4.03o Inverse 3x3 matrix

3 Let \(\mathbf { M } = \left( \begin{array} { r r r } 3 & 5 & 2 \\ 5 & 3 & - 2 \\ 2 & - 2 & - 4 \end{array} \right)\).

1. Show that the characteristic equation for \(\mathbf { M }\) is \(\lambda ^ { 3 } - 2 \lambda ^ { 2 } - 48 \lambda = 0\). You are given that \(\left( \begin{array} { r } 1 \\ - 1 \\ 1 \end{array} \right)\) is an eigenvector of \(\mathbf { M }\) corresponding to the eigenvalue 0 .
2. Find the other two eigenvalues of \(\mathbf { M }\), and corresponding eigenvectors.
3. Write down a matrix \(\mathbf { P }\), and a diagonal matrix \(\mathbf { D }\), such that \(\mathbf { P } ^ { - 1 } \mathbf { M } ^ { 2 } \mathbf { P } = \mathbf { D }\).
4. Use the Cayley-Hamilton theorem to find integers \(a\) and \(b\) such that \(\mathbf { M } ^ { 4 } = a \mathbf { M } ^ { 2 } + b \mathbf { M }\). Section B (18 marks)

3

1. Given the matrix \(\mathbf { Q } = \left( \begin{array} { r r r } 2 & - 1 & k \\ 1 & 0 & 1 \\ 3 & 1 & 2 \end{array} \right)\) (where \(k \neq 3\) ), find \(\mathbf { Q } ^ { - 1 }\) in terms of \(k\).
   Show that, when \(k = 4 , \mathbf { Q } ^ { - 1 } = \left( \begin{array} { r r r } - 1 & 6 & - 1 \\ 1 & - 8 & 2 \\ 1 & - 5 & 1 \end{array} \right)\). The matrix \(\mathbf { M }\) has eigenvectors \(\left( \begin{array} { l } 2 \\ 1 \\ 3 \end{array} \right) , \left( \begin{array} { r } - 1 \\ 0 \\ 1 \end{array} \right)\) and \(\left( \begin{array} { l } 4 \\ 1 \\ 2 \end{array} \right)\), with corresponding eigenvalues \(1 , - 1\) and 3 respectively.
2. Write down a matrix \(\mathbf { P }\) and a diagonal matrix \(\mathbf { D }\) such that \(\mathbf { P } ^ { - 1 } \mathbf { M P } = \mathbf { D }\), and hence find the matrix \(\mathbf { M }\).
3. Write down the characteristic equation for \(\mathbf { M }\), and use the Cayley-Hamilton theorem to find integers \(a , b\) and \(c\) such that \(\mathbf { M } ^ { 4 } = a \mathbf { M } ^ { 2 } + b \mathbf { M } + c \mathbf { I }\). Section B (18 marks)

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

1. Find \(\mathbf { D } ^ { - 1 }\).
2. Hence, or otherwise, solve the equations $$\begin{aligned} a x + 2 y & = 3 \\ 3 x + y + 2 z & = 4 \\ - y + z & = 1 \end{aligned}$$

10 The matrix \(\mathbf { A }\) is given by \(\mathbf { A } = \left( \begin{array} { r r r } a & 8 & 10 \\ 2 & 1 & 2 \\ 4 & 3 & 6 \end{array} \right)\). The matrix \(\mathbf { B }\) is such that \(\mathbf { A B } = \left( \begin{array} { l l l } a & 6 & 1 \\ 1 & 1 & 0 \\ 1 & 3 & 0 \end{array} \right)\).

1. Show that \(\mathbf { A B }\) is non-singular.
2. Find \(( \mathbf { A B } ) ^ { - 1 }\).
3. Find \(\mathbf { B } ^ { - 1 }\).

10 The matrix \(\mathbf { A }\) is given by \(\mathbf { A } = \left( \begin{array} { l l l } a & 2 & 1 \\ 1 & 3 & 2 \\ 4 & 1 & 1 \end{array} \right)\).

1. Find the value of \(a\) for which \(\mathbf { A }\) is singular.
2. Given that \(\mathbf { A }\) is non-singular, find \(\mathbf { A } ^ { - 1 }\) and hence solve the equations $$\begin{aligned} a x + 2 y + z & = 1 \\ x + 3 y + 2 z & = 2 \\ 4 x + y + z & = 3 \end{aligned}$$

8 The matrix \(\mathbf { M }\) is given by \(\mathbf { M } = \left( \begin{array} { r r r } a & 2 & - 1 \\ 2 & 3 & - 1 \\ 2 & - 1 & 1 \end{array} \right)\), where \(a\) is a constant.

1. Show that the determinant of \(\mathbf { M }\) is \(2 a\).
2. Given that \(a \neq 0\), find the inverse matrix \(\mathbf { M } ^ { - 1 }\).
3. Hence or otherwise solve the simultaneous equations $$\begin{array} { r } x + 2 y - z = 1 \\ 2 x + 3 y - z = 2 \\ 2 x - y + z = 0 \end{array}$$
4. Find the value of \(k\) for which the simultaneous equations $$\begin{array} { r } 2 y - z = k \\ 2 x + 3 y - z = 2 \\ 2 x - y + z = 0 \end{array}$$ have solutions.
5. Do the equations in part (iv), with the value of \(k\) found, have a solution for which \(x = z\) ? Justify your answer.

5 You are given that \(\mathbf { A } = \left( \begin{array} { l l l } 1 & 2 & 4 \\ 3 & 2 & 5 \\ 4 & 1 & 2 \end{array} \right)\) and \(\mathbf { B } = \left( \begin{array} { r r r } - 1 & 0 & 2 \\ 14 & - 14 & 7 \\ - 5 & 7 & - 4 \end{array} \right)\).

1. Calculate AB.
2. Write down \(\mathbf { A } ^ { - 1 }\).

3 Let \(\mathbf { P } = \left( \begin{array} { r r r } 4 & 2 & k \\ 1 & 1 & 3 \\ 1 & 0 & - 1 \end{array} \right) (\) where \(k \neq 4 )\) and \(\mathbf { M } = \left( \begin{array} { r r r } 2 & - 2 & - 6 \\ - 1 & 3 & 1 \\ 1 & - 2 & - 2 \end{array} \right)\).

1. Find \(\mathbf { P } ^ { - 1 }\) in terms of \(k\), and show that, when \(k = 2 , \mathbf { P } ^ { - 1 } = \frac { 1 } { 2 } \left( \begin{array} { r r r } - 1 & 2 & 4 \\ 4 & - 6 & - 10 \\ - 1 & 2 & 2 \end{array} \right)\).
2. Verify that \(\left( \begin{array} { l } 4 \\ 1 \\ 1 \end{array} \right) , \left( \begin{array} { l } 2 \\ 1 \\ 0 \end{array} \right)\) and \(\left( \begin{array} { r } 2 \\ 3 \\ - 1 \end{array} \right)\) are eigenvectors of \(\mathbf { M }\), and find the corresponding eigenvalues.
3. Show that \(\mathbf { M } ^ { n } = \left( \begin{array} { r r r } 4 & - 6 & - 10 \\ 2 & - 3 & - 5 \\ 0 & 0 & 0 \end{array} \right) + 2 ^ { n - 1 } \left( \begin{array} { r r r } - 2 & 4 & 4 \\ - 3 & 6 & 6 \\ 1 & - 2 & - 2 \end{array} \right)\). Section B (18 marks)

3

1. Find the inverse of the matrix $$\left( \begin{array} { r r r } 1 & 1 & a \\ 2 & - 1 & 2 \\ 3 & - 2 & 2 \end{array} \right)$$ where \(a \neq 4\).
   Show that when \(a = - 1\) the inverse is $$\frac { 1 } { 5 } \left( \begin{array} { r r r } 2 & 0 & 1 \\ 2 & 5 & - 4 \\ - 1 & 5 & - 3 \end{array} \right)$$
2. Solve, in terms of \(b\), the following system of equations. $$\begin{aligned} x + y - z & = - 2 \\ 2 x - y + 2 z & = b \\ 3 x - 2 y + 2 z & = 1 \end{aligned}$$
3. Find the value of \(b\) for which the equations $$\begin{aligned} x + y + 4 z & = - 2 \\ 2 x - y + 2 z & = b \\ 3 x - 2 y + 2 z & = 1 \end{aligned}$$ have solutions. Give a geometrical interpretation of the solutions in this case. Section B (18 marks)

3

1. Find the value of \(a\) for which the matrix $$\mathbf { M } = \left( \begin{array} { r r r } 1 & 2 & 3 \\ - 1 & a & 4 \\ 3 & - 2 & 2 \end{array} \right)$$ does not have an inverse.
   Assuming that \(a\) does not have this value, find the inverse of \(\mathbf { M }\) in terms of \(a\).
2. Hence solve the following system of equations. $$\begin{aligned} x + 2 y + 3 z & = 1 \\ - x + 4 z & = - 2 \\ 3 x - 2 y + 2 z & = 1 \end{aligned}$$
3. Find the value of \(b\) for which the following system of equations has a solution. $$\begin{aligned} x + 2 y + 3 z & = 1 \\ - x + 6 y + 4 z & = - 2 \\ 3 x - 2 y + 2 z & = b \end{aligned}$$ Find the general solution in this case and describe the solution geometrically.

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 }

3

1. Find the value of \(k\) for which the matrix $$\mathbf { M } = \left( \begin{array} { r r r } 1 & - 1 & k \\ 5 & 4 & 6 \\ 3 & 2 & 4 \end{array} \right)$$ does not have an inverse.
   Assuming that \(k\) does not take this value, find the inverse of \(\mathbf { M }\) in terms of \(k\).
2. In the case \(k = 3\), evaluate $$\mathbf { M } \left( \begin{array} { r } - 3 \\ 3 \\ 1 \end{array} \right)$$
3. State the significance of what you have found in part (ii).
4. Find the value of \(t\) for which the system of equations $$\begin{array} { r } x - y + 3 z = t \\ 5 x + 4 y + 6 z = 1 \\ 3 x + 2 y + 4 z = 0 \end{array}$$ has solutions. Find the general solution in this case and describe the solution geometrically.

4 Given that \(\mathbf { A }\) and \(\mathbf { B }\) are \(2 \times 2\) non-singular matrices and \(\mathbf { I }\) is the \(2 \times 2\) identity matrix, simplify $$\mathbf { B } ( \mathbf { A B } ) ^ { - 1 } \mathbf { A } - \mathbf { I } .$$

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 $$\begin{array} { r } 2 x - y + z = 1 \\ 3 y + z = 2 \\ x + y + a z = 2 \end{array}$$

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

1. Find the determinant of \(\mathbf { X }\) in terms of \(a\).
2. Hence find the values of \(a\) for which \(\mathbf { X }\) is singular.
3. Given that \(\mathbf { X }\) is non-singular, find \(\mathbf { X } ^ { - 1 }\) in terms of \(a\).

6 The matrix \(\mathbf { C }\) is given by \(\mathbf { C } = \left( \begin{array} { r r r } a & 1 & 0 \\ 1 & 2 & 1 \\ - 1 & 3 & 4 \end{array} \right)\), where \(a \neq 1\). Find \(\mathbf { C } ^ { - 1 }\).

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

1. Find \(\mathbf { A } ^ { - 1 }\). The matrix \(\mathbf { B }\) is given by \(\mathbf { B } = \left( \begin{array} { l l } 2 & a \\ 4 & 1 \end{array} \right)\).
2. Given that \(\mathbf { P A } = \mathbf { B }\), find the matrix \(\mathbf { P }\).

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 The matrix \(\mathbf { R }\) is \(\left( \begin{array} { r r } 0 & - 1 \\ 1 & 0 \end{array} \right)\).

1. Explain in terms of transformations why \(\mathbf { R } ^ { 4 } = \mathbf { I }\).
2. Describe the transformation represented by \(\mathbf { R } ^ { - 1 }\) and write down the matrix \(\mathbf { R } ^ { - 1 }\).
3. \(\mathbf { S }\) is the matrix representing rotation through \(60 ^ { \circ }\) anticlockwise about the origin. Find \(\mathbf { S }\).
4. Write down the smallest positive integers \(m\) and \(n\) such that \(\mathbf { S } ^ { m } = \mathbf { R } ^ { n }\), explaining your answer in terms of transformations.
5. Find \(\mathbf { R S }\) and explain in terms of transformations why \(\mathbf { R S } = \mathbf { S R }\). \section*{THERE ARE NO QUESTIONS WRITTEN ON THIS PAGE}

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 \(( \mathbf { A B } ) ^ { - 1 }\).

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. $$\begin{aligned} x + 3 y - z & = 25 \\ - x + 2 y - z & = 11 \\ - 2 x - y + 3 z & = - 23 \end{aligned}$$