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  • Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question15 Subquestion (i) Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question15 Subquestion (i) Maths Textbook Solution.

Answers (2)

Answer: x = 32\: and\: y = -32

Hint: Solve LHS and then equate with RHS.

Given:\begin{bmatrix} x-y &2 & -2\\ 4 &x & 6 \end{bmatrix} +\begin{bmatrix} 3 & -2 &2 \\ 1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0\\ 5 & 2x+y & 5 \end{bmatrix}

Here, we have to compute x  and y.

Solution:

  \begin{bmatrix} x-y &2 & -2\\ 4 &x & 6 \end{bmatrix} +\begin{bmatrix} 3 & -2 &2 \\ 1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0\\ 5 & 2x+y & 5 \end{bmatrix}

\begin{bmatrix} x-y+3 &2-2 & -2+2\\ 4+1 &x+0 & 6-1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0\\ 5 & 2x+y & 5 \end{bmatrix}

Equating this, we get:

x-y + 3 = 6
\Rightarrow x-y = 3                                                                                                       … (1)

Also,

x = 2x + y
\Rightarrow y = -x                                                                                                …. (2)

Applying the value of ‘y’ in equation (1), we get:

x-(-x) = 3
2x = 3
x = 32

Applying the value of ‘x ’ in equation (2), we get:

y = -32

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Answer:  x = \frac{3}{2} \: and\: y = -\frac{-3}{2}
Hint: Solve LHS and then equate with RHS.

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

\begin{bmatrix} x-y& 2 &-2 \\ 4 & x & 6 \end{bmatrix}+\begin{bmatrix} 3 &-2 &2 \\ 1&0 & -1 \end{bmatrix}= \begin{bmatrix} 6 & 0 &0 \\ 5 & 2x+y & 5 \end{bmatrix}

\begin{bmatrix} x-y+3& 2-2 &-2 +2\\ 4+3 & x +0& 6-1 \end{bmatrix}= \begin{bmatrix} 6 & 0 &0 \\ 5 & 2x+y & 5 \end{bmatrix}

  Equating this, we get:

x-y + 3 = 6
\Rightarrow x-y = 3                                                                                                               … (1)

Also,

x = 2x + y
\Rightarrow y = -x                                                                                                 …. (2)

Applying the value of ‘y’ in equation (1), we get:

x-(-x) = 3
2x = 3
x = \frac{3}{2}

Applying the value of ‘x ’ in equation (2), we get:

y = -\frac{3}{2}

Posted by

infoexpert27

View full answer

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