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(D) Let the matrix be $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$. The determinant is $|A| = ad - bc$.The total number of ways to choose $4$ nu

Vedclass · Accepted Answer

$\frac{43}{162}$

Vedclass · Answer

(D) Let the matrix be $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$. The determinant is $|A| = ad - bc$.The total number of ways to choose $4$ numbers from $6$ faces of a die is $6^4 = 1296$.We require all entries $a, b, c, d$ to be distinct. The number of ways to choose $4$ distinct numbers from ${1, 2, 3, 4, 5, 6}$ and arrange them in the matrix is $^6P_4 = 6 	imes 5 	imes 4 	imes 3 = 360$.For the matrix to be nonsingular,$|A| 
eq 0$,which means $ad 
eq bc$.We count the cases where $ad = bc$ with distinct $a, b, c, d$:$1$. $6 	imes 1 = 2 	imes 3$: The sets are ${1, 2, 3, 6}$. Possible arrangements $(a, b, c, d)$ such that $ad=bc$ are $(6, 2, 3, 1), (6, 3, 2, 1), (1, 2, 3, 6), (1, 3, 2, 6), (2, 6, 1, 3), (3, 6, 1, 2), (2, 1, 6, 3), (3, 1, 6, 2)$. Total $8$ cases.$2$. $6 	imes 2 = 3 	imes 4$: The sets are ${2, 3, 4, 6}$. Possible arrangements $(a, b, c, d)$ such that $ad=bc$ are $(6, 3, 4, 2), (6, 4, 3, 2), (2, 3, 4, 6), (2, 4, 3, 6), (3, 6, 2, 4), (4, 6, 2, 3), (3, 2, 6, 4), (4, 2, 6, 3)$. Total $8$ cases.Total cases where $ad = bc$ with distinct entries is $8 + 8 = 16$.Number of favorable cases = (Total ways with distinct entries) - (Cases where $ad = bc$) = $360 - 16 = 344$.Required probability = $\frac{344}{1296} = \frac{43}{162}$.

List-$I$	List-$II$
$A$. $\|k^{-1} A^{-1}\|$	$I$. $BA^k + A^kB$
$B$. $\|\text{Adj}(A^{-1})\|$	$II$. $\frac{B\text{Adj}(B)}{\|B\|}$
$C$. $BAB^{-1} = I \Rightarrow BA^kB^{-1} =$	$III$. $\frac{1}{\|B\|^3\|A\|}$
$D$. $\text{Adj}(\text{Adj}(A^{-1})) =$	$IV$. $\frac{1}{\|A\|}(A^{-1})$
	$V$. $\frac{1}{\|A\|^2}$

Four dice are thrown simultaneously and the numbers shown on these dice are recorded in $2 \times 2$ matrices. The probability that such formed matrices have all different entries and are nonsingular,is:

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