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Matrices and Determinants question

2021 · 17 Mar · Shift 2 · Q34
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Matrices and Determinants question

2021 · 17 Mar · Shift 2 · Q34

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
Let A=[abcd]A = \left[ {\begin{matrix} a & b \\ c & d \\ \end{matrix} } \right]A=[ac​bd​] and B=[αβ]e[00]B = \left[ {\begin{matrix} \alpha \\ \beta \\ \end{matrix} } \right] e \left[ {\begin{matrix} 0 \\ 0 \\ \end{matrix} } \right]B=[αβ​]e[00​] such that AB = B and a + d = 2021, then the value of ad −-− bc is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 2020

  1. Interpret the given condition.

    We have A=[abcd]A=\begin{bmatrix}a & b\\ c & d\end{bmatrix}A=[ac​bd​] and BBB is a non-zero column vector of the form B=[αβ],B≠[00]B=\begin{bmatrix}\alpha\\ \beta\end{bmatrix}, \qquad B\neq \begin{bmatrix}0\\0\end{bmatrix}B=[αβ​],B=[00​] such that AB=B.AB=B.AB=B.

  2. Rewrite the equation.

    From AB=B,AB=B,AB=B, we get AB−B=0AB-B=0AB−B=0 AB−IB=0AB-IB=0AB−IB=0 (A−I)B=0.(A-I)B=0.(A−I)B=0.

    Since B≠0B\neq 0B=0, this means the homogeneous system (A−I)X=0(A-I)X=0(A−I)X=0 has a non-trivial solution.

  3. Use the determinant condition.

    A homogeneous system has a non-trivial solution only when det⁡(A−I)=0.\det(A-I)=0.det(A−I)=0.

    Now A−I=[a−1bcd−1].A-I=\begin{bmatrix}a-1 & b\\ c & d-1\end{bmatrix}.A−I=[a−1c​bd−1​].

    Therefore, det⁡(A−I)=(a−1)(d−1)−bc=0.\det(A-I)=(a-1)(d-1)-bc=0.det(A−I)=(a−1)(d−1)−bc=0.

  4. Expand and simplify.

    ad−a−d+1−bc=0ad-a-d+1-bc=0ad−a−d+1−bc=0 ad−bc=a+d−1.ad-bc=a+d-1.ad−bc=a+d−1.

  5. Use the given trace condition.

    Given a+d=2021,a+d=2021,a+d=2021, so ad−bc=2021−1=2020.ad-bc=2021-1=2020.ad−bc=2021−1=2020.

  6. Final answer.

    2020\boxed{2020}2020​

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