Let A and B be two 3 $\times$ 3 real matrices such that (A2 $-$ B2) is invertible matrix. If A5 = B5 and A3B2 = A2B3, then the value of the determinant of the matrix A3 + B3 is equal to :
Solution
C = A<sup>2</sup> $-$ B<sup>2</sup>; | C | $\ne$ 0<br><br>A<sup>2</sup> = B<sup>5</sup> and A<sup>3</sup>B<sup>2</sup> = A<sup>2</sup>B<sup>2</sup><br><br>Now, A<sup>5</sup> $-$ A<sup>3</sup>B<sup>2</sup> = B<sup>5</sup> $-$ A<sup>2</sup>B<sup>3</sup><br><br>$\Rightarrow$ A<sup>3</sup> (A<sup>2</sup> $-$ B<sup>2</sup>) + B<sup>3</sup> (A<sup>2</sup> $-$ B<sup>2</sup>) = 0<br><br>$\Rightarrow$ (A<sup>3</sup> + B<sup>3</sup>(A<sup>2</sup> $-$ B<sup>2</sup>) = 0<br><br>$\Rightarrow$ A<sup>3</sup> + B<sup>3</sup> = 0 $\left( \because{\left| {{A^2} - {B^2} \ne 0} \right|} \right)$
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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