Verify $| a d j A | = {| A |}^{n - 1}$

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Solution

$\begin{matrix} | a d j A | = {| A |}^{n = 1} \Rightarrow A (a d j A) = | A | . I \Rightarrow | A a d j A | = | | A | . I | \Rightarrow | A | . | a d j A | = {| A |}^{n} & [∵ | A | . I n = {| A |}^{n} . | I n | = {| A |}^{n}] | a d j A | = {| A |}^{n - 1} \end{matrix}$

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Q. Verify A ( adj A ) = ( adj A ) A = I .

Let $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 1 - 2 & 3 & 1 1 & 1 & 5 \end{matrix} ⎤ ⎥ ⎦$ , verify that $(a) [a d j A]^{- 1} = a d j (A^{- 1})$

$(b) (A^{- 1})^{- 1} = A$

Let verify that

(i)

(ii)

Q. Paragraph for below question

नीचे दिए गए प्रश्न के लिए अनुच्छेद

If A is a square matrix of order n, then A(adjA) = I_n and |adjA| = |A|ⁿ^–1,

adj (k A) = k^{n - 1} (adj A)

and

| k A | = k^{n} \cdot | A |

यदि कोटि n वाला एक वर्ग आव्यूह A है, तब A(adjA) = I_n तथा |adjA| = |A|ⁿ^–1

adj (k A) = k^{n - 1} (adj A)

तथा

| k A | = k^{n} \cdot | A |

Q. Given

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 3 & 0 2 & 3 & 4 3 & 3 & 4 \end{matrix} ⎤ ⎥ ⎦,

then |adjA| is equal to

प्रश्न - दिया गया है

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 3 & 0 2 & 3 & 4 3 & 3 & 4 \end{matrix} ⎤ ⎥ ⎦,

तब |adjA| का मान बराबर है

Q. Let

A

be a

2 \times 2

matrix
Statement 1:

a d j (a d j A

)

= A

Statement 2:

| a d j A |

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

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Verify |adj A|=|A|n−1

|adjA|=|A|n=1⇒A(adjA)=|A|.I⇒|AadjA|=||A|.I|⇒|A|.|adjA|=|A|n[∵|A|.In=|A|n.|In|=|A|n]|adjA|=|A|n−1

Verify $| a d j A | = {| A |}^{n - 1}$

$\begin{matrix} | a d j A | = {| A |}^{n = 1} \Rightarrow A (a d j A) = | A | . I \Rightarrow | A a d j A | = | | A | . I | \Rightarrow | A | . | a d j A | = {| A |}^{n} & [∵ | A | . I n = {| A |}^{n} . | I n | = {| A |}^{n}] | a d j A | = {| A |}^{n - 1} \end{matrix}$