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2022 Paper 2 Q8
D: 1500.0 B: 1500.0
invariant lines matrices quadratic equation trigonometric identities angle between lines linear transformations

Let \(\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) be a real matrix with \(a \neq d\). The transformation represented by \(\mathbf{M}\) has exactly two distinct invariant lines through the origin.

  1. Show that, if neither invariant line is the \(y\)-axis, then the gradients of the invariant lines are the roots of the equation \[bm^2 + (a-d)m - c = 0.\] If one invariant line is the \(y\)-axis, what is the gradient of the other?
  2. Show that, if the angle between the two invariant lines is \(45^\circ\), then \[(a-d)^2 = (b-c)^2 - 4bc.\]
  3. Find a necessary and sufficient condition, on some or all of \(a\), \(b\), \(c\) and \(d\), for the two invariant lines to make equal angles with the line \(y = x\).
  4. Give an example of a matrix which satisfies both the conditions in parts (ii) and (iii).

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2020 Paper 2 Q6
D: 1500.0 B: 1500.0
matrices trace determinant proof matrix powers necessary and sufficient conditions involutory matrix

A \(2 \times 2\) matrix \(\mathbf{M}\) is real if it can be written as \(\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), where \(a\), \(b\), \(c\) and \(d\) are real. In this case, the \emph{trace} of matrix \(\mathbf{M}\) is defined to be \(\mathrm{tr}(\mathbf{M}) = a + d\) and \(\det(\mathbf{M})\) is the determinant of matrix \(\mathbf{M}\). In this question, \(\mathbf{M}\) is a real \(2 \times 2\) matrix.

  1. Prove that \[\mathrm{tr}(\mathbf{M}^2) = \mathrm{tr}(\mathbf{M})^2 - 2\det(\mathbf{M}).\]
  2. Prove that \[\mathbf{M}^2 = \mathbf{I} \text{ but } \mathbf{M} \neq \pm\mathbf{I} \iff \mathrm{tr}(\mathbf{M}) = 0 \text{ and } \det(\mathbf{M}) = -1,\] and that \[\mathbf{M}^2 = -\mathbf{I} \iff \mathrm{tr}(\mathbf{M}) = 0 \text{ and } \det(\mathbf{M}) = 1.\]
  3. Use part (ii) to prove that \[\mathbf{M}^4 = \mathbf{I} \iff \mathbf{M}^2 = \pm\mathbf{I}.\] Find a necessary and sufficient condition on \(\det(\mathbf{M})\) and \(\mathrm{tr}(\mathbf{M})\) so that \(\mathbf{M}^4 = -\mathbf{I}\).
  4. Give an example of a matrix \(\mathbf{M}\) for which \(\mathbf{M}^8 = \mathbf{I}\), but which does not represent a rotation or reflection. [Note that the matrices \(\pm\mathbf{I}\) are both rotations.]

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2019 Paper 2 Q8
D: 1500.0 B: 1638.7
matrices multiplicative function determinant 2x2 matrix identity matrix matrix properties proof function properties

The domain of the function f is the set of all \(2 \times 2\) matrices and its range is the set of real numbers. Thus, if \(M\) is a \(2 \times 2\) matrix, then \(f(M) \in \mathbb{R}\). The function f has the property that \(f(MN) = f(M)f(N)\) for any \(2 \times 2\) matrices \(M\) and \(N\).

  1. You are given that there is a matrix \(M\) such that \(f(M) \neq 0\). Let \(I\) be the \(2 \times 2\) identity matrix. By considering \(f(MI)\), show that \(f(I) = 1\).
  2. Let \(J = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\). You are given that \(f(J) \neq 1\). By considering \(J^2\), evaluate \(f(J)\). Using \(J\), show that, for any real numbers \(a\), \(b\), \(c\) and \(d\), $$.f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) = -f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right) = f\left(\begin{pmatrix} d & c \\ b & a \end{pmatrix}\right)$$
  3. Let \(K = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}\) where \(k \in \mathbb{R}\). Use \(K\) to show that, if the second row of the matrix \(A\) is a multiple of the first row, then \(f(A) = 0\).
  4. Let \(P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\). By considering the matrices \(P^2\), \(P^{-1}\), and \(K^{-1}PK\) for suitable values of \(k\), evaluate \(f(P)\).

Solution:

  1. Consider \(f(M) = f(MI) = f(M)f(I)\). Since \(f(M) \neq 0\) we can divide by \(f(M)\) to obtain \(f(I) = 1\)
  2. Let \(J = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\), then \(J^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I\). Therefore \(1 = f(I) = f(J^2) = f(J)f(J) \Rightarrow f(J) = \pm 1 \Rightarrow f(J) = -1\) since \(f(J) \neq 1\). \begin{align*} \begin{pmatrix} a & b \\ c & d \end{pmatrix}J &= \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\ &= \begin{pmatrix} b & a \\ d & c \end{pmatrix} \\ J\begin{pmatrix} a & b \\ c & d \end{pmatrix} &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} \\ &= \begin{pmatrix} c & d \\ a & b \end{pmatrix} \\ J\begin{pmatrix} a & b \\ c & d \end{pmatrix}J &=\begin{pmatrix} d & c \\ b & a \end{pmatrix} \end{align*} Therefore \(f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right) = f \left (J\begin{pmatrix} a & b \\ c & d \end{pmatrix} \right) = f(J) f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) = -f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)\) and \(f\left(\begin{pmatrix} d & c \\ b & a \end{pmatrix}\right) = f\left(J\begin{pmatrix} a & b \\ c & d \end{pmatrix}J \right) = f(J)f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)f(J) = f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)\) as required.
  3. First consider \(O\) the matrix of \(0\), then \begin{align*} && JO &= O \\ \Rightarrow && f(JO) &= f(O) \\ \Rightarrow && f(J)f(O) &= f(O) \\ \Rightarrow && -f(O) &= f(O) \\ \Rightarrow && f(O) &= 0 \end{align*} Now consider \(K_{k} = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}\). Suppose \(A = \begin{pmatrix} a & b \\ ka & kb \end{pmatrix}\) then \begin{align*} K_{\frac1k}A &= \begin{pmatrix} 1 & 0 \\ 0 & \frac1k \end{pmatrix} \begin{pmatrix} a & b \\ ka & kb \end{pmatrix} \\ &= \begin{pmatrix} a & b \\ a & b \end{pmatrix} \end{align*} And so \(f(K_{\frac1k}A) = f\left ( \begin{pmatrix} a & b \\ a & b \end{pmatrix} \right) = - f \left ( \begin{pmatrix} a & b \\ a & b \end{pmatrix} \right) = 0\), therefore either \(f(K_{\frac1k}) = 0\) or \(f(A) = 0\), but we know that \(f(I) \neq 0\) therefore \(f(A) = 0\).
  4. Let \(P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\), then \(P^2 = \begin{pmatrix} 1 &2 \\ 0 & 1 \end{pmatrix}\), \(P^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}\), \(K_k^{-1}PK_k = K_k^{-1}\begin{pmatrix} 1 & k \\ 0 & k \end{pmatrix} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\). If \(A\) has an inverse then \(f(A) \neq 0\) since \(1 = f(I) = f(A)f(A^{-1})\), in particular, \(f(A)f(A^{-1}) = 1\). Using this for \(K_2\) we have: \(f(P)^2 = f(P^2) = f(K_2^{-1}PK_2) = f(P)\) therefore \(f(P) = 0, 1\), but since \(f(P)\) has an inverse, \(f(P) \neq 0\) so \(f(P) = 1\)

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1993 Paper 2 Q6
D: 1600.0 B: 1516.0
matrices 2x2 matrix matrix multiplication determinant counterexample proof true or false commutativity singular matrix

In this question, \(\mathbf{A,\mathbf{B\) }}and \(\mathbf{X\) are non-zero \(2\times2\) real matrices.} Are the following assertions true or false? You must provide a proof or a counterexample in each case.

  1. If \(\mathbf{AB=0}\) then \(\mathbf{BA=0}.\)
  2. \((\mathbf{A-B)(A+B)=}\mathbf{A}^{2}-\mathbf{B}^{2}.\)
  3. The equation \(\mathbf{AX=0}\) has a non-zero solution \(\mathbf{X}\) if and only if \(\det\mathbf{A}=0.\)
  4. For any \(\mathbf{A}\) and \(\mathbf{B}\) there are at most two matrices \(\mathbf{X}\) such that \(\mathbf{X}^{2}+\mathbf{AX}+\mathbf{B}=\mathbf{0}.\)

Solution:

  1. This is false, for example let \(\mathbf{A} = \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\), then \begin{align*} \mathbf{AB} &= \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \\ &= \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix} \\ \mathbf{BA} &= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} \\ \end{align*}
  2. This is also false, using the same matrices from part (i), we find: \begin{align*} (\mathbf{A - B})(\mathbf{A + B}) &= \mathbf{A}^2-\mathbf{BA}+\mathbf{AB}-\mathbf{B}^2 \\ &= \mathbf{A}^2-\mathbf{B}^2+\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} \\ &\neq \mathbf{A}^2-\mathbf{B}^2 \end{align*}
  3. This is true. Claim: The equation \(\mathbf{AX=0}\) has a non-zero solution \(\mathbf{X}\) if and only if \(\det\mathbf{A}=0.\) Proof: \((\Rightarrow)\) Suppose \(\det\mathbf{A} \neq 0\) then \(\mathbf{A}\) has an inverse, and so we must have \(\mathbf{A}^{-1}\mathbf{AX} = \mathbf{0} \Rightarrow \mathbf{X} = \mathbf{0}\). \((\Leftarrow)\) Suppose \(\det \mathbf{A} = 0\) then \(ad-bc=0\), so consider the matrix \(\mathbf{X} = \begin{pmatrix} d & d\\ -c & -c\end{pmatrix}\) (or if this is zero, \(\mathbf{X} = \begin{pmatrix} a & a\\ -b & -b\end{pmatrix}\))
  4. This is false. Consider \(\mathbf{A} = \mathbf{B} = \mathbf{0}\), then \(\mathbf{X} = \begin{pmatrix} 0 & x \\ 0 & 0\end{pmatrix}\) has the property that \(\mathbf{X}^2 = \mathbf{0}\) for all \(x\), so at least more than 2 values

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1993 Paper 2 Q10
D: 1600.0 B: 1500.0
matrices diagonalisation quadratic form change of variables conic sections matrix multiplication transpose rectangular hyperbola

Verify that if \[ \mathbf{P}=\begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix}\qquad\mbox{ and }\qquad\mathbf{A}=\begin{pmatrix}-1 & 8\\ 8 & 11 \end{pmatrix} \] then \(\mathbf{PAP}\) is a diagonal matrix. Put $\mathbf{x}=\begin{pmatrix}x\\ y \end{pmatrix}\( and \)\mathbf{x}_{1}=\begin{pmatrix}x_{1}\\ y_{1} \end{pmatrix}.$ By writing \[ \mathbf{x}=\mathbf{P}\mathbf{x}_{1}+\mathbf{a} \] for a suitable vector \(\mathbf{a},\) show that the equation \[ \mathbf{x}^{\mathrm{T}}\mathbf{Ax}+\mathbf{b}^{\mathrm{T}}\mathbf{x}-11=0, \] where $\mathbf{b}=\begin{pmatrix}18\\ 6 \end{pmatrix}\( and \) \mathbf{x}^{\mathrm{T}} \( is the transpose of \)\mathbf{x},$ becomes \[ 3x_{1}^{2}-y_{1}^{2}=c \] for some constant \(c\) (which you should find).

Solution: \begin{align*} \mathbf{PAP} &= \begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix}\begin{pmatrix}-1 & 8\\ 8 & 11 \end{pmatrix}\begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix} \\ &= \begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix}\begin{pmatrix}15 & -10\\ 30 & 5 \end{pmatrix} \\ &= \begin{pmatrix}75 & 0\\ 0 & -25 \end{pmatrix} \end{align*} Which is diagonal as required. Letting \(\mathbf{x}=\mathbf{P}\mathbf{x}_{1}+\mathbf{a}\) \begin{align*} && \mathbf{x}^{\mathrm{T}}\mathbf{Ax}+\mathbf{b}^{\mathrm{T}}\mathbf{x}-11&=0 \\ \Leftrightarrow && (\mathbf{P}\mathbf{x}_{1}+\mathbf{a})^T\mathbf{A}(\mathbf{P}\mathbf{x}_{1}+\mathbf{a}) + \mathbf{b}^T(\mathbf{P}\mathbf{x}_{1}+\mathbf{a}) - 11 &= 0 \\ \Leftrightarrow && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1 + \mathbf{x}_{1}^T\mathbf{PAa} + \mathbf{a}^T\mathbf{AP}\mathbf{x}_{1}+\mathbf{a}^T\mathbf{Aa} + \mathbf{b}^T(\mathbf{P}\mathbf{x}_{1}+\mathbf{a}) - 11 &= 0 \\ \Leftrightarrow && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1 +(2\mathbf{a}^T\mathbf{A}+\mathbf{b}^T)\mathbf{P}\mathbf{x}_{1}+\mathbf{a}^T\mathbf{Aa} + \mathbf{b}^T\mathbf{a} - 11 &= 0 \\ \end{align*} It would be nice if we picked \(\mathbf{a}\) such that \(2\mathbf{a}^T\mathbf{A}+\mathbf{b}^T = 0\), if \(\mathbf{a} = \begin{pmatrix} a_1 \\a_2 \end{pmatrix}\) then this equation becomes: \begin{align*} && 2\begin{pmatrix}-a_1 + 8a_2 & 8a_1+11a_2 \end{pmatrix} + \begin{pmatrix}18 & 6 \end{pmatrix} &= 0 \\ \Rightarrow && a_1 = 1, a_2 = -1 \end{align*} So our equation is now \begin{align*} && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1 +(2\mathbf{a}^T\mathbf{A}+\mathbf{b}^T)\mathbf{P}\mathbf{x}_{1}+\mathbf{a}^T\mathbf{Aa} + \mathbf{b}^T\mathbf{a} - 11 &= 0 \\ \Leftrightarrow && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1-6 +12 - 11 &= 0 \\ \Leftrightarrow && 25(3x_1^2 - y_1^2) &= 5 \\ \Leftrightarrow && 3x_1^2 - y_1^2 &= \frac{1}{5} \end{align*}

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1992 Paper 3 Q2
D: 1700.0 B: 1540.7
matrices matrix powers proof by induction 2x2 matrix binomial expansion recurrence relation

The matrices \(\mathbf{I}\) and \(\mathbf{J}\) are \[ \mathbf{I}=\begin{pmatrix}1 & 0\\ 0 & 1 \end{pmatrix}\quad\mbox{ and }\quad\mathbf{J}=\begin{pmatrix}1 & 1\\ 1 & 1 \end{pmatrix} \] respectively and \(\mathbf{A}=\mathbf{I}+a\mathbf{J},\) where \(a\) is a non-zero real constant. Prove that \[ \mathbf{A}^{2}=\mathbf{I}+\tfrac{1}{2}[(1+2a)^{2}-1]\mathbf{J}\quad\mbox{ and }\quad\mathbf{A}^{3}=\mathbf{I}+\tfrac{1}{2}[(1+2a)^{3}-1]\mathbf{J} \] and obtain a similar form for \(\mathbf{A}^{4}.\) If \(\mathbf{A}^{k}=\mathbf{I}+p_{k}\mathbf{J},\) suggest a suitable form for \(p_{k}\) and prove that it is correct by induction, or otherwise.

Solution: If $\mathbf{J}=\begin{pmatrix}1 & 1\\ 1 & 1 \end{pmatrix}\(, them \)\mathbf{J}^2=\begin{pmatrix}2 & 2\\ 2 & 2 \end{pmatrix} = 2\mathbf{J}\(. Therefore \)\mathbf{J}^n = 2\mathbf{J}^{n-1} = 2^{n-1}\mathbf{J}$ Let \(\mathbf{A}=\mathbf{I}+a\mathbf{J}\) then \begin{align*} \mathbf{A}^2 &=\l \mathbf{I}+a\mathbf{J}\r^2 \\ &= \mathbf{I}+2a\mathbf{J} + a^2\mathbf{J}^2 \\ &= \mathbf{I}+2a\mathbf{J} + 2a^2\mathbf{J} \\ &= \mathbf{I}+(2a+ 2a^2)\mathbf{J} \\ &= \mathbf{I}+\frac12(1+4a+ 4a^2-1)\mathbf{J} \\ &= \mathbf{I}+\frac12((1+2a)^2-1)\mathbf{J} \\ \end{align*} \begin{align*} \mathbf{A}^3 &=\l \mathbf{I}+a\mathbf{J}\r^3 \\ &= \mathbf{I}+3a\mathbf{J} + a^2\mathbf{J} + a^3\mathbf{J}^3 \\ &= \mathbf{I}+3a\mathbf{J} + 6a^2\mathbf{J} + 4a^3\mathbf{J} \\ &= \mathbf{I}+(3a+ 6a^3+4a^3)\mathbf{J} \\ &= \mathbf{I}+\frac12(1+3\cdot2a+3\dot4a^2+ 8a^3-1)\mathbf{J} \\ &= \mathbf{I}+\frac12((1+2a)^3-1)\mathbf{J} \\ \end{align*} \begin{align*} \mathbf{A}^4 &=\l \mathbf{I}+a\mathbf{J}\r^4 \\ &= \mathbf{I}+4a\mathbf{J} + 6a^2\mathbf{J}^2 + 4a^3\mathbf{J}^3+a^4\mathbf{J}^4 \\ &= \mathbf{I}+4a\mathbf{J} + 12a^2\mathbf{J} + 16a^3\mathbf{J}+8a^4\mathbf{J}\\ &= \mathbf{I}+(4a+ 12a^3+16a^3+8a^4)\mathbf{J} \\ &= \mathbf{I}+\frac12(1+4\cdot2a+6\cdot4a^2+ 4\cdot8a^3+16a^4-1)\mathbf{J} \\ &= \mathbf{I}+\frac12((1+2a)^4-1)\mathbf{J} \\ \end{align*} Claim: \(\mathbf{A}^k = \mathbf{I} + \frac12 ((1+2a)^{k}-1)\mathbf{J}\) Proof: Firstly, note that \(\mathbf{I}\) commutes with everything, so we can just apply the binomial theorem as if we were using real numbers: \begin{align*} \mathbf{A}^k &=\l \mathbf{I}+a\mathbf{J}\r^k \\ &= \sum_{i=0}^k \binom{k}{i}a^i\mathbf{J}^i \\ &= \mathbf{I} + \sum_{i=1}^k \binom{k}{i}a^i2^{i-1}\mathbf{J} \\ &= \mathbf{I} + \frac12\l\sum_{i=1}^k \binom{k}{i}a^i2^{i}\r\mathbf{J} \\ &= \mathbf{I} + \frac12\l\sum_{i=0}^k \binom{k}{i}a^i2^{i} - 1\r\mathbf{J} \\ &= \mathbf{I} + \frac12\l(1+2a)^k - 1\r\mathbf{J} \end{align*} as required

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1990 Paper 2 Q10
D: 1600.0 B: 1496.1
matrices determinant matrix product zero matrix polynomial factorisation upper triangular matrix Cayley-Hamilton counterexample

Two square matrices \(\mathbf{A}\) and \(\mathbf{B}\) satisfies \(\mathbf{AB=0}.\) Show that either \(\det\mathbf{A}=0\) or \(\det\mathbf{B}=0\) or \(\det\mathbf{A}=\det\mathbf{B}=0\). If \(\det\mathbf{B}\neq0\), what must \(\mathbf{A}\) be? Give an example to show that the condition \(\det\mathbf{A}=\det\mathbf{B}=0\) is not sufficient for the equation \(\mathbf{AB=0}\) to hold. Find real numbers \(p,q\) and \(r\) such that \[ \mathbf{M}^{3}+2\mathbf{M}^{2}-5\mathbf{M}-6\mathbf{I}=(\mathbf{M}+p\mathbf{I})(\mathbf{M}+q\mathbf{I})(\mathbf{M}+r\mathbf{I}), \] where \(\mathbf{M}\) is any square matrix and \(\mathbf{I}\) is the appropriate identity matrix. Hence, or otherwise, find all matrices \(\mathbf{M}\) of the form $\begin{pmatrix}a & c\\ 0 & b \end{pmatrix}$ which satisfy the equation \[ \mathbf{M}^{3}+2\mathbf{M}^{2}-5\mathbf{M}-6\mathbf{I}=\mathbf{0}. \]

Solution: Since \(0 = \det \mathbf{0} = \det \mathbf{AB} = \det \mathbf{A} \det\mathbf{B}\) at least one of \(\det \mathbf{A}\) or \(\det \mathbf{B}\) is zero. If \(\det \mathbf{B} \neq 0\) then \(\mathbf{B}\) is invertible, and multiplying on the right by \(\mathbf{B}^{-1}\) gives us \(\mathbf{A} = \mathbf{0}\). If \(\mathbf{A} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} 1 & 0 \\1 & 0 \end{pmatrix}\), then \(\det \mathbf{A} = \det \mathbf{B} = 0\), but \(\mathbf{AB} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \neq \mathbf{0}\) Since \(\mathbf{M}\) commutes with itself and the identity matrix, this is equivalent to factorising the polynomial over the reals. Therefore $$\mathbf{M}^{3}+2\mathbf{M}^{2}-5\mathbf{M}-6\mathbf{I}=(\mathbf{M}-2\mathbf{I})(\mathbf{M}+\mathbf{I})(\mathbf{M}+3\mathbf{I}),$$ Since we now know at least one of \(\det (\mathbf{M}-2\mathbf{I})\), \(\det (\mathbf{M}+\mathbf{I})\), \(\det (\mathbf{M}+3\mathbf{I})\), we should look at cases: Since at least one of those must be non-zero, we must have the following cases: \((a,b) = (2,-1), (-1,2), (-1,-3), (-3,-1), (2,-3), (-3,2)\) In each of those cases, we will have: \(\begin{pmatrix} 0 & c \\ 0 & b+k \end{pmatrix}\begin{pmatrix} a+l & c \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}\) and so all of those solutions are valid. So \(c\) can be anything as long as \((a,b)\) are in that set of solutions

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1987 Paper 2 Q9
D: 1500.0 B: 1500.0
matrices matrix inverse transpose orthogonal matrix skew-symmetric matrix if and only if proof matrix algebra

For any square matrix \(\mathbf{A}\) such that \(\mathbf{I-A}\) is non-singular (where \(\mathbf{I}\) is the unit matrix), the matrix \(\mathbf{B}\) is defined by \[ \mathbf{B}=(\mathbf{I+A})(\mathbf{I-A})^{-1}. \] Prove that \(\mathbf{B}^{\mathrm{T}}\mathbf{B}=\mathbf{I}\) if and only if \(\mathbf{A+A}^{\mathrm{T}}=\mathbf{O}\) (where \(\mathbf{O}\) is the zero matrix), explaining clearly each step of your proof. {[}You may quote standard results about matrices without proof.{]}

Solution: We use the following properties: \((\mathbf{AB})^T = \mathbf{B}^T\mathbf{A}^T\), \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\), and \((\mathbf{A}^T)^{-1} = (\mathbf{A}^{-1})^T\) \begin{align*} &&\mathbf{I} &= \mathbf{B}^{\mathrm{T}}\mathbf{B} \\ \Leftrightarrow && {\mathbf{B}^{T}}^{-1} &= \mathbf{B} \\ \Leftrightarrow && (\mathbf{I+A})(\mathbf{I-A})^{-1} &= {((\mathbf{I+A})(\mathbf{I-A})^{-1})^{T}}^{-1} \\ &&&= {((\mathbf{I+A})(\mathbf{I-A})^{-1})^{-1}}^{T} \\ &&&= ((\mathbf{I-A})(\mathbf{I+A})^{-1})^{T} \\ &&&= {(\mathbf{I+A})^{-1}}^T(\mathbf{I-A})^T \\ &&&= {(\mathbf{I+A}^T)}^{-1}(\mathbf{I-A}^T) \\ \Leftrightarrow && (\mathbf{I+A}^T)(\mathbf{I+A}) &= (\mathbf{I-A}^T)(\mathbf{I-A}) \\ \Leftrightarrow && \mathbf{I}+\mathbf{A}+\mathbf{A}^T+\mathbf{A}^T\mathbf{A} &= \mathbf{I}-\mathbf{A}-\mathbf{A}^T+\mathbf{A}^T\mathbf{A} \\ \Leftrightarrow && 2( \mathbf{A}^T+\mathbf{A}) &= \mathbf{O} \\ \Leftrightarrow && \mathbf{A}+\mathbf{A}^T &= \mathbf{O} \end{align*}

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