If for any square matrix A 2- A- then I - A n is equal to where I is identity matrix A- 2-2 I -2n-2 AB- I -2 n -1 AC- 2n-1 I-AD- 2n-1 I-2n-1 A

The correct option is B I + (2^n 1) AA^2 = A ⇒ A^3 = A ⇒ A^4 = A ⇒ A^n = A (I+A)^n = ^nC_0 I^n + ^nC_1 I^n 1A + ⋯ + ^nC_n A^n = I + [ ^nC_1 + ^nC_2 + ^nC_ ...

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Byju's Answer

Standard XII

Mathematics

Multiplication of Matrices

If for any sq...

Question

If for any square matrix $A^{2}$ = A, then $(I + A)^{n}$ is equal to (where I is identity matrix)

I + (2^{n} - 1) A

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2^{n - 1} I + A

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2^{\frac{n}{2}} I + 2^{\frac{n}{2}} A

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2^{n - 1} I + 2^{n - 1} A

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Solution

The correct option is A $I + (2^{n} - 1) A$
$A^{2} = A \Rightarrow A^{3} = A \Rightarrow A^{4} = A \Rightarrow A^{n} = A (I + A)^{n} =^{n} C_{0} I^{n} +^{n} C_{1} I^{n - 1} A + \dots +^{n} C_{n} A^{n} = I + [^{n} C_{1} +^{n} C_{2} +^{n} C_{3} + \dots^{n} C_{n}] A = I + (2^{n} - 1) A$

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If for any square matrix A2 = A, then (I+A)n is equal to (where I is identity matrix)

The correct option is A I+(2n−1)AA2=A⇒A3=A⇒A4=A⇒An=A(I+A)n=nC0In+nC1In−1A+⋯+nCnAn=I+[nC1+nC2+nC3+⋯nCn]A=I+(2n−1)A

If for any square matrix $A^{2}$ = A, then $(I + A)^{n}$ is equal to (where I is identity matrix)

The correct option is A $I + (2^{n} - 1) A$
$A^{2} = A \Rightarrow A^{3} = A \Rightarrow A^{4} = A \Rightarrow A^{n} = A (I + A)^{n} =^{n} C_{0} I^{n} +^{n} C_{1} I^{n - 1} A + \dots +^{n} C_{n} A^{n} = I + [^{n} C_{1} +^{n} C_{2} +^{n} C_{3} + \dots^{n} C_{n}] A = I + (2^{n} - 1) A$