If a square matrix B satisfies B^2 = I - B and B^n = 5I - 8B, then the value of n is : Option 1: 4 Option 2: 6 Option 3: 8 Option 4: 10

Answer: 6. Solution: Using B^2 = I - B repeatedly: B^3 = B * B^2 = B(I - B) = B - B^2 = B - (I - B) = 2B - I. B^4 = B * B^3 = B(2B - I) = 2B^2 - B = 2(I - B) - B = 2I - 3B. B^5 = B * B^4 = 2B - 3B^2 = 2B - 3(I - B) = 5B - 3I. B^6 = B * B^5 = 5B^2 - 3B = 5(I - B) - 3B = 5I - 8B. So n = 6.

Applied Mathematics

Algebra

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If a square matrix B satisfies $B^2 = I - B$ and $B^n = 5I - 8B$ , then the value of n is :

✅ Correct Option: 2

Using $B^2 = I - B$ repeatedly:

$B^3 = B \cdot B^2 = B(I - B) = B - B^2 = B - (I - B) = 2B - I$ .

$B^4 = B \cdot B^3 = B(2B - I) = 2B^2 - B = 2(I - B) - B = 2I - 3B$ .

$B^5 = B \cdot B^4 = 2B - 3B^2 = 2B - 3(I - B) = 5B - 3I$ .

$B^6 = B \cdot B^5 = 5B^2 - 3B = 5(I - B) - 3B = 5I - 8B$ .

So $n = 6$ .

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(A) Break even point is at x = 10

(B) Break even point is at x = 8

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(E) Fixed cost is 0

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Match List - I with List - II.

	List - I (Matrix)		List - II (Type)
(A)	$\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$	(I)	Lower Triangular Matrix
(B)	$\begin{bmatrix} 1 & 5 & 0 & -1 \end{bmatrix}$	(II)	Row Matrix
(C)	$\begin{bmatrix} 3 & 0 & 0 \\ 1 & -1 & 0 \\ 2 & 5 & 4 \end{bmatrix}$	(III)	Diagonal Matrix
(D)	$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix}$	(IV)	Scalar Matrix

Choose the correct answer from the options given below :

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If a square matrix B satisfies B2=I−BB^2 = I - BB2=I−B and Bn=5I−8BB^n = 5I - 8BBn=5I−8B, then the value of n is :

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CUET Mathematics Past Year Questions (Topic-Wise)

If a square matrix B satisfies $B^2 = I - B$ and $B^n = 5I - 8B$ , then the value of n is :