1. Elementary Operations (Transformations)

These are the legal moves we can perform on a matrix. They are the foundation for solving systems of linear equations (like in Gaussian elimination) and for finding the inverse of a matrix. There are three basic operations for rows and three identical ones for columns.

  • (i) Interchange of any two rows or columns:

    • Notation: Ri↔RjR_i \leftrightarrow R_j (Row i is swapped with Row j).
    • Purpose: This is like reordering equations in a system. It helps to get a non-zero element into a pivot position (like a11a_{11}).

  • (ii) Multiplication of a row or column by a non-zero number:

    • Notation: Ri→kRiR_i \to kR_i (Row i is replaced by k times Row i).
    • Purpose: This is like multiplying an entire equation by a constant. It's used to create a '1' in a pivot position, which is a key step in creating an identity matrix.

  • (iii) Addition of a multiple of one row or column to another:

    • Notation: Ri→Ri+kRjR_i \to R_i + kR_j (Row i is replaced by the sum of itself and k times Row j).
    • Purpose: This is the workhorse operation, equivalent to adding a multiple of one equation to another to eliminate a variable. In matrix terms, this is used to create zeros in all other positions in a pivot's column.

2. Invertible Matrices

  • Definition: A square matrix A of order 'n' is called invertible (or non-singular) if we can find another square matrix B of the same order 'n' that acts as its multiplicative inverse.
    • Think about numbers: the inverse of 5 is 15\frac{1}{5} because 5×15=15 \times \frac{1}{5} = 1.
    • For matrices, the Identity Matrix (I) is the equivalent of the number 1. So, we are looking for a matrix B such that:

AB=BA=IAB = BA = I

- If such a matrix B exists, it is called the **inverse** of A and is denoted by A−1A^{-1}.
- A matrix that does not have an inverse is called **non-invertible** or **singular**.

  • Uniqueness: If a matrix has an inverse, that inverse is unique. There's only one undo matrix for any given invertible matrix.

  • Property (Reversal Law for Inverses): (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}.

    • Explanation (The Socks and Shoes Rule): To undo the action of putting on socks (A) and then shoes (B), you must first take off the shoes (B−1B^{-1}) and then take off the socks (A−1A^{-1}). The order of the undo operations is reversed.

3. Finding the Inverse using Elementary Operations

This is a standard algorithm to find the inverse of a matrix A. It works by solving for A.

  1. Write the equation A=IAA = IA.

    • This equation just says A is equal to itself. The 'I' on the right side is a placeholder that will record all the operations we do.
    • Think of it as: [Your Matrix]=[Starting Point]×A\text{[Your Matrix]} = \text{[Starting Point]} \times A.

  2. Apply a sequence of elementary row operations on the LHS and RHS.

    • Your goal is to turn the 'A' on the LHS into the identity matrix 'I'.
    • Every row operation you apply to the LHS, you must apply the exact same operation to the 'I' on the RHS.
    • The 'A' on the far right remains untouched throughout the process.

  3. Continue until the LHS becomes 'I'.

    • As you apply the row operations, the LHS matrix (which started as A) morphs into I. In the same way, the RHS matrix (which started as I) morphs into a new matrix, let's call it B.

  4. The equation will be in the form I=BAI = BA.

    • By the definition of an inverse, if I=BAI=BA, then the matrix B must be the inverse of A. The 'I' matrix has become A−1A^{-1}.
  • Note on Column Operations: You can do the same process with column operations, but you must start with the equation A=AIA = AI and only use column operations. You cannot mix row and column operations in the same problem.

Example 1: Applying Row Operations

Question: Apply the operation R2→R2−2R1R_2 \to R_2 - 2R_1 to the matrix A=[1221]A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}.

Explanation:

  1. Identify the Rows: R1=[12]R_1 = \begin{bmatrix} 1 & 2 \end{bmatrix} and R2=[21]R_2 = \begin{bmatrix} 2 & 1 \end{bmatrix}.
  2. State the Operation: The operation R2→R2−2R1R_2 \to R_2 - 2R_1 means "The new Row 2 will be the old Row 2 minus two times Row 1." Row 1 itself does not change.
  3. Calculate the New R2R_2: New R2=[21]−2×[12]R_2 = \begin{bmatrix} 2 & 1 \end{bmatrix} - 2 \times \begin{bmatrix} 1 & 2 \end{bmatrix} =[21]−[24]= \begin{bmatrix} 2 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 4 \end{bmatrix} =[2−21−4]=[0−3]= \begin{bmatrix} 2-2 & 1-4 \end{bmatrix} = \begin{bmatrix} 0 & -3 \end{bmatrix}
  4. Construct the New Matrix: The first row is unchanged, and we use the new second row.

Answer: The new matrix is [120−3]\begin{bmatrix} 1 & 2 \\ 0 & -3 \end{bmatrix}.

Example 2: Finding Inverse of a 2x2 Matrix

Question: Find the inverse of the matrix A=[122−1]A = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} using elementary row operations.

Explanation:

  1. Set up the Equation: Start with A=IAA = IA. [122−1]=[1001]A\begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A

  2. Goal: Turn the LHS matrix into I=[1001]I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.

  3. Step 1: Get a 0 in position (2,1). Apply the operation R2→R2−2R1R_2 \to R_2 - 2R_1. We do this to both sides.

    • LHS: R2=[2−1]−2[12]=[0−5]R_2 = \begin{bmatrix} 2 & -1 \end{bmatrix} - 2\begin{bmatrix} 1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & -5 \end{bmatrix}.
    • RHS: R2=[01]−2[10]=[−21]R_2 = \begin{bmatrix} 0 & 1 \end{bmatrix} - 2\begin{bmatrix} 1 & 0 \end{bmatrix} = \begin{bmatrix} -2 & 1 \end{bmatrix}.

    The equation becomes: [120−5]=[10−21]A\begin{bmatrix} 1 & 2 \\ 0 & -5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}A.

  4. Step 2: Get a 1 in position (2,2). Apply R2→−15R2R_2 \to -\frac{1}{5}R_2.

    • LHS: R2=−15[0−5]=[01]R_2 = -\frac{1}{5}\begin{bmatrix} 0 & -5 \end{bmatrix} = \begin{bmatrix} 0 & 1 \end{bmatrix}.
    • RHS: R2=−15[−21]=[2/5−1/5]R_2 = -\frac{1}{5}\begin{bmatrix} -2 & 1 \end{bmatrix} = \begin{bmatrix} 2/5 & -1/5 \end{bmatrix}.

    The equation becomes: [1201]=[102/5−1/5]A\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2/5 & -1/5 \end{bmatrix}A.

  5. Step 3: Get a 0 in position (1,2). Apply R1→R1−2R2R_1 \to R_1 - 2R_2.

    • LHS: R1=[12]−2[01]=[10]R_1 = \begin{bmatrix} 1 & 2 \end{bmatrix} - 2\begin{bmatrix} 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \end{bmatrix}.
    • RHS: R1=[10]−2[2/5−1/5]=[1−4/50−(−2/5)]=[1/52/5]R_1 = \begin{bmatrix} 1 & 0 \end{bmatrix} - 2\begin{bmatrix} 2/5 & -1/5 \end{bmatrix} = \begin{bmatrix} 1-4/5 & 0-(-2/5) \end{bmatrix} = \begin{bmatrix} 1/5 & 2/5 \end{bmatrix}.

  6. Final Result: The equation is now in the form I=BAI = BA. [1001]=[1/52/52/5−1/5]A\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1/5 & 2/5 \\ 2/5 & -1/5 \end{bmatrix}A.

Answer: A−1=[1/52/52/5−1/5]A^{-1} = \begin{bmatrix} 1/5 & 2/5 \\ 2/5 & -1/5 \end{bmatrix}.

Example 3: Inverse of a 3x3 Matrix

Question: Find the inverse of A=[012123311]A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}.

Explanation:

  1. Set up the Equation: Start with A=IAA = IA. [012123311]=[100010001]A\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}A.

  2. Step 1: Get a 1 in position (1,1). Swap Row 1 and Row 2 (R1↔R2R_1 \leftrightarrow R_2). [123012311]=[010100001]A\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}A.

  3. Step 2: Get a 0 in position (3,1). Apply R3→R3−3R1R_3 \to R_3 - 3R_1.

    • LHS: R3=[311]−3[123]=[0−5−8]R_3 = \begin{bmatrix} 3 & 1 & 1 \end{bmatrix} - 3\begin{bmatrix} 1 & 2 & 3 \end{bmatrix} = \begin{bmatrix} 0 & -5 & -8 \end{bmatrix}.
    • RHS: R3=[001]−3[010]=[0−31]R_3 = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix} - 3\begin{bmatrix} 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -3 & 1 \end{bmatrix}.

    The equation becomes: [1230120−5−8]=[0101000−31]A\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{bmatrix}A.

  4. Continue the Process: The goal is to create an identity matrix on the LHS. This involves more steps:

    • R3→R3+5R2R_3 \to R_3 + 5R_2 (to get a 0 in position 3,2)
    • R3→(1/2)R3R_3 \to (1/2)R_3 (to get a 1 in position 3,3)
    • R1→R1−2R2R_1 \to R_1 - 2R_2 (to get a 0 in position 1,2)
    • R1→R1−(...)R3R_1 \to R_1 - (...)R_3 and R2→R2−(...)R3R_2 \to R_2 - (...)R_3 (to get zeros in positions 1,3 and 2,3)

  5. Final Result: After all operations are completed, the equation becomes: I=[1/2−1/21/2−43−15/2−3/21/2]AI = \begin{bmatrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{bmatrix}A.

Answer: A−1=[1/2−1/21/2−43−15/2−3/21/2]A^{-1} = \begin{bmatrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{bmatrix}.

Example 4: A Matrix whose Inverse does not Exist

Question: Show that the matrix A=[2−6−13]A = \begin{bmatrix} 2 & -6 \\ -1 & 3 \end{bmatrix} is not invertible.

Explanation: We will attempt to find the inverse using row operations and show that the process fails.

  1. Set up the Equation: Start with A=IAA = IA. [2−6−13]=[1001]A\begin{bmatrix} 2 & -6 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A.

  2. Step 1: Get a 1 in position (1,1). Apply R1→12R1R_1 \to \frac{1}{2}R_1. [1−3−13]=[1/2001]A\begin{bmatrix} 1 & -3 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 1/2 & 0 \\ 0 & 1 \end{bmatrix}A.

  3. Step 2: Get a 0 in position (2,1). Apply R2→R2+R1R_2 \to R_2+R_1.

    • LHS: R2=[−13]+[1−3]=[00]R_2 = \begin{bmatrix} -1 & 3 \end{bmatrix} + \begin{bmatrix} 1 & -3 \end{bmatrix} = \begin{bmatrix} 0 & 0 \end{bmatrix}.
    • RHS: R2=[01]+[1/20]=[1/21]R_2 = \begin{bmatrix} 0 & 1 \end{bmatrix} + \begin{bmatrix} 1/2 & 0 \end{bmatrix} = \begin{bmatrix} 1/2 & 1 \end{bmatrix}.

    The equation becomes: [1−300]=[1/201/21]A\begin{bmatrix} 1 & -3 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1/2 & 0 \\ 1/2 & 1 \end{bmatrix}A.

  4. Conclusion: We have obtained a row of all zeros on the left-hand side. It is now impossible to continue the process to create the identity matrix. This failure indicates that the matrix A does not have an inverse. It is a singular (non-invertible) matrix.

Answer: The inverse does not exist.

Example 5: Using the Property (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}

Question: If A−1=[3−1−42]A^{-1} = \begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix} and B=[12−23]B = \begin{bmatrix} 1 & 2 \\ -2 & 3 \end{bmatrix}, find (AB)−1(AB)^{-1}.

Explanation:

  1. Recall the Reversal Law: The inverse of a product is the product of the inverses in the reverse order: (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}
  2. Find B−1B^{-1}: We need to find the inverse of BB. For a 2x2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}, the inverse is 1ad−bc[d−b−ca]\frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
    • Determinant of B: ad−bc=(1)(3)−(2)(−2)=3+4=7ad-bc = (1)(3) - (2)(-2) = 3+4=7
    • B−1=17[3−221]B^{-1} = \frac{1}{7}\begin{bmatrix} 3 & -2 \\ 2 & 1 \end{bmatrix}
  3. Multiply the Inverses: Now we compute B−1A−1B^{-1}A^{-1}. (AB)−1=B−1A−1=17[3−221][3−1−42](AB)^{-1} = B^{-1}A^{-1} = \frac{1}{7}\begin{bmatrix} 3 & -2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}
    • (Row 1, Col 1): (3)(3)+(−2)(−4)=9+8=17(3)(3) + (-2)(-4) = 9+8=17
    • (Row 1, Col 2): (3)(−1)+(−2)(2)=−3−4=−7(3)(-1) + (-2)(2) = -3-4=-7
    • (Row 2, Col 1): (2)(3)+(1)(−4)=6−4=2(2)(3) + (1)(-4) = 6-4=2
    • (Row 2, Col 2): (2)(−1)+(1)(2)=−2+2=0(2)(-1) + (1)(2) = -2+2=0 (AB)−1=17[17−720]=[17/7−12/70](AB)^{-1} = \frac{1}{7}\begin{bmatrix} 17 & -7 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 17/7 & -1 \\ 2/7 & 0 \end{bmatrix}

Answer: (AB)−1=[17/7−12/70](AB)^{-1} = \begin{bmatrix} 17/7 & -1 \\ 2/7 & 0 \end{bmatrix}.