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QJEEM24Mac2c5: JEE Main 2024 Mathematics - Class 12 JEEM24Mac2c5 Solution: Let . If , where is the identity matrix of order , then is…

Let A=[2amp;aamp;01amp;3amp;10amp;5amp;b]A=\left[\begin{array}{lll}2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b\end{array}\right] . If A3=4A2−A−21IA^3=4 A^2-A-21 I , where II is the identity matrix of order 3×33 \times 3 , then 2a+3b2 a+3 b is equal to

Multiple Choice Options

  • A −10-10
  • B −12-12
  • C −13-13
  • D −9-9

Step-by-Step Verified Solution

amp;∣A−λI∣=0amp;∣2−λamp;aamp;01amp;3−λamp;10amp;5amp;b−λ∣=0amp;(2−λ)[(3−λ)(b−λ)−5]−a[b−λ−0]+0=0amp;(2−λ)[3b−3λ−bλ+λ2−5]−ab+aλ=0amp;λ3−(b+5)λ2+(1−a+5b)λ+(10−6b+ab)=0amp;A3−(b+5)A2+(1−a+5b)A+(10−6b+ab)I=0amp;⇒b+5=4,1−a+5b=1,10−6b+ab=21amp;⇒a=−5,b=−1amp;⇒2a+3b=−13\begin{aligned}& |A-\lambda I|=0 \\& \left|\begin{array}{ccc}2-\lambda & a & 0 \\1 & 3-\lambda & 1 \\0 & 5 & b-\lambda\end{array}\right|=0 \\& (2-\lambda)[(3-\lambda)(b-\lambda)-5]-a[b-\lambda-0]+0=0 \\& (2-\lambda)\left[3 b-3 \lambda-b \lambda+\lambda^2-5\right]-a b+a \lambda=0 \\& \lambda^3-(b+5) \lambda^2+(1-a+5 b) \lambda+(10-6 b+a b)=0 \\& A^3-(b+5) A^2+(1-a+5 b) A+(10-6 b+a b) I=0 \\& \Rightarrow \mathrm{b}+5=4,1-a+5 b=1,10-6 b+a b=21 \\& \Rightarrow a=-5, b=-1 \\& \Rightarrow 2 a+3 b=-13\end{aligned}

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