If A is an invertible skew-symmetric matrix, then A -1 is a:
Skew-symmetric matrix.
The question asks us to identify the type of matrix that results from the inverse of an invertible skew-symmetric matrix (A-1). Let's break down the properties involved.
A matrix 'A' is called skew-symmetric if its transpose is equal to its negative. Mathematically, this is represented as:
$$ A^T = -A $$
For a matrix A to be skew-symmetric, all its diagonal elements must be zero. For example:
$$ A = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix} $$
When you transpose this matrix, you swap rows and columns. Notice how each element $A_{ij}$ becomes $A_{ji}$, and because it's skew-symmetric, $A_{ji} = -A_{ij}$. This includes the diagonal elements where $A_{ii} = -A_{ii}$, which implies $2A_{ii} = 0$, so $A_{ii} = 0$.
A square matrix is invertible if it has a multiplicative inverse, denoted as A-1. This means that when you multiply the matrix by its inverse, you get the identity matrix (I):
$$ A A^{-1} = A^{-1} A = I $$
An important condition for invertibility is that the determinant of the matrix must not be zero (det(A) ≠ 0).
We are given that A is an invertible skew-symmetric matrix. We need to find the property of A-1.
$$ A^T = -A $$
$$ (A^{-1})^T = (A^T)^{-1} $$
$$ (A^T)^{-1} = (-A)^{-1} $$
Proof for this property: Consider the product $(-A) \times -(A^{-1}) = (-1)A \times (-1)A^{-1} = (-1)(-1) \times A A^{-1} = 1 \times I = I$. Since multiplying $-A$ by $-(A^{-1})$ gives the identity matrix, $-(A^{-1})$ is indeed the inverse of $-A$.
$$ (A^{-1})^T = -(A^{-1}) $$
This final equation, $(A^{-1})^T = -(A^{-1})$, is the definition of a skew-symmetric matrix. Therefore, if A is an invertible skew-symmetric matrix, its inverse, A-1, must also be skew-symmetric.
Based on the derivation, the inverse of an invertible skew-symmetric matrix is a skew-symmetric matrix.
If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&1&{ - 1}\\ 2&{ - 3}&4\\ 3&{ - 2}&3 \end{array}} \right]{\rm{\;and\;\;B}} = \left[ {\begin{array}{*{20}{c}} { - 1}&{ - 2}&{ - 1}\\ 6&{12}&6\\ 5&{10}&5 \end{array}} \right]\) then which of the following is/are correct?
1. A and B commute.
2. AB is a null matrix.
Select the correct answer using the code given below:Which one of the following matrices is an elementary matrix?
The matrix is \(\left[ {\begin{array}{c} 0&{ - 4 + i}\\ {4 + i}&0 \end{array}} \right]\)
How many distinct matrices exist with all four entries taken from (1, 2)?
If A and B are square matrices of order 2 such that det(AB) = det(BA), then which one of the following is correct?