If A is a square matrix of order n then, ( adj adj adj … adj A ) ( n - 1 ) times ( adj adj adj … adj A n times is equal to:

Question

If $A$ is a square matrix of order $n$ then, $\underset{(n - 1) times}{\underset{⏟}{(adj adj adj \dots adj A)} \cdot} \underset{n times}{\underset{⏟}{(adj adj adj \dots adj A)}}$ is equal to:

Smart Achievers · Accepted Answer

(4)

$A \cdot Adj A =$ $| A |$ $I_{n}$

$(Adj A) (AdjAdj A) =$ $| Adj A |$ $I_{n} =$ $| A |^{n - 1}$ $I_{n}$

$(AdjAdj A) (AdjAdjAdj A) =$ $| A |^{{(n - 1)}^{2}}$ $I_{n}$

$\Rightarrow$ $(AdjAdj \dots A) (AdjAdjAdj \dots A) =$ $| A |^{{(n - 1)}^{(n - 1)}}$ $I_{n}$

Solution

Q.
If $A$ is a square matrix of order $n$ then, $\underset{(n - 1) times}{\underset{⏟}{(adj adj adj \dots adj A)} \cdot} \underset{n times}{\underset{⏟}{(adj adj adj \dots adj A)}}$ is equal to:

1 $| A^{n - 1} |^{{(n - 1)}^{2}}$

2 $| A^{- 1} |^{{(n - 1)}^{(n - 1)}}$

3 $\frac{1}{n} | A |^{{(n - 1)}^{(n - 1)}} \cdot I_{n \times n}$

4 $| A |^{{(n - 1)}^{(n - 1)}} \cdot I_{n \times n}$

Ans.
(4)

$A \cdot Adj A =$ $| A |$ $I_{n}$

$(Adj A) (AdjAdj A) =$ $| Adj A |$ $I_{n} =$ $| A |^{n - 1}$ $I_{n}$

$(AdjAdj A) (AdjAdjAdj A) =$ $| A |^{{(n - 1)}^{2}}$ $I_{n}$

$\Rightarrow$ $(AdjAdj \dots A) (AdjAdjAdj \dots A) =$ $| A |^{{(n - 1)}^{(n - 1)}}$ $I_{n}$

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Solution

Q. If A is a square matrix of order n then, (adjadjadj…adjA)⏟·(n-1) times(adjadjadj…adjA)⏟n times is equal to:

1 |A n-1|(n-1)2

2 |A-1|(n-1)(n-1)

3 1n|A|(n-1)(n-1)·In×n

4 |A|(n-1)(n-1)·In×n

Ans. (4) A·AdjA=|A|In (AdjA)(AdjAdjA)=|AdjA|In=|A|n-1In (AdjAdjA)(AdjAdjAdjA)=|A|(n-1)2In ⇒(AdjAdj…A)(AdjAdjAdj…A)=|A|(n-1)(n-1)In