Algebra of matrices

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Algebra of Matrices – Linear Algebra

Algebra of Matrices

😊 Hello, students! Now that we understand what a matrix is and its different types, it’s time to learn how to operate on them. Think of matrices as numbers with structure—just like you can add or multiply numbers, you can do similar (but more interesting!) things with matrices. But beware: matrix algebra has its own rules! Let’s explore them carefully.

1. Matrix Addition

Two matrices can be added only if they have the same order (same number of rows and columns). If \( A = (a_{ij}) \) and \( B = (b_{ij}) \) are both \( m \times n \) matrices, their sum \( A + B \) is the \( m \times n \) matrix defined by: \[ A + B = (a_{ij} + b_{ij}) \]

Let

\[ A = \begin{bmatrix} 1 & -2 \\ 3 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 4 & 1 \\ -1 & 5 \end{bmatrix} \]

Then

\[ A + B = \begin{bmatrix} 1+4 & -2+1 \\ 3+(-1) & 0+5 \end{bmatrix} = \begin{bmatrix} 5 & -1 \\ 2 & 5 \end{bmatrix} \]

Properties of Matrix Addition:

  • Commutative: \( A + B = B + A \)
  • Associative: \( (A + B) + C = A + (B + C) \)
  • Additive Identity: There exists a zero matrix \( O \) such that \( A + O = A \)
  • Additive Inverse: For every \( A \), there exists \( -A \) such that \( A + (-A) = O \)

2. Scalar Multiplication

If \( k \in \mathbb{R} \) and \( A = (a_{ij}) \) is an \( m \times n \) matrix, then the scalar multiple \( kA \) is defined by: \[ kA = (k \cdot a_{ij}) \]

\[ A = \begin{bmatrix} 2 & -1 \\ 0 & 3 \end{bmatrix}, \quad -2A = \begin{bmatrix} -4 & 2 \\ 0 & -6 \end{bmatrix} \]

Properties:

  • \( k(A + B) = kA + kB \)
  • \( (k + \ell)A = kA + \ell A \)
  • \( (k\ell)A = k(\ell A) \)
  • \( 1 \cdot A = A \)

3. Matrix Multiplication

⚠️ Crucial: Matrix multiplication is NOT component-wise! It’s defined via dot products of rows and columns.

Let \( A \) be an \( m \times n \) matrix and \( B \) be an \( n \times p \) matrix. Then the product \( AB \) is an \( m \times p \) matrix whose \( (i,j) \)-entry is: \[ (AB)_{ij} = \sum_{k=1}^{n} a_{ik} b_{kj} \] That is: row \( i \) of \( A \) dotted with column \( j \) of \( B \).

Let

\[ A = \begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} 3 & 0 & 1 \\ -2 & 1 & 0 \end{bmatrix} \]

Then \( AB \) is \( 2 \times 3 \):

\[ AB = \begin{bmatrix} (1)(3) + (2)(-2) & (1)(0) + (2)(1) & (1)(1) + (2)(0) \\ (0)(3) + (-1)(-2) & (0)(0) + (-1)(1) & (0)(1) + (-1)(0) \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 2 & -1 & 0 \end{bmatrix} \]
🤔 Quick check: Can you compute \( BA \)? No! \( B \) is \( 2 \times 3 \), \( A \) is \( 2 \times 2 \) → inner dimensions (3 vs 2) don’t match!

Key Properties of Matrix Multiplication:

  • Associative: \( (AB)C = A(BC) \)
  • Distributive: \( A(B + C) = AB + AC \), \( (A + B)C = AC + BC \)
  • Identity: If \( A \) is \( n \times n \), then \( AI_n = I_nA = A \)
  • NOT Commutative: In general, \( AB \ne BA \) (even if both exist!)

🙄 Common mistakes:

  • Assuming \( AB = 0 \Rightarrow A = 0 \) or \( B = 0 \) → False!
  • Assuming \( AB = AC \Rightarrow B = C \) → False!
\[ \text{Example: } A = \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix},\ B = \begin{bmatrix} 1 \\ 1 \end{bmatrix},\ AB = \begin{bmatrix} 0 \\ 0 \end{bmatrix}, \text{ but } A \ne 0,\ B \ne 0. \]

4. Transpose of a Matrix

The transpose of a matrix \( A = (a_{ij})_{m \times n} \), denoted \( A^T \), is the \( n \times m \) matrix obtained by interchanging rows and columns: \[ (A^T)_{ij} = a_{ji} \]

\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \quad \Rightarrow \quad A^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix} \]

Properties of Transpose:

  • \( (A^T)^T = A \)
  • \( (A + B)^T = A^T + B^T \)
  • \( (kA)^T = kA^T \)
  • \( (AB)^T = B^T A^T \) (note the reversal!)
😠 Remember: \( (AB)^T = B^T A^T \), **not** \( A^T B^T \)! This reversal reflects how multiplication order changes under duality.

5. Matrix Algebra and Linear Systems

Matrix multiplication lets us write systems of linear equations compactly.

Example:

\[ \begin{cases} 2x + 3y = 5 \\ x – y = 1 \end{cases} \quad \Longleftrightarrow \quad \begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 1 \end{bmatrix} \]

This is \( A\mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix.

Summary Table

Operation Condition Result Size
\( A + B \) Same order Same as \( A \)
\( kA \) Always defined Same as \( A \)
\( AB \) Columns of \( A \) = Rows of \( B \) Rows of \( A \) × Columns of \( B \)
\( A^T \) Always defined Columns of \( A \) × Rows of \( A \)

Self-Check: Test Your Understanding

  1. Let \( A = \begin{bmatrix} 1 & 2 \end{bmatrix} \), \( B = \begin{bmatrix} 3 \\ 4 \end{bmatrix} \). Compute \( AB \) and \( BA \).
  2. If \( A \) is \( 3 \times 4 \) and \( B \) is \( 4 \times 2 \), what is the size of \( AB \)?
  3. True or False: \( (A + B)^2 = A^2 + 2AB + B^2 \) for matrices.

Answers: (1) \( AB = [11] \), \( BA = \begin{bmatrix} 3 & 6 \\ 4 & 8 \end{bmatrix} \); (2) \( 3 \times 2 \); (3) False (unless \( AB = BA \))

Algebra of Matrices – Questions & Full Solutions

Algebra of Matrices – Questions & Full Solutions

😊 Hello students! Below are all matrix algebra questions from your documents, plus key supplementary exercises, each with a complete, detailed solution. Use this to master matrix addition, scalar multiplication, matrix multiplication, transpose, and more!

Q1. Let \[ A = \begin{bmatrix} 1 & 2 & 4 \\ 2 & 3 & 1 \\ 5 & 0 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} -2 & 1 & 3 \\ 2 & 4 & 2 \\ 3 & 6 & 1 \end{bmatrix} \] Compute \( A + B \) and \( 2A – 3B \).

Solution:

Part 1: \( A + B \)

\[ A + B = \begin{bmatrix} 1 + (-2) & 2 + 1 & 4 + 3 \\ 2 + 2 & 3 + 4 & 1 + 2 \\ 5 + 3 & 0 + 6 & 3 + 1 \end{bmatrix} = \begin{bmatrix} -1 & 3 & 7 \\ 4 & 7 & 3 \\ 8 & 6 & 4 \end{bmatrix} \]

Part 2: \( 2A – 3B \)

First compute \( 2A \) and \( 3B \):

\[ 2A = \begin{bmatrix} 2 & 4 & 8 \\ 4 & 6 & 2 \\ 10 & 0 & 6 \end{bmatrix}, \quad 3B = \begin{bmatrix} -6 & 3 & 9 \\ 6 & 12 & 6 \\ 9 & 18 & 3 \end{bmatrix} \]

Now subtract:

\[ 2A – 3B = \begin{bmatrix} 2 – (-6) & 4 – 3 & 8 – 9 \\ 4 – 6 & 6 – 12 & 2 – 6 \\ 10 – 9 & 0 – 18 & 6 – 3 \end{bmatrix} = \begin{bmatrix} 8 & 1 & -1 \\ -2 & -6 & -4 \\ 1 & -18 & 3 \end{bmatrix} \]

Q2. Let \[ A = \begin{bmatrix} 2 & 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 5 \\ 6 \\ 7 \end{bmatrix}, \quad C = \begin{bmatrix} 5 & 3 \\ 6 & 4 \\ 7 & 5 \end{bmatrix} \] Compute: (a) \( AB \), (b) \( AC \).

Solution:

(a) \( AB \): \( 1 \times 3 \) times \( 3 \times 1 \) ⇒ result is \( 1 \times 1 \)

\[ AB = [2\cdot5 + 3\cdot6 + 4\cdot7] = [10 + 18 + 28] = [56] \]

(b) \( AC \): \( 1 \times 3 \) times \( 3 \times 2 \) ⇒ result is \( 1 \times 2 \)

\[ AC = \begin{bmatrix} 2\cdot5 + 3\cdot6 + 4\cdot7 & 2\cdot3 + 3\cdot4 + 4\cdot5 \end{bmatrix} = \begin{bmatrix} 10 + 18 + 28 & 6 + 12 + 20 \end{bmatrix} = \begin{bmatrix} 56 & 38 \end{bmatrix} \]

Q3. Let \[ A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} a & 3 \\ 5 & b \end{bmatrix} \] Find \( a \) and \( b \) such that \( AB = BA \).

Solution:

Compute \( AB \):

\[ AB = \begin{bmatrix} 1\cdot a + 4\cdot5 & 1\cdot3 + 4\cdot b \\ 2\cdot a + 3\cdot5 & 2\cdot3 + 3\cdot b \end{bmatrix} = \begin{bmatrix} a + 20 & 3 + 4b \\ 2a + 15 & 6 + 3b \end{bmatrix} \]

Compute \( BA \):

\[ BA = \begin{bmatrix} a\cdot1 + 3\cdot2 & a\cdot4 + 3\cdot3 \\ 5\cdot1 + b\cdot2 & 5\cdot4 + b\cdot3 \end{bmatrix} = \begin{bmatrix} a + 6 & 4a + 9 \\ 5 + 2b & 20 + 3b \end{bmatrix} \]

Set \( AB = BA \) ⇒ equate entries:

  1. Top-left: \( a + 20 = a + 6 \) ⇒ \( 20 = 6 \) → **Contradiction!**

✅ **Conclusion:** No such \( a, b \) exist. Matrix multiplication is not commutative here.

Q4. Verify that \( (AB)^T = B^T A^T \) for \[ A = \begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} 3 & 0 & 1 \\ -2 & 1 & 0 \end{bmatrix} \]

Solution:

First, compute \( AB \) (from earlier example):

\[ AB = \begin{bmatrix} -1 & 2 & 1 \\ 2 & -1 & 0 \end{bmatrix} \Rightarrow (AB)^T = \begin{bmatrix} -1 & 2 \\ 2 & -1 \\ 1 & 0 \end{bmatrix} \]

Now compute \( B^T \) and \( A^T \):

\[ B^T = \begin{bmatrix} 3 & -2 \\ 0 & 1 \\ 1 & 0 \end{bmatrix}, \quad A^T = \begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix} \]

Then \( B^T A^T \) (3×2 × 2×2 = 3×2):

\[ B^T A^T = \begin{bmatrix} 3\cdot1 + (-2)\cdot2 & 3\cdot0 + (-2)\cdot(-1) \\ 0\cdot1 + 1\cdot2 & 0\cdot0 + 1\cdot(-1) \\ 1\cdot1 + 0\cdot2 & 1\cdot0 + 0\cdot(-1) \end{bmatrix} = \begin{bmatrix} -1 & 2 \\ 2 & -1 \\ 1 & 0 \end{bmatrix} = (AB)^T \]

✅ Verified!

Q5. Find non-zero matrices \( A \) and \( B \) such that \( AB = 0 \).

Solution:

Example:

\[ A = \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]

Then:

\[ AB = \begin{bmatrix} 1\cdot1 + (-1)\cdot1 \\ 1\cdot1 + (-1)\cdot1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]

Both \( A \ne 0 \), \( B \ne 0 \), but \( AB = 0 \). This shows matrices have **zero divisors**.

Q6. Express as \( AX = B \): \[ \begin{cases} 5x + 4y – 3z = 7 \\ 4x + 3y – 2z = 5 \\ 3x + 2y – z = 5 \end{cases} \]

Solution:

\[ A = \begin{bmatrix} 5 & 4 & -3 \\ 4 & 3 & -2 \\ 3 & 2 & -1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 7 \\ 5 \\ 5 \end{bmatrix} \]

So the system is \( AX = B \).

Q7. Let \[ A = \begin{bmatrix} 2 & 1 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} -1 & 2 \\ 0 & 1 \\ 1 & 0 \end{bmatrix}, \quad C = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ 2 & 1 \end{bmatrix} \] Verify \( A(B + C) = AB + AC \).

Solution:

Left side: \( B + C = \begin{bmatrix} -1 & 3 \\ 1 & 1 \\ 3 & 1 \end{bmatrix} \)

\[ A(B + C) = [2(-1) + 1(1) + 0(3),\ 2(3) + 1(1) + 0(1)] = [-1,\ 7] \]

Right side:

\[ AB = [2(-1) + 1(0) + 0(1),\ 2(2) + 1(1) + 0(0)] = [-2,\ 5] \\ AC = [2(0) + 1(1) + 0(2),\ 2(1) + 1(0) + 0(1)] = [1,\ 2] \\ AB + AC = [-2 + 1,\ 5 + 2] = [-1,\ 7] \]

✅ Both sides equal.

📌 Key Reminders:

  • Matrix addition: element-wise, same order only.
  • Scalar multiplication: distribute scalar to every entry.
  • Matrix multiplication: row × column; **not commutative**.
  • \( (AB)^T = B^T A^T \) — order reverses!
  • \( AB = 0 \) does **not** imply \( A = 0 \) or \( B = 0 \).

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