Matrices

2 Find the value of the constant \(k\) for which the system of equations $$\begin{aligned} 2 x - 3 y + 4 z & = 1 \\ 3 x - y & = 2 \\ x + 2 y + k z & = 1 \end{aligned}$$ does not have a unique solution. For this value of \(k\), solve the system of equations.

4

1. Find the value of \(k\) for which the set of linear equations $$\begin{aligned} x + 3 y + k z & = 4 \\ 4 x - 2 y - 10 z & = - 5 \\ x + y + 2 z & = 1 \end{aligned}$$ has no unique solution.
2. For this value of \(k\), find the set of possible solutions, giving your answer in the form $$\left( \begin{array} { c } x \\ y \\ z \end{array} \right) = \mathbf { a } + t \mathbf { b } ,$$ where \(\mathbf { a }\) and \(\mathbf { b }\) are vectors and \(t\) is a scalar.

2 The matrices \(\mathbf { A }\) and \(\mathbf { I }\) are given by \(\mathbf { A } = \left( \begin{array} { l l } 1 & 2 \\ 1 & 3 \end{array} \right)\) and \(\mathbf { I } = \left( \begin{array} { l l } 1 & 0 \\ 0 & 1 \end{array} \right)\) respectively.

1. Find \(\mathbf { A } ^ { 2 }\) and verify that \(\mathbf { A } ^ { 2 } = 4 \mathbf { A } - \mathbf { I }\).
2. Hence, or otherwise, show that \(\mathbf { A } ^ { - 1 } = 4 \mathbf { I } - \mathbf { A }\).

5 Find the value of \(a\) for which the system of equations $$\begin{aligned} & x - y + 2 z = 4 \\ & x + a y - 3 z = b \\ & x - y + 7 z = 13 \end{aligned}$$ where \(a\) and \(b\) are constants, has no unique solution. Taking \(a\) as the value just found,

1. find the general solution in the case \(b = - 5\),
2. interpret the situation geometrically in the case \(b \neq - 5\).

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.

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 .

1 You are given the matrix \(\mathbf { M } = \left( \begin{array} { r r } 2 & 3 \\ - 2 & 1 \end{array} \right)\).
Find the inverse of \(\mathbf { M }\).
The transformation associated with \(\mathbf { M }\) is applied to a figure of area 2 square units. What is the area of the transformed figure?