The Matrix Multiplication Solver

Example 1: Multiplying a 2×3 Matrix with a 3×2 Matrix

Multiply the matrices:
\[
A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}, \quad B = \begin{bmatrix} 7 & 8 \\ 9 & 10 \\ 11 & 12 \end{bmatrix}
\]

Solution

1. Check Dimensions:
\( A \) is \( 2 \times 3 \) and \( B \) is \( 3 \times 2 \). Since the number of columns in \( A \) equals the number of rows in \( B \), matrix multiplication is valid. The resulting matrix will be \( 2 \times 2 \).

2. Compute Each Element of \( C = AB \):
– \( C_{11} = (1 \times 7) + (2 \times 9) + (3 \times 11) = 7 + 18 + 33 = 58 \)
– \( C_{12} = (1 \times 8) + (2 \times 10) + (3 \times 12) = 8 + 20 + 36 = 64 \)
– \( C_{21} = (4 \times 7) + (5 \times 9) + (6 \times 11) = 28 + 45 + 66 = 139 \)
– \( C_{22} = (4 \times 8) + (5 \times 10) + (6 \times 12) = 32 + 50 + 72 = 154 \)

3. Resulting Matrix:
\[
C = \begin{bmatrix} 58 & 64 \\ 139 & 154 \end{bmatrix}
\]

Example 2: Multiplying a 2×2 Matrix with a 2×2 Matrix

Multiply the matrices:
\[
X = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix}, \quad Y = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}
\]

Solution

1. Check Dimensions:
Both \( X \) and \( Y \) are \( 2 \times 2 \). Matrix multiplication is valid, and the resulting matrix will also be \( 2 \times 2 \).

2. Compute Each Element of \( Z = XY \):
– \( Z_{11} = (2 \times 5) + (4 \times 7) = 10 + 28 = 38 \)
– \( Z_{12} = (2 \times 6) + (4 \times 8) = 12 + 32 = 44 \)
– \( Z_{21} = (1 \times 5) + (3 \times 7) = 5 + 21 = 26 \)
– \( Z_{22} = (1 \times 6) + (3 \times 8) = 6 + 24 = 30 \)

3. Resulting Matrix:
\[
Z = \begin{bmatrix} 38 & 44 \\ 26 & 30 \end{bmatrix}
\]

Example 3: Multiplying a 3×3 Matrix with a 3×3 Matrix

Multiply the matrices:
\[
P = \begin{bmatrix} 1 & 0 & 2 \\ 3 & 4 & 5 \\ 6 & 7 & 8 \end{bmatrix}, \quad Q = \begin{bmatrix} 9 & 8 & 7 \\ 6 & 5 & 4 \\ 3 & 2 & 1 \end{bmatrix}
\]

Solution

1. Check Dimensions:
Both \( P \) and \( Q \) are \( 3 \times 3 \). Matrix multiplication is valid, and the resulting matrix will also be \( 3 \times 3 \).

2. Compute Each Element of \( R = PQ \):
– Row 1:
– \( R_{11} = (1 \times 9) + (0 \times 6) + (2 \times 3) = 9 + 0 + 6 = 15 \)
– \( R_{12} = (1 \times 8) + (0 \times 5) + (2 \times 2) = 8 + 0 + 4 = 12 \)
– \( R_{13} = (1 \times 7) + (0 \times 4) + (2 \times 1) = 7 + 0 + 2 = 9 \)
– Row 2:
– \( R_{21} = (3 \times 9) + (4 \times 6) + (5 \times 3) = 27 + 24 + 15 = 66 \)
– \( R_{22} = (3 \times 8) + (4 \times 5) + (5 \times 2) = 24 + 20 + 10 = 54 \)
– \( R_{23} = (3 \times 7) + (4 \times 4) + (5 \times 1) = 21 + 16 + 5 = 42 \)
– Row 3:
– \( R_{31} = (6 \times 9) + (7 \times 6) + (8 \times 3) = 54 + 42 + 24 = 120 \)
– \( R_{32} = (6 \times 8) + (7 \times 5) + (8 \times 2) = 48 + 35 + 16 = 99 \)
– \( R_{33} = (6 \times 7) + (7 \times 4) + (8 \times 1) = 42 + 28 + 8 = 78 \)

3. Resulting Matrix:
\[
R = \begin{bmatrix} 15 & 12 & 9 \\ 66 & 54 & 42 \\ 120 & 99 & 78 \end{bmatrix}
\]