Question

Let $A = \{a_{ij}\}$ be a $3 \times 3$ matrix, where, $a_{ij} = \begin{cases} (-1)^{j-i} \quad \text{if } i < j \\ 2 \qquad \quad \text{if } i = j \\ (-1)^{i+j} \quad \text{if } i > j \end{cases}$ and $B$ be a matrix of order $3 \times 3$ and $\det(B)=2$.

Match each entry in List-I to the correct entries in List-II.

	List-I		List-II
(P)	If $\det(3 \operatorname{adj}(2A^{-1})) = 2^m 3^n$, then $m+n =$	1	21
(Q)	If $	\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2A))	= 2^p 3^q$, then$p+q =$
(R)	If $\det(\det(B) \operatorname{adj}(5 \operatorname{adj}(B^3))) = 2^r \cdot 5^s$, then $r+s =$	3	40
(S)	If $	3 \operatorname{adj}(	3B
		5	5

• A. If $\det(3 \operatorname{adj}(2A^{-1})) = 2^m 3^n$, then $m+n =$
• B. If $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2A))| = 2^p 3^q$, then $p+q =$
• C. If $\det(\det(B) \operatorname{adj}(5 \operatorname{adj}(B^3))) = 2^r \cdot 5^s$, then $r+s =$
• D. If $|3 \operatorname{adj}(|3B|B^2)| = 2^u \times 3^v$, then $u+v =$

Answer 1

P -> 5, Q -> 3, R -> 1, S -> 2

Answer 2

First, let's construct the matrix $A$ and calculate its determinant. The elements of the $3 \times 3$ matrix $A = \{a_{ij}\}$ are given by: $a_{ij} = \begin{cases} (-1)^{j-i} \quad \text{if } i < j \\ 2 \qquad \quad \text{if } i = j \\ (-1)^{i+j} \quad \text{if } i > j \end{cases}$

So, the matrix $A$ is: $A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} = \begin{pmatrix} 2 & (-1)^{2-1} & (-1)^{3-1} \\ (-1)^{2+1} & 2 & (-1)^{3-2} \\ (-1)^{3+1} & (-1)^{3+2} & 2 \end{pmatrix} = \begin{pmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix}$

Now, calculate $\det(A)$: $\det(A) = 2(2 \cdot 2 - (-1)(-1)) - (-1)((-1) \cdot 2 - (-1) \cdot 1) + 1((-1)(-1) - 2 \cdot 1)$ $\det(A) = 2(4 - 1) + 1(-2 + 1) + 1(1 - 2)$ $\det(A) = 2(3) + 1(-1) + 1(-1)$ $\det(A) = 6 - 1 - 1 = 4$.

We are given that $B$ is a $3 \times 3$ matrix and $\det(B)=2$. We will use the following properties for an $n \times n$ matrix $M$, where $n=3$:

1. $\det(kM) = k^n \det(M) = k^3 \det(M)$
2. $\det(\operatorname{adj}(M)) = (\det(M))^{n-1} = (\det(M))^2$
3. $\operatorname{adj}(\operatorname{adj}(M)) = (\det(M))^{n-2} M = \det(M) M$
4. $\det(M^{-1}) = (\det(M))^{-1}$
5. $\det(M^k) = (\det(M))^k$

**(P) If $\det(3 \operatorname{adj}(2A^{-1})) = 2^m 3^n$, then $m+n =$} Let $M = 2A^{-1}$. $\det(3 \operatorname{adj}(M)) = 3^3 \det(\operatorname{adj}(M))$ (using property 1) $= 3^3 (\det(M))^2$ (using property 2) Now, calculate $\det(M) = \det(2A^{-1})$: $\det(2A^{-1}) = 2^3 \det(A^{-1})$ (using property 1) $= 2^3 (\det(A))^{-1}$ (using property 4) $= 8 \cdot (4)^{-1} = 8 \cdot \frac{1}{4} = 2$. Substitute this back: $\det(3 \operatorname{adj}(2A^{-1})) = 3^3 (2)^2 = 27 \cdot 4 = 108$. To match $2^m 3^n$: $108 = 4 \cdot 27 = 2^2 \cdot 3^3$. So, $m=2$ and $n=3$. $m+n = 2+3 = 5$. (P) matches with 5.

**(Q) If $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2A))| = 2^p 3^q$, then $p+q =$} Let $M = 2A$. We need to calculate $\det(\operatorname{adj}(\operatorname{adj}(\operatorname{adj} M)))$. Using property 2: $\det(\operatorname{adj}(X)) = (\det(X))^2$. $\det(\operatorname{adj}(\operatorname{adj}(\operatorname{adj} M))) = (\det(\operatorname{adj}(\operatorname{adj} M)))^2$. Now, let's find $\det(\operatorname{adj}(\operatorname{adj} M))$. Using property 3: $\operatorname{adj}(\operatorname{adj} M) = \det(M) M$. So, $\det(\operatorname{adj}(\operatorname{adj} M)) = \det(\det(M) M)$. Using property 1: $\det(\det(M) M) = (\det(M))^3 \det(M) = (\det(M))^4$. Substitute this back: $\det(\operatorname{adj}(\operatorname{adj}(\operatorname{adj} M))) = ((\det(M))^4)^2 = (\det(M))^8$. Now, calculate $\det(M) = \det(2A)$: $\det(2A) = 2^3 \det(A) = 8 \cdot 4 = 32$. So, $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2A))| = (32)^8 = (2^5)^8 = 2^{40}$. To match $2^p 3^q$: $2^{40} = 2^p \cdot 3^q$. So, $p=40$ and $q=0$. $p+q = 40+0 = 40$. (Q) matches with 3.

**(R) If $\det(\det(B) \operatorname{adj}(5 \operatorname{adj}(B^3))) = 2^r \cdot 5^s$, then $r+s =$} Given $\det(B)=2$. Let $k = \det(B) = 2$. We need to calculate $\det(k \operatorname{adj}(5 \operatorname{adj}(B^3)))$. Using property 1: $\det(k \operatorname{adj}(5 \operatorname{adj}(B^3))) = k^3 \det(\operatorname{adj}(5 \operatorname{adj}(B^3)))$. Substitute $k=2$: $2^3 \det(\operatorname{adj}(5 \operatorname{adj}(B^3)))$. Using property 2: $\det(\operatorname{adj}(X)) = (\det(X))^2$. Let $X = 5 \operatorname{adj}(B^3)$. So, $2^3 (\det(5 \operatorname{adj}(B^3)))^2$. Now, calculate $\det(5 \operatorname{adj}(B^3))$. Let $Y = B^3$. $\det(5 \operatorname{adj}(Y)) = 5^3 \det(\operatorname{adj}(Y))$ (using property 1) $= 5^3 (\det(Y))^2$ (using property 2) $= 5^3 (\det(B^3))^2$. Now, calculate $\det(B^3)$: $\det(B^3) = (\det(B))^3$ (using property 5) $= (2)^3 = 8$. Substitute this back: $\det(5 \operatorname{adj}(B^3)) = 5^3 (8)^2 = 5^3 (2^3)^2 = 5^3 \cdot 2^6$. Substitute this back into the main expression: $2^3 (5^3 \cdot 2^6)^2 = 2^3 (5^{3 \cdot 2} \cdot 2^{6 \cdot 2}) = 2^3 (5^6 \cdot 2^{12}) = 2^{3+12} \cdot 5^6 = 2^{15} \cdot 5^6$. To match $2^r \cdot 5^s$: $r=15$ and $s=6$. $r+s = 15+6 = 21$. (R) matches with 1.

**(S) If $|3 \operatorname{adj}(|3B|B^2)| = 2^u \times 3^v$, then $u+v =$} Given $\det(B)=2$. First, evaluate $|3B| = \det(3B)$: $\det(3B) = 3^3 \det(B) = 27 \cdot 2 = 54$. So, we need to calculate $|3 \operatorname{adj}(54 B^2)|$. Let $k = 54$. $|3 \operatorname{adj}(k B^2)| = 3^3 \det(\operatorname{adj}(k B^2))$ (using property 1) $= 3^3 (\det(k B^2))^2$ (using property 2) Now, calculate $\det(k B^2)$: $\det(k B^2) = k^3 \det(B^2)$ (using property 1) $= k^3 (\det(B))^2$ (using property 5) Substitute $k=54$ and $\det(B)=2$: $\det(k B^2) = (54)^3 (2)^2 = (2 \cdot 3^3)^3 \cdot 2^2 = 2^3 \cdot (3^3)^3 \cdot 2^2 = 2^3 \cdot 3^9 \cdot 2^2 = 2^{3+2} \cdot 3^9 = 2^5 \cdot 3^9$. Substitute this back into the main expression: $3^3 (2^5 \cdot 3^9)^2 = 3^3 (2^{5 \cdot 2} \cdot 3^{9 \cdot 2}) = 3^3 (2^{10} \cdot 3^{18}) = 2^{10} \cdot 3^{3+18} = 2^{10} \cdot 3^{21}$. To match $2^u \times 3^v$: $u=10$ and $v=21$. $u+v = 10+21 = 31$. (S) matches with 2.

Matching List-I to List-II: (P) $\to$ 5 (Q) $\to$ 3 (R) $\to$ 1 (S) $\to$ 2