Question

Which of the following is not correct ? 

A. |A| = |A^T|, where A = [a_ij]_3×3 

B. |kA| = k³ |A|, where A = [a_ij]_3×3 

C. If A is a skew-symmetric matrix of odd order, then |A| = 0

D. \(\begin{vmatrix} a+b & c+d \\[0.3em] e+f & g+h \\[0.3em] \end{vmatrix}\)= \(\begin{vmatrix} a & c \\[0.3em] e & g \\[0.3em] \end{vmatrix}\) + \(\begin{vmatrix} b &d \\[0.3em] f & h \\[0.3em] \end{vmatrix}\)

Answer 1

We are given that, 

A = [a_ij]_3×3 

That is, 

order of matrix A = 3 

Example : 

Let,

The transpose of a matrix is a new matrix whose rows are the columns of the original. 

So,

So, 

We can conclude that, 

|A| = |A^T|, where A = [a_ij]_3×3. 

Option (B) is correct. 

|kA| = k³|A|, where A = [a_ij]_3×3 

Example : 

Let k = 2. 

And,

A = \(\begin{bmatrix} 2 &3& 4 \\[0.3em] 1& 2 &3 \\[0.3em] 3 &2 & 1 \end{bmatrix}\)

Take Left Hand Side of the equation : 

LHS = |kA|

⇒ LHS = 4(4 × 2 – 6 × 4) – 6(2 × 2 – 6 × 6) + 8(2 × 4 – 4 × 6) 

⇒ LHS = 4(8 – 24) – 6(4 – 36) + 8(8 – 24) 

⇒ LHS = 4(-16) – 6(-32) + 8(-16) 

⇒ LHS = - 64 + 192 – 128 

⇒ LHS = 0 

Take Right Hand Side of the equation : 

RHS = k³|A|

⇒ RHS = 8 [2(2 × 1 – 3 × 2) – 3(1 × 1 – 3 × 3) + 4(1 × 2 – 2 × 3)] 

⇒ RHS = 8 [2(2 – 6) – 3(1 – 9) + 4(2 – 6)] 

⇒ RHS = 8 [2(-4) – 3(-8) + 4(-4)] 

⇒ RHS = 8 [-8 + 24 – 16] 

⇒ RHS = 8 × 0 

⇒ RHS = 0 

Since, 

LHS = RHS. 

We can conclude that, 

|kA| = k³|A|, where A = [a_ij]_3×3 

Option (C) is also correct.

If A is a skew-symmetric matrix of odd order, then |A| = 0. 

If the transpose of a matrix is equal to the negative of itself, the matrix is said to be skew symmetric. In other words, A^T = -A. 

Example, 

Let a matrix of odd order 3×3 be,

A = \(\begin{bmatrix} 0 &-6 & 4 \\[0.3em] 6 & 0 & 7 \\[0.3em] -4 & -7 & 0 \end{bmatrix}\)

Take determinant of A.

Thus, 

We can conclude that If A is a skew-symmetric matrix of odd order, then |A| = 0. 

Option (D) is incorrect. 

Let a = 1, b = 3, c = 3, d = -4, 

e = -2, f = 5, g = 0 and h = 2. 

Take Left Hand Side,

⇒ LHS = 4 × 2 – (-1) × 3 

⇒ LHS = 8 + 3 

⇒ LHS = 11 

Take Right Hand Side,

⇒ RHS = (1 × 0 – 3 × (-2)) + (3 × 2 – (-4) × 5) 

⇒ RHS = (0 + 6) + (6 + 20) 

⇒ RHS = 6 + 26 

⇒ RHS = 32 

Since, 

LHS ≠ RHS. 

Then, 

We can conclude that,