Challenging +1.8 This is a Further Maths linear algebra question requiring row reduction to find rank, solving a non-homogeneous system, finding the null space, and applying a constraint. While systematic, it demands multiple techniques (rank calculation, particular/general solution structure, inequality analysis) and careful algebraic manipulation across several steps, placing it well above average difficulty but not at the extreme end for Further Maths content.
Find the rank of the matrix \(\mathbf{A}\), where
$$\mathbf{A} = \begin{pmatrix} 1 & 1 & 2 & 3 \\ 4 & 3 & 5 & 10 \\ 6 & 6 & 13 & 13 \\ 14 & 12 & 23 & 45 \end{pmatrix}.$$ [3]
Find vectors \(\mathbf{x_0}\) and \(\mathbf{e}\) such that any solution of the equation
$$\mathbf{A}\mathbf{x} = \begin{pmatrix} 0 \\ 2 \\ -1 \\ -3 \end{pmatrix} \quad (*)$$
can be expressed in the form \(\mathbf{x_0} + \lambda\mathbf{e}\), where \(\lambda \in \mathbb{R}\). [5]
Hence show that there is no vector which satisfies \((*)\) and has all its elements positive. [3]
Find the rank of the matrix $\mathbf{A}$, where
$$\mathbf{A} = \begin{pmatrix} 1 & 1 & 2 & 3 \\ 4 & 3 & 5 & 10 \\ 6 & 6 & 13 & 13 \\ 14 & 12 & 23 & 45 \end{pmatrix}.$$ [3]
Find vectors $\mathbf{x_0}$ and $\mathbf{e}$ such that any solution of the equation
$$\mathbf{A}\mathbf{x} = \begin{pmatrix} 0 \\ 2 \\ -1 \\ -3 \end{pmatrix} \quad (*)$$
can be expressed in the form $\mathbf{x_0} + \lambda\mathbf{e}$, where $\lambda \in \mathbb{R}$. [5]
Hence show that there is no vector which satisfies $(*)$ and has all its elements positive. [3]
\hfill \mbox{\textit{CAIE FP1 2005 Q11 [11]}}