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Chapter 3: Problem 58

Let \(A\) be an \(n \times n\) nonzero matrix satisfying \(A^{10}=O .\) Explain why \(A\) must be singular. What properties of determinants are you using in your argument?

Short Answer

Expert verified
The matrix \(A\) is singular because its determinant equals to zero. This conclusion is reached by using the properties of determinants, namely the fact that a matrix is singular if and only if its determinant is zero, and that the determinant of the product of two matrices equals the product of their determinants.

Step by step solution

01

Understanding Concepts

To start with, it's crucial to know that a matrix is singular if and only if its determinant is zero. This means that the matrix doesn't have an inverse. In addition, the determinant of a product of two matrices is the product of their determinants, written in mathematical symbols as \(\det(AB) = \det(A) \times \det(B)\). These properties are going to be central in figuring out our solution.
02

Compute the Determinant of Both Sides of the Equation

Given the equation \(A^{10}=O\), where \(O\) is a zero matrix and \(A^{10}\) means the matrix A multiplied by itself 10 times, calculate the determinant of both sides of the equation. For a zero matrix, the determinant is zero, \(\det(O) = 0\). For \(A^{10}\), based on the property of determinants, \(\det(A^{10})\) can be written as \((\det(A))^{10}\). Thus, our equation becomes \((\det(A))^{10} = 0\)
03

Conclude That A is Singular

In the equation from step 2, \((\det(A))^{10} = 0\), the only way for this equation to hold true is if \(\det(A)\) is zero. Hence, it's fair to say that the determinant of matrix A is zero. And from the proposition mentioned in step 1, if the determinant of a matrix is zero then the matrix is singular. Therefore, \(A\) must be singular.

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Most popular questions from this chapter

Illustrate the formula provided in Exercise 11 for the matrix \(A=\left[\begin{array}{rr}1 & 0 \\ 1 & -2\end{array}\right]\)

Cramer's Rule has been used to solve for one of the variables in a system of equations. Determine whether Cramer's Rule was used correctly to solve for the variable. If not, identify the mistake. System of Equations \\[ \begin{array}{rr} x-4 y-z= & -1 \\ 2 x-3 y+z= & 6 \\ x+y-4 z= & 1 \end{array} \\] Solve for \(z=\frac{\left|\begin{array}{rrr}-1 & -4 & -1 \\ 6 & -3 & 1 \\ 1 & 1 & -4\end{array}\right|}{\left|\begin{array}{rrr}1 & -4 & -1 \\ 2 & -3 & 1 \\\ 1 & 1 & -4\end{array}\right|}\)

Find (a) the characteristic equation, (b) the eigenvalues, and (c) the corresponding eigenvectors of the matrix. $$\left[\begin{array}{rrr} 1 & 0 & 1 \\ 0 & -1 & 0 \\ 2 & 1 & -1 \end{array}\right]$$

Use a graphing utility or computer software program with matrix capabilities to find the eigenvalues of the matrix. Then find the corresponding eigenvectors. $$\left[\begin{array}{rrr} 1 & 1 & 0 \\ 0 & -2 & 1 \\ 0 & -2 & 2 \end{array}\right]$$

Determine whether the points are collinear. $$(-2,5),(0,-1),(3,-9)$$
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