Scalar Multiple Property in Determinants

The scalar multiple property is an important concept in linear algebra, particularly when dealing with determinants of matrices. This property states that if each element of a row (or column) of a square matrix is multiplied by a scalar, the determinant of the new matrix is equal to the scalar multiplied by the determinant of the original matrix.

Understanding the Scalar Multiple Property

Let's denote a square matrix as $A$ and its determinant as $det(A)$ or $|A|$. If we multiply a row or column of $A$ by a scalar $k$, the new matrix will be denoted as $A'$, and its determinant as $det(A')$ or $|A'|$.

The scalar multiple property can be expressed as:

$$ det(A') = k \cdot det(A) $$

This means that the determinant of the new matrix $A'$ is $k$ times the determinant of the original matrix $A$.

Table of Differences and Important Points

Property	Description	Example
Original Matrix	The matrix before any scalar multiplication is applied to its rows or columns.	$A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$
Scalar Multiplication	A scalar $k$ is multiplied to all elements of a particular row or column.	$A' = \begin{bmatrix} ka & kb \ c & d \end{bmatrix}$
Determinant of Original Matrix	The determinant is calculated using the elements of the original matrix.	$det(A) = ad - bc$
Determinant after Scalar Multiplication	The determinant is calculated using the elements of the matrix after scalar multiplication.	$det(A') = (ka)d - bc = k(ad) - bc$
Scalar Multiple Property	The determinant of the new matrix is the scalar multiplied by the determinant of the original matrix.	$det(A') = k \cdot det(A)$

Formulas

For a 2x2 matrix:

$$ A = \begin{bmatrix} a & b \ c & d \end{bmatrix} $$

The determinant is:

$$ det(A) = ad - bc $$

After multiplying the first row by scalar $k$:

$$ A' = \begin{bmatrix} ka & kb \ c & d \end{bmatrix} $$

The determinant of the new matrix is:

$$ det(A') = (ka)d - (kb)c = k(ad - bc) = k \cdot det(A) $$

Examples

Example 1: 2x2 Matrix

Consider the matrix:

$$ A = \begin{bmatrix} 2 & 3 \ 4 & 5 \end{bmatrix} $$

The determinant is:

$$ det(A) = (2)(5) - (3)(4) = 10 - 12 = -2 $$

Now, let's multiply the first row by scalar $k = 3$:

$$ A' = \begin{bmatrix} 6 & 9 \ 4 & 5 \end{bmatrix} $$

The determinant of the new matrix is:

$$ det(A') = (6)(5) - (9)(4) = 30 - 36 = -6 $$

Using the scalar multiple property:

$$ det(A') = 3 \cdot det(A) = 3 \cdot (-2) = -6 $$

Example 2: 3x3 Matrix

Consider the matrix:

$$ B = \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{bmatrix} $$

The determinant is more complex for a 3x3 matrix, but for the sake of this example, let's assume it's calculated and equals $det(B) = 0$.

Now, let's multiply the second row by scalar $k = 2$:

$$ B' = \begin{bmatrix} 1 & 2 & 3 \ 8 & 10 & 12 \ 7 & 8 & 9 \end{bmatrix} $$

The determinant of the new matrix is:

$$ det(B') = 2 \cdot det(B) = 2 \cdot 0 = 0 $$

As expected, the determinant of the new matrix $B'$ is twice the determinant of the original matrix $B$, which in this case remains $0$.

In conclusion, the scalar multiple property is a fundamental concept in linear algebra that simplifies the computation of determinants after scalar multiplication of rows or columns. It is essential for students to understand and apply this property, especially in the context of solving systems of linear equations and finding eigenvalues of matrices.