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  • Explain Solution RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.5 Question 8 Maths.

Explain Solution RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.5 Question 8 Maths.

Answers (1)

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

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

Given: A = \begin{bmatrix} 3 &-2 &-4 \\ 3& -2 &-5 \\ -1 &1 & 2 \end{bmatrix}

Solution:

Step - 1\rightarrow A^{T} = \begin{bmatrix} 3 &3 &-1 \\ -2& -2 &1 \\ -4 &-5 & 2 \end{bmatrix}

Step - 2\rightarrow (A + A^{T}) = \begin{bmatrix} 6 &-1 &-5 \\ 1& -4 & -4\\ -5& -4 &4 \end{bmatrix}

Step - 3\rightarrow (A - A^{T}) = \begin{bmatrix} 0 & -5 & -3\\ 5 &0 &-6 \\ 3& 6 & 0 \end{bmatrix}

Step - 4\rightarrow P=\frac{1}{2} (A + A^{T}) =\begin{bmatrix} 3 &\frac{1}{2} &\frac{-5}{2} \\ \frac{1}{2} & -2 &-2 \\ \frac{-5}{2} & -2 &2 \end{bmatrix}

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

Verify:P+ Q = A whereP is symmetric and Q is skew-symmetric.

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

Thus, A is expressed as a sum of the symmetric and skew-symmetric matrix

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