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Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMatrices2 Marks
Question
Express matrix as the sum of a symmetric and a skew-symmetric matrix:$\left[\begin{array}{rr} {3} & {5} \\ {1} & {-1} \end{array}\right]$
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Answer

As per Theorem “Any square matrix can be expressed as the sum of a symmetric and skew-symmetric matrix.” So in order to prove this, we will be using Theorem which states that “For any square matrix A with real number entries, A + A’ is a symmetric matrix and A – A’ is a skew-symmetric matrix.”
Now, Let A = $\left[\begin{array}{cc} {3} & {5} \\ {1} & {-1} \end{array}\right]$
Therefore, A' = $\left[\begin{array}{cc} {3} & {1} \\ {5} & {-1} \end{array}\right]$
Now, on adding A and A’ we will get,
$A+A^{\prime}=\left[\begin{array}{cc} {3} & {5} \\ {1} & {-1} \end{array}\right]+\left[\begin{array}{cc} {3} & {1} \\ {5} & {-1} \end{array}\right]$
$\Rightarrow \mathrm{A}+\mathrm{A}^{\prime}=\left[\begin{array}{cc} {3+3} & {5+1} \\ {1+5} & {-1+(-1)} \end{array}\right]$
$\Rightarrow A+A^{\prime}=\left[\begin{array}{cc} {6} & {6} \\ {6} & {-2} \end{array}\right]$
Now, Let M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$
Therefore, $M=\frac{1}{2}\left[\begin{array}{cc} {6} & {6} \\ {6} & {-2} \end{array}\right] = \left[\begin{array}{cc} {3} & {3} \\ {3} & {-1} \end{array}\right]$
Now, $M^{\prime}=\left[\begin{array}{cc} {3} & {3} \\ {3} & {-1} \end{array}\right]$
$\Rightarrow$ M' = M Thus M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is a symmetric matrix as M' = M.
Now on subtracting A’ from A we will get,
$A-A^{\prime}=\left[\begin{array}{cc} {3} & {5} \\ {1} & {-1} \end{array}\right]-\left[\begin{array}{cc} {3} & {1} \\ {5} & {-1} \end{array}\right]$
$\Rightarrow A-A^{\prime}=\left[\begin{array}{cc} {0} & {4} \\ {-4} & {0} \end{array}\right]$
Now, Let N = $\frac{1}{2}\left(\mathrm{A}-\mathrm{A}^{\prime}\right)$
Therefore, $N=\frac{1}{2}\left[\begin{array}{cc} {0} & {4} \\ {-4} & {0} \end{array}\right]$
$\Rightarrow \mathrm{N}=\left[\begin{array}{cc} {0} & {2} \\ {-2} & {0} \end{array}\right]$
Now, N' =$\left[\begin{array}{cc} {0} & {-2} \\ {2} & {0} \end{array}\right]$
$\Rightarrow \mathrm{N}^{\prime}=(-1)\left[\begin{array}{cc} {0} & {2} \\ {-2} & {0} \end{array}\right]$
$\Rightarrow$ N' = - N
Thus M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is a skew-symmetric matrix as N' = -N
Now, Add M and N, we get,
$M+N=\left[\begin{array}{cc} {3} & {3} \\ {3} & {-1} \end{array}\right]+\left[\begin{array}{cc} {0} & {2} \\ {-2} & {0} \end{array}\right]$
$\Rightarrow \mathrm{M}+\mathrm{N}=\left[\begin{array}{cc} {3+0} & {3+2} \\ {3+(-2)} & {-1+0} \end{array}\right]$
$\Rightarrow \mathrm{M}+\mathrm{N}=\left[\begin{array}{cc} {3} & {5} \\ {1} & {-1} \end{array}\right]$
So we see here, M + N = $\left[\begin{array}{cc} {3} & {5} \\ {1} & {-1} \end{array}\right]$ = A
Thus, A is represented as the sum of a symmetric matrix M and a skew-symmetric matrix N.
Hence proved.

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