KCET · Maths · Matrices
If the matrix \(\left[\begin{array}{rr}2 & 3 \\ 5 & -1\end{array}\right]=A+B\), where \(A\) is symmetric and \(B\) is skew-symmetric, then \(B\) is equal to
- A \(\left[\begin{array}{cc}2 & 4 \\ 4 & -1\end{array}\right]\)
- B \(\left[\begin{array}{rr}0 & -2 \\ 2 & 0\end{array}\right]\)
- C \(\left[\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}\right]\)
- D \(\left[\begin{array}{rr}0 & -1 \\ 1 & 0\end{array}\right]\)
Answer & Solution
Correct Answer
(D) \(\left[\begin{array}{rr}0 & -1 \\ 1 & 0\end{array}\right]\)
Step-by-step Solution
Detailed explanation
Given,
\(\left[\begin{array}{rr}
2 & 3 \\
5 & -1
\end{array}\right]=A+B\)
We know that, every square matrix can be expressed uniquely as the sum of symmetric and skew-symmetric matrix.
Let
\(C=\left[\begin{array}{rr}
2 & 3 \\
5 & -1
\end{array}\right]\)
Since, \(A\) is symmetric and \(B\) is skew-symmetric matrix.
\(\begin{aligned}
\therefore \quad C &=\frac{1}{2}\left(C+C^{\prime}\right)+\frac{1}{2}\left(C-C^{\prime}\right)=A+B \\
\Rightarrow B &=\frac{1}{2}\left(C-C^{\prime}\right)=\frac{1}{2}\left\{\left(\begin{array}{rr}
2 & 3 \\
5 & -1
\end{array}\right)-\left(\begin{array}{rr}
2 & 5 \\
3 & -1
\end{array}\right)\right\} \\
&=\frac{1}{2}\left(\begin{array}{rr}
0 & -2 \\
2 & 0
\end{array}\right)=\left(\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}\right)
\end{aligned}\)
\(\left[\begin{array}{rr}
2 & 3 \\
5 & -1
\end{array}\right]=A+B\)
We know that, every square matrix can be expressed uniquely as the sum of symmetric and skew-symmetric matrix.
Let
\(C=\left[\begin{array}{rr}
2 & 3 \\
5 & -1
\end{array}\right]\)
Since, \(A\) is symmetric and \(B\) is skew-symmetric matrix.
\(\begin{aligned}
\therefore \quad C &=\frac{1}{2}\left(C+C^{\prime}\right)+\frac{1}{2}\left(C-C^{\prime}\right)=A+B \\
\Rightarrow B &=\frac{1}{2}\left(C-C^{\prime}\right)=\frac{1}{2}\left\{\left(\begin{array}{rr}
2 & 3 \\
5 & -1
\end{array}\right)-\left(\begin{array}{rr}
2 & 5 \\
3 & -1
\end{array}\right)\right\} \\
&=\frac{1}{2}\left(\begin{array}{rr}
0 & -2 \\
2 & 0
\end{array}\right)=\left(\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}\right)
\end{aligned}\)
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