Q63 / 150 · MHT CET 2007 · Mathematics

If $A$ is a square matrix of order $n \times n$, then $\operatorname{adj}(\operatorname{adj} A)$ is equal to

  • A
    | $\left.A\right|^{n} A$
  • B
    $|A|^{n-1} A$
  • Correct answer
    C
    $|A|^{n-2} A$
  • D
    $|A|^{n-3} A$

Solution

For any square matrix $B$, we have $B(\operatorname{adj} B)=|B| I_{n}$ On taking $B=\operatorname{adj} A$, we get $(\operatorname{adj} A)[\operatorname{adj}(\operatorname{adj} A)]=|\operatorname{adj} A| I_{n}$ $$ \Rightarrow \quad \operatorname{adj} A[\operatorname{adj}(\operatorname{adj} A)]=|A|^{n-1} I_{n} $$ $\left(\because|\operatorname{adj} A|=|A|^{n-1}\right)$ $\Rightarrow(A \operatorname{adj} A)[\operatorname{adj}(\operatorname{adj} A)]=|A|^{n-1} A$ $\Rightarrow\left(|A| I_{n}\right)[\operatorname{adj}(\operatorname{adj} A)]=|A|^{n-1} A$ $\Rightarrow$ $(\operatorname{adj} A)=|A|^{n-2} A$

Hence, the correct answer is option: 'C'.

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