Proof by induction(matrix version)

http://files.physicsandmathstutor.com/download/Maths/A-level/FP1/Papers-Edexcel/June%202015%20QP%20-%20FP1%20Edexcel.pdf
n=1
$\begin {pmatrix}1&0\\-\frac{1}{4}(5^1 -1)&5^1\end {pmatrix}$




Assume true for n=k that

$\begin {pmatrix}1&0\\-\frac{1}{4}(5^k-1)&5^k \end {pmatrix}$

what am i supposed to do next? it doesn't give me something to substitute something somewhere?

http://files.physicsandmathstutor.com/download/Maths/A-level/FP1/Papers-Edexcel/June%202015%20QP%20-%20FP1%20Edexcel.pdf
n=1
$\begin {pmatrix}1&0\\-\frac{1}{4}(5^1 -1)&5^1\end {pmatrix}$

$\therefore true\ for\ n=1$


Assume true for n=k that

$\begin {pmatrix}1&0\\-\frac{1}{4}(5^k-1)&5^k \end {pmatrix}$

what am i supposed to do next? it doesn't give me something to substitute something somewhere?

http://files.physicsandmathstutor.com/download/Maths/A-level/FP1/Papers-Edexcel/June%202015%20QP%20-%20FP1%20Edexcel.pdf
n=1
$\begin {pmatrix}1&0\\-\frac{1}{4}(5^1 -1)&5^1\end {pmatrix}$

$\therefore true\ for\ n=1$


Assume true for n=k that

$\begin {pmatrix}1&0\\-\frac{1}{4}(5^k-1)&5^k \end {pmatrix}$

what am i supposed to do next? it doesn't give me something to substitute something somewhere?

http://files.physicsandmathstutor.com/download/Maths/A-level/FP1/Papers-Edexcel/June%202015%20QP%20-%20FP1%20Edexcel.pdf
n=1
$\begin {pmatrix}1&0\\-\frac{1}{4}(5^1 -1)&5^1\end {pmatrix}$

$\therefore true\ for\ n=1$


Assume true for n=k that

$\begin {pmatrix}1&0\\-\frac{1}{4}(5^k-1)&5^k \end {pmatrix}$

what am i supposed to do next? it doesn't give me something to substitute something somewhere?

http://files.physicsandmathstutor.com/download/Maths/A-level/FP1/Papers-Edexcel/June%202015%20QP%20-%20FP1%20Edexcel.pdf
n=1
$\begin {pmatrix}1&0\\-\frac{1}{4}(5^1 -1)&5^1\end {pmatrix}$

$\therefore true\ for\ n=1$


Assume true for n=k that

$\begin {pmatrix}1&0\\-\frac{1}{4}(5^k-1)&5^k \end {pmatrix}$

what am i supposed to do next? it doesn't give me something to substitute something somewhere?

I might be wrong, but I have a feeling the kth term is the determinant...

But can you not do it for K+1, and use the result from the LHS of the equation in your assumption as a substitute?

Test for n=1
Assume true for n=k.
Prove for n=k+1. This would require you you to post-multiply both sides by the matrix 1, -1, 0, 5. (You know which one, cba to post it in the form) and factor every element to show it works for n=k+1.

Consider the matrix to the power of k + 1,and split that into $\begin {pmatrix} 1&0 \\-1&5 \end {pmatrix}^{k} \begin {pmatrix} 1&0 \\-1&5 \end {pmatrix}$ and sub that kth power matrix for the general formula we assumed was true for n = k.

Consider multiplying the RHS by $\begin {pmatrix}1&0\\-1&5 \end {pmatrix}$ on the right. Can you see why?

u sure i can do that? i considered it but i wasn't too sure if i was allowed to do that since it didn't state that matrix without to the power of n on it

Of course. Just like with summation of series when you add the (k+1)th term onto both sides, you multiply here by the same matrix as you're working with exponents.

Let $A=\begin {pmatrix}1 & 0 \\ {-1} & {5} \end {pmatrix}$

Then it follows that $A^{k+1}=(A^k)(A^1)$ (think of it as performing a transformation A a total amount of k times, then doing an additional one)

And you know what $A^k$ is because you assumed it to be true for $n=k$

i've got

$\begin {pmatrix}1&0\\{[-\frac{1}{4}(5^k -1)]-5^k}&5^k \times 5 \end {pmatrix}$

the $5^k \times 5=5^{k+1}$

but the other meaty term i've only gotten this far

$-\frac{1}{4}[5^k -1+(5^k \times 4)]$

Is it really meaty though? $5^k+4\cdot 5^k=5\cdot 5^k = 5^{k+1}$

Looks like we have another student (in the form of Mr or Mrs Huiop) who doesn't know their laws of indices. Typical.