Matrices, Sheet #3
Learning Targets
- Perform operations on matrices (add, subtract, multiply)
- Find the transpose, adjudicate, determinant of 2x2 and 3x3 matrices
- use matrices to find the solutions to linear equations
- Identify unit, symmetric, skew-symmetric, triangular, diagonal, singular and orthogonal matrices
Additional Resources
Tutorials
- Linear Algebra Playlist : The same playlist as mentioned last week.
Software
- Matlab Documentation : Getting familiar with matlab will be a great advantage - here is some documentation on what you can do with matrices and arrays.
Problem sheet
Skill Building Questions
Problem 1.
Given the matrices: $A=\begin{pmatrix}1&1 \newline 2&-1\end{pmatrix}$ and $B=\begin{pmatrix}0&1 \newline -2&3\end{pmatrix}$
Find the answers to the following operations:
(a) $A+B$
Solving with the Matlab terminal:
$\Rightarrow{}\quad \boxed{\begin{pmatrix}1&2 \newline 0&2\end{pmatrix}}$
a = [1 1 ; 2 -1]
b = [0 1 ; -2 3]
a + b
(b) $A-B$
Solving with the Matlab terminal:
$\Rightarrow{}\quad \boxed{\begin{pmatrix}1&0 \newline 4&-4\end{pmatrix}}$
a = [1 1 ; 2 -1]
b = [0 1 ; -2 3]
a - b
(c) $AB$
Solving with the Matlab terminal:
$\Rightarrow{}\quad \boxed{ \begin{pmatrix}-2&4\\2&-1\end{pmatrix}}$
a = [1 1 ; 2 -1]
b = [0 1 ; -2 3]
a * b
(d) $BA$
Solving with the Matlab terminal:
$\Rightarrow{}\quad \boxed{\begin{pmatrix}2&-1\\4&-5\end{pmatrix}}$
a = [1 1 ; 2 -1]
b = [0 1 ; -2 3]
b * a
(e) $4A+\frac{1}{2}B$
$\Rightarrow{}\quad 4\begin{pmatrix}1&1\\2&-1\end{pmatrix}+\frac{1}{2}\begin{pmatrix}0&1\\-2&3\end{pmatrix}$
$\Rightarrow{}\ \ \begin{pmatrix}4(1)+\frac{1}{2}(0)&4(1)+\frac{1}{2}(0)\\4(2)+\frac{1}{2}(-2)&4(-1)+\frac{1}{2}(3)\end{pmatrix}$
Otherwise solving with the Matlab terminal:
$\Rightarrow{} \quad \boxed{\begin{pmatrix}4&9/2\\7&-5/2\end{pmatrix}}$
$\Rightarrow{}\ \ \begin{pmatrix}4(1)+\frac{1}{2}(0)&4(1)+\frac{1}{2}(0)\\4(2)+\frac{1}{2}(-2)&4(-1)+\frac{1}{2}(3)\end{pmatrix}$
Otherwise solving with the Matlab terminal:
a = [1 1 ; 2 -1]
b = [0 1 ; -2 3]
(4 * a) + (0.5*b)
(f) $AB^T$
$\Rightarrow{}\quad$
$\begin{pmatrix}1&1\\2&-1\end{pmatrix}\begin{pmatrix}0&1\\-2&3\end{pmatrix}^T$
$\Rightarrow{} \quad \begin{pmatrix}1&1\\2&-1\end{pmatrix}\begin{pmatrix}0&-2\\1&3\end{pmatrix}$
$\Rightarrow{} \quad \begin{pmatrix}1(0)+1(1)&1(-2)+1(3)\\2(0)+-1(1)&2(-2)+-1(3)\end{pmatrix}$
Otherwise solving with the Matlab terminal:
$\Rightarrow{} \quad \boxed{\begin{pmatrix}1&1\\-1&-7\end{pmatrix}}$
$\Rightarrow{} \quad \begin{pmatrix}1&1\\2&-1\end{pmatrix}\begin{pmatrix}0&-2\\1&3\end{pmatrix}$
$\Rightarrow{} \quad \begin{pmatrix}1(0)+1(1)&1(-2)+1(3)\\2(0)+-1(1)&2(-2)+-1(3)\end{pmatrix}$
Otherwise solving with the Matlab terminal:
a = [1 1 ; 2 -1]
b = [0 1 ; -2 3]
b = transpose(b)
a * b
(g) $BA^T$
$\Rightarrow{}\quad$
$\begin{pmatrix}0&1\\-2&3\end{pmatrix}\begin{pmatrix}1&1\\2&-1\end{pmatrix}^T$
$\Rightarrow{} \quad \begin{pmatrix}0&1\\-2&3\end{pmatrix}\begin{pmatrix}1&2\\1&-1\end{pmatrix}$
$\Rightarrow{} \quad \begin{pmatrix}0(1)+1(1)&0(2)+1(-1)\\-2(1)+3(1)&-2(2)+3(-1)\end{pmatrix}$
Otherwise solving with the Matlab terminal:
$\Rightarrow{} \quad \boxed{\begin{pmatrix}1&-1\\1&-7\end{pmatrix}}$
$\Rightarrow{} \quad \begin{pmatrix}0&1\\-2&3\end{pmatrix}\begin{pmatrix}1&2\\1&-1\end{pmatrix}$
$\Rightarrow{} \quad \begin{pmatrix}0(1)+1(1)&0(2)+1(-1)\\-2(1)+3(1)&-2(2)+3(-1)\end{pmatrix}$
Otherwise solving with the Matlab terminal:
a = [1 1 ; 2 -1]
b = [0 1 ; -2 3]
a = transpose(a)
b * a
(h) $B^TA^T$
$\Rightarrow{}\quad$
$\begin{pmatrix}0&1\\-2&3\end{pmatrix}^T\begin{pmatrix}1&1\\2&-1\end{pmatrix}^T$
$\Rightarrow{} \quad \begin{pmatrix}0&-2\\1&3\end{pmatrix}\begin{pmatrix}1&2\\1&-1\end{pmatrix}$
$\Rightarrow{} \quad \begin{pmatrix}0(1)+-2(1)&0(2)+-2(-1)\\1(1)+3(1)&1(2)+3(-1)\end{pmatrix}$
Otherwise solving with the Matlab terminal:
$\Rightarrow{} \quad \boxed{\begin{pmatrix}-2&2\\4&-1\end{pmatrix}}$
$\Rightarrow{} \quad \begin{pmatrix}0&-2\\1&3\end{pmatrix}\begin{pmatrix}1&2\\1&-1\end{pmatrix}$
$\Rightarrow{} \quad \begin{pmatrix}0(1)+-2(1)&0(2)+-2(-1)\\1(1)+3(1)&1(2)+3(-1)\end{pmatrix}$
Otherwise solving with the Matlab terminal:
a = [1 1 ; 2 -1]
b = [0 1 ; -2 3]
a = transpose(a)
b = transpose(b)
b * a
(i) $det (A)$
$\Rightarrow{}\quad \begin{vmatrix}1&1\\2&-1\end{vmatrix}$
$\quad\Rightarrow{}\quad 1(-1)-2(1) \ \ \Rightarrow{}\ \boxed{ -3}$
Otherwise solving with the Matlab terminal:
Otherwise solving with the Matlab terminal:
a = [1 1 ; 2 -1]
det(a)
(j) $A^{-1}$
$\Rightarrow{}\quad$
$\frac{1}{det A}\begin{pmatrix}-1&-1\\-2&1\end{pmatrix}$
$\quad \Rightarrow{}\quad -\frac{1}{-3}\begin{pmatrix}-1&-1\\-2&1\end{pmatrix}$
$\Rightarrow{}\quad \boxed{\begin{pmatrix}\frac{1}{3}&\frac{1}{3}\\\frac{2}{3}&-\frac{1}{3}\end{pmatrix}}$
Otherwise solving with the Matlab terminal:
Otherwise solving with the Matlab terminal:
a = [1 1 ; 2 -1]
inv(a)
(k) Comment on any relationships between the results of question (a) to (j)
$\Rightarrow{} BA^T=(AB^T)^T$ and $(AB)^T=B^TA^T$
Problem 2
Given the matrices:
$A=\begin{pmatrix}2&0 \newline 3&-1 \newline 1&4\end{pmatrix}$, $B=\begin{pmatrix}-1&-3 \newline 5&2 \newline 7&1\end{pmatrix}$ and $C=\begin{pmatrix}1&3&5&7 \newline 2&4&6&8\end{pmatrix}$
Find the answers to the following operations, if they exist:
(a) $A+B$
Solving with the Matlab terminal:
$\Rightarrow{}\quad
\boxed{\begin{pmatrix}1&-3\\8&1\\8&5\end{pmatrix}}$
a = [2 0 ; 3 -1 ; 1 4]
b = [-1 -3 ; 5 2 ; 7 1]
c = [1 3 5 7 ; 2 4 6 8]
a + b
(b) $A-B$
Solving with the Matlab terminal:
$\Rightarrow{}\quad
\boxed{\begin{pmatrix}3&3\\-2&-3\\-6&3\end{pmatrix}}$
a - b
(c) $AC$
$\Rightarrow{} \ \ $
$\begin{pmatrix}2&0\\3&-1\\1&4\end{pmatrix}\begin{pmatrix}1&3&5&7\\2&4&6&8\end{pmatrix}$
$\Rightarrow{}\ \ \begin{pmatrix}2(1)+0(2)&2(3)+0(4)&2(5)+0(6)&2(7)+0(8)\\3(1)+-1(2)&3(3)+-1(4)&3(5)+-1(6)&3(7)+-1(8)\\1(1)+4(2)&1(3)+4(4)&1(5)+4(6)&1(7)+4(8)\end{pmatrix}$
Otherwise solving with the Matlab terminal:
$\Rightarrow{}\ \ \boxed{\begin{pmatrix}2&6&10&14\\1&5&9&13\\9&19&29&39\end{pmatrix}}$
$\Rightarrow{}\ \ \begin{pmatrix}2(1)+0(2)&2(3)+0(4)&2(5)+0(6)&2(7)+0(8)\\3(1)+-1(2)&3(3)+-1(4)&3(5)+-1(6)&3(7)+-1(8)\\1(1)+4(2)&1(3)+4(4)&1(5)+4(6)&1(7)+4(8)\end{pmatrix}$
Otherwise solving with the Matlab terminal:
a * c
(d) $A+C$
Operation is not possible. Two matrices must have an equal number of rows and columns to be added.
(e) $CA$
Operation not possible, $C_{2x4}A_{3x2}$
(f) $A^TB$
$\Rightarrow{}\ \ \begin{pmatrix}2&0\\3&-1\\1&4\end{pmatrix}^T\begin{pmatrix}-1&-3\\5&2\\7&1\end{pmatrix}\Rightarrow{}\begin{pmatrix}2&3&1\\0&-1&4\\\end{pmatrix}\begin{pmatrix}-1&-3\\5&2\\7&1\end{pmatrix}\\$
$\Rightarrow{}\ \ \begin{pmatrix}2(-1)+3(5)+1(7)&2(-3)+3(2)+1(1)\\0(-1)-1(5)+4(7)&0(-3)-1(2)+4(1)\end{pmatrix}$
$\Rightarrow{}\ \ \boxed{\begin{pmatrix}20&1\\23&2\end{pmatrix}}$
Otherwise solving with the Matlab terminal:
Otherwise solving with the Matlab terminal:
a = transpose(a)
a * b
(g) $AB^T$
$\Rightarrow{}\ \ \begin{pmatrix}2&0\\3&-1\\1&4\end{pmatrix} \begin{pmatrix}-1&-3\\5&2\\7&1\end{pmatrix}^T\ \Rightarrow{}\begin{pmatrix}2&0\\3&-1\\1&4\end{pmatrix} \begin{pmatrix}-1&5&7\\-3&2&1\end{pmatrix}$
$\Rightarrow{} \begin{pmatrix}2(-1)+0(-3)&2(5)+0(2)&2(7)+0(1)\\3(-1)-1(-3)&3(5)-1(2)&3(7)-1(1)\\1(-1)+4(-3)&1(5)+4(2)&1(7)+4(1)\end{pmatrix}\ \ \ \ \Rightarrow{}\ \ \boxed{\begin{pmatrix}-2&10&14\\0&13&20\\-13&13&11\end{pmatrix}}$
Otherwise solving with the Matlab terminal:
$\Rightarrow{} \begin{pmatrix}2(-1)+0(-3)&2(5)+0(2)&2(7)+0(1)\\3(-1)-1(-3)&3(5)-1(2)&3(7)-1(1)\\1(-1)+4(-3)&1(5)+4(2)&1(7)+4(1)\end{pmatrix}\ \ \ \ \Rightarrow{}\ \ \boxed{\begin{pmatrix}-2&10&14\\0&13&20\\-13&13&11\end{pmatrix}}$
Otherwise solving with the Matlab terminal:
b = transpose(b)
a * b
Problem 3.
Find the solution to the following linear equations:
\[\begin{align} 3x + 2y + z &= 10 \newline 3x + 5z &= 6 \newline 2x + 4y - 4z &= 2 \end{align}\]$\Rightarrow{} \begin{pmatrix}3 & 2 & 1\\3 & 0 & 5\\2&4&-4\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} =
\begin{pmatrix}10\\6\\2\end{pmatrix}$
$\Rightarrow\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}3&2&1\\3&0&5\\2&4&-4\end{pmatrix}^{-1} \ \begin{pmatrix}10\\6\\2\end{pmatrix}$
We find the inverse using the Matlab terminal:
$\Rightarrow\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}5&-3&-2.5\\-5.5&3.5&3\\-3&2&1.5\end{pmatrix} \ \begin{pmatrix}10\\6\\2\end{pmatrix}$
$\Rightarrow{} \quad\boxed{ x = 27 \\ y = -28 \\ z = -15}$
$\Rightarrow\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}3&2&1\\3&0&5\\2&4&-4\end{pmatrix}^{-1} \ \begin{pmatrix}10\\6\\2\end{pmatrix}$
We find the inverse using the Matlab terminal:
a = [3 2 1 ; 3 0 5 ; 2 4 -4]
inv(a)
$\Rightarrow\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}5&-3&-2.5\\-5.5&3.5&3\\-3&2&1.5\end{pmatrix} \ \begin{pmatrix}10\\6\\2\end{pmatrix}$
$\Rightarrow{} \quad\boxed{ x = 27 \\ y = -28 \\ z = -15}$
Problem 4.
Evaluate the following determinants and state whether the matrix is singular or non-singular:
(a) $\begin{vmatrix}20&6 \newline 1&2\end{vmatrix}$
$\Rightarrow{}$
$20(2)-6(1) \ \Rightarrow{}\ \boxed{34} \Rightarrow{}\ \boxed{\text{non-singular}}$
Solving with the Matlab terminal:
Solving with the Matlab terminal:
a = [20 6 ; 1 2]
det(a)
(b) $\begin{vmatrix}1&2&6 \newline 1&3&9 \newline 1&4&12\end{vmatrix}$
$\Rightarrow{}\ \ 1\begin{vmatrix}3&9\\4&12\end{vmatrix}-2\begin{vmatrix}1&9\\1&12\end{vmatrix}+6\begin{vmatrix}1&3\\1&4\end{vmatrix}$
$\Rightarrow{}\ \ 1[3(13)-9(4)]-2[1(12)-9(1)]+6[1(4)-3(1)] \ \ \Rightarrow{}\ \ \boxed{0} \Rightarrow{}\ \boxed{\text{singular}}$
Solving with the Matlab terminal:
$\Rightarrow{}\ \ 1[3(13)-9(4)]-2[1(12)-9(1)]+6[1(4)-3(1)] \ \ \Rightarrow{}\ \ \boxed{0} \Rightarrow{}\ \boxed{\text{singular}}$
Solving with the Matlab terminal:
a = [1 2 6 ; 1 3 9 ; 1 4 12]
det(a)
(c) $\begin{vmatrix}1&1&1 \newline \lambda & \mu & \nu \newline \lambda^3 & \mu^3 & \nu^3\end{vmatrix}$
$\Rightarrow{}\ \ 1\begin{vmatrix}\mu & \nu \\ \mu^3 & \nu^3 \end{vmatrix}-1\begin{vmatrix}\lambda & \nu \\ \lambda^3 & \nu^3 \end{vmatrix}+1\begin{vmatrix}\lambda & \mu \\ \lambda^3 & \mu^3 \end{vmatrix}$
$\Rightarrow{}\ \ (\mu\nu^3-\nu\mu^3) - (\lambda\nu^3-\nu\lambda^3) + (\lambda\mu^3-\mu\lambda^3)$ $\Rightarrow{}\ \ \boxed{\nu^3(\mu-\lambda)+\mu^3(\lambda-\nu)+\lambda^3(\nu-\mu)} \Rightarrow{}\ \boxed{\text{singularity is dependant on variables}}$
$\Rightarrow{}\ \ (\mu\nu^3-\nu\mu^3) - (\lambda\nu^3-\nu\lambda^3) + (\lambda\mu^3-\mu\lambda^3)$ $\Rightarrow{}\ \ \boxed{\nu^3(\mu-\lambda)+\mu^3(\lambda-\nu)+\lambda^3(\nu-\mu)} \Rightarrow{}\ \boxed{\text{singularity is dependant on variables}}$
(d) $\begin{vmatrix}4&9&12&52&84 \newline 2&28&30&28&7 \newline 2&5&8&92&34 \newline 14&2&37&56&9 \newline 12&35&82&2&28 \end{vmatrix}$
$\Rightarrow{}\ \ \boxed{-48100996} \Rightarrow{}\ \boxed{\text{non-singular}}$
It is only reasonable to solve this using a tool such as the Matlab terminal:
It is only reasonable to solve this using a tool such as the Matlab terminal:
a = [4 9 12 52 84 ; 2 28 30 28 7 ; 2 5 8 92 34 ; 14 2 37 56 9 ; 12 35 82 2 28]
det(a)
Problem 5.
Determine the elements of the matrix $M$, so that $AMB=C$, where
$A=\begin{pmatrix}2&1&1 \newline 1&1&0 \newline 0&0&1\end{pmatrix}$, $B=\begin{pmatrix}3&1 \newline 1&1\end{pmatrix}$ and $C=\begin{pmatrix}1&1 \newline 2&2 \newline 1&1\end{pmatrix}$
$\Rightarrow{}\quad M=A^{-1}CB^{-1}$
$\Rightarrow{}\quad det(a)=1$ and $det(b)=2$
$\Rightarrow{}\quad B^{-1}=\frac{1}{det(c)}adj(d) \quad\Rightarrow{}\quad B^{-1}=\frac{1}{2}\begin{pmatrix}1&-1\\-1&3\end{pmatrix} \quad\Rightarrow{}\quad B^{-1}=\begin{pmatrix}1/2&-1/2\\-1/2&3/2\end{pmatrix}$ $\Rightarrow{}\quad A^{-1}=\frac{1}{det(e)}adj(f)=\frac{1}{det(g)}C_A^T$, where $C_A$ is the cofactor matrix of A
$\Rightarrow{}\quad C_A=\begin{pmatrix} \begin{vmatrix}1&0\\0&1\end{vmatrix} & -\begin{vmatrix}1&0\\0&1\end{vmatrix} & \begin{vmatrix}1&1\\0&0\end{vmatrix} \\ -\begin{vmatrix}1&1\\0&1\end{vmatrix} & \begin{vmatrix}2&1\\0&1\end{vmatrix} & -\begin{vmatrix}2&1 \\0&0\end{vmatrix} \\ \begin{vmatrix}1&1\\1&0\end{vmatrix} & -\begin{vmatrix}2&1\\1&0\end{vmatrix} & \begin{vmatrix}2&1\\1&1\end{vmatrix} \end{pmatrix}$
$\Rightarrow{}\quad C_A=\begin{pmatrix}1&-1&0\\-1&2&0\\-1&1&1\end{pmatrix}$
$\Rightarrow{}\quad C_A^T=\begin{pmatrix}1&-1&-1\\-1&2&1\\0&0&1\end{pmatrix}$ $\Rightarrow{}\quad A^{-1}=\begin{pmatrix}1&-1&-1\\-1&2&1\\0&0&1\end{pmatrix}$
$\Rightarrow{} \quad M=\begin{pmatrix}1&-1&-1\\-1&2&1\\0&0&1\end{pmatrix} \begin{pmatrix}1&1\\2&2\\1&1\end{pmatrix}\begin{pmatrix}1/2&-1/2\\-1/2&3/2\end{pmatrix} \quad\Rightarrow{} \quad \boxed{M=\begin{pmatrix}0&-2\\0&4\\0&1\end{pmatrix}}$
Otherwise solving, simply, with Matlab:
$\Rightarrow{}\quad det(a)=1$ and $det(b)=2$
$\Rightarrow{}\quad B^{-1}=\frac{1}{det(c)}adj(d) \quad\Rightarrow{}\quad B^{-1}=\frac{1}{2}\begin{pmatrix}1&-1\\-1&3\end{pmatrix} \quad\Rightarrow{}\quad B^{-1}=\begin{pmatrix}1/2&-1/2\\-1/2&3/2\end{pmatrix}$ $\Rightarrow{}\quad A^{-1}=\frac{1}{det(e)}adj(f)=\frac{1}{det(g)}C_A^T$, where $C_A$ is the cofactor matrix of A
$\Rightarrow{}\quad C_A=\begin{pmatrix} \begin{vmatrix}1&0\\0&1\end{vmatrix} & -\begin{vmatrix}1&0\\0&1\end{vmatrix} & \begin{vmatrix}1&1\\0&0\end{vmatrix} \\ -\begin{vmatrix}1&1\\0&1\end{vmatrix} & \begin{vmatrix}2&1\\0&1\end{vmatrix} & -\begin{vmatrix}2&1 \\0&0\end{vmatrix} \\ \begin{vmatrix}1&1\\1&0\end{vmatrix} & -\begin{vmatrix}2&1\\1&0\end{vmatrix} & \begin{vmatrix}2&1\\1&1\end{vmatrix} \end{pmatrix}$
$\Rightarrow{}\quad C_A=\begin{pmatrix}1&-1&0\\-1&2&0\\-1&1&1\end{pmatrix}$
$\Rightarrow{}\quad C_A^T=\begin{pmatrix}1&-1&-1\\-1&2&1\\0&0&1\end{pmatrix}$ $\Rightarrow{}\quad A^{-1}=\begin{pmatrix}1&-1&-1\\-1&2&1\\0&0&1\end{pmatrix}$
$\Rightarrow{} \quad M=\begin{pmatrix}1&-1&-1\\-1&2&1\\0&0&1\end{pmatrix} \begin{pmatrix}1&1\\2&2\\1&1\end{pmatrix}\begin{pmatrix}1/2&-1/2\\-1/2&3/2\end{pmatrix} \quad\Rightarrow{} \quad \boxed{M=\begin{pmatrix}0&-2\\0&4\\0&1\end{pmatrix}}$
Otherwise solving, simply, with Matlab:
a = [2 1 1 ; 1 1 0 ; 0 0 1]
b = [3 1 ; 1 1]
c = [1 1 ; 2 2 ; 1 1]
a = inv(a)
b = inv(b)
a * c * b
Problem 6.
Given $A=\begin{pmatrix}1&2&3 \newline 1&3&5 \newline 1&5&12\end{pmatrix}$ find $|A|$, the adjoint of $A$ and $A^{-1}$. Verify that $AA^{-1}=I$, where $I$ is the corresponding unit matrix.
$\Rightarrow{}\quad det(A) = 1\begin{vmatrix}3&5\\5&12\end{vmatrix} - 2\begin{vmatrix}1&5\\1&12\end{vmatrix} + 3\begin{vmatrix}1&3\\1&5\end{vmatrix} \quad\Rightarrow{}\quad \boxed{det(A)=3}$
$\Rightarrow{}\quad adj(A) = C^T$, where C is the cofactor matrix of A.
$\Rightarrow{}\quad C=\begin{pmatrix} \begin{vmatrix}3&5\\5&12\end{vmatrix} & -\begin{vmatrix}1&5\\1&12\end{vmatrix} & \begin{vmatrix}1&3\\1&5\end{vmatrix}\\ -\begin{vmatrix}2&3\\5&12\end{vmatrix} & \begin{vmatrix}1&3\\1&12\end{vmatrix} & -\begin{vmatrix}1&2\\1&5\end{vmatrix}\\ \begin{vmatrix}2&3\\3&5\end{vmatrix} & -\begin{vmatrix}1&3\\1&5\end{vmatrix} & \begin{vmatrix}1&2\\1&3\end{vmatrix} \end{pmatrix} \quad\Rightarrow{}\quad C=\begin{pmatrix}11&-7&2\\-9&9&-3\\1&-2&1\end{pmatrix}$
$\Rightarrow{}\quad \boxed{ adj(A) = C^T = \begin{pmatrix}11&-9&1\\-7&9&-2\\2&-3&1\end{pmatrix}}$
$\Rightarrow{}\quad A^{-1} = \frac{1}{det(A)}adj(A) = \frac{1}{3} \begin{pmatrix}11&-9&1\\-7&9&-2\\2&-3&1\end{pmatrix}$
$\Rightarrow{}\quad \boxed{AA^{-1} = \begin{pmatrix}1&2&3\\1&3&5\\1&5&12\end{pmatrix} \begin{pmatrix}11/3 & -9/3 & 1/3 \\-7/3 & 9/3 & -2/3\\ 2/3 & -3/3 & 1/3 \end{pmatrix} = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}$
Otherwise solving, simply, with Matlab:
$\Rightarrow{}\quad adj(A) = C^T$, where C is the cofactor matrix of A.
$\Rightarrow{}\quad C=\begin{pmatrix} \begin{vmatrix}3&5\\5&12\end{vmatrix} & -\begin{vmatrix}1&5\\1&12\end{vmatrix} & \begin{vmatrix}1&3\\1&5\end{vmatrix}\\ -\begin{vmatrix}2&3\\5&12\end{vmatrix} & \begin{vmatrix}1&3\\1&12\end{vmatrix} & -\begin{vmatrix}1&2\\1&5\end{vmatrix}\\ \begin{vmatrix}2&3\\3&5\end{vmatrix} & -\begin{vmatrix}1&3\\1&5\end{vmatrix} & \begin{vmatrix}1&2\\1&3\end{vmatrix} \end{pmatrix} \quad\Rightarrow{}\quad C=\begin{pmatrix}11&-7&2\\-9&9&-3\\1&-2&1\end{pmatrix}$
$\Rightarrow{}\quad \boxed{ adj(A) = C^T = \begin{pmatrix}11&-9&1\\-7&9&-2\\2&-3&1\end{pmatrix}}$
$\Rightarrow{}\quad A^{-1} = \frac{1}{det(A)}adj(A) = \frac{1}{3} \begin{pmatrix}11&-9&1\\-7&9&-2\\2&-3&1\end{pmatrix}$
$\Rightarrow{}\quad \boxed{AA^{-1} = \begin{pmatrix}1&2&3\\1&3&5\\1&5&12\end{pmatrix} \begin{pmatrix}11/3 & -9/3 & 1/3 \\-7/3 & 9/3 & -2/3\\ 2/3 & -3/3 & 1/3 \end{pmatrix} = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}$
Otherwise solving, simply, with Matlab:
a = [1 2 3 ; 1 3 5 ; 1 5 12]
det(a)
adjoint(a)
inv(a)
b = inv(a)
a * b %should equal the identity matrix
Problem 7.
Solve the following sets of equations, if possible, using matrix methods. Otherwise, determine any relationships between the unknowns.
(a) $\begin{align} x + y + z &= 7 \newline 2x -y + 2z &= 8 \newline 3x + 2y -z &= 11 \end{align}$
$\Rightarrow{}\quad Ax = b \quad\Rightarrow{}\quad \begin{pmatrix}1&1&1\\2&-1&2\\3&2&-1\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}7\\8\\11\end{pmatrix}$
$\Rightarrow{}\quad A^{-1}Ax = A^{-1}b \quad\Rightarrow{}\quad Ix = A^{-1}b \quad\Rightarrow{}\quad x = A^{-1}b$
$\Rightarrow{}\quad A^{-1} = \frac{1}{detA}adj(A) \quad\Rightarrow{}\quad A^{-1} = \frac{1}{12} \begin{pmatrix}-3&3&3\\8&-4&0\\7&1&-3\end{pmatrix}$
$\Rightarrow{}\quad \begin{pmatrix}x \newline y \newline z\end{pmatrix} = \frac{1}{12} \begin{pmatrix}-3&3&3 \newline 8&-4&0 \newline 7&1&-3\end{pmatrix} \begin{pmatrix}7 \newline 8 \newline 11\end{pmatrix} \quad\Rightarrow{}\quad \begin{pmatrix}x \newline y \newline z\end{pmatrix} = \begin{pmatrix} 3 \newline 2 \newline 2\end{pmatrix}$
$\Rightarrow{}\quad \boxed{x=3, \quad y=2, \quad z=2}$
Solving using Matlab:
$\Rightarrow{}\quad A^{-1}Ax = A^{-1}b \quad\Rightarrow{}\quad Ix = A^{-1}b \quad\Rightarrow{}\quad x = A^{-1}b$
$\Rightarrow{}\quad A^{-1} = \frac{1}{detA}adj(A) \quad\Rightarrow{}\quad A^{-1} = \frac{1}{12} \begin{pmatrix}-3&3&3\\8&-4&0\\7&1&-3\end{pmatrix}$
$\Rightarrow{}\quad \begin{pmatrix}x \newline y \newline z\end{pmatrix} = \frac{1}{12} \begin{pmatrix}-3&3&3 \newline 8&-4&0 \newline 7&1&-3\end{pmatrix} \begin{pmatrix}7 \newline 8 \newline 11\end{pmatrix} \quad\Rightarrow{}\quad \begin{pmatrix}x \newline y \newline z\end{pmatrix} = \begin{pmatrix} 3 \newline 2 \newline 2\end{pmatrix}$
$\Rightarrow{}\quad \boxed{x=3, \quad y=2, \quad z=2}$
Solving using Matlab:
a = [1 1 1 ; 2 -1 2 ; 3 2 -1]
b = [7 ; 8 ; 11]
a = inv(a)
a * b %x, y, z are given by each row respectively
(b) $\begin{align} x + 3y + z &= 0 \newline 5x + y + 3z &= 0 \newline 4x - 2y + 2z &= 0 \end{align}$
$\Rightarrow{}\quad Ax = b \quad\Rightarrow{}\quad \begin{pmatrix}1&3&1\\5&1&3\\4&-2&2\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}$
$\Rightarrow{}\quad detA = 0 \quad\Rightarrow{}\quad$ singular (not invertible)
$\Rightarrow{}\quad \boxed{x/y=4, \quad y/z=1/7}$
$\Rightarrow{}\quad detA = 0 \quad\Rightarrow{}\quad$ singular (not invertible)
$\Rightarrow{}\quad \boxed{x/y=4, \quad y/z=1/7}$
(c) $\begin{align} x + 3y + 4z &= 0 \newline x + y + 3z &= 0 \newline 2x + 5y + z &= 0 \end{align}$
$\Rightarrow{}\quad Ax = b \quad\Rightarrow{}\quad \begin{pmatrix}1&3&4 \\ 1&1&3 \\ 2&5&1\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}$
$\Rightarrow{}\quad detA = 13 \quad\Rightarrow{}\quad$ nonsingular (invertible)
$\Rightarrow{}\quad A^{-1}$ can be obtained, but whatever the elements of it $\Rightarrow{}\quad \boxed{x=y=z=0}$
$\Rightarrow{}\quad detA = 13 \quad\Rightarrow{}\quad$ nonsingular (invertible)
$\Rightarrow{}\quad A^{-1}$ can be obtained, but whatever the elements of it $\Rightarrow{}\quad \boxed{x=y=z=0}$
(d) $\begin{align} 2v + 3w + 7x + y + z &= 41 \newline w + 8x + 11z &= 43 \newline 8v + w + 7y + 3z &= 75 \newline 9v + 2w + 2x + y &= 31 \newline 6w + 9y + 4z &= 110 \end{align}$
$\Rightarrow{}\quad Ax = b \quad\Rightarrow{}\quad \begin{pmatrix}2&3&7&1&1 \\ 0&1&8&0&11 \\ 8&1&0&7&3 \\ 9&2&2&1&0 \\ 0&6&0&9&4 \end{pmatrix} \begin{pmatrix}v\\w\\x\\y\\z\end{pmatrix} = \begin{pmatrix}41 \\ 43 \\ 75 \\ 31 \\ 110 \end{pmatrix}$
$\Rightarrow{}\quad detA \neq 0 \quad\Rightarrow{}\quad$ non-singular (invertible)
$\Rightarrow{}\quad \begin{pmatrix}v \newline w \newline x \newline y \newline z\end{pmatrix} = A^{-1} \begin{pmatrix}41 \newline 43 \newline 75 \newline 31 \newline 110\end{pmatrix} \quad\Rightarrow{}\quad \begin{pmatrix}v \newline w \newline x \newline y \newline z\end{pmatrix} = \begin{pmatrix} 1 \newline 5 \newline 2 \newline 8 \newline 2\end{pmatrix}$
$\Rightarrow{}\quad \boxed{v=1, \quad w=5, \quad x=2, \quad y=8, \quad z=2}$
It would be unreasonable to calculate this by hand, therefore a tool like Matlab is needed and can be used as shown:
$\Rightarrow{}\quad detA \neq 0 \quad\Rightarrow{}\quad$ non-singular (invertible)
$\Rightarrow{}\quad \begin{pmatrix}v \newline w \newline x \newline y \newline z\end{pmatrix} = A^{-1} \begin{pmatrix}41 \newline 43 \newline 75 \newline 31 \newline 110\end{pmatrix} \quad\Rightarrow{}\quad \begin{pmatrix}v \newline w \newline x \newline y \newline z\end{pmatrix} = \begin{pmatrix} 1 \newline 5 \newline 2 \newline 8 \newline 2\end{pmatrix}$
$\Rightarrow{}\quad \boxed{v=1, \quad w=5, \quad x=2, \quad y=8, \quad z=2}$
It would be unreasonable to calculate this by hand, therefore a tool like Matlab is needed and can be used as shown:
a = [2 3 7 1 1 ; 0 1 8 0 11 ; 8 1 0 7 3 ; 9 2 2 1 0 ; 0 6 0 9 4]
b = [41 ; 43 ; 75 ; 31 ; 110]
a = inverse(a)
a * b
Extension Questions
Problem 8.
Determine whether the following matrices are symmetric, skew-symmetric, triangular, diagonal or singular:
(a) $\begin{pmatrix}1&2 \newline 2&3\end{pmatrix}$
SS
A symmetric matrix is a square matrix that is equal to its transpose; that is, it satisfies the condition $A=A^T$.
A skew-symmetric matrix is a square matrix whose transpose equals its negative; that is, it satisfies the condition $A^T=-A$.
A triangular matrix is a square matrix that is either lower triangular or upper triangular. In a lower triangular matrix, all the entries above the main diagonal are zero. In a upper triangular matrix, all the entries below the main diagonal are zero.
A diagonal matrix is a square matrix that is both upper and lower triangular; that is all entries outside the main diagonal are zero.
A singular matrix is a square matrix that is not invertible. A square matrix is singular if and only if its determinant is 0.
$\quad\Rightarrow{}\quad$ $\begin{pmatrix}1&2 \newline 2&3\end{pmatrix}^T=\begin{pmatrix}1&2 \newline 2&3\end{pmatrix} \quad\Rightarrow{}\ \ \begin{vmatrix}1&2 \newline 2&3\end{vmatrix} = 1(3)-2(2) = 1 \quad\Rightarrow{}\ \ \boxed{\text{symmetric}}$
A skew-symmetric matrix is a square matrix whose transpose equals its negative; that is, it satisfies the condition $A^T=-A$.
A triangular matrix is a square matrix that is either lower triangular or upper triangular. In a lower triangular matrix, all the entries above the main diagonal are zero. In a upper triangular matrix, all the entries below the main diagonal are zero.
A diagonal matrix is a square matrix that is both upper and lower triangular; that is all entries outside the main diagonal are zero.
A singular matrix is a square matrix that is not invertible. A square matrix is singular if and only if its determinant is 0.
$\quad\Rightarrow{}\quad$ $\begin{pmatrix}1&2 \newline 2&3\end{pmatrix}^T=\begin{pmatrix}1&2 \newline 2&3\end{pmatrix} \quad\Rightarrow{}\ \ \begin{vmatrix}1&2 \newline 2&3\end{vmatrix} = 1(3)-2(2) = 1 \quad\Rightarrow{}\ \ \boxed{\text{symmetric}}$
(b) $\begin{pmatrix}2&1 \newline -1&4\end{pmatrix}$
$\quad\Rightarrow{}\quad$
$\begin{pmatrix}2&1\\-1&4\end{pmatrix}^T=\begin{pmatrix}2&-1\\1&4\end{pmatrix} \quad\Rightarrow{}\ \ \begin{vmatrix}2&1\\-1&4\end{vmatrix} = 2(4)-1(-1) = 9 \quad\Rightarrow{}\ \ \boxed{\text{none}}$
(c) $\begin{pmatrix}0&a \newline a&0\end{pmatrix}$
$\quad\Rightarrow{}\quad$
$\begin{pmatrix}0&a\\a&0\end{pmatrix}^T=\begin{pmatrix}0&a\\a&0\end{pmatrix} \quad\Rightarrow{}\ \ \begin{vmatrix}0&a\\a&0\end{vmatrix} = 0(0)-a(a) = -a^2 \quad\Rightarrow{}\ \ \boxed{\text{symmetric}}$
(d) $\begin{pmatrix}a&0 \newline 0&a\end{pmatrix}$
$\quad\Rightarrow{}\quad$
$\begin{pmatrix}a&0\\0&a\end{pmatrix}^T=\begin{pmatrix}a&0\\0&a\end{pmatrix} \quad\Rightarrow{}\ \ \begin{vmatrix}a&0\\0&a\end{vmatrix} = a(a)-0(0) = a^2$
$\quad\Rightarrow{}\ \ \boxed{\text{symmetric and diagonal (and triangular)}}$
(e) $\begin{pmatrix}1&1 \newline 1&1\end{pmatrix}$
$\quad\Rightarrow{}\quad$
$\begin{pmatrix}1&1\\1&1\end{pmatrix}^T=\begin{pmatrix}1&1\\1&1\end{pmatrix} \quad\Rightarrow{}\ \ \begin{vmatrix}1&1\\1&1\end{vmatrix} = 1(1)-1(1) = 0$
$\quad\Rightarrow{}\ \ \boxed{\text{symmetric and singular}}$
(f) $\begin{pmatrix}0&-1 \newline 1&0\end{pmatrix}$
$\quad\Rightarrow{}\quad$
$\begin{pmatrix}0&-1\\1&0\end{pmatrix}^T=\begin{pmatrix}0&1\\-1&0\end{pmatrix} \quad\Rightarrow{}\ \ \begin{vmatrix}0&-1\\1&0\end{vmatrix} = 0(0)--1(1) = 1$
$\quad\Rightarrow{}\ \ \boxed{\text{skew-symmetric}}$
Problem 9.
Determine whether the following matrices are singular or orthogonal:
(a) $\begin{pmatrix}1&0.5&0 \newline 0.5&0.75&0.5 \newline 0&0.5&0.5\end{pmatrix}$
A orthogonal matrix is a square matrix whose columns and rows are orthogonal unit vectors; that is, it satisfies the condition $Q^TQ=QQ^T=I$.
A singular matrix is a square matrix that is not invertible. A square matrix is singular if and only if its determinant is 0.
$\quad\Rightarrow{}\quad \begin{vmatrix}1&0.5&0 \newline 0.5&0.75&0.5 \newline 0&0.5&0.5\end{vmatrix} = 1\begin{vmatrix}0.75&0.5 \newline 0.5&0.5\end{vmatrix} -0.5\begin{vmatrix}0.5&0.5 \newline 0&0.5\end{vmatrix} +0\begin{vmatrix}0.5&0.75 \newline 0&0.5\end{vmatrix}= 0$
$\quad\Rightarrow{}\quad \boxed{\text{singular}}$
$\quad\Rightarrow{}\quad \begin{pmatrix}1&0.5&0 \newline 0.5&0.75&0.5 \newline 0&0.5&0.5\end{pmatrix}^T = \begin{pmatrix}1&0.5&0 \newline 0.5&0.75&0.5 \newline 0&0.5&0.5\end{pmatrix}$ $\quad\Rightarrow{}\quad \begin{pmatrix}1&0.5&0 \newline 0.5&0.75&0.5 \newline 0&0.5&0.5\end{pmatrix}\begin{pmatrix}1&0.5&0 \newline 0.5&0.75&0.5 \newline 0&0.5&0.5\end{pmatrix} \neq I \quad\Rightarrow{}\quad \boxed{\text{non orthogonal}} $
$\quad\Rightarrow{}\quad \begin{pmatrix}1&0.5&0 \newline 0.5&0.75&0.5 \newline 0&0.5&0.5\end{pmatrix}^T = \begin{pmatrix}1&0.5&0 \newline 0.5&0.75&0.5 \newline 0&0.5&0.5\end{pmatrix}$ $\quad\Rightarrow{}\quad \begin{pmatrix}1&0.5&0 \newline 0.5&0.75&0.5 \newline 0&0.5&0.5\end{pmatrix}\begin{pmatrix}1&0.5&0 \newline 0.5&0.75&0.5 \newline 0&0.5&0.5\end{pmatrix} \neq I \quad\Rightarrow{}\quad \boxed{\text{non orthogonal}} $
(b) $\begin{pmatrix}0.6&-0.8&0 \newline 0.8&0.6&0 \newline 0&0&1\end{pmatrix}$
$\quad\Rightarrow{}\quad \begin{vmatrix}0.6&-0.8&0\\0.8&0.6&0\\0&0&1\end{vmatrix} = 0.6\begin{vmatrix}0.6&0\\0&1\end{vmatrix} -0.8\begin{vmatrix}0.8&0\\0&1\end{vmatrix} +0\begin{vmatrix}0.8&0.6\\0&0\end{vmatrix} = 1$
$\quad\Rightarrow{}\quad \boxed{\text{non singular }}$
$\quad\Rightarrow{}\quad \begin{pmatrix}0.6&-0.8&0 \newline 0.8&0.6&0 \newline 0&0&1\end{pmatrix}^T = \begin{pmatrix}0.6&0.8&0 \newline -0.8&0.6&0 \newline 0&0&1\end{pmatrix}$
$\quad\Rightarrow{}\quad \begin{pmatrix}0.6&-0.8&0 \newline 0.8&0.6&0 \newline 0&0&1\end{pmatrix} \begin{pmatrix}0.6&0.8&0 \newline -0.8&0.6&0 \newline 0&0&1\end{pmatrix} = \begin{pmatrix}1&0&0 \newline 0&1&0 \newline 0&0&1\end{pmatrix} \quad\Rightarrow{}\quad \boxed{\text{orthogonal}}$
$\quad\Rightarrow{}\quad \boxed{\text{non singular }}$
$\quad\Rightarrow{}\quad \begin{pmatrix}0.6&-0.8&0 \newline 0.8&0.6&0 \newline 0&0&1\end{pmatrix}^T = \begin{pmatrix}0.6&0.8&0 \newline -0.8&0.6&0 \newline 0&0&1\end{pmatrix}$
$\quad\Rightarrow{}\quad \begin{pmatrix}0.6&-0.8&0 \newline 0.8&0.6&0 \newline 0&0&1\end{pmatrix} \begin{pmatrix}0.6&0.8&0 \newline -0.8&0.6&0 \newline 0&0&1\end{pmatrix} = \begin{pmatrix}1&0&0 \newline 0&1&0 \newline 0&0&1\end{pmatrix} \quad\Rightarrow{}\quad \boxed{\text{orthogonal}}$
(c) $\begin{pmatrix}0&2^{-\frac{1}{2}}&2^{-\frac{1}{2}} \newline 2^{-\frac{1}{2}}&-0.5&0.5 \newline 2^{-\frac{1}{2}}&0.5&-0.5\end{pmatrix}$
$\quad\Rightarrow{}\quad \begin{vmatrix}0&2^{-\frac{1}{2}}&2^{-\frac{1}{2}}\\2^{-\frac{1}{2}}&-0.5&0.5\\2^{-\frac{1}{2}}&0.5&-0.5\end{vmatrix} = 0\begin{vmatrix}-0.5&0.5\\0.5&-0.5\end{vmatrix} -2^{-\frac{1}{2}}\begin{vmatrix}2^{-\frac{1}{2}}&0.5\\2^{-\frac{1}{2}}&-0.5\end{vmatrix} +2^{-\frac{1}{2}}\begin{vmatrix}2^{-\frac{1}{2}}&-0.5\\2^{-\frac{1}{2}}&0.5\end{vmatrix}$
$\quad\Rightarrow{}\quad \begin{vmatrix}0&2^{-\frac{1}{2}}&2^{-\frac{1}{2}}\\2^{-\frac{1}{2}}&-0.5&0.5\\2^{-\frac{1}{2}}&0.5&-0.5\end{vmatrix} = 1 \quad\Rightarrow{}\quad \boxed{\text{no singular}}$
$\quad\Rightarrow{}\quad \begin{pmatrix}0&2^{-\frac{1}{2}}&2^{-\frac{1}{2}}\\2^{-\frac{1}{2}}&-0.5&0.5\\2^{-\frac{1}{2}}&0.5&-0.5\end{pmatrix}^T = \begin{pmatrix}0&2^{-\frac{1}{2}}&2^{-\frac{1}{2}}\\2^{-\frac{1}{2}}&-0.5&0.5\\2^{-\frac{1}{2}}&0.5&-0.5\end{pmatrix}$
$\quad\Rightarrow{}\quad \begin{pmatrix}0&2^{-\frac{1}{2}}&2^{-\frac{1}{2}}\\2^{-\frac{1}{2}}&-0.5&0.5\\2^{-\frac{1}{2}}&0.5&-0.5\end{pmatrix} \begin{pmatrix}0&2^{-\frac{1}{2}}&2^{-\frac{1}{2}}\\2^{-\frac{1}{2}}&-0.5&0.5\\2^{-\frac{1}{2}}&0.5&-0.5\end{pmatrix} = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\\ \quad\Rightarrow{}\quad \boxed{\text{orthogonal}}$
$\quad\Rightarrow{}\quad \begin{vmatrix}0&2^{-\frac{1}{2}}&2^{-\frac{1}{2}}\\2^{-\frac{1}{2}}&-0.5&0.5\\2^{-\frac{1}{2}}&0.5&-0.5\end{vmatrix} = 1 \quad\Rightarrow{}\quad \boxed{\text{no singular}}$
$\quad\Rightarrow{}\quad \begin{pmatrix}0&2^{-\frac{1}{2}}&2^{-\frac{1}{2}}\\2^{-\frac{1}{2}}&-0.5&0.5\\2^{-\frac{1}{2}}&0.5&-0.5\end{pmatrix}^T = \begin{pmatrix}0&2^{-\frac{1}{2}}&2^{-\frac{1}{2}}\\2^{-\frac{1}{2}}&-0.5&0.5\\2^{-\frac{1}{2}}&0.5&-0.5\end{pmatrix}$
$\quad\Rightarrow{}\quad \begin{pmatrix}0&2^{-\frac{1}{2}}&2^{-\frac{1}{2}}\\2^{-\frac{1}{2}}&-0.5&0.5\\2^{-\frac{1}{2}}&0.5&-0.5\end{pmatrix} \begin{pmatrix}0&2^{-\frac{1}{2}}&2^{-\frac{1}{2}}\\2^{-\frac{1}{2}}&-0.5&0.5\\2^{-\frac{1}{2}}&0.5&-0.5\end{pmatrix} = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\\ \quad\Rightarrow{}\quad \boxed{\text{orthogonal}}$
Problem 10.
Given that the matrix $\textbf{M}$ is given by: $\textbf{M} = \begin{bmatrix} 4 \ 12 \newline x \ 8 \end{bmatrix}$ and: $(\textbf{M}^{-1})^T = -\frac{2}{184}\begin{bmatrix} 4 \ {-9} \newline {-6} \ 2 \end{bmatrix}$ find $x$
$\quad\Rightarrow{}\quad \textbf{M}^{-1} = -\frac{2}{184}\begin{bmatrix}
4 \ {-6} \newline {-9} \ 2
\end{bmatrix} = -\frac{1}{184}\begin{bmatrix}
8 \ {-12} \newline {-18} \ 4
\end{bmatrix}$
(note: multiplying the matrix by the coefficient's numerator leaves the inverted matrix in a more recognizable form)
given that for matrix $\textbf{A}$:
$\quad\Rightarrow{} \quad \textbf{A} = \begin{bmatrix} a \ b \newline c \ d \end{bmatrix} \quad\Rightarrow{}\quad \textbf{A}^{-1} = \frac{1}{\text{det}(\textbf{A})}\begin{bmatrix} d \ {-b} \newline {-c} \ a \end{bmatrix}$
$\quad\Rightarrow{}\quad {-x}={-18} \quad\Rightarrow{}\quad \boxed{x = 18}$
(note: multiplying the matrix by the coefficient's numerator leaves the inverted matrix in a more recognizable form)
given that for matrix $\textbf{A}$:
$\quad\Rightarrow{} \quad \textbf{A} = \begin{bmatrix} a \ b \newline c \ d \end{bmatrix} \quad\Rightarrow{}\quad \textbf{A}^{-1} = \frac{1}{\text{det}(\textbf{A})}\begin{bmatrix} d \ {-b} \newline {-c} \ a \end{bmatrix}$
$\quad\Rightarrow{}\quad {-x}={-18} \quad\Rightarrow{}\quad \boxed{x = 18}$
Exam Style Questions
Problem 11.
Given the following matrices: $A = \begin{bmatrix} 1 \ 2 \ 0 \newline 3 \ 1 \ 0 \newline 5 \ 0 \ 1 \end{bmatrix}$ $B = \begin{bmatrix} 3 \ 3 \ 3 \newline 3 \ 3 \ 3 \end{bmatrix} $ $C = \begin{bmatrix} \ 2 \ 1 \ 9 \end{bmatrix}.$
(a) Find the determinant of $A$
$detA= 1(1)-2(3)+0 = \boxed{-5}$
Matlab:
Matlab:
a = [1 2 0 ; 3 1 0 ; 5 0 1]
det(a)
(b) Find the inverse of matrix $A$ multiplied by the determinant of $A$
$A^{-1}|A| = \boxed{\begin{bmatrix}
1 & -2 & 0 \\ -3 & 1 & 0 \\ -5 & 10 & -5
\end{bmatrix}}$
Matlab:
Matlab:
x = inv(a)
y = det(a)
x * y
(c) Find the inverse of the transpose of matrix $A$
$(A^{T})^{-1} = (A^{-1})^T = \boxed{\frac{1}{-5} \begin{bmatrix}
1 & -3 & -5 \\ -2 & 1 & 10 \\ 0 & 0 & -5
\end{bmatrix}}$
Matlab:
Matlab:
x = transpose(a)
inverse(x)
(d) Find $BA$
$BA = \boxed{\begin{bmatrix}
27 & 9 & 3 \\ 27 & 9 & 3
\end{bmatrix}}$
Matlab:
Matlab:
b = [3 3 3 ; 3 3 3]
b * a
(e) What are the dimensions of $D$ in the following operation? $D = ((BA)^T B - A)C^T$
$[(2\times3)(3\times3)^T(2\times3)-(3\times3)](3\times1)=\boxed{(3\times1)}$
Problem 12.
The matrix $\textbf{M}$ is given by: \(\textbf{M} = \begin{bmatrix} k & 2 & 1 \newline 3 & 0 & 4 \newline 1 & -2 & 1 \newline \end{bmatrix}, \quad k \neq 1\)
(a) Show that the determinant of $\textbf{M} = 8\textit{k}-4$.
$Det(\textbf{M}) = k(0+8)-2(3-4)+1(-6)= \boxed{8\textit{k}-4}$
(b) Find $\textbf{M}^{-1}$ in terms of k.
Matrix of minors:
$$
\begin{bmatrix}
8 & -1 & -6\\
4 & k-1 & -(2k+2)\\
8 & 4k-3 & -6\\
\end{bmatrix}
$$
Matrix of cofactors and transpose:
$$
\begin{bmatrix}
8 & -4 & 8\\
1 & k-1 & 3-4k\\
-6 & 2k+2 & -6\\
\end{bmatrix}
$$
Multiply by $\frac{1}{det(\textbf{M})}$:
$$
\boxed{\frac{1}{8\textit{k}-4}
\begin{bmatrix}
8 & -4 & 8\\
1 & k-1 & 3-4k\\
-6 & 2k+2 & -6\\
\end{bmatrix}}
$$
The straight line $l_{1}$ is mapped onto the straight line $l_{2}$ by the transformation represented by the matrix: \(N = \begin{bmatrix} 1 & 2 & 1 \newline 3 & 0 & 4 \newline 1 & -2 & 1 \newline \end{bmatrix}\) The equation of $l_{2}$ is $\textbf{r}=\textbf{a}+\mu\textbf{b}$, where $\textbf{a} = 6\textbf{i}+5\textbf{j}+2\textbf{k}$ and $\textbf{b} = 6\textbf{i}+5\textbf{j}+4\textbf{k}$
(c) Find a vector equation for the line $l_{1}$.
Let $(x,y,z)$ be on $l_{1}$.
Equation $l_{2}$ can be written as:
$$r = \begin{pmatrix}
6+6\mu \newline
5+5\mu \newline
2+4\mu \newline
\end{pmatrix}$$
Use $\textbf{N}^{-1}$ to map $l_{2}$ back to $l_{1}$ (matrix N is just matrix M with $k=1$):
$$\begin{pmatrix}
x \newline
y \newline
z \newline
\end{pmatrix} =
\textbf{N}^{-1}
\begin{pmatrix}
6+6\mu \newline
5+5\mu \newline
2+4\mu \newline
\end{pmatrix}$$
$$\begin{pmatrix}
x \newline
y \newline
z \newline
\end{pmatrix} =
\begin{pmatrix}
(12-5+4)+(12-5+8)\mu \newline
(1.5-0.5)+(1.5-1)\mu \newline
(-9+5-3)+(-9+5-6)\mu \newline
\end{pmatrix} =
\begin{pmatrix}
11+15\mu \newline
1+0.5\mu \newline
-7-10\mu \newline
\end{pmatrix}$$
Therefore the equation of $l_{1}$ is $\boxed{\textbf{r}=\textbf{a}+\mu\textbf{b}}$ where $\boxed{\textbf{a}=11i+j-7k}$ and $\boxed{\textbf{b}=15i+0.5j-10k}$
Answers
For Printing
Revision Questions
The questions included are optional, but here if you want some extra practice.
- Engineering Mathematics 7th edition, Stroud and Dexter : Pages 489-508, 515-517
- Matrix Introduction : The latter questions start to push on to the next couple of topics but the start keeps it quite basic.