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  • Explain Solution RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.5 Question 7 Maths.

Explain Solution RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.5 Question 7 Maths.

Answers (1)

Answer: Symmetric matrix \begin{bmatrix} 3 &\frac{-3}{2} \\ \frac{-3}{2} & -1 \end{bmatrix} and Skew-symmetric matrix =\begin{bmatrix} 0 &\frac{-5}{2} \\ \frac{5}{2} & 0 \end{bmatrix}

Hint: FindP= \frac{1}{2} (A + A^{T}) and\: Q=\frac{1}{2} (A - A^{T})

Given:A = \begin{bmatrix} 3 &-4 \\ 1 & -1 \end{bmatrix}

Solution:

Step - 1\rightarrow A^{T} =\begin{bmatrix} 3 &1 \\ -4 &-1 \end{bmatrix}

Step - 2\rightarrow (A + A^{T}) = \begin{bmatrix} 6 &-3 \\ -3& -2 \end{bmatrix}

Step - 3\rightarrow (A - A^{T}) = \begin{bmatrix} 0 & -5\\ 5& 0 \end{bmatrix}

Step - 4\rightarrow P=\frac{1}{2} (A + A^{T}) = \begin{bmatrix} 3 & \frac{-3}{2}\\ \frac{-3}{2} & -1 \end{bmatrix}

Q = \frac{1}{2} (A - A^{T}) = \begin{bmatrix} 0 &\frac{-5}{2} \\ \frac{5}{2}& 0 \end{bmatrix}

Step - 5\rightarrow P^{T} =\begin{bmatrix} 3 & \frac{-3}{2}\\ \frac{-3}{2}& -1 \end{bmatrix} = P \: and\: Q^{T} = \begin{bmatrix} 0 &\frac{5}{2} \\ \frac{-5}{2}& 0 \end{bmatrix}= -Q

Now, P+ Q = A whereP is symmetric and Q is skew-symmetric.

\begin{bmatrix} 3 &\frac{-3}{2} \\ \frac{-3}{2} & -1 \end{bmatrix}+\begin{bmatrix} 0 &\frac{-5}{2} \\ \frac{5}{2} & 0 \end{bmatrix}=\begin{bmatrix} 3 &-4\\ \1 & -1\end{bmatrix}

Thus, Ais expressed as a sum of symmetric and skew symmetric matrix

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