If B is a 3 × 3 matrix such that B 2 = 0 , then d e t . I + B 50 - 50 B is equal to :

B 2 = 0 ⇒ B 4 = B 6 = B 8 = . . . . = B 50 = 0 and B 3 = B 2 B = 0 B = 0 ⇒ I + B 50 = I + C 1 50 B (higher powers are zero) ∴ d e t . I + B 50 - 50 B = d e t . I + 5 0 C 1 B - 5 0 B = d e t . I = 1

Mathematics - Matrices Question with Solution | TestHub

MathematicsMatricesProduct of MatricesMedium2 minPYQ_2014

MathematicsMediumsingle choice

If $B$ is a $3 \times 3$ matrix such that $B^{2} = 0$ , then $d e t . [{(I + B)}^{50} - 50 B]$ is equal to :

Options:

Answer:

Solution:

$B^{2} = 0 \Rightarrow B^{4} = B^{6} = B^{8} = . . . . = B^{50} = 0$ and $B^{3} = B^{2} B = 0 B = 0$

$\Rightarrow {(I + B)}^{50} = I + {}^{50}C_{1} B$ (higher powers are zero)

$∴$ $d e t . [{(I + B)}^{50} - 50 B]$

$= d e t . [I +^{50} C_{1} B - 50 B]$

$= d e t . [I]$

$= 1$

Stream:JEESubject:MathematicsTopic:MatricesSubtopic:Product of Matrices

⏱ 2mℹ️ Source: PYQ_2014

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