Question Video: Determining the Power of a Given Matrix | Nagwa Question Video: Determining the Power of a Given Matrix | Nagwa

# Question Video: Determining the Power of a Given Matrix Mathematics • First Year of Secondary School

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If 𝐴 = [0, 2 and 2, 0], which of the following is 𝐴⁹⁹? [A] 2⁹⁸ [0, 2 and 2, 0] [B] 2⁹⁹ [0, 2 and 2, 0] [C] 2¹⁰⁰ [0, 2 and 2, 0] [D] 2⁹⁸ [1, 0 and 0, 1] [E]2⁹⁹ [1, 0 and 0, 1]

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### Video Transcript

If 𝐴 is the two-by-two matrix zero, two, two, zero, which of the following is 𝐴 to the 99th power? Is it option (A) two to the power of 98 multiplied by the two-by-two matrix zero, two, two, zero? Is it option (B) two to the power of 99 multiplied by the two-by-two matrix zero, two, two, zero? Is it option (C) two to the power of 100 multiplied by the two-by-two matrix zero, two, two, zero? Is it option (D) two to the power of 98 multiplied by the two-by-two matrix one, zero, zero, one? Or is it option (E) two to the power of 99 multiplied by the two-by-two matrix one, zero, zero, one?

In this question, we’re given a two-by-two matrix 𝐴. And we need to determine which of five given expressions represents the matrix 𝐴 to the power of 99. To do this, let’s start by recalling what it means to raise a matrix to an exponent. If 𝐴 is a square matrix and the exponent is a positive integer, then this means we multiply the matrix by itself where the number of times 𝐴 appears in the product is equal to the exponent. In particular, if the exponent is one, then our matrix is just equal to itself. And when the exponent is 99, we need to multiply 𝐴 by itself where 𝐴 appears in the product 99 times.

Of course, working out this product directly would be very difficult. This is a product with 99 two-by-two matrices. Instead, let’s try and find a pattern that this product follows. We’ll do this by using the associativity property of matrix multiplication. Let’s just find 𝐴 squared. And remember, 𝐴 squared is 𝐴 multiplied by 𝐴. That’s the two-by-two matrix zero, two, two, zero multiplied by the matrix zero, two, two, zero.

Recall to multiply two matrices together, we need to find the sum of the products of the rows of the first matrix and the columns of the second matrix. So the element in row one, column one of the product of these two matrices will be the product of the elements of the first row in the first matrix and the first column in the second matrix. That’s zero multiplied by zero plus two multiplied by two, which we can calculate is equal to four. So the entry in row one, column one of 𝐴 squared is four. We can then do the same for the element in row one, column two. We need to multiply zero by two and then add on to this the product of two and zero. And both of these terms have a factor of zero. So the answer is zero.

We get a very similar story in the entry in row one, column two of our matrix. It’s equal to two times zero plus zero times two which is equal to zero. Finally, we can calculate the element in row two, column two. It’s equal to two times two plus zero times zero, which we can calculate is equal to four. Therefore, the matrix 𝐴 squared is the two-by-two matrix four, zero, zero, four. And this is a very useful result because we can see this is very similar to the identity matrix of order two. In particular, if we take a scale factor of four outside of this matrix, we can see that 𝐴 squared is equal to four multiplied by the identity matrix of order two. And this is a very useful result because multiplying a matrix by the identity matrix is very easy to calculate.

We can use this to simplify our expression for 𝐴 to the power of 99. To do this, we’re going to use the associativity of matrix multiplication. This will allow us to replace any product of 𝐴 and 𝐴 with 𝐴 squared. And we have to note there are only 99 factors of 𝐴 in this expression. This is an odd number. So when we rewrite these factors of 𝐴 times 𝐴 as 𝐴 squared, we will only have 49 pairs of this form and one last factor of 𝐴. And before we start evaluating this expression, there’s one thing worth noting. We can write the factor of 𝐴 anywhere in this expression. And this is because we can choose the 49 pairs of 𝐴 times 𝐴 in any way we want and this doesn’t matter. It won’t change the final answer of our question.

We’re now ready to evaluate this expression. We’ll do this by substituting 𝐴 squared is equal to four multiplied by the two-by-two identity matrix into this expression. Doing this gives us 49 factors of four multiplied by the identity matrix of order two and then we multiply by 𝐴. And now we can simplify this expression. Let’s start with the scalar factors. We have 49 factors of four, which simplifies to four to the power of 49. Next, we have 49 factors of the identity matrix multiplied by 𝐴. But remember, multiplying a matrix by the identity matrix doesn’t change its value. So this is just equal to 𝐴.

Therefore, we’ve shown that 𝐴 to the power of 99 is four to the power of 49 multiplied by 𝐴. And this is not any of our options. Instead, we need to simplify slightly. We need to write our scalar factor as a power of two, and we can do this by using the laws of exponents. Four is two squared, and two squared all raised to the power of 49 is two to the power of two times 49 which is two to the power of 98. Substituting this into our expression and writing our matrix 𝐴 out in four gives us our final answer. If 𝐴 is the two-by-two matrix zero, two, two, zero, then 𝐴 to the power of 99 is equal to two to the power of 98 multiplied by 𝐴, which is option (A).

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