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Chapter 2: Problem 11

Let \(A\) be an \(n \times n\) matrix. Is it possible for \(A^{2}+I=\) \(O\) in the case where \(n\) is odd? Answer the same question in the case where \(n\) is even.

Short Answer

Expert verified
In summary, if n is odd, it is not possible for the matrix equation \(A^2 + I = O\) to hold true, since the matrix A will always be singular and non-invertible. If n is even, the matrix equation \(A^2 + I = O\) can hold true, but it is not guaranteed, as there could be cases where the determinant of A is non-zero, resulting in A being invertible, or when the determinant of A is zero, resulting in A being non-invertible.

Step by step solution

01

Case 1: n is odd

For this case, we assume n is odd, and we have the matrix equation A^2 + I = O. First, we will find the determinant of both sides of the equation. Let's denote the determinant of A as det(A). Take the determinant of both sides: \[\det(A^2 + I) = \det(O)\] Now, we can rewrite the left side of the equation using properties of determinants: \[\det(A^2) \cdot \det(I) = 0\] Since det(I) = 1, we have: \[\det(A^2) = 0\] Now, recall that \(\det(A^2) = (\det(A))^2\). So, we can rewrite the equation as: \[(\det(A))^2 = 0\] In this case, since n is odd, the only solution for this equation is \(\det(A) = 0\), which means that A is singular and non-invertible.
02

Case 2: n is even

For this case, we assume n is even, and we have the matrix equation A^2 + I = O. Again, we find the determinant of both sides of the equation, and we can rewrite it: \[\det(A^2) \cdot \det(I) = 0\] \[(\det(A))^2 = 0\] In this case, since n is even, there could be possibilities that the matrix A has a determinant equal to 0 or not. Take for instance a 2x2 matrix: \[A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\] The equation A^2 + I = O holds true, and its determinant is non-zero. However, when the determinant of A is equal to 0, the same as in the odd case, A would be singular and non-invertible.
03

Conclusion

In summary, if n is odd, it is not possible for the matrix equation A^2 + I = O to hold true, since the matrix A will always be singular and non-invertible. If n is even, the matrix equation A^2 + I = O can hold true, but it is not guaranteed, as there could be cases where the determinant of A is non-zero, resulting in A being invertible, or when the determinant of A is zero, resulting in A being non-invertible.

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Most popular questions from this chapter

Find all values of \(\lambda\) for which the following determinant will equal 0 $$\left|\begin{array}{cc} 2-\lambda & 4 \\ 3 & 3-\lambda \end{array}\right|$$

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

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) \(A=\left(\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right)\)

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

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