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Consider the following in respect of matrices $\mathrm{A}$ and $\mathrm{B}$ of same order: $\quad[2018-I I]$
$1.$ $\mathrm{A}^{2}-\mathrm{B}^{2}=(\mathrm{A}+\mathrm{B})(\mathrm{A}-\mathrm{B})$
$2.$ $(A-I)(I+A)=0 \Leftrightarrow A^{2}=I$
Where I is the identity matrix and $\mathrm{O}$ is the null matrix.
Which of the above is/are correct?
Options:
$1.$ $\mathrm{A}^{2}-\mathrm{B}^{2}=(\mathrm{A}+\mathrm{B})(\mathrm{A}-\mathrm{B})$
$2.$ $(A-I)(I+A)=0 \Leftrightarrow A^{2}=I$
Where I is the identity matrix and $\mathrm{O}$ is the null matrix.
Which of the above is/are correct?
Solution:
2885 Upvotes
Verified Answer
The correct answer is:
2 only
Matrix product is commutative if both are diagonal matrices of same order.
$\Rightarrow \quad A^{2}-B^{2}=(A+B)(A-B)$ is not true.
Next, $(\mathrm{A}-\mathrm{I})(\mathrm{A}+\mathrm{I})=0$
$\Rightarrow \quad \mathrm{A}^{2}+\mathrm{AI}-\mathrm{IA}-\mathrm{I}^{2}=0(\because \mathrm{AI}=\mathrm{IA})$
$\Rightarrow \mathrm{A}^{2}=\mathrm{I}$ is correct.
$\Rightarrow \quad A^{2}-B^{2}=(A+B)(A-B)$ is not true.
Next, $(\mathrm{A}-\mathrm{I})(\mathrm{A}+\mathrm{I})=0$
$\Rightarrow \quad \mathrm{A}^{2}+\mathrm{AI}-\mathrm{IA}-\mathrm{I}^{2}=0(\because \mathrm{AI}=\mathrm{IA})$
$\Rightarrow \mathrm{A}^{2}=\mathrm{I}$ is correct.
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