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Matrices and Determinants question

2003 · Shift 0 · Q79
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  5. /2003 · Shift 0 · Q79

Matrices and Determinants question

2003 · Shift 0 · Q79

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If A=[abba]A = \left[ {\begin{matrix} a & b \\ b & a \\ \end{matrix} } \right]A=[ab​ba​] and A2=[αββα]{A^2} = \left[ {\begin{matrix} \alpha & \beta \\ \beta & \alpha \\ \end{matrix} } \right]A2=[αβ​βα​], then
  1. A
    α=2ab, β=a2+b2\alpha = 2ab,\,\beta = {a^2} + {b^2}α=2ab,β=a2+b2
  2. B
    α=a2+b2, β=ab\alpha = {a^2} + {b^2},\,\beta = abα=a2+b2,β=ab
  3. C
    α=a2+b2, β=2ab\alpha = {a^2} + {b^2},\,\beta = 2abα=a2+b2,β=2ab
  4. D
    α=a2+b2, β=a2−b2\alpha = {a^2} + {b^2},\,\beta = {a^2} - {b^2}α=a2+b2,β=a2−b2
View written solutionFree

Correct answer: C

  1. We are given A=[abba]A=\begin{bmatrix}a&b\\ b&a\end{bmatrix}A=[ab​ba​] and A2=[αββα].A^2=\begin{bmatrix}\alpha&\beta\\ \beta&\alpha\end{bmatrix}.A2=[αβ​βα​].

  2. Compute A2A^2A2 by matrix multiplication:

\begin{bmatrix}a&b\\ b&a\end{bmatrix} \begin{bmatrix}a&b\\ b&a\end{bmatrix}.$$ 3. Multiply entry-wise: - First row, first column: $$a\cdot a+b\cdot b=a^2+b^2$$ - First row, second column: $$a\cdot b+b\cdot a=ab+ab=2ab$$ - Second row, first column: $$b\cdot a+a\cdot b=ba+ab=2ab$$ - Second row, second column: $$b\cdot b+a\cdot a=b^2+a^2=a^2+b^2$$ So, $$A^2=\begin{bmatrix}a^2+b^2&2ab\\ 2ab&a^2+b^2\end{bmatrix}.$$ 4. Comparing with $$A^2=\begin{bmatrix}\alpha&\beta\\ \beta&\alpha\end{bmatrix},$$ we get $$\alpha=a^2+b^2,\qquad \beta=2ab.$$ 5. Hence the correct option is: $$\boxed{\text{C}}$$
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