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  • Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question 7 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question 7 Maths Textbook Solution.

Answers (1)

Answer: X= \begin{bmatrix} 4 & 4\\ 0 & 4 \end{bmatrix} and \: Y=\begin{bmatrix} 1 &-2 \\ 0 & 5 \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} 5 & 2\\ 0 & 9 \end{bmatrix} and\: X - Y = \begin{bmatrix} 3&6 \\ 0 & -1 \end{bmatrix}
Here, we have to compute X  and Y.
Solution:

\left ( X + Y \right )+\left ( X - Y \right ) = \begin{bmatrix} 5 & 2\\ 0 & 9 \end{bmatrix} + \begin{bmatrix} 3&6 \\ 0 & -1 \end{bmatrix}

\left ( X + Y \right )+\left ( X - Y \right ) = \begin{bmatrix} 5+3 & 2+6\\ 0+0 & 9-1 \end{bmatrix}

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

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

Now,

X+Y = \begin{bmatrix} 5 & 2\\ 0 & 9 \end{bmatrix}

\begin{bmatrix} 4 &4 \\ 0 & 4 \end{bmatrix}+Y = \begin{bmatrix} 5 & 2\\ 0 & 9 \end{bmatrix}

Y = \begin{bmatrix} 5 & 2\\ 0 & 9 \end{bmatrix}-\begin{bmatrix} 4 &4 \\ 0 & 4 \end{bmatrix}

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

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

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