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Matrices and Determinants question

2012 · Shift 2 · Q36
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  5. /2012 · Shift 2 · Q36

Matrices and Determinants question

2012 · Shift 2 · Q36

JEE AdvancedMathematicsMatrices and DeterminantsMCQ+3 / −1
If P is a 3 ×\times× 3 matrix such that PT = 2P + I, where PT is the transpose of P and I is the 3 ×\times× 3 identity matrix, then there exists a column matrix X=[xyz]e[000]X = \left[ {\begin{matrix} x \\ y \\ z \\ \end{matrix} } \right] e \left[ {\begin{matrix} 0 \\ 0 \\ 0 \\ \end{matrix} } \right]X=​xyz​​e​000​​ such that
  1. A
    PX=[000]PX = \left[ {\begin{matrix} 0 \\ 0 \\ 0 \\ \end{matrix} } \right]PX=​000​​
  2. B
    PX = X
  3. C
    PX = 2X
  4. D
    PX = −-− X
View written solutionFree

Correct answer: D

  1. Let P=[abcdefghi].P=\begin{bmatrix}a&b&c\\ d&e&f\\ g&h&i\end{bmatrix}.P=​adg​beh​cfi​​. Then PT=[adgbehcfi].P^T=\begin{bmatrix}a&d&g\\ b&e&h\\ c&f&i\end{bmatrix}.PT=​abc​def​ghi​​. Given PT=2P+I.P^T=2P+I.PT=2P+I. So,
\begin{bmatrix}2a+1&2b&2c\\ 2d&2e+1&2f\\ 2g&2h&2i+1\end{bmatrix}.$$ 2. Compare corresponding entries: - Diagonal entries: $$a=2a+1 \implies a=-1,$$ $$e=2e+1 \implies e=-1,$$ $$i=2i+1 \implies i=-1.$$ - Off-diagonal entries: $$d=2b,\quad b=2d,$$ $$g=2c,\quad c=2g,$$ $$h=2f,\quad f=2h.$$ From $$d=2b$$ and $$b=2d$$, $$b=2(2b)=4b \implies 3b=0 \implies b=0,$$ so $$d=0.$$ Similarly, $$c=0,\ g=0,\ f=0,\ h=0.$$ Hence, $$P=\begin{bmatrix}-1&0&0\\ 0&-1&0\\ 0&0&-1\end{bmatrix}=-I.$$ 3. Therefore for every non-zero column matrix $$X=\begin{bmatrix}x\\y\\z\end{bmatrix}\neq \begin{bmatrix}0\\0\\0\end{bmatrix},$$ we have $$PX=(-I)X=-X.$$ So there certainly exists such a non-zero column matrix satisfying $$PX=-X.$$ 4. Check options: - A: $$PX=0$$ is false for non-zero $X$ since $PX=-X\neq 0$. - B: $$PX=X$$ is false. - C: $$PX=2X$$ is false. - D: $$PX=-X$$ is true. Therefore, the correct option is $$\boxed{D}.$$
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