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Question: Which of the following statements is/are true about square matrix A of order n? (This question has...

Which of the following statements is/are true about square matrix A of order n?
(This question has multiple correct options)
A. (−A)−1{{\left( -A \right)}^{-1}} is equal to −A−1-{{A}^{-1}} when n is odd only
B. If An=O{{A}^{n}}=O, then I+A+A2+...+An−1=(I−A)−1I+A+{{A}^{2}}+...+{{A}^{n-1}}={{\left( I-A \right)}^{-1}}
C. If A is a skew symmetric matrix of odd order, then its inverse does not exist
D. (AT)−1=(A−1)T{{\left( {{A}^{T}} \right)}^{-1}}={{\left( {{A}^{-1}} \right)}^{T}} holds exist

Explanation

Solution

We here have been given that A is a square matrix and then a one or more correct options type questions. Since, it is one or more option base problem, we will check for all the options till the last option. We will try to solve every option by using different properties of matrices such as X−1=adj(X)∣X∣{{X}^{-1}}=\dfrac{adj\left( X \right)}{\left| X \right|}, adj(kA)=kn−1adj(A)adj\left( kA \right)={{k}^{n-1}}adj\left( A \right), ∣kA∣=kn∣A∣|kA|={{k}^{n}}|A| and ∣AT∣=∣A∣|{{A}^{T}}|=|A|. We will then see which of the options satisfy both the condition and the result. The options which do will be our required answers.

Complete step-by-step solution:
Here, we have been given that A is a square matrix.
Now, we have also been mentioned in the question that this question has multiple correct options, which means we will have to check for all the options.
Now, we will check for all the given options.
Option-A:
Here, we have been given that (−A)−1=−A−1{{\left( -A \right)}^{-1}}=-{{A}^{-1}} when n is odd only.
Now, we know that for any matrix X, its inverse is given as:
X−1=adj(X)∣X∣{{X}^{-1}}=\dfrac{adj\left( X \right)}{\left| X \right|}
Here, if we keep X=-A, we will get:
(−A)−1=adj(−A)∣−A∣{{\left( -A \right)}^{-1}}=\dfrac{adj\left( -A \right)}{|-A|}
Now, we know that adj(kA)=kn−1adj(A)adj\left( kA \right)={{k}^{n-1}}adj\left( A \right) and ∣kA∣=kn∣A∣|kA|={{k}^{n}}|A| where ‘n’ is the order of matrix A.
Thus, we get:
(−A)−1=(−1)n−1adj(−A)(−1)n∣A∣ ⇒(−A)−1=adj(A)(−1)∣A∣ ⇒(−A)−1=−adj(A)∣A∣ \begin{aligned} & {{\left( -A \right)}^{-1}}=\dfrac{{{\left( -1 \right)}^{n-1}}adj\left( -A \right)}{{{\left( -1 \right)}^{n}}|A|} \\\ & \Rightarrow {{\left( -A \right)}^{-1}}=\dfrac{adj\left( A \right)}{\left( -1 \right)|A|} \\\ & \Rightarrow {{\left( -A \right)}^{-1}}=-\dfrac{adj\left( A \right)}{|A|} \\\ \end{aligned}
Now, we know that adj(A)∣A∣=A−1\dfrac{adj\left( A \right)}{|A|}={{A}^{-1}}
Thus, we get:
(−A)−1=−adj(A)∣A∣ ∴(−A)−1=−A−1 \begin{aligned} & {{\left( -A \right)}^{-1}}=-\dfrac{adj\left( A \right)}{|A|} \\\ & \therefore {{\left( -A \right)}^{-1}}=-{{A}^{-1}} \\\ \end{aligned}
Thus, we can see that this statement is true for all values of n and not only for odd values of n and in the statement we have been asked only for the odd values of n.
Thus, this statement is false.
Option-B:
Here we have been given that An=O{{A}^{n}}=O and have been asked if I+A+A2+...+An−1=(I−A)−1I+A+{{A}^{2}}+...+{{A}^{n-1}}={{\left( I-A \right)}^{-1}} is true or not.
Now, if we consider the LHS, we have:
I+A+A2+...+An−1I+A+{{A}^{2}}+...+{{A}^{n-1}}
Now, multiplying this by (I-A), we get:
(I+A+A2+...+An−1)(I−A)=(I+A+A2+...+An−1)−(A+A2+A3+...+An−1+An) ⇒(I+A+A2+...+An−1)(I−A)=I+(A−A)+(A2−A2)+...(An−1−An−1)−An ⇒(I+A+A2+...+An−1)(I−A)=I−An \begin{aligned} & \left( I+A+{{A}^{2}}+...+{{A}^{n-1}} \right)\left( I-A \right)=\left( I+A+{{A}^{2}}+...+{{A}^{n-1}} \right)-\left( A+{{A}^{2}}+{{A}^{3}}+...+{{A}^{n-1}}+{{A}^{n}} \right) \\\ & \Rightarrow \left( I+A+{{A}^{2}}+...+{{A}^{n-1}} \right)\left( I-A \right)=I+\left( A-A \right)+\left( {{A}^{2}}-{{A}^{2}} \right)+...\left( {{A}^{n-1}}-{{A}^{n-1}} \right)-{{A}^{n}} \\\ & \Rightarrow \left( I+A+{{A}^{2}}+...+{{A}^{n-1}} \right)\left( I-A \right)=I-{{A}^{n}} \\\ \end{aligned}
Now, we have been given that An=O{{A}^{n}}=O
Thus, we get the above obtained equation as:
(I+A+A2+...+An−1)(I−A)=I−An ⇒(I+A+A2+...+An−1)(I−A)=I−O ⇒(I+A+A2+...+An−1)(I−A)=I \begin{aligned} & \left( I+A+{{A}^{2}}+...+{{A}^{n-1}} \right)\left( I-A \right)=I-{{A}^{n}} \\\ & \Rightarrow \left( I+A+{{A}^{2}}+...+{{A}^{n-1}} \right)\left( I-A \right)=I-O \\\ & \Rightarrow \left( I+A+{{A}^{2}}+...+{{A}^{n-1}} \right)\left( I-A \right)=I \\\ \end{aligned}
Now, if we right operate the equation with (I−A)−1{{\left( I-A \right)}^{-1}} on both sides, we will get:
(I+A+A2+...+An−1)(I−A)=I ⇒(I+A+A2+...+An−1)(I−A)(I−A)−1=I(I−A)−1 ⇒(I+A+A2+...+An−1)I=(I−A)−1 ∴(I+A+A2+...+An−1)=(I−A)−1 \begin{aligned} & \left( I+A+{{A}^{2}}+...+{{A}^{n-1}} \right)\left( I-A \right)=I \\\ & \Rightarrow \left( I+A+{{A}^{2}}+...+{{A}^{n-1}} \right)\left( I-A \right){{\left( I-A \right)}^{-1}}=I{{\left( I-A \right)}^{-1}} \\\ & \Rightarrow \left( I+A+{{A}^{2}}+...+{{A}^{n-1}} \right)I={{\left( I-A \right)}^{-1}} \\\ & \therefore \left( I+A+{{A}^{2}}+...+{{A}^{n-1}} \right)={{\left( I-A \right)}^{-1}} \\\ \end{aligned}
This is the statement given in the question which we have proved to be true.
Hence, option (B) is correct.
Option-C:
Here, we have been given that A is a skew-symmetric matrix of odd order and we have to see if it is invertible or not.
We know that only those matrices are invertible which have their determinant is non-zero.
Now, we know that a skew symmetric matrix is represented as:
AT=−A{{A}^{T}}=-A
Now, if we take determinant on both sides, we will get:
∣AT∣=∣−A∣|{{A}^{T}}|=|-A|
Now, we know that ∣AT∣=∣A∣|{{A}^{T}}|=|A|
Thus, we get:

& |{{A}^{T}}|=|-A| \\\ & \Rightarrow |A|=|-A| \\\ & \Rightarrow |A|={{\left( -1 \right)}^{n}}|A| \\\ \end{aligned}$$ Now, we know that ${{\left( -1 \right)}^{n}}=-1$ when n is odd and ${{\left( -1 \right)}^{n}}=1$ when n is even. Here, n is odd, thus we get: $$\begin{aligned} & |A|={{\left( -1 \right)}^{n}}|A| \\\ & \Rightarrow |A|=-|A| \\\ & \Rightarrow 2|A|=0 \\\ & \therefore |A|=0 \\\ \end{aligned}$$ Now, since the matrix A is a singular matrix ( $$|A|=0$$ ), we can say that the inverse of A does not exist. Hence, option (C) is also true. Option-D: Here we have been asked if ${{\left( {{A}^{T}} \right)}^{-1}}={{\left( {{A}^{-1}} \right)}^{T}}$ is true or not. Here, we can see that the inverse of matrix A is involved and we know that not all square matrices are invertible. Here, we have not given for sure if A is invertible or not. Thus, this option is only true for non-singular values of A not all values of it. Hence, this option is also false. **Thus, only options (B) and (C) are correct.** **Note:** Here we have been given many terms related to matrices. We should know the different types of matrices which will come in handy: 1\. Symmetric matrix: This is a square matrix that is characterized as: ${{A}^{T}}=A$ 2\. Skew symmetric matrix: This is a square matrix that is characterized as: ${{A}^{T}}=-{{A}^{T}}$ 3\. Identity matrix: This is a square matrix in which the diagonal elements are 1 and all others are 0. 4\. Diagonal matrix: This is a matrix in which all the elements except for the diagonal elements are 0. 5\. Null matrix: This is a matrix in which all the elements of the matrix are 0.