Matrices#
Solution to Exercise 1.1
(a) \( A = \begin{pmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{pmatrix} \)
(b) \( B = \begin{pmatrix} 1 & -1 & 1 & -1 \\ -1 & 1 & -1 & 1 \\ 1 & -1 & 1 & -1 \\ -1 & 1 & -1 & 1 \end{pmatrix} \)
(c) \( C = \begin{pmatrix} 0 & 1 & 1 & 1 \\ -1 & 0 & 1 & 1 \\ -1 & -1 & 0 & 1 \\ -1 & -1 & -1 & 0 \end{pmatrix} \)
Solution to Exercise 1.2
(a) The \(4 \times 4\) Hilbert matrix is
(b) Since addition of two numbers is commutative, i.e., \(i + j = j + i\) then \(h_{ij} = h_{ji}\) for all \(i, j = 1, 2, \ldots, n\) then the \(n \times n\) Hilbert matrix is symmetric.
Solution to Exercise 1.3
(a) \(A + B = \begin{pmatrix} 1 + 3 & -3 + 0 \\ 4 + (-1) & 2 + 5 \end{pmatrix} = \begin{pmatrix} 4 & -3 \\ 3 & 7 \end{pmatrix}\)
(b) \(B + C\) is undefined since \(B\) is \(2\times 2\) and \(C\) is \(2\times 1\)
(c) \(A^\mathsf{T} = \begin{pmatrix} 1 & 4 \\ -3 & 2 \end{pmatrix}\)
(d) \(C^\mathsf{T} = \begin{pmatrix} 5 & 9 \end{pmatrix}\)
(e) \(3 B - A = \begin{pmatrix} 9 & 0 \\ -3 & 15 \end{pmatrix} - \begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} 8 & 3 \\ -7 & 13 \end{pmatrix}\)
(f) \((F^\mathsf{T})^\mathsf{T} = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}^\mathsf{T} = \begin{pmatrix} 1 & -2 & 4 \end{pmatrix} = F\)
(g) \(A^\mathsf{T} + B^\mathsf{T} = \begin{pmatrix} 1 & 4 \\ -3 & 2 \end{pmatrix} + \begin{pmatrix} 3 & -1 \\ 0 & 5 \end{pmatrix} = \begin{pmatrix} 4 & 3 \\ -3 & 7 \end{pmatrix}\)
(h) \((A + B)^\mathsf{T} = \begin{pmatrix} 4 & -3 \\ 3 & 7 \end{pmatrix}^\mathsf{T} = \begin{pmatrix} 4 & 3 \\ -3 & 7 \end{pmatrix}\)
Solution to Exercise 1.4
(a) \(AB = \begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ -1 & 5 \end{pmatrix} = \begin{pmatrix} 3 + 3 & 0 - 15 \\ 7 - 2 & 0 + 10 \end{pmatrix} = \begin{pmatrix}6 & -15 \\ 10 & 10 \end{pmatrix}\)
(b) \(BA = \begin{pmatrix} 3 & 0 \\ -1 & 5 \end{pmatrix}\begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} 3 + 0 & -9 + 0 \\ -1 + 20 & 3 + 10 \end{pmatrix} = \begin{pmatrix} 3 & -9 \\ 19 & 13 \end{pmatrix}\)
(c) \(AC = \begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix}\begin{pmatrix} 5 \\ 9 \end{pmatrix} = \begin{pmatrix} 15 - 27 \\ 20 + 18 \end{pmatrix} = \begin{pmatrix} -22 \\ 38 \end{pmatrix}\)
(d) \(CA\) is undefined since \(C\) has 1 column and \(A\) has 2 rows
(e) \(C^\mathsf{T}C = \begin{pmatrix} 5 & 9 \end{pmatrix} \begin{pmatrix} 5 \\ 9 \end{pmatrix} = 25 + 81 = 106\)
(f) \(CC^\mathsf{T} = \begin{pmatrix} 5 \\ 9 \end{pmatrix}\begin{pmatrix} 5 & 9 \end{pmatrix} = \begin{pmatrix} 25 & 45 \\ 45 & 81 \end{pmatrix}\)
(g)
(h)
(i)
(j)
(k)
(l)
Solution to Exercise 1.5
(a) \(\det(A) = \begin{vmatrix} 1 & -3 \\ 4 & 2 \end{vmatrix} = 1(2) - (-3)(4) = 14\)
(b) \(\det(B) = \begin{vmatrix} 3 & 0 \\ -1 & 5 \end{vmatrix} = 3(5) - 0 (-1) = 15\)
(c) \(\begin{align*} \det(G) &= \begin{vmatrix} 4 & 2 & 3 \\ -2 & 6 & 0 \\ 0 & 7 & 1 \end{vmatrix} = 4 \begin{vmatrix} 6 & 0 \\ 7 & 1 \end{vmatrix} - (-2) \begin{vmatrix} 2 & 3 \\ 7 & 1 \end{vmatrix} = 4(6 - 0) + 2(2 - 21) = -14 \end{align*} \)
(d) \(\begin{align*} \det(H) &= \begin{vmatrix} 1 & 0 & 1 \\ 5 & 2 & -2 \\ 2 & -3 & 4 \end{vmatrix} = \begin{vmatrix} 2 & -2 \\ -3 & 4 \end{vmatrix} + \begin{vmatrix} 5 & 2 \\ 2 & -3 \end{vmatrix} = (8 - 6) + (-15 - 4) = -17 \end{align*} \)
Solution to Exercise 1.6
\(A\):
Check:
\(B\):
Check:
\(G\):
Check:
\(H\):
Check:
Solution to Exercise 1.7
For \(AA^\mathsf{T}\) to be symmetric we need to show that \(AA^\mathsf{T} = (AA^\mathsf{T})^\mathsf{T}\)
So \(AA^\mathsf{T}\) is symmetric
Solution to Exercise 1.8
To prove that \((AB)^{-1} = B^{-1}A^{-1}\) we need to show that \((AB)(B^{-1}A^{-1}) = I\)
Therefore \((AB)^{-1} = B^{-1}A^{-1}\).
Solution to Exercise 1.9
Expanding out \((A + B)^2\)
therefore \((A + B)^2 \neq A^2 + 2AB + B^2\).
This would be true if \(AB = BA\), i.e., \(A = B\) or \(A = I\) or \(B = I\).
Solution to Exercise 1.10
We need to show that \(A^2 = I\)
so \(a^2 + bc = 1\) for \(A\) to be an involutory matrix.
Solution to Exercise 1.11
(a) True
Proof: Right multiply both sides of \(A = B\) by \(C\).
(b) False, e.g., if \(C = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\) then this would hold for any two matrices.
(c) False, e.g., if \(A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\) and \(B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}\) then \(AB = O\).
(d) True
Proof: Add \(C\) to both sides of \(A = B\).
(e) False, e.g., if \(A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\) then \(A^2 = I\).
(f) True
Proof: If \(B = A^2\) then \(b_{ii} = \sum_{k=1}^n a_{ik}a_{ki}\) and since \(A\) is symmetric then \(a_{ik} = a_{ki}\) for \(i = 1, 2, \ldots, n\) so \(b_{ii} = \sum_{k=1}^n = a_{ij}^2\) which is always greater than or equal to 0.
(g) True
Proof:
Let \(A\) and \(B\) be two \(n \times n\) matrices then \(C\) is also an \(n \times n\) matrix.
Let \(A\) be an \(n \times n\) matrix and \(C\) be an \(m \times m\) matrix and \(B\) is a \(p \times q\) matrix. \(p = n\) and \(q = m\) but \(m = n\) so \(p = q = m\) and \(B\) is a square matrix.
Ley \(B\) be an \(m \times m\) matrix and \(C\) be an \(n \times n\) matrix and \(A\) is a \(p \times q\) matrix. \(p = n\) and \(q = m\) but \(m = n\) so \(p = q = m\) and \(A\) is a square matrix.
(h) True
Proof: Let \(B\) be a \(p \times q\) matrix then since \(C\) is an \(m \times 1\) matrix then \(q = 1\).
(i) False, e.g., if \(A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\) then \(A^3 = A\).
Solution to Exercise 1.12
(a)
(b)
(c)
(d)
(e)
(f)
(g) We begin by setting \(X = \begin{pmatrix}a & b \\ c & d\end{pmatrix}\), and then \(X^2 = B\) gives
Now we can solve the quadratic equations over \(\mathbb{R}\), since \([X^2]_{12} = 0\) then
For the case when \(b = 0\)
So we have four possible solutions
For the case when \(a = -d\)
which is a contradiction so \(a = -d\) yields no solutions.
(h)