Verified Solution • 2025–26 Session moderate

QJEEM23Mada95: JEE Main 2023 Mathematics - Class 12 JEEM23Mada95 Solution: Let be the adjoint of a matrix and . Then is equal to :

Let B=\left[\begin{array}{lll}1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4\end{array}\right], \alpha > 2 be the adjoint of a matrix AA and ∣A∣=2|A|=2 . Then [αamp;−2αamp;α]B[α−2αα]\left[\begin{array}{ccc}\alpha & -2 \alpha & \alpha\end{array}\right] B\left[\begin{array}{c}\alpha \\ -2 \alpha \\ \alpha\end{array}\right] is equal to :

Multiple Choice Options

  • A 32
  • B −- 16
  • C 0
  • D 16

Step-by-Step Verified Solution

B=\left[\begin{array}{lll}1 & 3 & \alpha \\1 & 2 & 3 \\\alpha & \alpha & 4\end{array}\right], \alpha>2

And adj⁡(A)=B,∣A∣=2\operatorname{adj}(A)=B,|A|=2

amp;⇒∣adj⁡(A)∣=∣B∣amp;⇒22=(8−3α)−3(4−3α)+α(−α)amp;⇒α2−6α+8=0\begin{aligned}& \Rightarrow|\operatorname{adj}(A)|=|B| \\\\& \Rightarrow 2^2=(8-3 \alpha)-3(4-3 \alpha)+\alpha(-\alpha) \\\\& \Rightarrow \alpha^2-6 \alpha+8=0\end{aligned}

⇒amp;(α−4)(α−2)=0amp;α=4,2 but αgt;2 so α=4\begin{aligned}\Rightarrow & (\alpha-4)(\alpha-2)=0 \\\\& \alpha=4,2 \text { but } \alpha>2 \text { so } \alpha=4\end{aligned}

Now

amp;[αamp;−2αamp;α]B[α−2αα]=[4−8amp;4][1amp;3amp;41amp;2amp;34amp;4amp;4][4−84]amp;=[12amp;12amp;8][4−84]amp;=48−96+32=−16\begin{aligned}& {\left[\begin{array}{ccc}\alpha & -2 \alpha & \alpha\end{array}\right] B\left[\begin{array}{c}\alpha \\-2 \alpha \\\alpha\end{array}\right]=\left[\begin{array}{lll}4-8 & 4\end{array}\right]\left[\begin{array}{lll}1 & 3 & 4 \\1 & 2 & 3 \\4 & 4 & 4\end{array}\right]\left[\begin{array}{c}4 \\-8 \\4\end{array}\right]} \\\\& =\left[\begin{array}{lll}12 & 12 & 8\end{array}\right]\left[\begin{array}{c}4 \\-8 \\4\end{array}\right] \\\\& = {48-96+32=-16}\end{aligned}

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