Matrix Magic

Unlock the secrets of matrices with interactive learning!

Exercise 1: Matrix Basics

Given the matrix:

\[ A = \begin{pmatrix} 8 & 9 & 4 & 3 \\ -1 & \sqrt{7} & \frac{\sqrt{3}}{2} & 5 \\ 1 & 4 & 3 & 0 \\ 6 & 8 & -11 & 1 \end{pmatrix} \]

Find:

  1. The number of elements
  2. The order of the matrix
  3. The elements \( a_{22}, a_{23}, a_{24}, a_{34}, a_{43}, a_{44} \)
1. Number of elements: Count all individual numbers/symbols in the matrix.
There are 4 rows × 4 columns = 16 elements.
2. Order of the matrix: Given as (rows × columns).
This matrix has 4 rows and 4 columns, so its order is 4×4.
3. Specific elements:
\( a_{22} \): 2nd row, 2nd column → \( \sqrt{7} \)
\( a_{23} \): 2nd row, 3rd column → \( \frac{\sqrt{3}}{2} \)
\( a_{24} \): 2nd row, 4th column → 5
\( a_{34} \): 3rd row, 4th column → 0
\( a_{43} \): 4th row, 3rd column → -11
\( a_{44} \): 4th row, 4th column → 1

Exercise 2: Possible Matrix Orders

If a matrix has:

  1. 18 elements, what are the possible orders it can have?
  2. 6 elements, what are the possible orders?
The possible orders are all the factor pairs of the number of elements.
1. For 18 elements:
Factor pairs of 18: (1,18), (2,9), (3,6), (6,3), (9,2), (18,1)
Possible orders: 1×18, 2×9, 3×6, 6×3, 9×2, 18×1
2. For 6 elements:
Factor pairs of 6: (1,6), (2,3), (3,2), (6,1)
Possible orders: 1×6, 2×3, 3×2, 6×1
1×6
2×3
3×2
6×1

Exercise 3: Constructing Matrices

Construct a 3×3 matrix whose elements are given by:

  1. \( a_{ij} = |i - 2j| \)
  2. \( a_{ij} = \frac{(i + j)^3}{3} \)
1. For \( a_{ij} = |i - 2j| \):
1
3
5
0
2
4
1
1
3
Final matrix: \[ \begin{pmatrix} 1 & 3 & 5 \\ 0 & 2 & 4 \\ 1 & 1 & 3 \end{pmatrix} \]
2. For \( a_{ij} = \frac{(i + j)^3}{3} \):
8/3
9
64/3
9
64/3
125/3
64/3
125/3
72
Final matrix: \[ \begin{pmatrix} \frac{8}{3} & 9 & \frac{64}{3} \\ 9 & \frac{64}{3} & \frac{125}{3} \\ \frac{64}{3} & \frac{125}{3} & 72 \end{pmatrix} \]

Exercise 4: Matrix Transpose

If \[ A = \begin{pmatrix} 5 & 4 & 3 \\ 1 & -7 & 9 \\ 3 & 8 & 2 \end{pmatrix} \], find the transpose of A.

The transpose of a matrix is obtained by interchanging its rows and columns.
Original matrix rows become columns in the transpose.
5
1
3
4
-7
8
3
9
2
Transpose of A: \[ A^T = \begin{pmatrix} 5 & 1 & 3 \\ 4 & -7 & 8 \\ 3 & 9 & 2 \end{pmatrix} \]

Exercise 5: Negative Matrix Transpose

If \[ A = \begin{pmatrix} \sqrt{7} & -3 \\ -\sqrt{5} & 2 \\ \sqrt{3} & -5 \end{pmatrix} \], find the transpose of -A.

First, find -A by multiplying each element by -1:
\[ -A = \begin{pmatrix} -\sqrt{7} & 3 \\ \sqrt{5} & -2 \\ -\sqrt{3} & 5 \end{pmatrix} \]
Now, find the transpose by swapping rows and columns:
-√7
√5
-√3
3
-2
5
Transpose of -A: \[ (-A)^T = \begin{pmatrix} -\sqrt{7} & \sqrt{5} & -\sqrt{3} \\ 3 & -2 & 5 \end{pmatrix} \]

Exercise 6: Double Transpose Property

If \[ A = \begin{pmatrix} 5 & 2 & 2 \\ -\sqrt{17} & 0.7 & \frac{5}{2} \\ 8 & 3 & 1 \end{pmatrix} \], verify that \( (A^T)^T = A \).

1. First, find \( A^T \):
\[ A^T = \begin{pmatrix} 5 & -\sqrt{17} & 8 \\ 2 & 0.7 & 3 \\ 2 & \frac{5}{2} & 1 \end{pmatrix} \]
2. Now, find \( (A^T)^T \) by transposing again:
\[ (A^T)^T = \begin{pmatrix} 5 & 2 & 2 \\ -\sqrt{17} & 0.7 & \frac{5}{2} \\ 8 & 3 & 1 \end{pmatrix} \]
3. Compare with original A:
We can see that \( (A^T)^T \) is identical to the original matrix A.
This verifies the property that the transpose of a transpose returns the original matrix.
5
2
2
-√17
0.7
5/2
8
3
1

Exercise 7: Solving Matrix Equations

Find the values of x, y, and z from the following equations:

  1. \[ \begin{pmatrix} 12 & 3 \\ x & 5 \end{pmatrix} = \begin{pmatrix} y & z \\ 3 & 5 \end{pmatrix} \]
  2. \[ \begin{pmatrix} x + y & 2 \\ 5 + z & xy \end{pmatrix} = \begin{pmatrix} 6 & 2 \\ 5 & 8 \end{pmatrix} \]
  3. \[ \begin{pmatrix} x + y + z \\ x + z \\ y + z \end{pmatrix} = \begin{pmatrix} 9 \\ 5 \\ 7 \end{pmatrix} \]
1. For the first equation:
Set corresponding elements equal:
12 = y → \( y = 12 \)
3 = z → \( z = 3 \)
x = 3 → \( x = 3 \)
2. For the second equation:
x + y = 6 → Equation 1
5 + z = 5 → z = 0 → \( z = 0 \)
xy = 8 → Equation 2
From Equation 1: y = 6 - x
Substitute into Equation 2: x(6 - x) = 8 → 6x - x² = 8 → x² - 6x + 8 = 0
Solving: (x-2)(x-4) = 0 → x = 2 or x = 4
If x = 2, y = 4 → xy = 8 ✔
If x = 4, y = 2 → xy = 8 ✔
So two solutions: (x=2, y=4, z=0) or (x=4, y=2, z=0)
3. For the third equation:
x + y + z = 9 → Equation 1
x + z = 5 → Equation 2
y + z = 7 → Equation 3
From Equation 2: x = 5 - z
From Equation 3: y = 7 - z
Substitute into Equation 1: (5 - z) + (7 - z) + z = 9 → 12 - z = 9 → z = 3
Then x = 5 - 3 = 2
And y = 7 - 3 = 4
Solution: x = 2, y = 4, z = 3