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Home  >>  CBSE XII  >>  Math  >>  Matrices

If $A =\begin{bmatrix} x & -2 \\ 3 & 7 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} \frac{7}{34} & \frac{1}{17} \\ \frac{3}{34} & \frac{2}{17} \end{bmatrix}$ then find the value of $x $.


$(A)\; x = 4$
$(B)\; x = -4$
$(C)\; x = 1$
$(D) \; x = -1$

1 Answer

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$AA^{-1} = I$
$\begin{bmatrix} x & -2 \\ 3 & 7 \end{bmatrix} \begin{bmatrix} \frac{7}{34} & \frac{1}{17} \\ \frac{-3}{34} & \frac{2}{17} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\implies \begin{bmatrix} \frac{7x}{34} + \frac{6}{34} & \frac{x}{17} - \frac{4}{17} \\ \frac{21}{34} - \frac{21}{34} & \frac{3}{17} + \frac{14}{17} \end{bmatrix}= \begin{bmatrix} 1& 0 \\ 0 & 1 \end{bmatrix}$
$ \implies \begin{bmatrix} \frac{7x+6}{34} & \frac{x-4}{17} \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\implies \frac{x-4}{17} = 0$
$\implies x = 4$
answered Dec 20, 2016 by priyanka.c
 

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