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  • Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question 10 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question 10 Maths Textbook Solution.

Answers (1)

Answer: X = \begin{bmatrix} 2 &3 &1 \\ 0 &1 &2 \\ 6 &4 &0 \end{bmatrix} and Y = \begin{bmatrix} 1 & 2 &0 \\ -1 & 0 &2 \\ 5 & 4 & 0 \end{bmatrix}

Hint: Try to add two matrices and find variable ‘X ’. Then substitute ‘X ’ value in any matrix to find ‘Y ’.

Given:X+Y= \begin{bmatrix} 3 &5 &1 \\ -1 &1 &4 \\ 11 & 8 & 0 \end{bmatrix} and X-Y = \begin{bmatrix} 1 &1 &1 \\ 1 &1 &0 \\ 1&0 &0 \end{bmatrix}

Here, we have to compute X  and Y .

Solution:

X+Y+X-Y = \begin{bmatrix} 3 & 5 &1 \\ -1 &1 &4 \\ 11 & 8 &0 \end{bmatrix} +\begin{bmatrix} 1 & 1 & 1\\ 1 & 1& 0\\ 1 &0 & 0 \end{bmatrix}

X+Y+X-Y= \begin{bmatrix} 3+1 & 5+1& 1+1\\ -1+1 &1+1 &4+0 \\ 11+1 & 8+0 & 0+0 \end{bmatrix}

2X= \begin{bmatrix} 4 & 6& 2\\ 0 &2 &4\\ 12 & 8 & 0 \end{bmatrix}

X= \frac{1}{2}\begin{bmatrix} 4 & 6& 2\\ 0 &2 &4\\ 12 & 8 & 0 \end{bmatrix} 

X= \begin{bmatrix} 2 & 3& 1\\ 0 &1 &2\\ 6 & 4 & 0 \end{bmatrix}

Now,

X+Y= \begin{bmatrix} 3 &5 &1 \\ -1 &1 &4 \\ 11 & 8 & 0 \end{bmatrix} 

\begin{bmatrix} 2 & 3& 1\\ 0 &1 &2\\ 6 & 4 & 0 \end{bmatrix}+Y  = \begin{bmatrix} 3 &5 &1 \\ -1 &1 &4 \\ 11 & 8 & 0 \end{bmatrix}= \begin{bmatrix} 3 &5 &1 \\ -1 &1 &4 \\ 11 & 8 & 0 \end{bmatrix}

Y = \begin{bmatrix} 3-2 & 5-3 & 1-1\\ -1-0 & 1-1 &4-2 \\ 11-6 &8-4 & 0-0 \end{bmatrix}

Y = \begin{bmatrix} 1 & 2 &0 \\ -1 & 0 &2 \\ 5 & 4 & 0 \end{bmatrix} 

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