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Home  >>  CBSE XII  >>  Math  >>  Matrices
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Which of the given values of x and y make the following pair of matrices equal: \[\begin{bmatrix}3x+7 & 5\\y+1 & 2-3x\end{bmatrix},\begin{bmatrix}0 & y-2\\8 & 4\end{bmatrix}\]

\begin{array}{1 1}(A)\;x=\frac{-1}{3},\;y=7 & (B)\;\text{Not possible to find}\\(C)\;y=7\;x=\frac{-2}{3} & (D)\;x=\frac{1}{3},y=\frac{2}{3}\end{array}


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  • If the order of 2 matrices are equal, their corresponding elements are equal, i.e, if $A_{ij}=B_{ij}$, then any element $a_{ij}$ in matrix A is equal to corresponding element $b_{ij}$ in matrix B.
  • We can then match the corresponding elements and solve the resulting equations to find the values of x and y.
Given $\begin{bmatrix}3x+7 & 5\\y+1 & 2-3x\end{bmatrix}= \begin{bmatrix}0 & y-2\\8 & 4\end{bmatrix}$ Since the order matrices are equal Since these matrices are equal, we can an obtain the value of x and y by comparing the matrices' corresponding elements.
By comparing the given two matrices of equal order, we can see that:
$3x+7=0$ (i)
$5=y-2$ (ii)
$y+1=8$ (iii)
$2-3x=4$ (iv)
Solving for $x$ in (i), $3x=-7 \rightarrow x = \frac{-7}{3}$.
Solving for $x$ in (iv), $-3x = 4-2 \rightarrow -3x = 2 \rightarrow x = \frac{-2}{3}$.
By solving for x in (i) and (iv) we see that $x$ has two values $\frac{-7}{3}$ and $\frac{-2}{3}$ which is not possible since $x$ and $y$ can have only one value for which the matrices can be equal.
Therefore, the correct answer is $(B):$ Not possible to find.
answered Feb 27, 2013 by balaji.thirumalai

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