Question
Express matrix as the sum of a symmetric and a skew-symmetric matrix: $\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$

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 = $\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$
Therefore, A' = $\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$
Now, on adding A and A’ we will get,
$A+A^{\prime}=\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]+\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$
$\Rightarrow A+A^{\prime}=\left[\begin{array}{rrr} {6+6} & {-2+(-2)} & {2+2} \\ {-2+(-2)} & {3+3} & {-1+(-1)} \\ {2+2} & {-1+(-1)} & {3+3} \end{array}\right]$
$\Rightarrow \mathrm{A}+\mathrm{A}^{\prime}=\left[\begin{array}{ccc} {12} & {-4} & {4} \\ {-4} & {6} & {-2} \\ {4} & {-2} & {6} \end{array}\right]$
Let, M = $\frac{1}{2}\left[\begin{array}{ccc} {12} & {-4} & {4} \\ {-4} & {6} & {-2} \\ {4} & {-2} & {6} \end{array}\right]$
$\Rightarrow M=\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$
Now, $M^{\prime}=\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$

$\Rightarrow$ 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,
$A-A^{\prime}=\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]-\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$
$\Rightarrow \mathrm{A}-\mathrm{A}^{\prime}=\left[\begin{array}{ccc} {6-6} & {-2-(-2)} & {2-2} \\ {-2} & {3-3} & {-1-(-1)} \\ {2-2} & {-1-(-1)} & {3-3} \end{array}\right]$
$\Rightarrow A-A^{\prime}=\left[\begin{array}{lll} {0} & {0} & {0} \\ {0} & {0} & {0} \\ {0} & {0} & {0} \end{array}\right]$
Now, Let N = $\frac{1}{2}\left(\mathrm{A}-\mathrm{A}^{\prime}\right)$
Therefore, N = $\frac{1}{2}\left[\begin{array}{lll} {0} & {0} & {0} \\ {0} & {0} & {0} \\ {0} & {0} & {0} \end{array}\right]$
$\Rightarrow \mathrm{N}=\left[\begin{array}{lll} {0} & {0} & {0} \\ {0} & {0} & {0} \\ {0} & {0} & {0} \end{array}\right]$
Now, $N^{\prime}=\left[\begin{array}{lll} {0} & {0} & {0} \\ {0} & {0} & {0} \\ {0} & {0} & {0} \end{array}\right]$
$\Rightarrow \mathrm{N}^{\prime}=(-1)\left[\begin{array}{lll} {0} & {0} & {0} \\ {0} & {0} & {0} \\ {0} & {0} & {0} \end{array}\right]$
$\Rightarrow$ N’ = - N
Thus M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is a symmetric matrix as N' = - N.
Now, Add M and N, we get,
$M+N=\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]+\left[\begin{array}{lll} {0} & {0} & {0} \\ {0} & {0} & {0} \\ {0} & {0} & {0} \end{array}\right]$
$\Rightarrow \mathrm{M}+\mathrm{N}=\left[\begin{array}{ccc} {6+0} & {-2+0} & {2+0} \\ {-2+0} & {3+0} & {-1+0} \\ {2+0} & {-1+0} & {3+0} \end{array}\right]$
$\Rightarrow \mathrm{M}+\mathrm{N}=\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$ So we see her, M + N = $\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$ = A

Thus, A is represented as the sum of a symmetric matrix M and a skew-symmetric matrix N.

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