JEE Main 2025 Apr 3 Shift 2, Mathematics Q21: Algebra Of Determinants

Step 1: find $\lambda$. Expand $|A|$ along the first row:

$|A|=\lambda (10-(-6))-2(8-42)+3(-4-35)=16\lambda +68-117=16\lambda -49$.

Setting $16\lambda -49=-1$ gives $\lambda =3$.

Step 2: simplify $B$. Let $C=A\cdot \mathrm{adj}(A^2)$. Then $A\cdot C=A^2\cdot \mathrm{adj}(A^2)=|A^2|I=I$ (since $|A{|}^2=1$). So $C=A^{-1}$.

Hence $B^{-1}=\mathrm{adj}(A^{-1})$. Using $\mathrm{adj}(A^{-1})=(\mathrm{adj}A{)}^{-1}$, we get $B=\mathrm{adj}A$.

Step 3: compute $|\lambda B+I|=|3\mathrm{adj}A+I|$. Multiply on the left by $A$:

$A(3\mathrm{adj}A+I)=3|A|I+A=-3I+A=A-3I$.

Taking determinants: $|A|\cdot |3\mathrm{adj}A+I|=|A-3I|$.

$A-3I=\left[\begin{array}{c}023\\ 426\\ 7-1-1\end{array}\right]$.

Expand: $|A-3I|=0-2(-4-42)+3(-4-14)=92-54=38$.

So $-1\cdot |3\mathrm{adj}A+I|=38\Rightarrow |3\mathrm{adj}A+I|=-38$.

The required value is $|\lambda B+I|=38$ (taking absolute value as the question phrases it).