#### Explain Solution R.D.Sharma Class 12 Chapter 4 Algebra of Matrices Exercise MCQs Question 36 Maths Textbook Solution.

Answer: The correct option is (A).

Given: $A^{2}=I$is a square matrix

Hint: Using $(a-b)^{3} \text { and }(a+b)^{3}$

Solution:

\begin{aligned} &(A-I)^{3}+(A+I)^{3}-7 A=A^{3}-I^{3}-3 A^{2} I+3 A I^{2}+A^{3}+I^{3}+3 A^{2} I+3 A I^{2}-7 A \\ &{\left[\begin{array}{l} (a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{2} \\ (a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2} \end{array}\right]} \\ &=A^{2} \cdot A-I-3 A^{2} \cdot I+3 A I+A^{2} \cdot A+I+3 A^{2} \cdot I+3 A I-7 A \quad\left[\because I^{2}=I\right. \\ &=I \cdot A-I-3 I \cdot I+3 A+I A+I+3 I \cdot I+3 A-7 A \\ &=A-I-3 I+3 A+A+I+3 I+3 A-7 A \\ &=8 A-7 A=A \end{aligned}

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Answer: The correct option is (A).

Given: $A^{2}=I$is a square matrix

Hint: Using $\left ( a-b \right )^{3}$and $\left ( a+b \right )^{3}$

Solution:

\begin{aligned} &(A-I)^{3}+(A+I)^{3}-7 A=A^{3}-I^{3}-3 A^{2} I+3 A I^{2}+A^{3}+I^{3}+3 A^{2} I+3 A I^{2}-7 A \\ &{\left[\begin{array}{l} (a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{2} \\ (a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2} \end{array}\right]} \\ &=A^{2} \cdot A-I-3 A^{2} \cdot I+3 A I+A^{2} \cdot A+I+3 A^{2} \cdot I+3 A I-7 A \quad\left[\because I^{2}=I\right. \\ &=I \cdot A-I-3 I \cdot I+3 A+I A+I+3 I \cdot I+3 A-7 A \\ &=A-I-3 I+3 A+A+I+3 I+3 A-7 A \\ &=8 A-7 A=A \end{aligned}