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Chapter 2: Problem 11

Show that if \(A\) is singular, then adj \(A\) is also singular.

Short Answer

Expert verified
Given that A is singular, we have \(\det(A) = 0\). The relationship between the determinant of a matrix and its adjugate is given by \(A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A)\). Since \(\det(A) = 0\), this leads to \(A^{-1} = \frac{1}{0} \cdot \text{adj}(A)\), which is undefined. Assuming adj A is not singular leads to a contradiction, so we conclude that adj A is singular as well.

Step by step solution

01

Find the determinant of matrix A

Given that matrix A is singular, it means that the determinant of A is equal to zero. Let's denote this as: \(\det(A) = 0\)
02

Recall the relationship between the determinant of a matrix and its adjugate

The adjugate of a matrix A can be used to find its inverse, if it exists, using the formula: \(A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A)\)
03

Prove that adj A is singular

Knowing that \(\det(A) = 0\) and using the formula for the inverse of a matrix, we can see that: \(A^{-1} = \frac{1}{0} \cdot \text{adj}(A)\) Since division by zero is undefined, this means that the inverse of matrix A does not exist. Now, let's assume that the adjugate of A is not singular, meaning that it has an inverse, denoted by (adj A)^{-1}. We can then multiply both sides of the equation by (adj A)^{-1}, which gives: \((\text{adj}(A))^{-1} A^{-1} = (\text{adj}(A))^{-1} \cdot \frac{1}{0} \cdot \text{adj}(A)\) But we know that this is not possible, as A is singular and does not have an inverse, and therefore the equation cannot hold. Since our assumption that adj A is not singular leads to a contradiction, we can conclude that adj A must be singular as well.

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Most popular questions from this chapter

Let \(A\) be a nonsingular matrix. Show that \\[\operatorname{det}\left(A^{-1}\right)=\frac{1}{\operatorname{det}(A)}\\]

Show that the elimination method of computing the value of the determinant of an \(n \times n\) matrix involves \([n(n-1)(2 n-1)] / 6\) additions and \(\left[(n-1)\left(n^{2}+n+3\right)\right] / 3\) multiplications and divisions. [Hint: At the ith step of the reduction process, it takes \(n-i\) divisions to calculate the multiples of the ith row that are to be subtracted from the remaining rows below the pivot. We must then calculate new values for the \((n-i)^{2}\) entries in rows \(i+1 \text { through } n \text { and columns } i+1 \text { through } n .]\)

For each of the following, compute the determinant and state whether the matrix is singular or nonsingular: (a) \(\left(\begin{array}{ll}3 & 1 \\ 6 & 2\end{array}\right)\) (b) \(\left(\begin{array}{ll}3 & 1 \\ 4 & 2\end{array}\right)\) (c) \(\left(\begin{array}{lll}3 & 3 & 1 \\ 0 & 1 & 2 \\ 0 & 2 & 3\end{array}\right)\) (d) \(\left(\begin{array}{lll}2 & 1 & 1 \\ 4 & 3 & 5 \\ 2 & 1 & 2\end{array}\right)\) (e) \(\left(\begin{array}{rrr}2 & -1 & 3 \\ -1 & 2 & -2 \\ 1 & 4 & 0\end{array}\right)\) (f) \(\left(\begin{array}{rrrr}1 & 1 & 1 & 1 \\ 2 & -1 & 3 & 2 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 7 & 3\end{array}\right)\)

Let \(A\) be a \(4 \times 4\) matrix. If \(\operatorname{adj} A=\left(\begin{array}{rrrr}2 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 \\\ 0 & 4 & 3 & 2 \\ 0 & -2 & -1 & 2\end{array}\right)\) (a) calculate the value of det(adj \(A\) ). What should the value of \(\operatorname{det}(A)\) be? \([\) Hint: Use the result from Exercise \(8 .]\) (b) find \(A\)

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 \(A\)
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