4.03r Solve simultaneous equations: using inverse matrix

3 You are given the matrix \(\mathbf { A } = \left( \begin{array} { r r r } k & - 7 & 4 \\ 2 & - 2 & 3 \\ 1 & - 3 & - 2 \end{array} \right)\).

1. Show that when \(k = 5\) the determinant of \(\mathbf { A }\) is zero. Obtain an expression for the inverse of \(\mathbf { A }\) when \(k \neq 5\).
2. Solve the matrix equation $$\left( \begin{array} { r r r }

9 The matrix \(\mathbf { A }\) is given by \(\mathbf { A } = \left( \begin{array} { r r r } 2 & - 1 & 1 \\ 0 & 3 & 1 \\ 1 & 1 & a \end{array} \right)\), where \(a \neq 1\).

1. Find \(\mathbf { A } ^ { - 1 }\).
2. Hence, or otherwise, solve the equations $$\begin{array} { r } 2 x - y + z = 1 \\ 3 y + z = 2 \\ x + y + a z = 2 \end{array}$$

10 You are given that \(\mathbf { A } = \left( \begin{array} { r r r } 3 & 4 & - 1 \\ 1 & - 1 & k \\ - 2 & 7 & - 3 \end{array} \right)\) and \(\mathbf { B } = \left( \begin{array} { r r c } 11 & - 5 & - 7 \\ 1 & 11 & 5 + k \\ - 5 & 29 & 7 \end{array} \right)\) and that \(\mathbf { A B }\) is of the form \(\mathbf { A B } = \left( \begin{array} { c c c } 42 & \alpha & 4 k - 8 \\ 10 - 5 k & - 16 + 29 k & - 12 + 6 k \\ 0 & 0 & \beta \end{array} \right)\).

1. Show that \(\alpha = 0\) and \(\beta = 28 + 7 k\).
2. Find \(\mathbf { A B }\) when \(k = 2\).
3. For the case when \(k = 2\) write down the matrix \(\mathbf { A } ^ { - 1 }\).
4. Use the result from part (iii) to solve the following simultaneous equations. $$\begin{aligned} 3 x + 4 y - z & = 1 \\ x - y + 2 z & = - 9 \\ - 2 x + 7 y - 3 z & = 26 \end{aligned}$$

4 You are given that if \(\mathbf { M } = \left( \begin{array} { r r r } 4 & 0 & 1 \\ - 6 & 1 & 1 \\ 5 & 2 & 5 \end{array} \right)\) then \(\mathbf { M } ^ { - 1 } = \frac { 1 } { k } \left( \begin{array} { r r r } - 3 & - 2 & 1 \\ - 35 & - 15 & 10 \\ 17 & 8 & - 4 \end{array} \right)\).
Find the value of \(k\). Hence solve the following simultaneous equations. $$\begin{aligned} 4 x + z & = 9 \\ - 6 x + y + z & = 32 \\ 5 x + 2 y + 5 z & = 81 \end{aligned}$$

\(\mathbf { 9 }\) You are given that \(\mathbf { A } = \left( \begin{array} { r r r } - 2 & 1 & - 5 \\ 3 & a & 1 \\ 1 & - 1 & 2 \end{array} \right)\) and \(\mathbf { B } = \left( \begin{array} { c c c } 2 a + 1 & 3 & 1 + 5 a \\ - 5 & 1 & - 13 \\ - 3 - a & - 1 & - 2 a - 3 \end{array} \right)\).

1. Show that \(\mathbf { A B } = ( 8 + a ) \mathbf { I }\).
2. State the value of \(a\) for which \(\mathbf { A } ^ { - 1 }\) does not exist. Write down \(\mathbf { A } ^ { - 1 }\) in terms of \(a\), when \(\mathbf { A } ^ { - 1 }\) exists.
3. Use \(\mathbf { A } ^ { - 1 }\) to solve the following simultaneous equations. $$\begin{aligned} - 2 x + y - 5 z & = - 55 \\ 3 x + 4 y + z & = - 9 \\ x - y + 2 z & = 26 \end{aligned}$$
4. What can you say about the solutions of the following simultaneous equations? $$\begin{aligned} - 2 x + y - 5 z & = p \\ 3 x - 8 y + z & = q \\ x - y + 2 z & = r \end{aligned}$$

9 You are given that \(\mathbf { A } = \left( \begin{array} { r r r } 8 & - 7 & - 12 \\ - 10 & 5 & 15 \\ - 9 & 6 & 6 \end{array} \right)\) and \(\mathbf { A } ^ { - 1 } = k \left( \begin{array} { r r r } 4 & 2 & 3 \\ 5 & 4 & 0 \\ 1 & - 1 & 2 \end{array} \right)\).

1. Find the exact value of \(k\).
2. Using your answer to part (i), solve the following simultaneous equations. $$\begin{aligned} 8 x - 7 y - 12 z & = 14 \\ - 10 x + 5 y + 15 z & = - 25 \\ - 9 x + 6 y + 6 z & = 3 \end{aligned}$$ You are also given that \(\mathbf { B } = \left( \begin{array} { r r r } - 7 & 5 & 15 \\ a & - 8 & - 21 \\ 2 & - 1 & - 3 \end{array} \right)\) and \(\mathbf { B } ^ { - 1 } = \frac { 1 } { 3 } \left( \begin{array} { r r r } 1 & 0 & 5 \\ - 4 & - 3 & 1 \\ 2 & 1 & b \end{array} \right)\).
3. Find the values of \(a\) and \(b\).
4. Write down an expression for \(( \mathbf { A B } ) ^ { - 1 }\) in terms of \(\mathbf { A } ^ { - 1 }\) and \(\mathbf { B } ^ { - 1 }\). Hence find \(( \mathbf { A B } ) ^ { - 1 }\).

1

1. Find the inverse of the matrix \(\mathbf { M } = \left( \begin{array} { r r } 4 & - 1 \\ 3 & 2 \end{array} \right)\).
2. Use this inverse to solve the simultaneous equations $$\begin{aligned} & 4 x - y = 49 \\ & 3 x + 2 y = 100 \end{aligned}$$ showing your working clearly.

2 You are given that \(\mathbf { M } = \left( \begin{array} { r r } 2 & - 5 \\ 3 & 7 \end{array} \right)\). \(\mathbf { M } \binom { x } { y } = \binom { 9 } { - 1 }\) represents two simultaneous equations.

1. Write down these two equations.
2. Find \(\mathbf { M } ^ { - 1 }\) and use it to solve the equations.

9 The simultaneous equations $$\begin{aligned} & 2 x - y = 1 \\ & 3 x + k y = b \end{aligned}$$ are represented by the matrix equation \(\mathbf { M } \binom { x } { y } = \binom { 1 } { b }\).

1. Write down the matrix \(\mathbf { M }\).
2. State the value of \(k\) for which \(\mathbf { M } ^ { - 1 }\) does not exist and find \(\mathbf { M } ^ { - 1 }\) in terms of \(k\) when \(\mathbf { M } ^ { - 1 }\) exists. Use \(\mathbf { M } ^ { - 1 }\) to solve the simultaneous equations when \(k = 5\) and \(b = 21\).
3. What can you say about the solutions of the equations when \(k = - \frac { 3 } { 2 }\) ?
4. The two equations can be interpreted as representing two lines in the \(x - y\) plane. Describe the relationship between these two lines
   (A) when \(k = 5\) and \(b = 21\),
   (B) when \(k = - \frac { 3 } { 2 }\) and \(b = 1\),
   (C) when \(k = - \frac { 3 } { 2 }\) and \(b = \frac { 3 } { 2 }\). RECOGNISING ACHIEVEMENT

9 You are given that \(\mathbf { A } = \left( \begin{array} { r r r } - 3 & - 4 & 1 \\ 2 & 1 & k \\ 7 & - 1 & - 1 \end{array} \right) , \mathbf { B } = \left( \begin{array} { r r c } - 4 & - 5 & 11 \\ - 19 & - 4 & - 7 \\ - 9 & - 31 & 2 - k \end{array} \right)\) and \(\mathbf { A B } = \left( \begin{array} { c c c } 79 & 0 & - 3 - k \\ - 9 k - 27 & - 31 k - 14 & q \\ p & 0 & 82 + k \end{array} \right)\) where \(p\) and \(q\) are to be determined.

1. Show that \(p = 0\) and \(q = 15 + 2 k - k ^ { 2 }\). It is now given that \(k = - 3\).
2. Find \(\mathbf { A B }\) and hence write down the inverse matrix \(\mathbf { A } ^ { - 1 }\).
3. Use a matrix method to find the values of \(x , y\) and \(z\) that satisfy the equation \(\mathbf { A } \left( \begin{array} { l } x \\ y \\ z \end{array} \right) = \left( \begin{array} { r } 14 \\ - 23 \\ 9 \end{array} \right)\). \section*{THERE ARE NO QUESTIONS WRITTEN ON THIS PAGE}

3 You are given that \(\mathbf { N } = \left( \begin{array} { r r r } - 9 & - 2 & - 4 \\ 3 & 2 & 2 \\ 5 & 1 & 2 \end{array} \right)\) and \(\mathbf { N } ^ { - 1 } = \left( \begin{array} { r r r } 1 & 0 & 2 \\ 2 & 1 & 3 \\ - \frac { 7 } { 2 } & p & - 6 \end{array} \right)\).

1. Find the value of \(p\).
2. Solve the equation \(\mathbf { N } \left( \begin{array} { c } x \\ y \\ z \end{array} \right) = \left( \begin{array} { r } - 39 \\ 5 \\ 22 \end{array} \right)\).

9 You are given that \(\mathbf { A } = \left( \begin{array} { r r r } 1 & 3 & - 1 \\ - 1 & \alpha & - 1 \\ - 2 & - 1 & 3 \end{array} \right) , \mathbf { B } = \left( \begin{array} { c c c } 3 \alpha - 1 & - 8 & \alpha - 3 \\ 5 & 1 & 2 \\ 2 \alpha + 1 & - 5 & \alpha + 3 \end{array} \right)\) and \(\mathbf { A B } = \left( \begin{array} { c c c } \gamma & 0 & 0 \\ \beta & \gamma & 0 \\ 0 & 0 & \gamma \end{array} \right)\).

1. Show that \(\beta = 0\).
2. Find \(\gamma\) in terms of \(\alpha\).
3. Write down \(\mathbf { A } ^ { - 1 }\) for the case when \(\alpha = 2\). State the value of \(\alpha\) for which \(\mathbf { A } ^ { - 1 }\) does not exist.
4. Use your answer to part (iii) to solve the following simultaneous equations. $$\begin{aligned} x + 3 y - z & = 25 \\ - x + 2 y - z & = 11 \\ - 2 x - y + 3 z & = - 23 \end{aligned}$$

1 Given that \(\mathbf { M } \binom { x } { y } = \binom { 1 } { 3 }\), where \(\mathbf { M } = \left( \begin{array} { r r } 4 & - 3 \\ 8 & 21 \end{array} \right)\), find \(x\) and \(y\).

The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M } = \left( \begin{array} { r r r r } 1 & 1 & 5 & 7 \\ 3 & 9 & 17 & 25 \\ 1 & 7 & 7 & 11 \\ 3 & 6 & 16 & 23 \end{array} \right)\).

1. In either order,
   1. show that the dimension of \(R\), the range space of T , is equal to 2 ,
   2. obtain a basis for \(R\).
   3. Show that the vector \(\left( \begin{array} { r } 1 \\ - 15 \\ - 17 \\ - 6 \end{array} \right)\) belongs to \(R\).
   4. It is given that \(\left\{ \mathbf { e } _ { 1 } , \mathbf { e } _ { 2 } \right\}\) is a basis for the null space of T , where \(\mathbf { e } _ { 1 } = \left( \begin{array} { r } 14 \\ 1 \\ - 3 \\ 0 \end{array} \right)\) and \(\mathbf { e } _ { 2 } = \left( \begin{array} { r } 19 \\ 2 \\ 0 \\ - 3 \end{array} \right)\). Show that, for all \(\lambda\) and \(\mu\), $$\mathbf { x } = \left( \begin{array} { r } 4 \\ - 3 \\ 0 \\ 0 \end{array} \right) + \lambda \mathbf { e } _ { 1 } + \mu \mathbf { e } _ { 2 }$$ is a solution of $$\mathbf { M x } = \left( \begin{array} { r } 1 \\ - 15 \\ - 17 \\ - 6 \end{array} \right)$$
   5. Hence find a solution of \(( * )\) of the form \(\left( \begin{array} { c } \alpha \\ 0 \\ \gamma \\ \delta \end{array} \right)\).

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 \(\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } + \delta ^ { 2 } = \frac { 11 } { 4 }\).

10 Find the set of values of \(a\) for which the system of equations $$\begin{aligned} x - 2 y - 2 z & = - 7 \\ 2 x + ( a - 9 ) y - 10 z & = - 11 \\ 3 x - 6 y + 2 a z & = - 29 \end{aligned}$$ has a unique solution. Show that the system has no solution in the case \(a = - 3\). Given that \(a = 5\),

1. show that the number of solutions is infinite,
2. find the solution for which \(x + y + z = 2\).

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 .

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.

has the form \(\left( \begin{array} { r } 1 \\ - 2 \\ 2 \\ - 1 \end{array} \right) + \lambda \mathbf { e } _ { 1 } + \mu \mathbf { e } _ { 2 }\), where \(\lambda\) and \(\mu\) are scalars and \(\left\{ \mathbf { e } _ { 1 } , \mathbf { e } _ { 2 } \right\}\) is a basis for \(K\). Hence obtain a solution \(\mathbf { x } ^ { \prime }\) of ( \(*\) ) such that the sum of the components \(\mathbf { x } ^ { \prime }\) is 6 and the sum of the squares of the components of \(\mathbf { x } ^ { \prime }\) is 26 . {www.cie.org.uk} after the live examination series. }

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.

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. \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\)

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 \(K _ { 2 }\).