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Mathematics Question on Matrices

For I=[10 01]I = \begin{bmatrix} 1 & 0 \\\ 0 & 1 \end{bmatrix}, if A=[ab c−a]A = \begin{bmatrix} a & b \\\ c & -a \end{bmatrix} be such that A2=IA^2 = I, then:

A

1+a2+bc=01 + a^2 + bc = 0

B

1−a2−bc=01 - a^2 - bc = 0

C

1−a2+bc=01 - a^2 + bc = 0

D

1+a2−bc=01 + a^2 - bc = 0

Answer

1−a2−bc=01 - a^2 - bc = 0

Explanation

Solution

The matrix A is given as:

A=[ab c−a].A = \begin{bmatrix} a & b \\\ c & -a \end{bmatrix}.

To compute A2A^2:

A2=[ab c−a]⋅[ab c−a].A^2 = \begin{bmatrix} a & b \\\ c & -a \end{bmatrix} \cdot \begin{bmatrix} a & b \\\ c & -a \end{bmatrix}.

A2=[a2+bcab−ab ac−acbc+(−a)2]=[a2+bc0 0bc+a2].A^2 = \begin{bmatrix} a^2 + bc & ab - ab \\\ ac - ac & bc + (-a)^2 \end{bmatrix} = \begin{bmatrix} a^2 + bc & 0 \\\ 0 & bc + a^2 \end{bmatrix}.

Since A2=IA^2 = I, we have:

[a2+bc0 0bc+a2]=[10 01].\begin{bmatrix} a^2 + bc & 0 \\\ 0 & bc + a^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\\ 0 & 1 \end{bmatrix}.

Equating the diagonal elements:

a2+bc=1.a^2 + bc = 1.

Rewriting this equation:

1−a2−bc=0.1 - a^2 - bc = 0.

Thus, the correct answer is Option (B).