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Chapter 3: Problem 72

If \(A\) is an idempotent matrix \(\left(A^{2}=A\right),\) then prove that the determinant of \(A\) is either 0 or 1

Short Answer

Expert verified
The determinant of an idempotent matrix \(A\) must be either 0 or 1.

Step by step solution

01

Use the property of determinants

Applying the property of determinants \(\text{det}(AB) = \text{det}(A)\times\text{det}(B)\) to the given idempotent matrix equation \(A^{2}=A\), we obtain \(\text{det}(A^{2})=\text{det}(A)\). Since \(A^{2}\) is simply \(A \times A\), this equation can be rewritten as \(\text{det}(A)\times\text{det}(A)=\text{det}(A)\).
02

Factorize the equation

Now, factorize the equation. This results in \(\text{det}(A)(\text{det}(A)-1)=0\). In order for this equation to be true, either \(\text{det}(A) = 0\) or \(\text{det}(A) - 1 = 0\), i.e. \(\text{det}(A) = 1\).
03

Conclude proof

Given that determinant of \(A\) must satisfy this equation and the only two possible solutions to this equation are 0 and 1, it can be concluded that the determinant of an idempotent matrix \(A\) is either 0 or 1 as required.

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Most popular questions from this chapter

Use Cramer's Rule to solve the system of linear equations, if possible. $$\begin{array}{l} 20 x_{1}+8 x_{2}=11 \\ 12 x_{1}-24 x_{2}=21 \end{array}$$

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}{l} 5 x-2 y+z=15 \\ 3 x-3 y-z=-7 \\ 2 x-y-7 z=-3 \end{array} \\] Solve for \(x\) \(x=\frac{\left|\begin{array}{rrr}15 & -2 & 1 \\ -7 & -3 & -1 \\\ -3 & -1 & -7\end{array}\right|}{\left|\begin{array}{rrr}5 & -2 & 1 \\ 3 & -3 & -1 \\ 2 & -1 & -7\end{array}\right|}\)

Use a graphing utility or a computer software program with matrix capabilities and Cramer's Rule to solve for \(x_{1}\) if possible. $$\begin{aligned} -x_{1}-x_{2} \quad+x_{4}=-8 \\ 3 x_{1}+5 x_{2}+5 x_{3} \quad=24 \\ 2 x_{3}+x_{4} =-6 \\ -2 x_{1}-3 x_{2}-3 x_{3} \quad=-15 \end{aligned}$$

Use a graphing utility or a computer software program with matrix capabilities and Cramer's Rule to solve for \(x_{1}\) if possible. $$\begin{array}{l} 4 x_{1}-x_{2}+x_{3}=-5 \\ 2 x_{1}+2 x_{2}+3 x_{3}=10 \\ 5 x_{1}-2 x_{2}+6 x_{3}=1 \end{array}$$

Use a graphing utility or computer software program with matrix capabilities to find the eigenvalues of the matrix. Then find the corresponding eigenvectors. $$\left[\begin{array}{rr} 4 & 3 \\ -3 & -2 \end{array}\right]$$
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