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Which of the following statements are correct?

(A) If A is a square matrix, then ∣A2∣=∣A∣2|A^2| = |A|^2.

(B) If A and B are square matrices of the same order, then det (AB) = det (A) + det (B).

(C) If A is a square matrix of order 3 and ∣A∣=2|A|=2, then the value of ∣−3A∣|-3A| is 54.

(D) If the matrix [5−xx−135]\begin{bmatrix} 5 -x & x -1 \\ 3 &5 \end{bmatrix} is singular, then the value of x is 7/2.

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 3

Let's check each statement one by one.


Statement (A): If AA is a square matrix, then ∣A2∣=∣A∣2|A^2| = |A|^2

We know that for any two square matrices of the same order, det⁡(AB)=det⁡(A)⋅det⁡(B)\det(AB) = \det(A) \cdot \det(B)

So,

∣A2∣=∣A⋅A∣=∣A∣⋅∣A∣=∣A∣2|A^2| = |A \cdot A| = |A| \cdot |A| = |A|^2

✅ Statement (A) is correct.


Statement (B): det⁡(AB)=det⁡(A)+det⁡(B)\det(AB) = \det(A) + \det(B)

This is wrong. The correct property is:

det⁡(AB)=det⁡(A)⋅det⁡(B)\det(AB) = \det(A) \cdot \det(B)

It is multiplication, not addition.

❌ Statement (B) is incorrect.


Statement (C): If AA is a square matrix of order 3 and ∣A∣=2|A| = 2, then ∣−3A∣=54|-3A| = 54

When you multiply a matrix of order nn by a scalar kk, each row gets multiplied by kk. So the determinant gets multiplied by knk^n:

∣kA∣=kn⋅∣A∣|kA| = k^n \cdot |A|

Here, k=−3k = -3, n=3n = 3, ∣A∣=2|A| = 2:

∣−3A∣=(−3)3⋅∣A∣|-3A| = (-3)^3 \cdot |A|

=−27×2= -27 \times 2

=−54= -54

The statement claims the value is 5454 (positive), but the actual value is −54-54.

❌ Statement (C) is incorrect.


Statement (D): If the matrix [5−xx−135]\begin{bmatrix} 5-x & x-1 \\ 3 & 5 \end{bmatrix} is singular, then x=72x = \dfrac{7}{2}

A singular matrix has its determinant equal to zero.

For a 2×22 \times 2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}, the determinant =ad−bc= ad - bc

So,

(5−x)(5)−(x−1)(3)=0(5 - x)(5) - (x - 1)(3) = 0

25−5x−3x+3=025 - 5x - 3x + 3 = 0

28−8x=028 - 8x = 0

x=288=72x = \dfrac{28}{8} = \dfrac{7}{2}

✅ Statement (D) is correct.


Statements (A) and (D) are correct.

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