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2003 Paper 2 Q1
D: 1600.0 B: 1516.0
3x3 matrices simultaneous equations linear systems Gaussian elimination consistency conditions parametric solution optimisation minimisation

Consider the equations \begin{alignat*}{2} ax-&y- \ z && =3 \;,\\ 2ax -&y -3z && = 7 \;,\\ 3ax-&y-5z && =b \;, \end{alignat*} where \(a\) and \(b\) are given constants.

  1. In the case \(a=0\,\), show that the equations have a solution if and only if \(b=11\,\).
  2. In the case \(a\ne0\) and \(b=11\,\) show that the equations have a solution with \(z=\lambda\) for any given number \(\lambda\,\).
  3. In the case \(a=2\) and \(b=11\,\) find the solution for which \(x^2+y^2+z^2\) is least.
  4. Find a value for \(a\) for which there is a solution such that \(x>10^6\) and \(y^2+z^2<1\,\).

Solution:

  1. If \(a = 0\), then then the LHS second equation is the average of the first and last equations, ie \(7 = \frac{b+3}{2}\) so \(b = 11\). This clearly has solutions, say \(x = 0, y = -1, z = -2\).
  2. If \(a \neq 0\) and \(b = 11\), it is still the case that the third equation a linear combination of the first two. Therefore we can consider the linear system: \begin{cases} ax - y &= 3 + \lambda \\ 2ax - y &= 7 + 3\lambda \end{cases} and since \(-a+2a = a \neq 0\) the solution has a unique solution for \(x\) and \(y\).
  3. \begin{align*} \begin{cases} 2x - y &= 3 + \lambda \\ 4x - y &= 7 + 3\lambda \end{cases} \Rightarrow x = 2 +\lambda, y = 1 + \lambda \\ x^2 + y^2 + z^2 &= (2 + \lambda)^2 + (1+\lambda)^2 + \lambda^2 \\ &= (4 + 1) + (4+2)\lambda + 3\lambda^2 \\ &= 5 + 3((\lambda+1)^2 - 1) \\ &= 3(\lambda + 1)^2 + 2 \end{align*} Therefore the solution is minimized when \(\lambda = -1, x = 1, y = 0, z = -1\)
  4. \begin{align*} \begin{cases} ax - y &= 3 + \lambda \\ 2ax - y &= 7 + 3\lambda \end{cases} \Rightarrow x = \frac{4 +2\lambda}{a}, y = 1 + \lambda \end{align*} We want say \(\lambda = -\frac12\) then we have \(y^2 + z^2 = \frac12\) and \(x = \frac{3}{a}\), so choose \(a < \frac{3}{10^6}\)

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1996 Paper 3 Q2
D: 1700.0 B: 1516.0
3x3 matrices simultaneous equations linear systems parameter analysis Gaussian elimination consistency conditions

For all values of \(a\) and \(b,\) either solve the simultaneous equations \begin{alignat*}{1} x+y+az & =2\\ x+ay+z & =2\\ 2x+y+z & =2b \end{alignat*} or prove that they have no solution.

Solution: Consider the matrix system: \begin{align*} \left(\begin{array}{ccc|c} 1 & 1 & a & 2 \\ 1 & a & 1 & 2 \\ 2 & 1 & 1 & 2b \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & 1 & a & 2 \\ 0 & a-1 & 1-a & 0 \\ 0 & -1 & 1-2a & 2b-4 \\ \end{array}\right)\\ \left(\begin{array}{ccc|c} 1 & 1 & a & 2 \\ 0 & a-1 & 1-a & 0 \\ 0 & 0 & -2a & 2b-4 \\ \end{array}\right) \\ \end{align*} Assuming that \(a \neq 1, 0\) all steps are fine and: \(z = \frac{2-b}{a}, y = \frac{2-b}{a}, x +(1+a)y = 2, x = 2 - \frac{(2-b)(1+a)}{a} = \frac{ab+b-2}{a}\) If \(a = 0\), \(y = z\) and \(\begin{cases} x + y &= 2 \\ 2x + 2y &= 2b \end{cases} \Rightarrow b= 2, x = t, y = 2-t, z = 2-t\) If \(a = 1\), \(x = 2b-2, y = t, z = 4-t-2b\), where \(t \in \mathbb{R}\)

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