Which of the following is not correct ?

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Which of the following is not correct ?

A. |A| = |AT|, where A = [aij]3×3

B. |kA| = k3 |A|, where A = [aij]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}$ + $\begin{vmatrix} b &d \\[0.3em] f & h \\[0.3em] \end{vmatrix}$

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We are given that,

A = [aij]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| = |AT|, where A = [aij]3×3

Option (B) is correct.

|kA| = k3|A|, where A = [aij]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 = [aij]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, AT = -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,