Problem Solving With Matrices

Matrix Theory provides a powerful set of tools for solving complex problems in various fields, including network analysis, machine learning, physics, and more. Whether it is solving systems of equations, representing complex networks, or finding the inverse of matrices, understanding matrix operations is essential in modern problem-solving techniques.

Matrices are widely used in real-world applications, helping to solve complex problems efficiently. Some key applications include:

  • Computer Graphics: Transformations like rotation, scaling, and translation
  • Physics & Engineering: Solving equations in circuits, structures, and mechanics
  • Machine Learning & AI: Processing large datasets in neural networks
  • Cryptography: Secure data encryption using matrix operations
  • Network Analysis: Representing and analyzing connections in social networks, transportation, and communication systems

Matrix Multiplication

Matrix multiplication combines two matrices to produce a new one. The rule is that the number of columns in the first matrix must match the number of rows in the second.

Steps to Multiply Matrices

  1. Ensure the dimensions allow multiplication
  2. Multiply corresponding elements from rows and columns
  3. \[C = A \times B \]

    OR

    \[ \begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \times \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} \]


  4. Sum up the products to form the resulting matrix
  5. \[ \begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{bmatrix} = \begin{bmatrix} a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \\ a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22} \end{bmatrix} \]


Inverse Matrices

The inverse of a matrix \(\textbf{A}\) is another matrix \(\textbf{A⁻¹}\) such that:

\( A \times A^{-1} = I \) (Identity Matrix)

Steps to Find the Inverse (for a 2×2 matrix)

  1. Check if the determinant is nonzero:

    \[ \det(A) = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{12}a_{21} \]

  2. If invertible, compute the inverse using:

    \[ A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{bmatrix} \]


Representing Networks as Matrices

Networks, such as social connections, transportation systems, and web links, can be represented using adjacency matrices.

Steps to Represent a Network

  1. Create a matrix where each row and column represent a node.
  2. Use 1 to indicate a direct connection and 0 otherwise.

Multipy the following matrices:

\( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} , \quad B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} \)

First, we can calculate \(C_{11}\):

\(C_{11} = a_{11}b_{11} + a_{12}b_{12}\)

\(C_{11} = (1 \times 5) + (2 \times 7)\)

\(C_{11} = 5 + 14\)

\(C_{11} = 19\)

Next, we can calculate \( C_{12} \):

\(C_{12} = a_{11}b_{12} + a_{12}b_{22}\)

\(C_{12} = (1 \times 6) + (2 \times 8)\)

\(C_{12} = 6 + 16\)

\(C_{12} = 22\)

Then, we can calculate \(C_{21}\):

\(C_{21} = a_{21}b_{11} + a_{22}b_{21}\)

\(C_{21} = (3 \times 5) + (4 \times 7)\)

\(C_{21} = 15 + 28\)

\(C_{21} = 43\)

After, we can calculate \(C_{22}\):

\(C_{22} = a_{21}b_{12} + a_{22}b_{22}\)

\(C_{22} = (3 \times 6) + (4 \times 8)\)

\(C_{22} = 18 + 32\)

\(C_{22} = 50 \)

Finally, we can put all the values together to get the resulting matrix:

\(C = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix} \)

Therefore, we can determine the resulting matrix, \(C\), is \(C = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix} \).


\(A = \begin{bmatrix} 2 & 5 \\ 7 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 4 & 8 \\ 6 & 1 \end{bmatrix} \)

First, we can calculate \(C_{11}\):

\(C_{11} = a_11b_11 + a_12b_12\)

\(C_{11} = (2 \times 4) + (5 \times 6)\)

\(C_{11} = 8 + 30\)

\(C_{11} = 38\)

Next, we can calculate \(C_{12}\):

\(C_{12} = a_{11}b_{12} + a_{12}b_{22}\)

\(C_{12} = (2 \times 8) + (5 \times 1)\)

\(C_{12} = 16 + 5\)

\(C_{12} = 21\)

Then, we can calculate \(C_{21}\):

\(C_{21} = a_{21}b_{11} + a_{22}b_{21}\)

\(C_{21} = (7 \times 4) + (3 \times 6)\)

\(C_{21} = 28 + 18\)

\(C_{21} = 46\)

After, we can calculate \(C_{22}\):

\(C_{22} = a_{21}b_{12} + a_{22}b_{22}\)

\(C_{22} = (7 \times 8) + (3 \times 1)\)

\(C_{22} = 56 + 3\)

\(C_{22} = 59\)

Finally, we can put all the values together to get the resulting matrix:

\(C = \begin{bmatrix} 38 & 21 \\ 46 & 59 \end{bmatrix} \)

Therefore, we can determine the resulting matrix, \(C\), is \(C = \begin{bmatrix} 38 & 21 \\ 46 & 59 \end{bmatrix}\).


Find the inverse of the following matrices:

\( A = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}\)

First, we can calculate the determinant of \(A\):

\(\det(A) = a_{11}a_{22} - a_{12}a_{21}\)

\(\det(A) = (4 \times 6) - (7 \times 2)\)

\(\det(A) = 24 - 14\)

\(\det(A) = 10\)

Next, we can apply the inverse matrix formula by substituting the determinant and matrix:

\[A^{-1} = \frac{1}{\det(A)} \begin{bmatrix}a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{bmatrix}\]

\[A^{-1} = \frac{1}{10} \begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix}\]

Then, we can simplify the matrix by multiplying each element by \(\cfrac{1}{10}\). We can then substitute the values:

\[A^{-1} = \begin{bmatrix} \frac{6}{10} & \frac{-7}{10} \\ \frac{-2}{10} & \frac{4}{10} \end{bmatrix}\]

\[A^{-1} = \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix}\]

Therefore, we can determine the inverse of matrix \(A\) is \(A^{-1} = \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix}\).


\(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\)

First, we can calculate the determinant of \(A\):

\(\det(A) = a_{11}a_{22} - a_{12}a_{21}\)

\(\det(A) = (1 \times 4) - (2 \times 3)\)

\(\det(A) = 4 - 6\)

\(\det(A) = -2\)

Next, we can apply the inverse matrix formula by substituting the determinant and matrix:

\[A^{-1} = \frac{1}{\det(A)} \begin{bmatrix}a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{bmatrix}\]

\[A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}\]

Then, we can simplify the matrix by multiplying each element by \(-\cfrac{1}{2}\). We can then simplify the values:

\[A^{-1} = \begin{bmatrix} \frac{4}{-2} & \frac{-2}{-2} \\ \frac{-3}{-2} & \frac{1}{-2} \end{bmatrix}\]

\[A^{-1} = \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}\]

Therefore, we can determine the inverse of matrix \( A \) is \(A^{-1} = \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}\).