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If [10b5]+2[a01−2]=I\begin{bmatrix} 1 & 0 \\ b & 5 \end{bmatrix} + 2\begin{bmatrix} a & 0 \\ 1 & -2 \end{bmatrix} = I, where II is a unit matrix of order 2, then the value of (a−b)(a - b) is:

Solution

✅ Correct Option: 3

The unit matrix II of order 2 is:

I=[1001]I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}


2[a01−2]=[2a02−4]2\begin{bmatrix} a & 0 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} 2a & 0 \\ 2 & -4 \end{bmatrix}


[10b5]+[2a02−4]=[1+2a0b+21]\begin{bmatrix} 1 & 0 \\ b & 5 \end{bmatrix} + \begin{bmatrix} 2a & 0 \\ 2 & -4 \end{bmatrix} = \begin{bmatrix} 1+2a & 0 \\ b+2 & 1 \end{bmatrix}


[1+2a0b+21]=[1001]\begin{bmatrix} 1+2a & 0 \\ b+2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}


Comparing corresponding elements:

1+2a=11 + 2a = 1

2a=02a = 0

a=0a = 0


b+2=0b + 2 = 0

b=−2b = -2


a−b=0−(−2)a - b = 0 - (-2)

a−b=2a - b = 2

Therefore, (a−b)=2(a - b) = 2

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