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If A=[14202x−3]A = \begin{bmatrix}\frac{1}{4} & 2\\0 & 2x-3\end{bmatrix}, B=[480−1]B = \begin{bmatrix}4 & 8\\0 & -1\end{bmatrix} and AB = I2I_2 (Identity matrix of order 2), then the value of xx is

Solution

✅ Correct Option: 1

AB=[18⋅14−20−(2x−3)]=[1003−2x]AB=\begin{bmatrix}1&8\cdot\frac14-2\\0&-(2x-3)\end{bmatrix}=\begin{bmatrix}1&0\\0&3-2x\end{bmatrix}.

For this to equal I2I_2, we need 3−2x=13-2x=1, so x=1x=1.

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