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In Exercises $7-12$ , assume the signals listed are solutions of the given difference equation. Determine if the signals form a basis for the solution space of the equation. Justify your answers using appropriate theorems. $$1^{k}, 2^{k},(-2)^{k} ; y_{k+3}-y_{k+2}-4 y_{k+1}+4 y_{k}=0$$

the signals $\mathrm { I } ^ { k } , 2 ^ { k } ,$ and $( - 2 ) ^ { k }$ are linearly independent.

Calculus 3

Chapter 4

Vector Spaces

Section 8

Applications to Difference Equations

Vectors

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in this example. We have a difference equation that's provided here. And we're told that one k to the power k and negative to to the para que are all solutions to this homogeneous difference equation. Next, we're going to let h do you know this up space of s that contains all solutions and determine if this set is a basis for age. To do this, we need to determine the Katsuragi Matrix for this particular difference equation. It's going to be a matrix. We can do no by see if we like. And in the first column we have one to the Power K, then one of the power of K plus one, then one of the power of K plus two. So, likewise, for the second column, we go to the second solution and right in to the power K to the Power K plus one, and then to the power of K plus two. Likewise, the less column is negative to power of K that negative two to the power of K plus one and then finally negative two to the power of K plus two. So for this particular matrix, we want to determine if it is potentially in vertebral for some choice of K, and the most nicest choice to work with is the following. Let's let K be equal to zero, then our cause. Roddy Matrix C will be equal to the following. Placing K a zero on the first row gives us 11 and one. If K is zero, then on the second row we have the bases all to power one. So we have 12 negative to then raise thes same basis to the power of to we have 14 and four. So this is our matrix, and our next goal is to determine if it's in vertebral. Well, this can be assessed through either taking a determinant or through row reduction. Going the road reduction route. I found that this matrix ISRO equivalent to 100 going column wise 11 zero one negative three and 12. And because there is a pivot in every room of this matrix, it follows that see inverse exists. So it's pause for a moment. We found that if we take K two b zero, then the Katsuragi matrix becomes in vertebral. That tells us that this set is linearly independent. Let's write that down as our next step so we can say this shows that let's call this set Maybe well, we could leave it without a name. It's one of the K to the power of K and negative to the power of K is leaned early independent. We also know that the dimension off the solution set H is equal to three and that becomes from the order of this equation which is order three. So we now know that we have three linearly independent solutions and the dimension of the spaces h Then we found a basis. So our conclusion is that this set meaning that one right above forms a basis four h and this completes our solution.

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