Let A be a square matrix of order n, then which of the following are TRUE?

(A) $|adj A| = |A|^{n-1}$

(B) $|A. adj A| = |A|^n$

(C) $A. (adj A) = |A|$

(D) $|KA| = K|A|$

(E) $|A^{-1}| = \frac{1}{|A|}, |A| \neq 0$

Choose the correct answer from the options given below:

For a square matrix A of order n:

The determinant of the adjoint equals the determinant of the original matrix raised to power (n-1):

$|adj A| = |A|^{n-1}$

This is a standard formula in matrix algebra. The adjoint matrix is constructed from cofactors, and its determinant relationship follows from the properties of cofactor expansion.

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The fundamental property states:

$A \cdot adj(A) = |A| \cdot I_n$

Taking determinant of both sides:

$|A \cdot adj(A)| = ||A| \cdot I_n|$

Applying determinant properties:

$|A| \cdot |adj(A)| = |A|^n \cdot |I_n|$

$|A| \cdot |adj(A)| = |A|^n \cdot 1$

$|A| \cdot |adj(A)| = |A|^n$

Using $|adj(A)| = |A|^{n-1}$:

$|A| \cdot |A|^{n-1} = |A|^n$

Therefore, $|A. adj A| = |A|^n$ is true.

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The correct formula is:

$A \cdot adj(A) = |A| \cdot I_n$

The right side is $|A|$ multiplied by the identity matrix, not just the scalar $|A|$. Since $|A|$ is a scalar and $|A| \cdot I_n$ is a matrix, these are not the same.

Therefore, $A. (adj A) = |A|$ is false.

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When multiplying matrix A by scalar K, every entry gets multiplied by K. This is equivalent to multiplying each of the n rows by K. When one row is multiplied by K, the determinant gets multiplied by K. Since there are n rows:

$|KA| = K^n|A|$

Therefore, $|KA| = K|A|$ is false.

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For an invertible matrix:

$A \cdot A^{-1} = I$

Taking determinant of both sides:

$|A \cdot A^{-1}| = |I|$

Applying the property $|AB| = |A| \cdot |B|$:

$|A| \cdot |A^{-1}| = 1$

$|A^{-1}| = \frac{1}{|A|}$

Therefore, $|A^{-1}| = \frac{1}{|A|}, |A| \neq 0$ is true.

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Statements (A), (B), and (E) are true.

Statements (C) and (D) are false.

Answer: (A), (B) and (E) only