4.03s Consistent/inconsistent: systems of equations

80 questions

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CAIE Further Paper 2 2020 November Q3
9 marks Standard +0.8
3
  1. Show that the system of equations $$\begin{array} { r } x - 2 y - 4 z = 1 \\ x - 2 y + k z = 1 \\ - x + 2 y + 2 z = 1 \end{array}$$ where \(k\) is a constant, does not have a unique solution.
  2. Given that \(k = - 4\), show that the system of equations in part (a) is consistent. Interpret this situation geometrically.
  3. Given instead that \(k = - 2\), show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically.
  4. For the case where \(k \neq - 2\) and \(k \neq - 4\), show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically. \includegraphics[max width=\textwidth, alt={}, center]{7da7fa35-1b97-4708-a1a2-cba9e35c8bf0-06_894_841_260_612} The diagram shows the curve with equation \(\mathrm { y } = 1 - \mathrm { x } ^ { 3 }\) for \(0 \leqslant x \leqslant 1\), together with a set of \(n\) rectangles of width \(\frac { 1 } { n }\).
CAIE Further Paper 2 2020 November Q9
16 marks Standard +0.8
9 It is given that \(a\) is a positive constant.
  1. Show that the system of equations $$\begin{aligned} a x + ( 2 a + 5 ) y + ( a + 1 ) z & = 1 \\ - 4 y & = 2 \\ 3 y - z & = 3 \end{aligned}$$ has a unique solution and interpret this situation geometrically.
    The matrix \(\mathbf { A }\) is given by $$\mathbf { A } = \left( \begin{array} { c c c } a & 2 a + 5 & a + 1 \\ 0 & - 4 & 0 \\ 0 & 3 & - 1 \end{array} \right)$$
  2. Show that the eigenvalues of \(\mathbf { A }\) are \(a , - 1\) and - 4 .
  3. Find a matrix \(\mathbf { P }\) such that $$\mathbf { A } = \mathbf { P } \left( \begin{array} { r r r } a & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 4 \end{array} \right) \mathbf { P } ^ { - 1 } .$$
  4. Use the characteristic equation of \(\mathbf { A }\) to find \(\mathbf { A } ^ { - 1 }\).
    If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
CAIE Further Paper 2 2020 November Q3
9 marks Standard +0.8
3
  1. Show that the system of equations $$\begin{array} { r } x - 2 y - 4 z = 1 \\ x - 2 y + k z = 1 \\ - x + 2 y + 2 z = 1 \end{array}$$ where \(k\) is a constant, does not have a unique solution.
  2. Given that \(k = - 4\), show that the system of equations in part (a) is consistent. Interpret this situation geometrically.
  3. Given instead that \(k = - 2\), show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically.
  4. For the case where \(k \neq - 2\) and \(k \neq - 4\), show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically. \includegraphics[max width=\textwidth, alt={}, center]{23c7189f-850d-4745-a8ce-46a140ed0176-06_894_841_260_612} The diagram shows the curve with equation \(\mathrm { y } = 1 - \mathrm { x } ^ { 3 }\) for \(0 \leqslant x \leqslant 1\), together with a set of \(n\) rectangles of width \(\frac { 1 } { n }\).
CAIE Further Paper 2 2022 November Q2
7 marks Standard +0.8
2
  1. Show that the system of equations $$\begin{aligned} & x - y + 2 z = 4 \\ & x - y - 3 z = a \\ & x - y + 7 z = 13 \end{aligned}$$ where \(a\) is a constant, does not have a unique solution.
  2. Given that \(a = - 5\), show that the system of equations in part (a) is consistent. Interpret this situation geometrically.
  3. Given instead that \(a \neq - 5\), show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically.
CAIE Further Paper 2 2022 November Q1
5 marks Standard +0.3
1
  1. Find the set of values of \(k\) for which the system of equations $$\begin{aligned} x + 2 y + 3 z & = 1 \\ k x + 4 y + 6 z & = 0 \\ 7 x + 8 y + 9 z & = 3 \end{aligned}$$ has a unique solution.
  2. Interpret the situation geometrically in the case where the system of equations does not have a unique solution.
CAIE Further Paper 2 2023 November Q1
4 marks Standard +0.8
1 Show that the system of equations $$\begin{aligned} 14 x - 4 y + 6 z & = 5 \\ x + y + k z & = 3 \\ - 21 x + 6 y - 9 z & = 14 \end{aligned}$$ where \(k\) is a constant, does not have a unique solution and interpret this situation geometrically.
CAIE Further Paper 2 2024 November Q1
4 marks Standard +0.8
1 Find the set of values of \(k\) for which the system of equations $$\begin{array} { r } x + 5 y + 6 z = 1 \\ k x + 2 y + 2 z = 2 \\ - 3 x + 4 y + 8 z = 3 \end{array}$$ has a unique solution and interpret this situation geometrically.
Edexcel FP3 2015 June Q3
12 marks Standard +0.3
3. $$\mathbf { A } = \left( \begin{array} { l l l } 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{array} \right)$$
  1. Find the eigenvalues of \(\mathbf { A }\).
  2. Find a normalised eigenvector for each of the eigenvalues of \(\mathbf { A }\).
  3. Write down a matrix \(\mathbf { P }\) and a diagonal matrix \(\mathbf { D }\) such that \(\mathbf { P } ^ { \mathrm { T } } \mathbf { A P } = \mathbf { D }\).
Edexcel FP3 2016 June Q6
11 marks Standard +0.8
6. $$\mathbf { M } = \left( \begin{array} { r r r } p & - 2 & 0 \\ - 2 & 6 & - 2 \\ 0 & - 2 & q \end{array} \right)$$ where \(p\) and \(q\) are constants.
Given that \(\left( \begin{array} { r } 2 \\ - 2 \\ 1 \end{array} \right)\) is an eigenvector of the matrix \(\mathbf { M }\),
  1. find the eigenvalue corresponding to this eigenvector,
  2. find the value of \(p\) and the value of \(q\). Given that 6 is another eigenvalue of \(\mathbf { M }\),
  3. find a corresponding eigenvector. Given that \(\left( \begin{array} { l } 1 \\ 2 \\ 2 \end{array} \right)\) is a third eigenvector of \(\mathbf { M }\) with eigenvalue 3
  4. find a matrix \(\mathbf { P }\) and a diagonal matrix \(\mathbf { D }\) such that $$\mathbf { P } ^ { \mathrm { T } } \mathbf { M } \mathbf { P } = \mathbf { D }$$
OCR MEI FP2 2006 June Q3
18 marks Standard +0.8
3
  1. Find the inverse of the matrix \(\left( \begin{array} { r r r } 4 & 1 & k \\ 3 & 2 & 5 \\ 8 & 5 & 13 \end{array} \right)\), where \(k \neq 5\).
  2. Solve the simultaneous equations $$\begin{aligned} & 4 x + y + 7 z = 12 \\ & 3 x + 2 y + 5 z = m \\ & 8 x + 5 y + 13 z = 0 \end{aligned}$$ giving \(x , y\) and \(z\) in terms of \(m\).
  3. Find the value of \(p\) for which the simultaneous equations $$\begin{aligned} & 4 x + y + 5 z = 12 \\ & 3 x + 2 y + 5 z = p \\ & 8 x + 5 y + 13 z = 0 \end{aligned}$$ have solutions, and find the general solution in this case.
OCR FP1 2007 June Q7
8 marks Standard +0.3
7 The matrix \(\mathbf { M }\) is given by \(\mathbf { M } = \left( \begin{array} { l l l } a & 4 & 0 \\ 0 & a & 4 \\ 2 & 3 & 1 \end{array} \right)\).
  1. Find, in terms of \(a\), the determinant of \(\mathbf { M }\).
  2. In the case when \(a = 2\), state whether \(\mathbf { M }\) is singular or non-singular, justifying your answer.
  3. In the case when \(a = 4\), determine whether the simultaneous equations $$\begin{aligned} a x + 4 y \quad = & 6 \\ a y + 4 z & = 8 \\ 2 x + 3 y + z & = 1 \end{aligned}$$ have any solutions.
OCR FP1 Specimen Q8
14 marks Standard +0.8
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.
OCR MEI FP2 2011 June Q3
18 marks Challenging +1.2
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.
OCR FP1 2009 January Q5
5 marks Standard +0.3
5 By using the determinant of an appropriate matrix, or otherwise, find the value of \(k\) for which the simultaneous equations $$\begin{aligned} 2 x - y + z & = 7 \\ 3 y + z & = 4 \\ x + k y + k z & = 5 \end{aligned}$$ do not have a unique solution for \(x , y\) and \(z\).
OCR FP1 2011 January Q9
9 marks Standard +0.3
9 The matrix \(\mathbf { M }\) is given by \(\mathbf { M } = \left( \begin{array} { r r r } a & - a & 1 \\ 3 & a & 1 \\ 4 & 2 & 1 \end{array} \right)\).
  1. Find, in terms of \(a\), the determinant of \(\mathbf { M }\).
  2. Hence find the values of \(a\) for which \(\mathbf { M } ^ { - 1 }\) does not exist.
  3. Determine whether the simultaneous equations $$\begin{aligned} & 6 x - 6 y + z = 3 k \\ & 3 x + 6 y + z = 0 \\ & 4 x + 2 y + z = k \end{aligned}$$ where \(k\) is a non-zero constant, have a unique solution, no solution or an infinite number of solutions, justifying your answer.
  4. Show that \(\frac { 1 } { r } - \frac { 2 } { r + 1 } + \frac { 1 } { r + 2 } \equiv \frac { 2 } { r ( r + 1 ) ( r + 2 ) }\).
  5. Hence find an expression, in terms of \(n\), for $$\sum _ { r = 1 } ^ { n } \frac { 2 } { r ( r + 1 ) ( r + 2 ) }$$
  6. Show that \(\sum _ { r = n + 1 } ^ { \infty } \frac { 2 } { r ( r + 1 ) ( r + 2 ) } = \frac { 1 } { ( n + 1 ) ( n + 2 ) }\).
OCR FP1 2009 June Q9
10 marks Standard +0.3
9 The matrix \(\mathbf { A }\) is given by \(\mathbf { A } = \left( \begin{array} { l l l } a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & 2 \end{array} \right)\).
  1. Find, in terms of \(a\), the determinant of \(\mathbf { A }\).
  2. Hence find the values of \(a\) for which \(\mathbf { A }\) is singular.
  3. State, giving a brief reason in each case, whether the simultaneous equations $$\begin{aligned} a x + y + z & = 2 a \\ x + a y + z & = - 1 \\ x + y + 2 z & = - 1 \end{aligned}$$ have any solutions when
    1. \(a = 0\),
    2. \(a = 1\).
OCR FP1 2012 June Q10
10 marks Standard +0.3
10 The matrix \(\mathbf { D }\) is given by \(\mathbf { D } = \left( \begin{array} { r r r } a & 2 & - 1 \\ 2 & a & 1 \\ 1 & 1 & a \end{array} \right)\).
  1. Find the determinant of \(\mathbf { D }\) in terms of \(a\).
  2. Three simultaneous equations are shown below. $$\begin{array} { r } a x + 2 y - z = 0 \\ 2 x + a y + z = a \\ x + y + a z = a \end{array}$$ For each of the following values of \(a\), determine whether or not there is a unique solution. If the solution is not unique, determine whether the equations are consistent or inconsistent.
    1. \(\quad a = 3\)
    2. \(a = 2\)
    3. \(\quad a = 0\) \section*{THERE ARE NO QUESTIONS WRITTEN ON THIS PAGE}
OCR FP1 2015 June Q9
10 marks Standard +0.3
9 The matrix \(\mathbf { D }\) is given by \(\mathbf { D } = \left( \begin{array} { l l l } 1 & 3 & 4 \\ 2 & a & 3 \\ 0 & 1 & a \end{array} \right)\).
  1. Find the values of \(a\) for which \(\mathbf { D }\) is singular.
  2. Three simultaneous equations are shown below. $$\begin{array} { r } x + 3 y + 4 z = 3 \\ 2 x + a y + 3 z = 2 \\ y + a z = 0 \end{array}$$ For each of the following values of \(a\), determine whether or not there is a unique solution. If a unique solution does not exist, determine whether the equations are consistent or inconsistent.
    1. \(a = 3\)
    2. \(a = 1\)
OCR FP1 2016 June Q9
6 marks Standard +0.3
9
  1. The matrix \(\mathbf { X }\) is given by \(\mathbf { X } = \left( \begin{array} { r r r } a & 3 & - 2 \\ 0 & a & 5 \\ 1 & 2 & 1 \end{array} \right)\). Show that the determinant of \(\mathbf { X }\) is \(a ^ { 2 } - 8 a + 15\).
  2. Explain briefly why the equations $$\begin{array} { r } 3 x + 3 y - 2 z = 1 \\ 3 y + 5 z = 5 \\ x + 2 y + z = 2 \end{array}$$ do not have a unique solution and determine whether these equations are consistent or inconsistent.
  3. Use an algebraic method to find the square roots of the complex number \(9 + 40 \mathrm { i }\).
  4. Show that \(9 + 40 \mathrm { i }\) is a root of the quadratic equation \(z ^ { 2 } - 18 z + 1681 = 0\).
  5. By using the substitution \(z = \frac { 1 } { u ^ { 2 } }\), find the roots of the equation \(1681 u ^ { 4 } - 18 u ^ { 2 } + 1 = 0\). Give your answers in the form \(x + \mathrm { i } y\), where \(x\) and \(y\) are real.
CAIE FP1 2008 June Q12 OR
Challenging +1.8
The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix $$\left( \begin{array} { r r r r } 1 & 2 & - 1 & - 1 \\ 1 & 3 & - 1 & 0 \\ 1 & 0 & 3 & 1 \\ 0 & 3 & - 4 & - 1 \end{array} \right) .$$ The range space of T is denoted by \(V\).
  1. Determine the dimension of \(V\).
  2. Show that the vectors \(\left( \begin{array} { l } 1 \\ 1 \\ 1 \\ 0 \end{array} \right) , \left( \begin{array} { l } 2 \\ 3 \\ 0 \\ 3 \end{array} \right) , \left( \begin{array} { r } - 1 \\ - 1 \\ 3 \\ - 4 \end{array} \right)\) are linearly independent.
  3. Write down a basis of \(V\). The set of elements of \(\mathbb { R } ^ { 4 }\) which do not belong to \(V\) is denoted by \(W\).
  4. State, with a reason, whether \(W\) is a vector space.
  5. Show that if the vector \(\left( \begin{array} { l } x \\ y \\ z \\ t \end{array} \right)\) belongs to \(W\) then \(y - z - t \neq 0\).
CAIE FP1 2009 June Q12 OR
Challenging +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 $$\mathbf { M } _ { 1 } = \left( \begin{array} { r r r r } 1 & 1 & 1 & 2 \\ 1 & 4 & 7 & 8 \\ 1 & 7 & 11 & 13 \\ 1 & 2 & 5 & 5 \end{array} \right) , \quad \mathbf { M } _ { 2 } = \left( \begin{array} { r r r r } 2 & 0 & - 1 & - 1 \\ 5 & 1 & - 3 & - 3 \\ 3 & - 1 & - 1 & - 1 \\ 13 & - 1 & - 6 & - 6 \end{array} \right) .$$
  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 \(\mathrm { T } _ { 3 }\).
CAIE FP1 2010 June Q10
11 marks Challenging +1.2
10 Find the set of values of \(a\) for which the system of equations $$\begin{aligned} x + 4 y + 12 z & = 5 \\ 2 x + a y + 12 z & = a - 1 \\ 3 x + 12 y + 2 a z & = 10 \end{aligned}$$ has a unique solution. Show that the system does not have any solution in the case \(a = 18\). Given that \(a = 8\), show that the number of solutions is infinite and find the solution for which \(x + y + z = 1\).
CAIE FP1 2010 June Q12 OR
Challenging +1.2
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 \(\left( \begin{array} { c } \alpha \\ 0 \\ \gamma \\ \delta \end{array} \right)\).
CAIE FP1 2011 June Q11 OR
Challenging +1.8
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 \(\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } + \delta ^ { 2 } = \frac { 11 } { 4 }\).
CAIE FP1 2012 June Q10
11 marks Challenging +1.2
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 \(x + y + z = 2\).