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  • Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question 9 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question 9 Maths Textbook Solution.

Answers (2)

Answer: X  = 3-21-21-1  and Y  = 02200-3

Hint: Try to take the one variable out by using Addition and Multiplication.

Given: 2X-Y  = 6-60-421  and X+2Y  = 325-21-7

Here, we have to compute X  and Y .

Solution:

2X-Y  =6-60-421                                                                                    … (1)

X+2Y   = 325-21-7                                                                                  … (2)

Multiplying equation (1) with 2, we get:

2(2X-Y ) = 26-60-421

4X-2 Y = 12-120-842                                                                            … (3)

Adding equation (2) with equation (3), we get:

(4X-2 Y) + (X+2Y  ) = 12-120-842  + 325-21-7

4X-2 Y + X+2Y   = 12+3-12+20+5-8-24+12-7

5X  = 15-105-105-5

X  = 15  15-105-105-5

X  = 3-21-21-1

Applying ‘X ’ value in equation (2), we get:

X+2Y   = 325-21-7

3-21-21-1  + 2Y  = 325-21-7

2Y  = 3-32+25-1-2+21-1- 7+1

Y  = 12  04400-6

Y  = 02200-3

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Answer: X = \begin{bmatrix} 3 &-2 & 1\\ -2& 1& -1 \end{bmatrix} and \: Y = \begin{bmatrix} 0 & 2 & 2\\ 0 & 0 & -3 \end{bmatrix}
Hint: Try to take the one variable out by using Addition and Multiplication.

Given:2X-Y = \begin{bmatrix} 6 & -6 &0 \\ -4 &2 & 1 \end{bmatrix} and\: X+2Y = \begin{bmatrix} 3 & 2 &5 \\ -2 &1 & -7 \end{bmatrix}
Here, we have to compute X  and Y .
Solution:

2X-Y =\begin{bmatrix} 6 & -6&0 \\ -4 & 2& 1 \end{bmatrix}                                                         … (1)

X+2Y = \begin{bmatrix} 3 & 2&5 \\ -2 & 1 & -7 \end{bmatrix}                                                                   … (2)

Multiplying equation (1) with 2, we get:

 2\left (2X-Y \right ) =2\begin{bmatrix} 6 & -6&0 \\ -4 & 2& 1 \end{bmatrix}

4X-Y =\begin{bmatrix} 12 & -12&0 \\ -8 & 4& 2 \end{bmatrix}                                                                  … (3)

Adding equation (2) with equation (3), we get:

(4X-2 Y) + (X+2Y ) = \begin{bmatrix} 12 & -12 &0 \\ -8 & 4 &2 \end{bmatrix}+\begin{bmatrix} 3 & 2 &5 \\ -2 &1 & -7 \end{bmatrix}

4X-2 Y + X+2Y = \begin{bmatrix} 12+3 & -12+2 &0+5 \\ -8-2 & 4+1 &2-7 \end{bmatrix}

5X = \begin{bmatrix} 15 & -10 &5 \\ -10 & 5 &-5 \end{bmatrix}

X =\frac{1}{5} \begin{bmatrix} 15 & -10 &5 \\ -10 & 5 &-5 \end{bmatrix}

X = \begin{bmatrix} 3 & -2 &1 \\ -2 & 1 &-1 \end{bmatrix}

Applying ‘X ’ value in equation (2), we get:

X+2Y = \begin{bmatrix} 3 & -2 &1 \\ -2 & 1 &-1 \end{bmatrix}

\begin{bmatrix} 3 & -2 &1 \\ -2 & 1 &-1 \end{bmatrix}+2Y = \begin{bmatrix} 3 & 2 & 5\\ -2 & 1 & -7 \end{bmatrix}

2Y = \begin{bmatrix} 3 -3& 2+2 & 5-1\\ -2+2 & 1-1 & -7+1 \end{bmatrix}

Y =\frac{1}{2} \begin{bmatrix} 0& 4 & 4\\ 0 & 0 & -6 \end{bmatrix}

Y = \begin{bmatrix} 0& 2 & 2\\ 0 & 0 & -3 \end{bmatrix}

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