Matrices Test - 16

We have matrices A and B of same order.

Let P=(AB' - BA')

Then, P' = (AB' - BA')' = (AB')' - (BA')'

= (B')'(A)' - (A')'B' = BA' - AB'

= -(AB' - BA') = - P

Hence, (AB' - BA') is a skew-symmetric matrix.

We have, \(A^2 = I\)

\(\therefore\) \((A - I)^3 + (A + I)^3 - 7A\)

= [(A - I) + (A + I){\((A - I)^2 + (A + I)^2 \) - (A - I)(A + I)}] - 7A

\([\because a^3 + b^3 = (a + b)(a^2 + b^2 - ab)]\)

= \([(2A)\) {\(A^2 + I^2 - 2AI + A^2 + I^2 +2 AI -(A^2 - I^2)\)}] - 7A

= 2A[ I + \(I^2 + I + I^2 - A^2 + I^2] - 7A [\because A^2 = AI]\)

= 2A[5I - I] - 7A

= 8AI - 7AI [\(\because\) A = AI]

= AI = A

For any two matrices A and B, we may have AB = BA = I, AB \(\neq\) BA and AB = 0 but it is not always true.

Given that, \(\begin{bmatrix} 1 &-3 \\[0.3em] 2 &4 \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix} 1 &-1 \\[0.3em] 0 &1 \\[0.3em] \end{bmatrix}\) \(\begin{bmatrix} 3 & 1 \\[0.3em] 2 &4 \\[0.3em] \end{bmatrix}\)

On using \(C_2 \rightarrow C_2 - 2C_1\), \(\begin{bmatrix} 1 &-5 \\[0.3em] 2 &0 \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix} 1 &-1 \\[0.3em] 0 &1 \\[0.3em] \end{bmatrix}\)\(\begin{bmatrix} 3 &-5 \\[0.3em] 2 &0 \\[0.3em] \end{bmatrix}\)

Since, on using elementary column operation on X = AB, we apply these operations simultaneously on X and on the second matrix B of the product AB on RHS.

We have, \(\begin{bmatrix} 4 & 2 \\[0.3em] 3 & 3 \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix} 1 & 2 \\[0.3em] 0 & 3 \\[0.3em] \end{bmatrix}\) \(\begin{bmatrix} 2 & 0 \\[0.3em] 1 & 1 \\[0.3em] \end{bmatrix}\)

Using elementary row operation \(R_1 \) → \(R_1 - 3R_2\),

\(\begin{bmatrix} -5 & -7 \\[0.3em] 3 & 3 \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix} 1 & -7 \\[0.3em] 0 & 3 \\[0.3em] \end{bmatrix}\) \(\begin{bmatrix} 2 & 0 \\[0.3em] 1 & 1 \\[0.3em] \end{bmatrix}\)

Since, on using elementary row operation on X = AB, we apply these operation simultaneously on X and on the first matrix A of the product AB on RHS.

\(A^2 = A.A = \) \(\begin{bmatrix} 1 &0 & 0 \\[0.3em] 0 & 1 & 0 \\[0.3em] a &b &-1 \end{bmatrix}\) \(\begin{bmatrix} 1 &0 & 0 \\[0.3em] 0 & 1 & 0 \\[0.3em] a &b &-1 \end{bmatrix}\)

= \(\begin{bmatrix} 1 &0 & 0 \\[0.3em] 0 & 1 & 0 \\[0.3em] 0 &0 & 1 \end{bmatrix}\) = I

\(A^2\) = \(\begin{bmatrix} 2& 2 \\[0.3em] a & b \\[0.3em] \end{bmatrix}\) \(\begin{bmatrix} 2& 2 \\[0.3em] a & b \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix} 4 + 2a & 4 + 2b \\[0.3em] 2a + ab & 2a + b^2 \\[0.3em] \end{bmatrix}\) = 0

= \(\begin{bmatrix} 0& 0 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}\)

⇒ 4 + 2a = 0, 4 + 2b = 0, 2a + ab = 0, 2a + \(b^2 \) = 0 must be consistent.

⇒ a = -2, b = -2

A = \(\begin{bmatrix} i & 0 \\[0.3em] 0 & i \\[0.3em] \end{bmatrix}\); \(A^2 = A.A = \)  \(\begin{bmatrix} i & 0 \\[0.3em] 0 & i \\[0.3em] \end{bmatrix}\) \(\begin{bmatrix} i & 0 \\[0.3em] 0 & i \\[0.3em] \end{bmatrix}\)

\(A^2\) = \(\begin{bmatrix} -1 & 0 \\[0.3em] 0 & -1 \\[0.3em] \end{bmatrix}\) \([\because i^2 = -1]\)

A = \(\begin{bmatrix} x & 1 \\[0.3em] 1 & 0 \\[0.3em] \end{bmatrix}\), \(\therefore \) \(A^2 = I\) 

⇒ \(\begin{bmatrix} x & 1 \\[0.3em] 1 & 0 \\[0.3em] \end{bmatrix}\) \(\begin{bmatrix} x & 1 \\[0.3em] 1 & 0 \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}\)

⇒ \(\begin{bmatrix} x^2 + 1 & x \\[0.3em] x & 1 \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}\)

⇒ \(x^2 + 1 = 1\) 

⇒ x = 0

It is a property of matrix multiplication.