Let a \in R and A be a matrix of order 3 \times 3 such that \operatornamedet(A)=-4 and A+I=[\beginarraylll1 & a & 1 2 & 1 & 0 a & 1 & 2\endarray] , where I is the identity matrix of order 3 \times 3 . If \operatornamedet((a+1) \operatornameadj((a-1) A)) is 2^m 3^n, m , n \in\0,1,2, \ldots, 20\ , then m+n is equal to :

Matrices and Determinants PYQ — JEE Main 2025 · 2 Apr · Shift 1 · Q28 with Solution

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Correct answer: D

Given data

We have $\det(A)=-4$ and $A+I=\begin{bmatrix}1&a&1\\2&1&0\\a&1&2\end{bmatrix}.$

So,

=\begin{bmatrix}0&a&1\\2&0&0\\a&1&1\end{bmatrix}.$$ 2. **Use the determinant condition to find $a$** Compute $$\det(A)=\begin{vmatrix}0&a&1\\2&0&0\\a&1&1\end{vmatrix}.$$ Expanding along the first row: $$\det(A)=0\cdot\begin{vmatrix}0&0\\1&1\end{vmatrix}-a\cdot\begin{vmatrix}2&0\\a&1\end{vmatrix}+1\cdot\begin{vmatrix}2&0\\a&1\end{vmatrix}.$$ Now, $$\begin{vmatrix}2&0\\a&1\end{vmatrix}=2.$$ Hence, $$\det(A)=-a(2)+1(2)=2-2a.$$ Given $\det(A)=-4$, so $$2-2a=-4$$ $$-2a=-6$$ $$a=3.$$ 3. **Find the required determinant** We need $$\det\big((a+1)\,\operatorname{adj}((a-1)A)\big).$$ Since $a=3$, $$a+1=4,\qquad a-1=2.$$ Thus the expression becomes $$\det\big(4\,\operatorname{adj}(2A)\big).$$ 4. **Use determinant properties** For a $3\times 3$ matrix $M$: - $$\det(kM)=k^3\det(M),$$ - $$\det(\operatorname{adj}(M))=(\det M)^{3-1}=(\det M)^2.$$ Now, $$\det\big(4\,\operatorname{adj}(2A)\big)=4^3\det(\operatorname{adj}(2A)).$$ Also, $$\det(2A)=2^3\det(A)=8(-4)=-32.$$ Therefore, $$\det(\operatorname{adj}(2A))=(\det(2A))^2=(-32)^2=1024=2^{10}.$$ And $$4^3=(2^2)^3=2^6.$$ So, $$\det\big(4\,\operatorname{adj}(2A)\big)=2^6\cdot 2^{10}=2^{16}.$$ Hence it is of the form $$2^m3^n$$ with $$m=16,\qquad n=0.$$ Therefore, $$m+n=16.$$ 5. **Option check** The correct option is: $$\boxed{\text{D: }16}$$

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View written solutionFree

Correct answer: D

Given data

We have $\det(A)=-4$ and $A+I=\begin{bmatrix}1&a&1\\2&1&0\\a&1&2\end{bmatrix}.$

So,

=\begin{bmatrix}0&a&1\\2&0&0\\a&1&1\end{bmatrix}.$$ 2. **Use the determinant condition to find $a$** Compute $$\det(A)=\begin{vmatrix}0&a&1\\2&0&0\\a&1&1\end{vmatrix}.$$ Expanding along the first row: $$\det(A)=0\cdot\begin{vmatrix}0&0\\1&1\end{vmatrix}-a\cdot\begin{vmatrix}2&0\\a&1\end{vmatrix}+1\cdot\begin{vmatrix}2&0\\a&1\end{vmatrix}.$$ Now, $$\begin{vmatrix}2&0\\a&1\end{vmatrix}=2.$$ Hence, $$\det(A)=-a(2)+1(2)=2-2a.$$ Given $\det(A)=-4$, so $$2-2a=-4$$ $$-2a=-6$$ $$a=3.$$ 3. **Find the required determinant** We need $$\det\big((a+1)\,\operatorname{adj}((a-1)A)\big).$$ Since $a=3$, $$a+1=4,\qquad a-1=2.$$ Thus the expression becomes $$\det\big(4\,\operatorname{adj}(2A)\big).$$ 4. **Use determinant properties** For a $3\times 3$ matrix $M$: - $$\det(kM)=k^3\det(M),$$ - $$\det(\operatorname{adj}(M))=(\det M)^{3-1}=(\det M)^2.$$ Now, $$\det\big(4\,\operatorname{adj}(2A)\big)=4^3\det(\operatorname{adj}(2A)).$$ Also, $$\det(2A)=2^3\det(A)=8(-4)=-32.$$ Therefore, $$\det(\operatorname{adj}(2A))=(\det(2A))^2=(-32)^2=1024=2^{10}.$$ And $$4^3=(2^2)^3=2^6.$$ So, $$\det\big(4\,\operatorname{adj}(2A)\big)=2^6\cdot 2^{10}=2^{16}.$$ Hence it is of the form $$2^m3^n$$ with $$m=16,\qquad n=0.$$ Therefore, $$m+n=16.$$ 5. **Option check** The correct option is: $$\boxed{\text{D: }16}$$

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