# Prove the matrix satisfies the equation $A^2 -4A-5I=0$ [closed]

How to prove that $$A=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$$

satisfies the equation $A^2 -4A-5I=0$?

• If nothing else, just compute the expression.. Apr 20, 2014 at 18:49
• In such easy example I would simple calculate that out. Apr 20, 2014 at 18:49
• I know that but i wonder how ease as is.Thnx very much Apr 20, 2014 at 18:50
• @Amzoti yes written correctly. Apr 20, 2014 at 18:52
• What's $5I$?${}$ Apr 20, 2014 at 18:55

$$A^2 = \left( \begin{array}{ccc} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \\ \end{array} \right)$$

$$-4A = \left( \begin{array}{ccc} -4 & -8 & -8 \\ -8 & -4 & -8 \\ -8 & -8 & -4 \\ \end{array} \right)$$

$$A^2 - 4A = \left( \begin{array}{ccc} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \\ \end{array} \right)$$

This is just as easy as any other way for this problem.

• Thnx i got what you are talking about . Apr 20, 2014 at 19:03
• Nice explication! Apr 22, 2014 at 11:19

$A=2J-I$, where $J$ is the all-one matrix with $J^2=3J$. Therefoer $$A^2-4A-5I=(2J-I)^2-4(2J-I)-5I=4J^2-12J=0.$$

• Excellent idea thnx. Apr 20, 2014 at 19:36

If you would like to sped up the calculation (supposing you are doing it on your own) you can use this approach.

Let's take the equation $$A^2 - 4A - 5I = 0.$$ And now forget about the matrices. What we have is simple quadratic equation in $x$ written as $$x^2 - 4x - 5 = 0,$$ now just find the roots $$(x-5)(x+1) = 0.$$ Let's get back to that matrix $$(A-5I)(A+I) = 0.$$ Simple calculations give $$\begin{pmatrix} -4 & 2 &2\\ 2 & -4 &2\\ 2 & 2 &-4 \end{pmatrix}\cdot \begin{pmatrix} 2 & 2 &2\\ 2 & 2 &2\\ 2 & 2 &2 \end{pmatrix} = 0.$$

And now it is very simple since you divide by 2 $$\begin{pmatrix} -4 & 2 &2\\ 2 & -4 &2\\ 2 & 2 &-4 \end{pmatrix}\cdot \begin{pmatrix} 1 & 1 &1\\ 1 & 1 &1\\ 1 & 1 &1 \end{pmatrix} = 0.$$ Multiplying those matrices is just adding all the elements in each row of the first matrix. So it is $0$ everywhere. Such tricks might come in handy in more complicated examples.

• Excellent idea thnx. Apr 20, 2014 at 19:36
• I am glad I could be of some help. Apr 20, 2014 at 19:37

Another way to do this is to note that $x^2 - 4x - 5 = (x-5)(x+1)$ so if we wanted to calculate this matrix $A^2 - 4A + 5I$ we can just calculate $(A-5I)(A+I)$.

This might not be much easier, especially just for a quadratic, but it does reduce the number of times you have to calculate a product of matrices, and the number of times you have to add them up.

The other ways given are perfectly good too, I just happen to prefer solving this sort of problem this way.