\operatorname{rank}(\mathbf{M})=2\).

To find the null space, solve \(\mathbf{Mx}=\mathbf{0}\). From the echelon form:

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

\(y+2z-3t=0\)

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

\(y=3\mu-2\lambda\)

and

\(x=3y+z-2t=3(3\mu-2\lambda)+\lambda-2\mu=7\mu-5\lambda\).

\(\mathbf{x}=\lambda\begin{pmatrix}-5\\-2\\1\\0\end{pmatrix}+\mu\begin{pmatrix}7\\3\\0\\1\end{pmatrix}.\)

Hence a basis for \(K\) is \(\{(-5,-2,1,0)^T,(7,3,0,1)^T\}\).

Now evaluate the given product:

\(\mathbf{M}\begin{pmatrix}1\\-2\\-3\\-4\end{pmatrix}=\begin{pmatrix}1&-3&-1&2\\4&-10&0&2\\1&-1&3&-4\\5&-12&1&1\end{pmatrix}\begin{pmatrix}1\\-2\\-3\\-4\end{pmatrix}=\begin{pmatrix}2\\16\\10\\22\end{pmatrix}.\)

Therefore the vector \(\begin{pmatrix}1\\-2\\-3\\-4\end{pmatrix}\) is one particular solution of \(\mathbf{Mx}=\begin{pmatrix}2\\16\\10\\22\end{pmatrix}\).

If \(\mathbf{x}\) is any other solution, then

\(\mathbf{M}\left(\mathbf{x}-\begin{pmatrix}1\\-2\\-3\\-4\end{pmatrix}\right)=\mathbf{Mx}-\mathbf{M}\begin{pmatrix}1\\-2\\-3\\-4\end{pmatrix}=\mathbf{0},\)

so \(\mathbf{x}-\begin{pmatrix}1\\-2\\-3\\-4\end{pmatrix}\in K\).

Since \(K\) has basis \(\{\mathbf{e}_1,\mathbf{e}_2\}\), every vector in \(K\) can be written as \(\lambda \mathbf{e}_1+\mu \mathbf{e}_2\). Hence every solution has the form

\(\mathbf{x}=\begin{pmatrix}1\\-2\\-3\\-4\end{pmatrix}+\lambda \mathbf{e}_1+\mu \mathbf{e}_2.\)