Question

Let $A$ and $B$be $3\times 3$real matrices such that is a $A$ symmetric matrix and $B$is a skew-symmetric matrix. Then the system of linear equations $\left(A^2{B}^2-B^2{A}^2\right)x=O$ where $X$ is a $3\times 1$ column matrix of unknown variables and $O$ is a $3\times 1$ null matrix has:

• A. Unique Solution
• B. exactly two solutions
• C. infinitely many solutions
• D. No solution

Answer 1

The correct option is C

infinitely many solutions

Explanation for the correct answer:

Step 1: Identifying the type of matrix using given conditions

Given, $A$ is symmetric matrix

$\Rightarrow A=A^T$

$B$is a skew-symmetric matrix

$\Rightarrow B=-B^T$

Let $C=\left(A^2{B}^2-B^2{A}^2\right)$

$$\begin{gathered} C^T={\left(A^2{B}^2-B^2{A}^2\right)}^T \\ ={\left(A^2{B}^2\right)}^T-{\left(B^2{A}^2\right)}^T \\ ={\left(B^2\right)}^T{\left(A^2\right)}^T-{\left(B^2\right)}^T{\left(A^2\right)}^T\left[\because {\left(AB\right)}^T=B^T{A}^T\right] \\ =B^2{A}^2-A^2{B}^2 \end{gathered}$$

$\Rightarrow C=-C^T$

Therefore, $C$ is a skew-symmetric matrix.

Step 2: Find the system of equations by defining variables

Given $X$ is a $3\times 1$ column matrix of unknown variables and $O$ is a $3\times 1$ null matrix

$\Rightarrow$$C=\left[\begin{array}{ccc}0& a& b\\ -a& 0& c\\ -b& -c& 0\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],O=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

$$\begin{gathered} \left(A^2{B}^2-B^2{A}^2\right)X=O \\ CX=O \\ \left[\begin{array}{ccc}0& a& b\\ -a& 0& c\\ -b& -c& 0\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \end{gathered}$$

$C\to 3\times 3$ matrix $X\to 3\times 1$ matrix multiplication is possible and the resultant matrix be $3\times 1$matrix

$\left[\begin{array}{c}ay+bz\\ -ax+cz\\ -bx-cy\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

$$\begin{gathered} ay+bz=0\to \left(i\right) \\ -ax+cz=0\to \left(ii\right) \\ -bx-cy=0\to \left(iii\right) \end{gathered}$$

Step 3: Find the nature of solutions of the equation using crammers rule

From this equation

$$\begin{gathered} ∆=\left|\begin{array}{ccc}0& a& b\\ -a& 0& c\\ -b& -c& 0\end{array}\right|=0-a\left(bc\right)+b\left(ac\right)=0 \\ {∆}_1=\left|\begin{array}{ccc}0& a& b\\ 0& 0& c\\ 0& -c& 0\end{array}\right|=0 \\ {∆}_2=\left|\begin{array}{ccc}0& 0& b\\ -a& 0& c\\ -b& 0& 0\end{array}\right|=0 \\ {∆}_3=\left|\begin{array}{ccc}0& a& b\\ -a& 0& c\\ -b& -c& 0\end{array}\right|=0 \end{gathered}$$

By crammers rule, the system of the equation has an infinite number of solutions.

Hence, the correct answer is option (C).