If A = {bmatrix} a & b \\ c & -a {bmatrix} is such that A² = I, then . . . . . . .

Correct Answer is : 1−a2−bc=01 - a² - bc = 01−a2−bc=0

Correct Answer is : 1−a2−bc=01 - a² - bc = 01−a2−bc=0

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GUJCET 2026 Math Paper

Question : 11 of 40

Marks: +1, -0

A = \begin{bmatrix} a & b \\ c & -a \end{bmatrix}

A^2 = I

, then . . . . . . .

Solution:

Concept:

For a matrix,

A^2=I

means multiplying

A

by itself gives the identity matrix.

Explanation:

Given

A=\begin{bmatrix} a & b \\ c & -a \end{bmatrix}

Compute

A^2

A^2=\begin{bmatrix} a & b \\ c & -a \end{bmatrix}\begin{bmatrix} a & b \\ c & -a \end{bmatrix}

=\begin{bmatrix} a^2+bc & ab-ab \\ ac-ac & bc+a^2 \end{bmatrix}

=\begin{bmatrix} a^2+bc & 0 \\ 0 & a^2+bc \end{bmatrix}

Since

A^2=I

, compare the diagonal elements:

a^2+bc=1

Rearranging gives:

1-a^2-bc=0

So the correct option is C.

Answer:

1-a^2-bc=0

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