If A = $(1 0 0 1 0 1 0 1 0)$ then A50 equals to ________.

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(1 0 0 25 1 0 25 0 0)

(1 25 25 1 0 0 0 0 1)

A2 + 24(A2 − I)

None of these

answer is A.

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Detailed Solution

Given matrix A is a 3×3 matrix having 3 rows and 3 columns.
A =

(1 0 0 1 0 1 0 1 0)

\times

3)
We need to compute A50 if it was given to calculate A2 ⇒ A3 we could have just multiplied A×A and A×A×A respectively but here A50 is to be calculated, so, simply multiplication of A×A up to 50 times is not possible.
We can use the hit on trial method which is just a simple method.
A =

(1 0 0 1 0 1 0 1 0)

Now, find A×A i.e. A2 multiplying a given matrix with itself.
By multiplication property, two matrices can only be multiplied if columns of first matrix is equal to rows of the second matrix. Here, A = 3×3 matrix
so, when A will be multiplied with itself it would satisfy the multiplication condition.
Multiplying A×A
A × A =

(1 0 0 1 0 1 0 1 0) \times (1 0 0 1 0 1 0 1 0)

1st 2nd
Elements of the row of the first matrix will get multiplied with all the elements of all the columns of the second matrix and will add up.
A × A =

(1 \times 1 + 0 \times 1 + 0 \times 0 1 \times 0 + 0 \times 0 + 0 \times 1 1 \times 0 + 0 \times 1 + 0 \times 0 1 \times 1 + 0 \times 1 + 1 \times 0 1 \times 0 + 0 \times 0 + 1 \times 1 1 \times 0 + 0 \times 1 + 1 \times 0 0 \times 1 + 1 \times 1 - 0 \times 0 0 \times 0 + 1 \times 0 - 0 \times 1 0 \times 0 + 1 \times 1 - 0 \times 0)

A2 =

(1 0 0 1 1 0 1 0 1)

Similarly, A3 =

(1 0 0 2 0 1 1 1 0)

Similarly, A4 =

(1 0 0 2 1 0 2 0 1)

Can you observe the pattern in A1 ↔ A3 with odd powers and A2 ↔ A4 with even powers.
First row always remains = 1, 0, 0. Observe column first, second and third elements are getting increased by 1 in A with even powers A2 ↔ A4 So, by hit and trial we see A50 as:
A50 is with even power, will follow A2, A4 type
First row = 1, 0, 0
First column = Second and third elements
∴ A2 ↔ A4 : From 1 → 2
⇒ A4 ↔ A6 : From 2 → 3
⇒ A6 ↔ A8 : From 3 → 4
Thus, A2 contains second and third elements of first column as