Question:
If $A = \begin{bmatrix} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{bmatrix}$ is such that $A^2 = I$, then
If $A = \begin{bmatrix} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{bmatrix}$ is such that $A^2 = I$, then
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When verifying $A^2 = I$, matrix multiplication and equating corresponding entries is a key technique.
Updated On: May 19, 2025
- $b = \dfrac{ac}{2}$
- $b = -\dfrac{ac}{2}$
- $b = \dfrac{a + c}{2}$
- $b = \sqrt{ac}$
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The Correct Option is B
Solution and Explanation
Given $A^2 = I$, compute $A^2$ using the given matrix.
Multiply $A$ with itself and equate with the identity matrix.
Comparing the third row, first column entry: $b + ac + b = 0 \Rightarrow 2b + ac = 0 \Rightarrow b = -\dfrac{ac}{2}$
Multiply $A$ with itself and equate with the identity matrix.
Comparing the third row, first column entry: $b + ac + b = 0 \Rightarrow 2b + ac = 0 \Rightarrow b = -\dfrac{ac}{2}$
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