Multiplicación de Matrices: Guía Completa, Método Paso a Paso y Regla del Determinante

Direct Answer / AI Overview: Matrix multiplication is a binary operation that takes a pair of matrices and produces another matrix by calculating the dot product of rows from the first matrix and columns from the second. For multiplication $C = AB$ to be valid, the number of columns in $A$ ($m \times k$) must equal the number of rows in $B$ ($k \times n$), yielding a product matrix of dimension $m \times n$. For square matrices, one of the most powerful theorems in linear algebra is the Determinant Product Rule: $\det(AB) = \det(A) \cdot \det(B)$.

When Can You Multiply Matrices? (Dimension Compatibility)

Before performing matrix multiplication, verify whether the two matrices are compatible.

Matrix multiplication is defined if and only if the inner dimensions match:

$$\text{Matrix } A_{(m \times k)} \times \text{Matrix } B_{(k \times n)} = \text{Matrix } C_{(m \times n)}$$

• The number of columns in $A$ must equal the number of rows in $B$.
• The resulting matrix $C$ takes its row count from $A$ and its column count from $B$.

    A (m × k)         B (k × n)         C (m × n)
┌              ┐   ┌              ┐   ┌              ┐
│              │   │              │   │              │
│   m rows     │ × │   k rows     │ = │   m rows     │
│              │   │              │   │              │
└              ┘   └              ┘   └              ┘
    k cols             n cols             n cols
        ▲                 ▲
        └───── MATCH ─────┘

[!WARNING] Matrix multiplication is not commutative: in general, $AB \neq BA$. Even when both products exist (e.g., when multiplying square $n \times n$ matrices), the resulting matrices are rarely identical.

The Step-by-Step Row $\times$ Column Method

To find the entry $c_{ij}$ in the $i$-th row and $j$-th column of product $C$, compute the scalar dot product of Row $i$ of matrix $A$ and Column $j$ of matrix $B$:

$$c_{ij} = \sum_{p=1}^k a_{ip} b_{pj} = a_{i1}b_{1j} + a_{i2}b_{2j} + \dots + a_{ik}b_{kj}$$

2x2 Worked Example

Let $A = \begin{bmatrix} 2 & 3 \ 1 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 1 \ 0 & 2 \end{bmatrix}$.

1. Calculate $c_{11}$ (Row 1 of $A$ $\cdot$ Column 1 of $B$):
   $c_{11} = (2 \cdot 5) + (3 \cdot 0) = 10 + 0 = 10$

2. Calculate $c_{12}$ (Row 1 of $A$ $\cdot$ Column 2 of $B$):
   $c_{12} = (2 \cdot 1) + (3 \cdot 2) = 2 + 6 = 8$

3. Calculate $c_{21}$ (Row 2 of $A$ $\cdot$ Column 1 of $B$):
   $c_{21} = (1 \cdot 5) + (4 \cdot 0) = 5 + 0 = 5$

4. Calculate $c_{22}$ (Row 2 of $A$ $\cdot$ Column 2 of $B$):
   $c_{22} = (1 \cdot 1) + (4 \cdot 2) = 1 + 8 = 9$

Resulting product:

$$AB = \begin{bmatrix} 10 & 8 \ 5 & 9 \end{bmatrix}$$

The Determinant Product Theorem: $\det(AB) = \det(A)\det(B)$

One of the cornerstone properties in linear algebra states that for any two $n \times n$ square matrices $A$ and $B$:

$$\det(AB) = \det(A) \cdot \det(B)$$

Mathematical Proof Verification using our Example:

Let us evaluate each determinant using our Determinant Calculator 2x2:

• $\det(A) = (2 \cdot 4) - (3 \cdot 1) = 8 - 3 = 5$
• $\det(B) = (5 \cdot 2) - (1 \cdot 0) = 10 - 0 = 10$
• Product of individual determinants: $\det(A) \cdot \det(B) = 5 \cdot 10 = 50$
• Direct determinant of $AB$:
  $\det(AB) = (10 \cdot 9) - (8 \cdot 5) = 90 - 40 = 50$

The equality $\det(AB) = \det(A)\det(B) = 50$ holds true.

Crucial Algebraic Properties of Matrix Multiplication

When manipulating algebraic expressions or matrices, keep these rules in mind:

For an in-depth breakdown of all 10 fundamental determinant rules, check our tutorial on Matrix Determinant Properties.

Why Isn’t Matrix Determinant Additive?

A very common mistake students make is writing $\det(A + B) = \det(A) + \det(B)$. This is false.

Consider: $$A = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix}$$

• $\det(A) = 0$
• $\det(B) = 0$
• $A + B = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I_2 \implies \det(A + B) = 1 \neq 0 + 0$

While determinants distribute over matrix multiplication ($\det(AB) = \det(A)\det(B)$), they do not distribute over matrix addition.

Frequently Asked Questions (FAQs)

What happens if you multiply a matrix by its inverse?

By definition, $A \cdot A^{-1} = I$. Applying the determinant product rule: $$\det(A \cdot A^{-1}) = \det(A) \cdot \det(A^{-1}) = \det(I) = 1$$ This yields the classic identity $\det(A^{-1}) = \frac{1}{\det(A)}$. This is also why a matrix is invertible if and only if $\det(A) \neq 0$, as explained in our Invertible Matrix Theorem guide.

Can you multiply two non-square matrices and take a determinant?

If $A$ is $m \times k$ and $B$ is $k \times m$ with $m \neq k$, the product $AB$ is an $m \times m$ square matrix, so its determinant $\det(AB)$ exists. Under the Cauchy-Binet Formula, if $m > k$, $\det(AB) = 0$. If $m \leq k$, $\det(AB)$ equals the sum of products of corresponding minors.