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Chapter 2: Problem 5

Let \(A\) be an \(n \times n\) matrix and \(\alpha\) a scalar. Show that \\[ \operatorname{det}(\alpha A)=\alpha^{n} \operatorname{det}(A) \\]

Short Answer

Expert verified
The proof involves expanding the determinant of the scalar multiplied matrix using the Laplace expansion, factoring out the scalar 伪, and generalizing for the nth power of the scalar. We obtain the expression \(\operatorname{det}(\alpha A) = \alpha^n \operatorname{det}(A)\), which verifies the given relation.

Step by step solution

01

Define Variables and Rewrite the Expression

Define the square matrix A with elements a_{ij}, where i and j refer to the row and column indices, respectively. The scalar value is represented by 伪. The expression we want to prove is \[ \operatorname{det}(\alpha A) = \alpha^n \operatorname{det}(A) \] First, let's rewrite the expression for the scalar multiplied matrix, 伪A: \[ (\alpha A)_{ij} = \alpha a_{ij} \]
02

Expand the Determinant using the Laplace Expansion

Now, let's expand the determinant using the Laplace expansion, which states that the determinant of a matrix A can be calculated as follows: \[ \operatorname{det}(A)=\sum_{j=1}^n a_{1j} C_{1j} \] where C_{1j} is the cofactor of element a_{1j}. By substituting 伪A in place of A in the Laplace expansion, we have: \[ \operatorname{det}(\alpha A)=\sum_{j=1}^n (\alpha A)_{1j} C_{1j} = \sum_{j=1}^n (\alpha a_{1j}) C_{1j} \]
03

Factor Out the Scalar

Since 伪 is a scalar, we can factor it out of the summation: \[ \operatorname{det}(\alpha A) = \alpha \sum_{j=1}^n a_{1j} C_{1j} \] Now, the expression inside the summation is the determinant of matrix A. Thus, we have: \[ \operatorname{det}(\alpha A) = \alpha \operatorname{det}(A) \]
04

Generalize for the nth Power of the Scalar

In order to prove the given expression, we need to show that the scalar 伪 is raised to the nth power, where n is the order of the matrix. As we expand the Laplace expansion for matrix A with increasing order, we will have n independent instances of the scalar 伪 present. Therefore, after factoring out the scalar 伪 for each Laplace expansion, we will have: \[ \operatorname{det}(\alpha A) = (\alpha)^n \operatorname{det}(A) \] Hence, the proof is complete, and we have shown the given expression to be true.

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Most popular questions from this chapter

Let \\[ A=\left(\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array}\right) \\] (a) Compute the determinant of \(A\). Is \(A\) nonsingular? (b) Compute adj \(A\) and the product \(A\) adj \(A\)

Prove that if a row or a column of an \(n \times n\) matrix \(A\) consists entirely of zeros, then \(\operatorname{det}(A)=0\)

Let \(A\) and \(B\) be \(n \times n\) matrices. Prove that if \(A B=I,\) then \(B A=I .\) What is the significance of this result in terms of the definition of a nonsingular matrix?

Let \(E_{1}, E_{2},\) and \(E_{3}\) be \(3 \times 3\) elementary matrices of types I, II, and III, respectively, and let \(A\) be a \(3 \times 3\) matrix with \(\operatorname{det}(A)=6 .\) Assume, additionally, that \(E_{2}\) was formed from \(I\) by multiplying its second row by 3. Find the values of each of the following: (a) \(\operatorname{det}\left(E_{1} A\right)\) (b) \(\operatorname{det}\left(E_{2} A\right)\) (c) \(\operatorname{det}\left(E_{3} A\right)\) (d) \(\operatorname{det}\left(A E_{1}\right)\) (e) \(\operatorname{det}\left(E_{1}^{2}\right)\) (f) \(\operatorname{det}\left(E_{1} E_{2} E_{3}\right)\)

For each of the following, compute (i) \(\operatorname{det}(A)\) (ii) adj \(A,\) and (iii) \(A^{-1}:\) (a) \(A=\left(\begin{array}{rr}1 & 2 \\ 3 & -1\end{array}\right)\) (b) \(A=\left(\begin{array}{ll}3 & 1 \\ 2 & 4\end{array}\right)\) (c) \(A=\left(\begin{array}{rrr}1 & 3 & 1 \\ 2 & 1 & 1 \\ -2 & 2 & -1\end{array}\right)\) (d) \(A=\left(\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right)\)
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