Rank and null space basis

The linear transformations \(\mathrm { T } _ { 1 } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) and \(\mathrm { T } _ { 2 } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) are represented by the matrices \(\mathbf { M } _ { 1 }\) and \(\mathbf { M } _ { 2 }\), respectively, where $$\mathbf { M } _ { 1 } = \left( \begin{array} { r r r r } 1 & 1 & 1 & 2 \\ 1 & 4 & 7 & 8 \\ 1 & 7 & 11 & 13 \\ 1 & 2 & 5 & 5 \end{array} \right) , \quad \mathbf { M } _ { 2 } = \left( \begin{array} { r r r r } 2 & 0 & - 1 & - 1 \\ 5 & 1 & - 3 & - 3 \\ 3 & - 1 & - 1 & - 1 \\ 13 & - 1 & - 6 & - 6 \end{array} \right) .$$

1. Find a basis for \(R _ { 1 }\), the range space of \(\mathrm { T } _ { 1 }\).
2. Find a basis for \(K _ { 2 }\), the null space of \(\mathrm { T } _ { 2 }\), and hence show that \(K _ { 2 }\) is a subspace of \(R _ { 1 }\). The set of vectors which belong to \(R _ { 1 }\) but do not belong to \(K _ { 2 }\) is denoted by \(W\).
3. State whether \(W\) is a vector space, justifying your answer. The linear transformation \(\mathrm { T } _ { 3 } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is the result of applying \(\mathrm { T } _ { 1 }\) and then \(\mathrm { T } _ { 2 }\), in that order.
4. Find the dimension of the null space of \(\mathrm { T } _ { 3 }\).

3 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M } = \left( \begin{array} { r r r r } 1 & 3 & - 2 & 4 \\ 5 & 15 & - 9 & 19 \\ - 2 & - 6 & 3 & - 7 \\ 3 & 9 & - 5 & 11 \end{array} \right)\).

1. Find the rank of \(\mathbf { M }\).
2. Obtain a basis for the null space of T .

Determine the rank of the matrix $$\mathbf { A } = \left( \begin{array} { l l l l } 1 & - 1 & - 1 & 1 \\ 2 & - 1 & - 4 & 3 \\ 3 & - 3 & - 2 & 2 \\ 5 & - 4 & - 6 & 5 \end{array} \right)$$ Show that if $$\mathbf { A x } = p \left( \begin{array} { l } 1 \\ 2 \\ 3 \\ 5 \end{array} \right) + q \left( \begin{array} { l } - 1 \\ - 1 \\ - 3 \\ - 4 \end{array} \right) + r \left( \begin{array} { l } - 1 \\ - 4 \\ - 2 \\ - 6 \end{array} \right)$$ where \(p , q\) and \(r\) are given real numbers, then $$\mathbf { x } = \left( \begin{array} { c } p + \lambda \\ q + \lambda \\ r + \lambda \\ \lambda \end{array} \right) ,$$ where \(\lambda\) is real. Find the values of \(p , q\) and \(r\) such that $$p \left( \begin{array} { l } 1 \\ 2 \\ 3 \\ 5 \end{array} \right) + q \left( \begin{array} { l } - 1 \\ - 1 \\ - 3 \\ - 4 \end{array} \right) + r \left( \begin{array} { l } - 1 \\ - 4 \\ - 2 \\ - 6 \end{array} \right) = \left( \begin{array} { r } 3 \\ 7 \\ 8 \\ 15 \end{array} \right) .$$ Find the solution \(\mathbf { x } = \left( \begin{array} { l } \alpha \\ \beta \\ \gamma \\ \delta \end{array} \right)\) of the equation \(\mathbf { A } \mathbf { x } = \left( \begin{array} { r } 3 \\ 7 \\ 8 \\ 15 \end{array} \right)\) for which \(\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } + \delta ^ { 2 } = \frac { 11 } { 4 }\).

7 The linear transformations \(\mathrm { T } _ { 1 } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) and \(\mathrm { T } _ { 2 } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) are represented by the matrices $$\mathbf { M } _ { 1 } = \left( \begin{array} { r r r r } 1 & 1 & 1 & 4 \\ 2 & 1 & 4 & 11 \\ 3 & 4 & 1 & 9 \\ 4 & - 3 & 18 & 37 \end{array} \right) \quad \text { and } \quad \mathbf { M } _ { 2 } = \left( \begin{array} { r r r r } 1 & 1 & 1 & - 1 \\ 2 & 3 & 0 & 1 \\ 3 & 4 & 1 & 0 \\ 4 & 5 & 2 & 0 \end{array} \right)$$ respectively. The null space of \(\mathrm { T } _ { 1 }\) is denoted by \(K _ { 1 }\) and the null space of \(\mathrm { T } _ { 2 }\) is denoted by \(K _ { 2 }\). Show that the dimension of \(K _ { 1 }\) is 2 and that the dimension of \(K _ { 2 }\) is 1 . Find the basis of \(K _ { 1 }\) which has the form \(\left\{ \left( \begin{array} { c } p \\ q \\ 1 \\ 0 \end{array} \right) , \left( \begin{array} { c } r \\ s \\ 0 \\ 1 \end{array} \right) \right\}\) and show that \(K _ { 2 }\) is a subspace of \(K _ { 1 }\).

7 The linear transformations \(\mathrm { T } _ { 1 } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) and \(\mathrm { T } _ { 2 } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) are represented by the matrices $$\mathbf { M } _ { 1 } = \left( \begin{array} { r r r r } 1 & 1 & 1 & 4 \\ 2 & 1 & 4 & 11 \\ 3 & 4 & 1 & 9 \\ 4 & - 3 & 18 & 37 \end{array} \right) \quad \text { and } \quad \mathbf { M } _ { 2 } = \left( \begin{array} { r r r r } 1 & 1 & 1 & - 1 \\ 2 & 3 & 0 & 1 \\ 3 & 4 & 1 & 0 \\ 4 & 5 & 2 & 0 \end{array} \right)$$ respectively. The null space of \(\mathrm { T } _ { 1 }\) is denoted by \(K _ { 1 }\) and the null space of \(\mathrm { T } _ { 2 }\) is denoted by \(K _ { 2 }\). Show that the dimension of \(K _ { 1 }\) is 2 and that the dimension of \(K _ { 2 }\) is 1 . Find the basis of \(K _ { 1 }\) which has the form \(\left\{ \left( \begin{array} { c } p \\ q \\ 1 \\ 0 \end{array} \right) , \left( \begin{array} { l } r \\ s \\ 0 \\ 1 \end{array} \right) \right\}\) and show that \(K _ { 2 }\) is a subspace of \(K _ { 1 }\).

8 The linear transformations \(\mathrm { T } _ { 1 } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) and \(\mathrm { T } _ { 2 } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) are represented by the matrices \(\mathbf { M } _ { 1 }\) and \(\mathbf { M } _ { 2 }\) respectively, where $$\mathbf { M } _ { 1 } = \left( \begin{array} { r r r r } 1 & - 2 & 3 & 5 \\ 3 & - 4 & 17 & 33 \\ 5 & - 9 & 20 & 36 \\ 4 & - 7 & 16 & 29 \end{array} \right) \quad \text { and } \quad \mathbf { M } _ { 2 } = \left( \begin{array} { r r r r } 1 & - 2 & 0 & - 3 \\ 2 & - 1 & 0 & 0 \\ 4 & - 7 & 1 & - 9 \\ 6 & - 10 & 0 & - 14 \end{array} \right) .$$ The null spaces of \(\mathrm { T } _ { 1 }\) and \(\mathrm { T } _ { 2 }\) are denoted by \(K _ { 1 }\) and \(K _ { 2 }\) respectively. Find a basis for \(K _ { 1 }\) and a basis for \(K _ { 2 }\). It is given that \(\mathbf { a } = \left( \begin{array} { l } 1 \\ 2 \\ 3 \\ 4 \end{array} \right)\). The vectors \(\mathbf { x } _ { 1 }\) and \(\mathbf { x } _ { 2 }\) are such that \(\mathbf { M } _ { 1 } \mathbf { x } _ { 1 } = \mathbf { M } _ { 1 } \mathbf { a }\) and \(\mathbf { M } _ { 2 } \mathbf { x } _ { 2 } = \mathbf { M } _ { 2 } \mathbf { a }\). Given that \(\mathbf { x } _ { 1 } - \mathbf { x } _ { 2 } = \left( \begin{array} { c } p \\ 5 \\ 7 \\ q \end{array} \right)\), find \(p\) and \(q\).

6 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { r r r r } - 2 & 5 & 3 & - 1 \\ 0 & 1 & - 4 & - 2 \\ 6 & - 14 & - 13 & 1 \\ \alpha & \alpha & - 2 \alpha & - 11 \alpha \end{array} \right)$$ and \(\alpha\) is a constant. The null space of T is denoted by \(K _ { 1 }\) when \(\alpha \neq 0\), and by \(K _ { 2 }\) when \(\alpha = 0\). Find a basis for \(K _ { 1 }\) and a basis for \(K _ { 2 }\).

6 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { r r r r } 2 & - 1 & 1 & 3 \\ 2 & 0 & 0 & 5 \\ 6 & - 2 & 2 & 11 \\ 10 & - 3 & 3 & 19 \end{array} \right)$$

1. Find the rank of \(\mathbf { M }\) and state a basis for the range space of T .
2. Obtain a basis for the null space of T .

The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { r r r r } 1 & - 2 & 3 & - 4 \\ 2 & - 4 & 7 & - 9 \\ 4 & - 8 & 14 & - 18 \\ 5 & - 10 & 17 & - 22 \end{array} \right)$$ Find the rank of \(\mathbf { M }\). Obtain a basis for the null space \(K\) of T . Evaluate $$\mathbf { M } \left( \begin{array} { r } 1 \\ - 2 \\ 2 \\ - 1 \end{array} \right)$$ and hence show that any solution of $$\mathbf { M x } = \left( \begin{array} { l }

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

1. Find the rank of \(\mathbf { A }\) and show that \(\left\{ \left( \begin{array} { r } 2 \\ 2 \\ - 1 \\ 0 \end{array} \right) , \left( \begin{array} { l } 1 \\ 3 \\ 0 \\ 1 \end{array} \right) \right\}\) is a basis for the null space of the transformation. \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\)
2. Show that if $$\mathbf { A x } = p \left( \begin{array} { r } 1 \\ 3 \\ 5 \\ - 2 \end{array} \right) + q \left( \begin{array} { r } - 1 \\ - 1 \\ - 8 \\ 3 \end{array} \right) ,$$ where \(p\) and \(q\) are given real numbers, then $$\mathbf { x } = \left( \begin{array} { c } p + 2 \lambda + \mu \\ q + 2 \lambda + 3 \mu \\ - \lambda \\ \mu \end{array} \right) ,$$ where \(\lambda\) and \(\mu\) are real numbers. \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\)
3. Find the values of \(p\) and \(q\) such that $$p \left( \begin{array} { r } 1 \\ 3 \\ 5 \\ - 2 \end{array} \right) + q \left( \begin{array} { r } - 1 \\ - 1 \\ - 8 \\ 3 \end{array} \right) = \left( \begin{array} { r } 3 \\ 7 \\ 18 \\ - 7 \end{array} \right)$$ \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\)
4. Find the solution of the equation \(\mathbf { A x } = \left( \begin{array} { r } 3 \\ 7 \\ 18 \\ - 7 \end{array} \right)\) of the form \(\mathbf { x } = \left( \begin{array} { l } 4 \\ 9 \\ \alpha \\ \beta \end{array} \right)\), where \(\alpha\) and \(\beta\) are positive integers to be found. \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \footnotetext{

8 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 3 }\) is represented by the matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { r r c r } 1 & 2 & \alpha & - 1 \\ 2 & 6 & - 3 & - 3 \\ 3 & 10 & - 6 & - 5 \end{array} \right)$$ and \(\alpha\) is a constant. When \(\alpha \neq 0\) the null space of T is denoted by \(K _ { 1 }\).

1. Find a basis for \(K _ { 1 }\).
   When \(\alpha = 0\) the null space of T is denoted by \(K _ { 2 }\).
2. Find a basis for \(K _ { 2 }\).
3. Determine, justifying your answer, whether \(K _ { 1 }\) is a subspace of \(K _ { 2 }\).

5 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { r r r r } 1 & 2 & 0 & 4 \\ 5 & 2 & 1 & - 3 \\ 4 & 0 & 1 & - 7 \\ - 2 & 4 & - 1 & \alpha \end{array} \right)$$ It is given that the rank of \(\mathbf { M }\) is 2 .

1. Find the value of \(\alpha\) and state a basis for the range space of T .
2. Obtain a basis for the null space of T .

10 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { H }\), where $$\mathbf { H } = \left( \begin{array} { r r r r } 1 & 2 & - 3 & - 5 \\ - 1 & 4 & 5 & 1 \\ 2 & 3 & 0 & - 3 \\ - 3 & 5 & 7 & 2 \end{array} \right)$$

1. Find the dimension of the range space of T .
2. Find a basis for the null space of \(T\).
3. It is given that \(\mathbf { x }\) satisfies the equation $$\mathbf { H } \mathbf { x } = \left( \begin{array} { r } 2 \\ - 10 \\ - 1 \\ - 15 \end{array} \right)$$ Using the fact that $$\mathbf { H } \left( \begin{array} { r } 1 \\ - 3 \\ 1 \\ - 2 \end{array} \right) = \left( \begin{array} { r } 2 \\ - 10 \\ - 1 \\ - 15 \end{array} \right) ,$$ find the least possible value of \(| \mathbf { x } |\).
   [0pt] [For the vector \(\mathbf { x } = \left( \begin{array} { c } x _ { 1 } \\ x _ { 2 } \\ x _ { 3 } \\ x _ { 4 } \end{array} \right) , | \mathbf { x } | = \sqrt { } \left( x _ { 1 } ^ { 2 } + x _ { 2 } ^ { 2 } + x _ { 3 } ^ { 2 } + x _ { 4 } ^ { 2 } \right)\).]

1 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix $$\left( \begin{array} { r r r r } 1 & 5 & 2 & 6 \\ 2 & 0 & - 1 & 7 \\ 3 & - 1 & - 2 & 10 \\ 4 & 10 & 13 & 29 \end{array} \right)$$ Find the dimension of the null space of T .

6 The matrix \(\mathbf { A }\) is defined by $$\mathbf { A } = \left( \begin{array} { r r r r } 1 & - 1 & - 2 & - 3 \\ - 2 & 1 & 7 & 2 \\ - 3 & 3 & 6 & \alpha \\ 7 & - 6 & - 17 & - 17 \end{array} \right) .$$

1. Show that if \(\alpha = 9\) then the rank of \(\mathbf { A }\) is 2, and find a basis for the null space of \(\mathbf { A }\) in this case.
2. Find the rank of \(\mathbf { A }\) when \(\alpha \neq 9\).

6 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { r r r r } 1 & - 3 & - 1 & 2 \\ 4 & - 10 & 0 & 2 \\ 1 & - 1 & 3 & - 4 \\ 5 & - 12 & 1 & 1 \end{array} \right)$$ 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 $$\mathbf { M x } = \left( \begin{array} { r } 2 \\ 16 \\ 10 \\ 22 \end{array} \right)$$ 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 \(K\).

5 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { A }\), where $$\mathbf { A } = \left( \begin{array} { r r r r } 1 & 3 & 5 & 7 \\ 2 & 8 & 7 & 9 \\ 3 & 13 & 9 & 11 \\ 6 & 24 & 21 & 27 \end{array} \right)$$ Find

1. the rank of \(\mathbf { A }\),
2. a basis for the range space of T ,
3. a basis for the null space of T .

7 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { A }\), where $$\mathbf { A } = \left( \begin{array} { r r r r } 1 & - 1 & - 2 & 3 \\ 5 & - 3 & - 4 & 25 \\ 6 & - 4 & - 6 & 28 \\ 7 & - 5 & - 8 & 31 \end{array} \right)$$

1. Find the rank of \(\mathbf { A }\) and a basis for the null space of T .
2. Find the matrix product \(\mathbf { A } \left( \begin{array} { r } - 1 \\ 1 \\ - 1 \\ 1 \end{array} \right)\) and hence find the general solution of the equation \(\mathbf { A } \mathbf { x } = \left( \begin{array} { r } 3 \\ 21 \\ 24 \\ 27 \end{array} \right)\). \(8 \quad\) Let \(I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sec ^ { n } x \mathrm {~d} x\) for \(n > 0\).

5 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { r r r r } 3 & 2 & 0 & 1 \\ 6 & 5 & - 1 & 3 \\ 9 & 8 & - 2 & 5 \\ - 3 & - 2 & 0 & - 1 \end{array} \right)$$

1. Find the rank of \(\mathbf { M }\).
   Let \(K\) be the null space of T .
2. Find a basis for \(K\).
3. Find the general solution of $$\mathbf { M } \mathbf { x } = \left( \begin{array} { r } 2 \\ 5 \\ 8 \\ - 2 \end{array} \right) .$$

7 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { r r r r } 1 & - 2 & - 3 & 1 \\ 3 & - 5 & - 7 & 7 \\ 5 & - 9 & - 13 & 9 \\ 7 & - 13 & - 19 & 11 \end{array} \right)$$

1. Find the rank of \(\mathbf { M }\) and a basis for the null space of T .
2. The vector \(\left( \begin{array} { l } 1 \\ 2 \\ 3 \\ 4 \end{array} \right)\) is denoted by \(\mathbf { e }\). Show that there is a solution of the equation \(\mathbf { M x } = \mathbf { M e }\) of the form \(\mathbf { x } = \left( \begin{array} { c } a \\ b \\ - 1 \\ - 1 \end{array} \right)\), where the constants \(a\) and \(b\) are to be found.

4 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { r r r r } 3 & 4 & 2 & 5 \\ 6 & 7 & 5 & 8 \\ 9 & 9 & 9 & 9 \\ 15 & 16 & 14 & 17 \end{array} \right)$$ Find

1. the rank of \(\mathbf { M }\) and a basis for the range space of T ,
2. a basis for the null space of T .

6 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { r r r r } 1 & - 3 & - 1 & 2 \\ 4 & - 10 & 0 & 2 \\ 1 & - 1 & 3 & - 4 \\ 5 & - 12 & 1 & 1 \end{array} \right)$$ 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 $$\mathbf { M x } = \left( \begin{array} { r } 2 \\ 16 \\ 10 \\ 22 \end{array} \right)$$ 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 \(K\).

7 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { r r r r } 1 & - 2 & - 3 & 1 \\ 3 & - 5 & - 7 & 7 \\ 5 & - 9 & - 13 & 9 \\ 7 & - 13 & - 19 & 11 \end{array} \right)$$ Find the rank of \(\mathbf { M }\) and a basis for the null space of T . The vector \(\left( \begin{array} { l } 1 \\ 2 \\ 3 \\ 4 \end{array} \right)\) is denoted by \(\mathbf { e }\). Show that there is a solution of the equation \(\mathbf { M x } = \mathbf { M e }\) of the form \(\mathbf { x } = \left( \begin{array} { c } a \\ b \\ - 1 \\ - 1 \end{array} \right)\), where the constants \(a\) and \(b\) are to be found.