Question

a. Find \(\operatorname{det} A\) if \(A\) is \(3 \times 3\) and \(\operatorname{det}(2 A)=6\). b. Under what conditions is \(\operatorname{det}(-A)=\operatorname{det} A\) ?

Short answer

a. \( \operatorname{det} A = \frac{3}{4} \); b. \( \operatorname{det} A = 0 \).

Step-by-step solution

Understanding the Property of Determinant Scaling

When a matrix \( A \) is multiplied by a scalar \( c \), the determinant of the resulting matrix is scaled by \( c^{n} \), where \( n \) is the order of the matrix. For a \( 3 \times 3 \) matrix \( A \), this means: \( \operatorname{det}(c A) = c^3 \cdot \operatorname{det}(A) \).

Applying the Property

Given \( \operatorname{det}(2A) = 6 \), apply the property from Step 1: \( \operatorname{det}(2A) = 2^3 \cdot \operatorname{det}(A) = 8 \cdot \operatorname{det}(A) = 6 \).

Solving for \( \operatorname{det}(A) \)

Rearrange the equation from the previous step to find \( \operatorname{det}(A) \): \( 8 \cdot \operatorname{det}(A) = 6 \). Divide both sides by 8 to obtain \( \operatorname{det}(A) = \frac{6}{8} = \frac{3}{4} \).

Exploring Condition for \( \operatorname{det}(-A) = \operatorname{det}(A) \)

When \( A \) is an odd-order \( n \times n \) matrix, \( \operatorname{det}(-A) = (-1)^n \cdot \operatorname{det}(A) \). For a \( 3 \times 3 \) matrix, \( n = 3 \), so \( (-1)^3 = -1 \). Therefore, \( \operatorname{det}(-A) = -\operatorname{det}(A) \). For \( \operatorname{det}(-A) = \operatorname{det}(A) \), \( \operatorname{det}(A) \) must equal zero.

Key concepts

Scalar Multiplication and Determinants

When you multiply a matrix by a scalar, the effect on its determinant is quite systematic. Suppose you have a matrix, denoted as \( A \), and you choose to scale it by a scalar value \( c \). This results in a new matrix \( cA \). To find out how this transformation affects the determinant, you apply a key formula: \( \operatorname{det}(cA) = c^n \cdot \operatorname{det}(A) \). Here, \( n \) stands for the order of the matrix, which is basically the number of rows or columns since the matrix is square.

For example, consider a \( 3 \times 3 \) matrix (as in the given exercise), we get \( \operatorname{det}(cA) = c^3 \cdot \operatorname{det}(A) \). This means you raise the scalar to the third power before multiplying it with the determinant of \( A \).

Given the property, if we know that \( \operatorname{det}(2A) = 6 \) for \( 3 \times 3 \) matrix \( A \), we follow the formula by substituting the values: \( 2^3 \cdot \operatorname{det}(A) = 6 \). Solving this equation gives us \( \operatorname{det}(A) = \frac{3}{4} \).

Thus, understanding this property helps us calculate determinants after scalar multiplication without diving into complex computations.

Properties of Determinants

Determinants come with a delightful assortment of properties that offer valuable shortcuts for matrix algebra. One interesting property is how determinants behave under multiplication with negative scalars. If you multiply every element of matrix \( A \) by \(-1\), creating \(-A\), the new determinant is calculated as \( \operatorname{det}(-A) = (-1)^n \cdot \operatorname{det}(A) \).

Here, \( n \) denotes the order of the matrix. This property is crucial because it illustrates that the sign of the determinant depends on the order of the matrix. Specifically:

• For odd-order matrices, \((-1)^n\) multiplies the determinant of \( A \) by \(-1\), effectively flipping its sign.
• For even-order matrices, \((-1)^n = 1\), hence the determinant remains unchanged when multiplied by \(-1\).

Applying this rule helps in figuring out circumstances where \( \operatorname{det}(-A) = \operatorname{det}(A) \). For odd-order matrices like a \( 3 \times 3 \), this equal condition arises only when \( \operatorname{det}(A) = 0 \). Thus, for it to hold, \( A \) must be singular, meaning it doesn't have an inverse.

Odd-Order Matrices

Mathematicians classify matrices by their size into orders. Odd-order matrices have an interesting characteristic with respect to determinants. Specifically, when dealing with a matrix of odd order, its determinant undergoes a sign change when each entry is multiplied by \(-1\).

As discussed in the properties of determinants, when matrix \( A \) is a \( 3 \times 3 \) matrix (or any odd number \( n \)), \( \operatorname{det}(-A) = -\operatorname{det}(A) \). This scenario occurs because \((-1)^n = -1\) for odd \( n \).

A practical implication of this property is visible when you explore conditions like \( \operatorname{det}(-A) = \operatorname{det}(A) \). Since multiplying \( A \) by \(-1\) isn't affecting the determinant positively or negatively due to cancellation, the determinant must be zero.

• This means that the given matrix \( A \) is singular (i.e., not invertible).
• It connects to scenarios where the equations represent dependent or redundant systems.

Overall, understanding how odd-order matrices interact with these properties provides keys to unlocking more complex algebraic conclusions involving determinants.