linear Algebra Matrix Quadratic

[cgossman]

this is a homework question that I've been struggling with for a few days :/ any help would be much appreciated:

Use the method of completing the square to find the four 2x2 matrices 'X' which satisfy the quadratic:

[2 1] X² + [6 3] X = [-15/4 -15/4]
[1 2] [3 6] [ 0 0 ]


and prove that a matrix quadratic AX² +BX+C=0 where A,B,C and 0 are 2x2, can be solved by completing the square only if A is invertible and B is a multiple of A.

 

[HallsofIvy]

this is a homework question that I've been struggling with for a few days :/ any help would be much appreciated:

Use the method of completing the square to find the four 2x2 matrices 'X' which satisfy the quadratic:

[2 1] X² + [6 3] X = [-15/4 -15/4]
[1 2] [3 6] [ 0 0 ]


and prove that a matrix quadratic AX² +BX+C=0 where A,B,C and 0 are 2x2, can be solved by completing the square only if A is invertible and B is a multiple of A.

Using LaTeX that would be \(\displaystyle \begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix}X^2+ \begin{bmatrix}6 & 3 \\ 3 & 6\end{bmatrix}X= \begin{bmatrix}-\frac{15}{4} & -\frac{15}{4} \\ 0 & 0 \end{bmatrix}\).

Now, think about how you would complete the square if those we numbers. First, factor out the leading coefficient:
\(\displaystyle A(X^2+ A^{-1}BX= A^{-1}C\)

You know that \(\displaystyle (x- p)^2= x^2- 2px+ p^2\) so that the the last term, in a perfect square, is the square of half the coefficient of x. Here, the coefficient of X is \(\displaystyle A^{-1}B\) so half of that is \(\displaystyle \frac{A^{-1}B}{2}\) and the square of that is \(\displaystyle \frac{A^{-2}B^2}{4}\). Adding that to both sides, gives \(\displaystyle A(X^2+ A^{-1}BX+ A^{-2}B^2/4= (X+ A^{-1}B/2)^2= \frac{A^{-1}B}{2}- C\)

 

[cgossman]

Wow, thanks a lot!