Let \alpha \beta eq 0 and A=[\beginarrayrrr\beta & \alpha & 3 \alpha & \alpha & \beta -\beta & \alpha & 2 \alpha\endarray] . If B=[\beginarrayrrr3 \alpha & -9 & 3 \alpha -\alpha & 7 & -2 \alpha -2 \alpha & 5 & -2 \beta\endarray] is the matrix of cofactors of the elements of A , then \operatornamedet(A B) is equal to :

Matrices and Determinants PYQ — JEE Main 2024 · 5 Apr · Shift 2 · Q45 with Solution

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Correct answer: D

Since $B$ is the matrix of cofactors of $A$ , we use the standard identity

$A\,(\text{adj }A)= (\det A)I.$

The adjugate matrix is the transpose of the cofactor matrix. Hence, whether $B$ is given as cofactor matrix or its transpose, we have

$\det(AB)=\det(A)\det(B).$

Also, for a $3\times 3$ matrix,

$\det(\text{cofactor matrix})=\det(\text{adj }A)= (\det A)^{2}.$

Therefore,

$\det(AB)=\det(A)\det(B)=\det(A)\cdot (\det A)^2=(\det A)^3.$

So we only need $\det A$ .

Compute $\det A$ using the first row:

\beta & \alpha & 3\\ \alpha & \alpha & \beta\\ -\beta & \alpha & 2\alpha \end{bmatrix}.$$ Thus,

\det A= \beta\begin{vmatrix} \alpha & \beta\ \alpha & 2\alpha \end{vmatrix} -\alpha\begin{vmatrix} \alpha & \beta\ -\beta & 2\alpha \end{vmatrix} +3\begin{vmatrix} \alpha & \alpha\ -\beta & \alpha \end{vmatrix}.

Now evaluate each minor:

\begin{vmatrix} \alpha & \beta\ \alpha & 2\alpha \end{vmatrix} =2\alpha^2-\alpha\beta=\alpha(2\alpha-\beta),

\begin{vmatrix} \alpha & \beta\ -\beta & 2\alpha \end{vmatrix} =2\alpha^2+\beta^2,

\begin{vmatrix} \alpha & \alpha\ -\beta & \alpha \end{vmatrix} =\alpha^2+\alpha\beta=\alpha(\alpha+\beta).

Hence,

\det A=eta\alpha(2\alpha-\beta)-\alpha(2\alpha^2+\beta^2)+3\alpha(\alpha+\beta).

This looks messy, so instead we use the given cofactor matrix $B$ to determine $\alpha,\beta$. --- 3. Since $B$ is the cofactor matrix of $A$, compare entries of $B$ with actual cofactors of $A$. Given $$B=\begin{bmatrix} 3\alpha & -9 & 3\alpha\\ -\alpha & 7 & -2\alpha\\ -2\alpha & 5 & -2\beta \end{bmatrix}.$$ Now compute some cofactors of $A$. ### Cofactor $C_{12}$

C_{12}=(-1)^{1+2}\begin{vmatrix} \alpha & \beta\ -\beta & 2\alpha \end{vmatrix} =-(2\alpha^2+\beta^2).

From $B$, this equals $-9$. So, $$2\alpha^2+\beta^2=9. \qquad (1)$$ ### Cofactor $C_{22}$

C_{22}=\begin{vmatrix} \beta & 3\ -\beta & 2\alpha \end{vmatrix} =2\alpha\beta+3\beta=\beta(2\alpha+3).

From $B$, this equals $7$. So, $$\beta(2\alpha+3)=7. \qquad (2)$$ ### Cofactor $C_{32}$

C_{32}=(-1)^{3+2}\begin{vmatrix} \beta & 3\ \alpha & \beta \end{vmatrix} =-(\beta^2-3\alpha)=3\alpha-\beta^2.

From $B$, this equals $5$. Thus, $$3\alpha-\beta^2=5. \qquad (3)$$ --- 4. Solve for $\alpha,\beta$. From (3), $$\beta^2=3\alpha-5.$$ Substitute into (1): $$2\alpha^2+(3\alpha-5)=9$$ $$2\alpha^2+3\alpha-14=0.$$ Factor: $$2\alpha^2+3\alpha-14=(2\alpha+7)(\alpha-2)=0.$$ So, $$\alpha=2 \quad \text{or} \quad \alpha=-\frac72.$$ Now use (2). If $\alpha=2$, then from (3): $$\beta^2=6-5=1 \Rightarrow \beta=\pm1.$$ Using (2): $$\beta(4+3)=7 \Rightarrow 7\beta=7 \Rightarrow \beta=1.$$ So one solution is $$\alpha=2,\quad \beta=1.$$ If $\alpha=-\frac72$, then $$\beta^2=3\left(-\frac72\right)-5=-\frac{31}{2},$$ impossible over reals. Hence rejected. Therefore, $$\alpha=2,\quad \beta=1.$$ --- 5. Now substitute into $A$: $$A=\begin{bmatrix} 1 & 2 & 3\\ 2 & 2 & 1\\ -1 & 2 & 4 \end{bmatrix}.$$ Compute its determinant:

\det A = 1\begin{vmatrix}2&1\2&4\end{vmatrix} -2\begin{vmatrix}2&1\-1&4\end{vmatrix} +3\begin{vmatrix}2&2\-1&2\end{vmatrix}.

That is,

\det A = 1(8-2)-2(8+1)+3(4+2) =6-18+18=6.

Thus, $$\det A=6.$$ --- 6. Therefore, $$\det(AB)=(\det A)^3=6^3=216.$$ So the correct option is $$\boxed{216}.$$

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View written solutionFree

Correct answer: D

Since $B$ is the matrix of cofactors of $A$ , we use the standard identity

$A\,(\text{adj }A)= (\det A)I.$

The adjugate matrix is the transpose of the cofactor matrix. Hence, whether $B$ is given as cofactor matrix or its transpose, we have

$\det(AB)=\det(A)\det(B).$

Also, for a $3\times 3$ matrix,

$\det(\text{cofactor matrix})=\det(\text{adj }A)= (\det A)^{2}.$

Therefore,

$\det(AB)=\det(A)\det(B)=\det(A)\cdot (\det A)^2=(\det A)^3.$

So we only need $\det A$ .

Compute $\det A$ using the first row:

\beta & \alpha & 3\\ \alpha & \alpha & \beta\\ -\beta & \alpha & 2\alpha \end{bmatrix}.$$ Thus,

Now evaluate each minor:

\begin{vmatrix} \alpha & \beta\ \alpha & 2\alpha \end{vmatrix} =2\alpha^2-\alpha\beta=\alpha(2\alpha-\beta),

\begin{vmatrix} \alpha & \beta\ -\beta & 2\alpha \end{vmatrix} =2\alpha^2+\beta^2,

\begin{vmatrix} \alpha & \alpha\ -\beta & \alpha \end{vmatrix} =\alpha^2+\alpha\beta=\alpha(\alpha+\beta).

Hence,

\det A=eta\alpha(2\alpha-\beta)-\alpha(2\alpha^2+\beta^2)+3\alpha(\alpha+\beta).

This looks messy, so instead we use the given cofactor matrix $B$ to determine $\alpha,\beta$. --- 3. Since $B$ is the cofactor matrix of $A$, compare entries of $B$ with actual cofactors of $A$. Given $$B=\begin{bmatrix} 3\alpha & -9 & 3\alpha\\ -\alpha & 7 & -2\alpha\\ -2\alpha & 5 & -2\beta \end{bmatrix}.$$ Now compute some cofactors of $A$. ### Cofactor $C_{12}$

C_{12}=(-1)^{1+2}\begin{vmatrix} \alpha & \beta\ -\beta & 2\alpha \end{vmatrix} =-(2\alpha^2+\beta^2).

From $B$, this equals $-9$. So, $$2\alpha^2+\beta^2=9. \qquad (1)$$ ### Cofactor $C_{22}$

C_{22}=\begin{vmatrix} \beta & 3\ -\beta & 2\alpha \end{vmatrix} =2\alpha\beta+3\beta=\beta(2\alpha+3).

From $B$, this equals $7$. So, $$\beta(2\alpha+3)=7. \qquad (2)$$ ### Cofactor $C_{32}$

C_{32}=(-1)^{3+2}\begin{vmatrix} \beta & 3\ \alpha & \beta \end{vmatrix} =-(\beta^2-3\alpha)=3\alpha-\beta^2.

From $B$, this equals $5$. Thus, $$3\alpha-\beta^2=5. \qquad (3)$$ --- 4. Solve for $\alpha,\beta$. From (3), $$\beta^2=3\alpha-5.$$ Substitute into (1): $$2\alpha^2+(3\alpha-5)=9$$ $$2\alpha^2+3\alpha-14=0.$$ Factor: $$2\alpha^2+3\alpha-14=(2\alpha+7)(\alpha-2)=0.$$ So, $$\alpha=2 \quad \text{or} \quad \alpha=-\frac72.$$ Now use (2). If $\alpha=2$, then from (3): $$\beta^2=6-5=1 \Rightarrow \beta=\pm1.$$ Using (2): $$\beta(4+3)=7 \Rightarrow 7\beta=7 \Rightarrow \beta=1.$$ So one solution is $$\alpha=2,\quad \beta=1.$$ If $\alpha=-\frac72$, then $$\beta^2=3\left(-\frac72\right)-5=-\frac{31}{2},$$ impossible over reals. Hence rejected. Therefore, $$\alpha=2,\quad \beta=1.$$ --- 5. Now substitute into $A$: $$A=\begin{bmatrix} 1 & 2 & 3\\ 2 & 2 & 1\\ -1 & 2 & 4 \end{bmatrix}.$$ Compute its determinant:

\det A = 1\begin{vmatrix}2&1\2&4\end{vmatrix} -2\begin{vmatrix}2&1\-1&4\end{vmatrix} +3\begin{vmatrix}2&2\-1&2\end{vmatrix}.

That is,

\det A = 1(8-2)-2(8+1)+3(4+2) =6-18+18=6.

Thus, $$\det A=6.$$ --- 6. Therefore, $$\det(AB)=(\det A)^3=6^3=216.$$ So the correct option is $$\boxed{216}.$$

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