If A=2x0xx and A−1=10−12, then the value of x is

A=2x 0x x⇒|A|=2x2Also, adj⁡A=x 0−x 2x⇒A−1=adj⁡A|A|⇒A−1=12x2x0−x2x=12x0−12x1x⇒10−12=12x0−12x1x⇒12x=1⇒x=12∴Option (d) is the correct answer.

If A=2x0xx and A−1=10−12, then the value of x is

A=2x 0x x⇒|A|=2x2Also, adj⁡A=x 0−x 2x⇒A−1=adj⁡A|A|⇒A−1=12x2x0−x2x=12x0−12x1x⇒10−12=12x0−12x1x⇒12x=1⇒x=12∴Option (d) is the correct answer.

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If $A = [\begin{array}{cc} 2 x & 0 \\ x & x \end{array}]$ and $A^{- 1} = [\begin{array}{cc} 1 & 0 \\ - 1 & 2 \end{array}],$ then the value of x is

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$- \frac{1}{2}$

$\frac{1}{2}$

answer is D.

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Detailed Solution

$A = [\begin{array}{rr} 2 x 0 \\ x x \end{array}] \Rightarrow | A | = 2 x^{2}$

Also, $adj A = [\begin{array}{rr} x 0 \\ - x 2 x \end{array}] \Rightarrow A^{- 1} = \frac{adj A}{| A |}$

$\begin{array}{l} \Rightarrow A^{- 1} = \frac{1}{2 x^{2}} [\begin{array}{rr} x & 0 \\ - x & 2 x \end{array}] = [\begin{array}{cc} \frac{1}{2 x} & 0 \\ - \frac{1}{2 x} & \frac{1}{x} \end{array}] \\ \Rightarrow [\begin{array}{rr} 1 & 0 \\ - 1 & 2 \end{array}] = [\begin{array}{cc} \frac{1}{2 x} & 0 \\ - \frac{1}{2 x} & \frac{1}{x} \end{array}] \\ \Rightarrow \frac{1}{2 x} = 1 \Rightarrow x = \frac{1}{2} \end{array}$