Matrices II

Inverse of a Matrix

Lesson Objectives

Understand different types of matrices.
Calculate the determinant of 3x3 matrices.
Solve linear equations using inverse matrices.

Lesson Introduction

A matrix is a rectangular array of numbers. It can be used to represent and solve systems of linear equations. In this lesson, we will focus on 3x3 matrices.

Types of Matrices

Example 1: Identify the type of matrix: \( A = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)
This is a zero matrix because all entries are 0.

Example 2: Identify the matrix type: \( B = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \)
This is a scalar matrix because it's a diagonal matrix with all diagonal elements equal.

Example 3: Identify: \( C = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
This is an identity matrix.

Example 4: \( D = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \)
This is a rectangular matrix because it has 2 rows and 3 columns.

Example 5: \( E = \begin{bmatrix} 2 & 4 & 6 \\ 4 & 8 & 12 \\ 6 & 12 & 18 \end{bmatrix} \)
This is a square matrix and also a symmetric matrix if \( E = E^T \).

Determinant of a 3x3 Matrix

Example 1: Find the determinant of:
\( A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{bmatrix} \)

Using the formula:
\( \text{det}(A) = 1(4 \cdot 6 - 5 \cdot 0) - 2(0 \cdot 6 - 5 \cdot 1) + 3(0 \cdot 0 - 4 \cdot 1) \)
= \( 1(24) - 2(-5) + 3(-4) = 24 + 10 - 12 = 22 \)

Example 2:
\( B = \begin{bmatrix} 2 & 3 & 1 \\ 1 & 0 & 4 \\ 5 & 6 & 0 \end{bmatrix} \)
Expand along the first row:
\( \text{det}(B) = 2(0 \cdot 0 - 4 \cdot 6) - 3(1 \cdot 0 - 4 \cdot 5) + 1(1 \cdot 6 - 0 \cdot 5) \)
= \( 2(-24) - 3(-20) + 6 = -48 + 60 + 6 = 18 \)

Example 3:
\( C = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 9 \end{bmatrix} \)
Apply the determinant rule:
= \( 1(2 \cdot 9 - 3 \cdot 4) - 1(1 \cdot 9 - 3 \cdot 1) + 1(1 \cdot 4 - 2 \cdot 1) \)
= \( 1(18 - 12) - 1(9 - 3) + 1(4 - 2) = 6 - 6 + 2 = 2 \)

Example 4:
\( D = \begin{bmatrix} 4 & 0 & 1 \\ 2 & 3 & 0 \\ 1 & 2 & 1 \end{bmatrix} \)
Expand along the first row:
\( 4(3 \cdot 1 - 0 \cdot 2) - 0(2 \cdot 1 - 0 \cdot 1) + 1(2 \cdot 2 - 3 \cdot 1) \)
= \( 4(3) + 0 + 1(4 - 3) = 12 + 1 = 13 \)

Example 5:
\( E = \begin{bmatrix} 2 & -1 & 0 \\ -3 & 4 & 1 \\ 1 & 2 & 3 \end{bmatrix} \)
Expand:
\( 2(4 \cdot 3 - 1 \cdot 2) + 1(-3 \cdot 3 - 1 \cdot 1) + 0 = 2(12 - 2) + 1(-9 - 1) = 2(10) -10 = 10 \)

Inverse of a 2×2 Matrix

If \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), then the inverse is:

\( A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \), provided \( ad - bc \ne 0 \).

Example 1: Find the inverse of \( A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \).
Step 1: Calculate the determinant:
\( \text{det}(A) = (2)(4) - (3)(1) = 8 - 3 = 5 \)
Step 2: Swap the diagonal and change signs:
\( A^{-1} = \frac{1}{5} \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix} \)

Example 2: Find the inverse of \( A = \begin{bmatrix} 5 & 2 \\ 1 & 1 \end{bmatrix} \).
\( \text{det} = 5(1) - 2(1) = 5 - 2 = 3 \)
\( A^{-1} = \frac{1}{3} \begin{bmatrix} 1 & -2 \\ -1 & 5 \end{bmatrix} \)

Example 3: \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)
\( \text{det} = 1(4) - 2(3) = 4 - 6 = -2 \)
\( A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} \)

Example 4: \( A = \begin{bmatrix} 3 & 0 \\ 5 & 2 \end{bmatrix} \)
\( \text{det} = 3(2) - 0(5) = 6 \)
\( A^{-1} = \frac{1}{6} \begin{bmatrix} 2 & 0 \\ -5 & 3 \end{bmatrix} \)

Example 5: \( A = \begin{bmatrix} 7 & 2 \\ 3 & 1 \end{bmatrix} \)
\( \text{det} = 7(1) - 2(3) = 7 - 6 = 1 \)
\( A^{-1} = \begin{bmatrix} 1 & -2 \\ -3 & 7 \end{bmatrix} \)

Inverse of a 3×3 Matrix

The inverse of a 3×3 matrix \( A \), if it exists, is given by:

\[ A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) \]

Example 1: Find the inverse of:
\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix} \] Step 1: Compute the determinant of A
\[ \det(A) = 1 \cdot (1 \cdot 0 - 4 \cdot 6) - 2 \cdot (0 \cdot 0 - 4 \cdot 5) + 3 \cdot (0 \cdot 6 - 1 \cdot 5) \] \[ \det(A) = 1(0 - 24) - 2(0 - 20) + 3(0 - 5) \] \[ \det(A) = -24 + 40 - 15 = 1 \] Step 2: Find the matrix of minors
Minor of element (1,1): remove row 1 and column 1: \[ \begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} = 1(0) - 4(6) = -24 \] Minor of element (1,2): \[ \begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} = 0(0) - 4(5) = -20 \] Minor of element (1,3): \[ \begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix} = 0(6) - 1(5) = -5 \] Minor of element (2,1): \[ \begin{vmatrix} 2 & 3 \\ 6 & 0 \end{vmatrix} = 2(0) - 3(6) = -18 \] Minor of element (2,2): \[ \begin{vmatrix} 1 & 3 \\ 5 & 0 \end{vmatrix} = 1(0) - 3(5) = -15 \] Minor of element (2,3): \[ \begin{vmatrix} 1 & 2 \\ 5 & 6 \end{vmatrix} = 1(6) - 2(5) = -4 \] Minor of element (3,1): \[ \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} = 2(4) - 3(1) = 5 \] Minor of element (3,2): \[ \begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix} = 1(4) - 3(0) = 4 \] Minor of element (3,3): \[ \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1(1) - 2(0) = 1 \] Matrix of minors:
\[ \begin{bmatrix} -24 & -20 & -5 \\ -18 & -15 & -4 \\ 5 & 4 & 1 \end{bmatrix} \] Step 3: Apply signs to get cofactor matrix
\[ \text{Cofactor} = \begin{bmatrix} -24 & 20 & -5 \\ 18 & -15 & 4 \\ 5 & -4 & 1 \end{bmatrix} \] Step 4: Transpose to get adjoint
\[ \text{adj}(A) = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix} \] Step 5: Inverse of A
\[ A^{-1} = \frac{1}{1} \cdot \text{adj}(A) = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix} \]

Example 2: Find the inverse of:
\[ B = \begin{bmatrix} 2 & 0 & 1 \\ 3 & 0 & 0 \\ 5 & 1 & 1 \end{bmatrix} \] Step 1: Find determinant
\[ \det(B) = 2 \cdot \begin{vmatrix} 0 & 0 \\ 1 & 1 \end{vmatrix} - 0 \cdot \begin{vmatrix} 3 & 0 \\ 5 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 3 & 0 \\ 5 & 1 \end{vmatrix} \] \[ = 2(0 - 0) + 1(3 - 0) = 3 \] Step 2: Find minors
Minor(1,1): \[ \begin{vmatrix} 0 & 0 \\ 1 & 1 \end{vmatrix} = 0 \] Minor(1,2): \[ \begin{vmatrix} 3 & 0 \\ 5 & 1 \end{vmatrix} = 3(1) - 0(5) = 3 \] Minor(1,3): \[ \begin{vmatrix} 3 & 0 \\ 5 & 1 \end{vmatrix} = 3(1) - 0(5) = 3 \] Minor(2,1): \[ \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} = 0(1) - 1(1) = -1 \] Minor(2,2): \[ \begin{vmatrix} 2 & 1 \\ 5 & 1 \end{vmatrix} = 2(1) - 1(5) = -3 \] Minor(2,3): \[ \begin{vmatrix} 2 & 0 \\ 5 & 1 \end{vmatrix} = 2(1) - 0(5) = 2 \] Minor(3,1): \[ \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} = 0(0) - 1(0) = 0 \] Minor(3,2): \[ \begin{vmatrix} 2 & 1 \\ 3 & 0 \end{vmatrix} = 2(0) - 1(3) = -3 \] Minor(3,3): \[ \begin{vmatrix} 2 & 0 \\ 3 & 0 \end{vmatrix} = 2(0) - 0(3) = 0 \] Matrix of minors:
\[ \begin{bmatrix} 0 & 3 & 3 \\ -1 & -3 & 2 \\ 0 & -3 & 0 \end{bmatrix} \] Step 3: Cofactor matrix
\[ \begin{bmatrix} 0 & -3 & 3 \\ 1 & -3 & -2 \\ 0 & 3 & 0 \end{bmatrix} \] Step 4: Adjoint = Transpose of cofactor
\[ \text{adj}(B) = \begin{bmatrix} 0 & 1 & 0 \\ -3 & -3 & 3 \\ 3 & -2 & 0 \end{bmatrix} \] Step 5: Inverse
\[ B^{-1} = \frac{1}{3} \begin{bmatrix} 0 & 1 & 0 \\ -3 & -3 & 3 \\ 3 & -2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & \frac{1}{3} & 0 \\ -1 & -1 & 1 \\ 1 & -\frac{2}{3} & 0 \end{bmatrix} \]

Solving Equations Using Matrices (3x3)

Example 1: Solve the system:
\( x + y + z = 6 \)
\( 2x + 3y + z = 14 \)
\( x + 2y + 3z = 14 \)

Matrix form: \( AX = B \)
\( A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 1 \\ 1 & 2 & 3 \end{bmatrix}, B = \begin{bmatrix} 6 \\ 14 \\ 14 \end{bmatrix} \)
Find \( A^{-1} \), then compute \( X = A^{-1}B \)
After computation: \( X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \)

Solving Equations Using Matrices (3x3)

Example 1: Solve the system:
\( x + y + z = 6 \)
\( 2x + 3y + z = 14 \)
\( x + 2y + 3z = 14 \)

Matrix form: \( AX = B \)
\( A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 1 \\ 1 & 2 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 6 \\ 14 \\ 14 \end{bmatrix} \)
Find \( A^{-1} \) and compute \( X = A^{-1}B \)
Final solution: \( X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \)

Example 2: Solve the system:
\( 2x + y - z = 1 \)
\( x - y + 2z = 3 \)
\( 3x + 2y + z = 10 \)

Matrix form:
\( A = \begin{bmatrix} 2 & 1 & -1 \\ 1 & -1 & 2 \\ 3 & 2 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 3 \\ 10 \end{bmatrix} \)
Compute \( A^{-1} \) and then \( X = A^{-1}B \)
Final answer: \( X = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} \)

Example 3: Solve the system:
\( x + 2y + 3z = 14 \)
\( 2x + 3y + 2z = 15 \)
\( 3x + y + z = 10 \)

Matrix form:
\( A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 2 \\ 3 & 1 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 14 \\ 15 \\ 10 \end{bmatrix} \)
Compute \( X = A^{-1}B \)
Solution: \( X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \)

Example 4: Solve:
\( x + y + z = 3 \)
\( 2x + y + 3z = 7 \)
\( x + 4y + 9z = 17 \)

Matrix form:
\( A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 1 & 3 \\ 1 & 4 & 9 \end{bmatrix}, \quad B = \begin{bmatrix} 3 \\ 7 \\ 17 \end{bmatrix} \)
Use inverse matrix to find \( X \)
Answer: \( X = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} \)

Example 5: Solve:
\( x + 2y + z = 4 \)
\( 3x - y + 2z = 10 \)
\( 2x + y + 2z = 7 \)

Matrix form:
\( A = \begin{bmatrix} 1 & 2 & 1 \\ 3 & -1 & 2 \\ 2 & 1 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 4 \\ 10 \\ 7 \end{bmatrix} \)
Find \( X = A^{-1}B \)
Final result: \( X = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \)

Exercises

Identify the type of matrix: \( \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)
Determine the determinant of: \( \begin{bmatrix} 2 & 1 & 3 \\ 4 & 1 & 2 \\ 3 & 0 & 1 \end{bmatrix} \)
Solve: \( x + y + z = 6; 2x + y + 3z = 14; x + 4y + 3z = 20 \)
Find the determinant of a matrix with rows [1, 2, 3], [4, 5, 6], [7, 8, 9]
[NECO] What type of matrix is \( \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \)? (Past Question)
Find inverse of \( \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \)
Calculate determinant of \( \begin{bmatrix} 6 & 1 & 1 \\ 4 & -2 & 5 \\ 2 & 8 & 7 \end{bmatrix} \)
Solve the system using matrices:
2x + y + z = 5; x + 3y + z = 10; x + y + 4z = 13
Classify: \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} \)
Find the determinant of identity matrix \( 3 \times 3 \)

Conclusion/Recap

We explored different types of matrices and learned to compute determinants and solve systems of equations using 3x3 matrices. Understanding matrix operations is essential for advanced math and applications in science and engineering.