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

\[\begin{split} \begin{align*} H = \begin{pmatrix} 1 & 1/2 & 1/3 & 1/4 \\ 1/2 & 1/3 & 1/4 & 1/5 \\ 1/3 & 1/4 & 1/5 & 1/6 \\ 1/4 & 1/5 & 1/6 & 1/7 \end{pmatrix}. \end{align*} \end{split}\]

(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)  

\[\begin{split} \begin{align*} DE &= \begin{pmatrix} 1 & 1 & 3 \\ 4 & -2 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 6 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} 1 + 0 - 6 & 2 + 6 + 9 \\ 4 + 0 - 6 & 8 - 12 + 9 \end{pmatrix} \\ &= \begin{pmatrix} -5 & 17 \\ -2 & 5 \end{pmatrix} \end{align*} \end{split}\]

(h)  

\[\begin{split} \begin{align*} GH &= \begin{pmatrix} 4 & 2 & 3 \\ -2 & 6 & 0 \\ 0 & 7 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 1 \\ 5 & 2 & -2 \\ 2 & -3 & 4 \end{pmatrix} \\ &= \begin{pmatrix} 4 + 10 + 6 & 0 + 4 - 9 & 4 - 4 + 12 \\ -2 + 30 + 0 & 0 + 12 + 0 & -2 - 12 + 0 \\ 0 + 35 + 2 & 0 + 14 - 3 & 0 - 14 + 4 \end{pmatrix} \\ &= \begin{pmatrix} 20 & -5 & 12 \\ 28 & 12 & -14 \\ 37 & 11 & -10 \end{pmatrix} \end{align*} \end{split}\]

(i)  

\[\begin{split} \begin{align*} A(DE) &= \begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} \left( \begin{pmatrix} 1 & 1 & 3 \\ 4 & -2 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 6 \\ -2 & 3 \end{pmatrix} \right) \\ &= \begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} -5 & 17 \\ -2 & 5 \end{pmatrix} \\ &= \begin{pmatrix} 1 & 2 \\ -24 & 78 \end{pmatrix} \end{align*} \end{split}\]

(j)  

\[\begin{split} \begin{align*} (AD)E &= \left( \begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} 1 & 1 & 3 \\ 4 & -2 & 3 \end{pmatrix} \right) \begin{pmatrix} 1 & 2 \\ 0 & 6 \\ -2 & 3 \end{pmatrix} \\ &= \begin{pmatrix} -11 & 7 & -6 \\ 12 & 0 & 18 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 6 \\ -2 & 3 \end{pmatrix} \\ &= \begin{pmatrix} 1 & 2 \\ -24 & 78 \end{pmatrix} \end{align*} \end{split}\]

(k)  

\[\begin{split} \begin{align*} A^3 &= \begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} \\ &= \begin{pmatrix} -11 & -9 \\ 12 & -8 \end{pmatrix} \begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} \\ &= \begin{pmatrix} -47 & 15 \\ -20 & -52 \end{pmatrix} \end{align*} \end{split}\]

(l)  

\[\begin{split} \begin{align*} G^2 &= \begin{pmatrix} 4 & 2 & 3 \\ -2 & 6 & 0 \\ 0 & 7 & 1 \end{pmatrix} \begin{pmatrix} 4 & 2 & 3 \\ -2 & 6 & 0 \\ 0 & 7 & 1 \end{pmatrix} \\ &= \begin{pmatrix} 12 & 41 & 15 \\ -20 & 32 & -6 \\ -14 & 49 & 1 \end{pmatrix} \\ \therefore G^4 &= G^2G^2 = \begin{pmatrix} 12 & 41 & 15 \\ -20 & 32 & -6 \\ -14 & 49 & 1 \end{pmatrix} \begin{pmatrix} 12 & 41 & 15 \\ -20 & 32 & -6 \\ -14 & 49 & 1 \end{pmatrix} \\ &= \begin{pmatrix} -886 & 2539 & -51 \\ -796 & -90 & -498 \\ -1162 & 1043 & -503 \end{pmatrix} \end{align*} \end{split}\]

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

\(A\):

\[\begin{split} \begin{align*} \adj(A) &= \begin{pmatrix} 2 & -4 \\ 3 & 1 \end{pmatrix}^\mathsf{T} = \begin{pmatrix} 2 & 3 \\ -4 & 1 \end{pmatrix}, \\ \therefore A^{-1} &= \frac{1}{14} \begin{pmatrix} 2 & 3 \\ -4 & 1 \end{pmatrix} = \begin{pmatrix} 1/7 & 3/14 \\ -2/7 & 1/14 \end{pmatrix}. \end{align*} \end{split}\]

Check:

\[\begin{split} \begin{align*} AA^{-1} &= \begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} 1/7 & 3/14 \\ -2/7 & 1/14 \end{pmatrix} = \begin{pmatrix} 1/7 + 6/7 & 3/14 - 3/14 \\ 4/7 - 4/7 & 12/14 + 2/14 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \end{align*} \end{split}\]

\(B\):

\[\begin{split} \begin{align*} \adj{B} &= \begin{pmatrix} 5 & 1 \\ 0 & 3 \end{pmatrix}^\mathsf{T} = \begin{pmatrix} 5 & 0 \\ 1 & 3 \end{pmatrix}, \\ \therefore B^{-1} &= \frac{1}{15} \begin{pmatrix} 5 & 0 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 1/3 & 0 \\ 1/15 & 1/5 \end{pmatrix}. \end{align*} \end{split}\]

Check:

\[\begin{split} \begin{align*} BB^{-1} &= \begin{pmatrix} 3 & 0 \\ -1 & 5 \end{pmatrix} \begin{pmatrix} 1/3 & 0 \\ 1/15 & 1/5 \end{pmatrix} = \begin{pmatrix} 1 + 0 & 0 + 0 \\ -1/3 + 5/15 & 0 + 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \end{align*} \end{split}\]

\(G\):

\[\begin{split} \begin{align*} \adj{G} &= \begin{pmatrix} \begin{vmatrix} 6 & 0 \\ 7 & 1 \end{vmatrix} & - \begin{vmatrix} -2 & 0 \\ 0 & 1 \end{vmatrix} & \begin{vmatrix} -2 & 6 \\ 0 & 7 \end{vmatrix} \\ - \begin{vmatrix} 2 & 3 \\ 7 & 1 \end{vmatrix} & \begin{vmatrix} 4 & 3 \\ 0 & 1 \end{vmatrix} & - \begin{vmatrix} 4 & 2 \\ 0 & 7 \end{vmatrix} \\ \begin{vmatrix} 2 & 3 \\ 6 & 0 \end{vmatrix} & - \begin{vmatrix} 4 & 3 \\ -2 & 0 \end{vmatrix} & \begin{vmatrix} 4 & 2 \\ -2 & 6 \end{vmatrix} \end{pmatrix}^\mathsf{T} \\ &= \begin{pmatrix} 6 & 2 & -14 \\ 19 & 4 & -28 \\ -18 & -6 & 28 \end{pmatrix}^\mathsf{T} = \begin{pmatrix} 6 & 19 & -18 \\ 2 & 4 & -6 \\ -14 & -28 & 28 \end{pmatrix}, \\ \therefore G^{-1} &= -\frac{1}{14} \begin{pmatrix} 6 & 19 & -18 \\ 2 & 4 & -6 \\ -14 & -28 & 28 \end{pmatrix} = \begin{pmatrix} -3/7 & -19/14 & 9/7 \\ -1/7 & -2/7 & 3/7 \\ 1 & 2 & -2 \end{pmatrix}. \end{align*} \end{split}\]

Check:

\[\begin{split} \begin{align*} GG^{-1} &= \begin{pmatrix} 4 & 2 & 3 \\ -2 & 6 & 0 \\ 0 & 7 & 1 \end{pmatrix} \begin{pmatrix} -3/7 & -19/14 & 9/7 \\ -1/7 & -2/7 & 3/7 \\ 1 & 2 & -2 \end{pmatrix} \\ &= \begin{pmatrix} -12/7 - 2/7 + 3 & -76/14 - 4/7 + 6 & 36/7 + 6/7 - 6 \\ 6/7 - 6/7 + 0 & 38/14 - 12/7 + 0 & -18/7 + 18/7 + 0 \\ 0 - 7/7 + 1 & 0 - 14/7 + 2 & 0 + 21/7 - 2 \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{align*} \end{split}\]

\(H\):

\[\begin{split} \begin{align*} \adj(H) &= \begin{pmatrix} \begin{vmatrix} 2 & -2 \\ -3 & 4 \end{vmatrix} & - \begin{vmatrix} 5 & -2 \\ 2 & 4 \end{vmatrix} & \begin{vmatrix} 5 & 2 \\ 2 & -3 \end{vmatrix} \\ - \begin{vmatrix} 0 & 1 \\ -3 & 4 \end{vmatrix} & \begin{vmatrix} 1 & 1 \\ 2 & 4 \end{vmatrix} & - \begin{vmatrix} 1 & 0 \\ 2 & -3 \end{vmatrix} \\ \begin{vmatrix} 0 & 1 \\ 2 & -2 \end{vmatrix} & - \begin{vmatrix} 1 & 1 \\ 5 & -2 \end{vmatrix} & \begin{vmatrix} 1 & 0 \\ 5 & 2 \end{vmatrix} \end{pmatrix}^\mathsf{T} \\ &= \begin{pmatrix} 2 & -24 & -19 \\ -3 & 2 & 3 \\ -2 & 7 & 2 \end{pmatrix}^\mathsf{T} = \begin{pmatrix} 2 & -3 & -2 \\ -24 & 2 & 7 \\ -19 & 3 & 2 \end{pmatrix}, \\ \therefore H^{-1} &= -\frac{1}{17} \begin{pmatrix} 2 & -3 & -2 \\ -24 & 2 & 7 \\ -19 & 3 & 2 \end{pmatrix} = \begin{pmatrix} -2/17 & 3/17 & 2/17 \\ 24/17 & -2/17 & -7/17 \\ 19/17 & -3/17 & -2/17 \end{pmatrix} . \end{align*} \end{split}\]

Check:

\[\begin{split} \begin{align*} HH^{-1} &= \begin{pmatrix} 1 & 0 & 1 \\ 5 & 2 & -2 \\ 2 & -3 & 4 \end{pmatrix} \begin{pmatrix} -2/17 & 3/17 & 2/17 \\ 24/17 & -2/17 & -7/17 \\ 19/17 & -3/17 & -2/17 \end{pmatrix} \\ &= \begin{pmatrix} -2/17 + 0 + 19/17 & 3/17 - 3/17 & 2/17 - 2/17 \\ -5/17 + 48/17 - 38/17 & 15/17 - 4/17 + 6/17 & 10/17 - 14/17 + 4/17 \\ -4/17 - 72/17 + 76/17 & 6/17 + 6/17 - 12/17 & 4/17 + 21/17 - 8/17 \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{align*} \end{split}\]

Solution to Exercise 1.7

For \(AA^\mathsf{T}\) to be symmetric we need to show that \(AA^{T} = (AA^\mathsf{T})^\mathsf{T}\)

\[\begin{split} \begin{align*} (AA^\mathsf{T})^\mathsf{T} &= (A^\mathsf{T})^\mathsf{T}A^\mathsf{T} && \textsf{since $(AB)^\mathsf{T} = B^\mathsf{T}A^\mathsf{T}$} \\ &= AA^\mathsf{T} && \textsf{since $(A^\mathsf{T})^\mathsf{T}$.} \end{align*} \end{split}\]

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 \((AB)(B^{-1}A^{-1}) = I\)

\[\begin{split} \begin{align*} (AB)(B^{-1}A^{-1}) &= A(BB^{-1}A^{-1}) && \textsf{associativity law} \\ &= A(IA^{-1}) && \textsf{since $BB^{-1} = I$} \\ &= AA^{-1} && \textsf{since $AI = IA = A$}\\ &= I && \textsf{definition of $A^{-1}.$} \end{align*} \end{split}\]

Therefore \((AB)^{-1} = B^{-1}A^{-1}\).

Solution to Exercise 1.9

Expanding out \((A + B)^2\)

\[ (A + B)^2 = (A + B)(A + B) = A^2 + AB + BA + B^2, \]

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 \(A^2 = I\)

\[\begin{split} \begin{align*} A^2 = \begin{pmatrix} a & b \\ c & -a \end{pmatrix} \begin{pmatrix} a & b \\ c & -a \end{pmatrix} = \begin{pmatrix} a^2 + bc & 0 \\ 0 & a^2 + bc \end{pmatrix}, \end{align*} \end{split}\]

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)

\[\begin{split} \begin{align*} 5X &= A \\ X &= \frac{1}{5} A \\ &= \dfrac{1}{5}\begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{5} & -\frac{3}{5} \\ \frac{4}{5} & \frac{2}{5} \end{pmatrix} \end{align*} \end{split}\]

(b)

\[\begin{split} \begin{align*} X + A &= I \\ X &= I - A \\ &= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} 0 & 3 \\ -4 & -1 \end{pmatrix} \end{align*} \end{split}\]

(c)

\[\begin{split} \begin{align*} 2X - B &= A \\ 2X &= A + B \\ X &= \dfrac{1}{2}(A + B) = \dfrac{1}{2}\left( \begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} + \begin{pmatrix} 3 & 0 \\ -1 & 5 \end{pmatrix} \right) \\ &= \dfrac{1}{2} \begin{pmatrix} 4 & -3 \\ 3 & 7 \end{pmatrix} = \begin{pmatrix} 2 & -\frac{3}{2} \\ \frac{3}{2} & \frac{7}{2} \end{pmatrix} \end{align*} \end{split}\]

(d)

\[\begin{split} \begin{align*} XA &= I \\ X &= IA^{-1} = A^{-1}I \\ &= \dfrac{1}{14} \begin{pmatrix} 2 & 3 \\ -4 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{7} & \frac{3}{14} \\ -\frac{2}{7} & \frac{1}{14} \end{pmatrix} \end{align*} \end{split}\]

(e)

\[\begin{split} \begin{align*} BX &= A \\ X &= B^{-1}A \\ &= \dfrac{1}{15}\begin{pmatrix} 5 & 0 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} \\ &= \dfrac{1}{15} \begin{pmatrix} 5 & -15 \\ 13 & 3\end{pmatrix} = \begin{pmatrix} \frac{1}{3} & -1 \\ \frac{13}{15} & \frac{1}{5} \end{pmatrix} \end{align*} \end{split}\]

(f)

\[\begin{split} \begin{align*} A^2 &= X \\ X &= \begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix}\begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} -11 & -9 \\ 12 & -8 \end{pmatrix} \end{align*} \end{split}\]

(g)   We begin by setting \(X = \begin{pmatrix}a & b \\ c & d\end{pmatrix}\), and then \(X^2 = B\) gives

\[\begin{split} \begin{align*} \begin{pmatrix}a^2 + bc & ab + bd \\ ca+dc & cb+d^2\end{pmatrix} = \begin{pmatrix} a^2 + bc & b(a + d) \\ c(a + d) & cb + d^2 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ -1 & 5 \end{pmatrix}. \end{align*} \end{split}\]

Now we can solve the quadratic equations over \(\mathbb{R}\), since \([X^2]_{12} = 0\) then

\[\begin{split} \begin{align*} b(a + d) &= 0 \\ \therefore b &= 0 \text{ or } a = -d. \end{align*} \end{split}\]

For the case when \(b = 0\)

\[\begin{split} \begin{align*} a^2 + bc &= 3 \implies a = \pm \sqrt{3}, \\ cb + d^2 &= 5 \implies d = \pm \sqrt{5}. \end{align*} \end{split}\]

So we have four possible solutions

\[\begin{split} \begin{align*} X &= \begin{pmatrix} \pm \sqrt{3} & 0 \\ - 1 / (\pm \sqrt{3} \pm \sqrt{5}) & \pm\sqrt{5} \end{pmatrix}, \text{ or }\\ X &= \begin{pmatrix} \pm \sqrt{3} & 0 \\ - 1 / (\pm \sqrt{3} \mp \sqrt{5}) & \mp\sqrt{5} \end{pmatrix}. \end{align*} \end{split}\]

For the case when \(a = -d\)

\[ \begin{align*} c(a + d) = 0c = -1, \end{align*} \]

which is a contradiction so \(a = -d\) yields no solutions.

(h)

\[\begin{split} \begin{align*} (X + A)B &= I \\ X + A &= I B^{-1} \\ X &= B^{-1} - A \\ &= \frac{1}{15} \begin{pmatrix} 5 & 0 \\ 1 & 3 \end{pmatrix} - \begin{pmatrix} 1 & -3 \\ 4 & 2 \end{pmatrix} \\ &= \begin{pmatrix} -2/3 & 3 \\ -59/15 & -9/5 \end{pmatrix} \\ &= \frac{1}{15} \begin{pmatrix} -10 & 45 \\ -59 & -27 \end{pmatrix} \end{align*} \end{split}\]