# 2nd degree matrix equation

Let $X$ be a matrix with 2 rows and 2 columns.

Solve the following equation: $$X^2 = \begin{pmatrix} 3 & 5\\ -5 & 8 \end{pmatrix}$$

Here is what I did:

Let $X = \begin{pmatrix} a & b\\ c & d \end{pmatrix}$. After multiplying I got the following system:

$$\left\{\begin{matrix} a^2 + bc = 3\\ ab + bd = 5\\ ac + cd = -5\\ d^2 + bc = 8 \end{matrix}\right.$$

At this point I got stucked.

• Do you happen to know anything about eigenvalues/eigenvectors? – Omnomnomnom Oct 21 '15 at 13:17
• – Emilio Novati Oct 21 '15 at 13:19
• I haven't studied eigenvalues yet. – George R. Oct 21 '15 at 13:32
• @GeorgeR. the soultion with eigenvalues etc. becomes sort of necessary for bigger matrices, in your case you would get something which isn't so easy to solve either, so the straight forward way using all equations (you missed one crucial one, check my answer) is definitely the better choice here – user190080 Oct 21 '15 at 14:49

We have the following criteria which you already stated correctly, but you missed one more information $(5)$ - still, you can solve this root problem without this additional knowledge by plugging in recursively - which comes from the determinant, we get then \begin{align} a^2 + bc &= 3 \tag1\\ ab + bd &= 5\tag2 \\ ac + cd &= -5\tag3 \\ d^2 + bc &= 8\tag4 \\ \det(X)=ad-bc&=7=\sqrt{\det(M)} \tag{5a} \end{align} Remark remember that we have $\det(AB)=\det(A)\det(B)$

This gives us

\begin{align} (1)-(4)&=a^2-d^2=(a-d)(a+d)=-5\\ (2)&=b(a+d)=5 \\ (3)&=c(a+d)=-5 \\ (1)+(5a)&=a(a+d)=10 \end{align} so we get from $(2)\wedge(3)$ $b=-c$ and further $a-d=c$ and $a=-2(a-d)\iff\frac32a=d$ therefore \begin{align} a(a+d)=10=a(a+\frac32a)=\frac52a^2\iff4=a^2 \end{align}

and thus we get for $a=2$ \begin{align} a=2,d=3,c=-1,b=1 \end{align} so $$X_1=\begin{pmatrix}2 &1\\ -1&3\end{pmatrix}$$ and for $a=-2$ \begin{align} a=-2,d=-3,c=1,b=-1 \end{align} so $$X_2=\begin{pmatrix}-2 &-1\\ 1&-3\end{pmatrix}=-X_1$$ Remark due to Robert Israel:

Indeed we have to investigate the other possible determinant solution \begin{align} \det(X)=ad-bc&=-7 \tag{5b} \end{align} then we get \begin{align} (1)+(5b)&=a(a+d)=-4 \\ (1)-(4)&=a^2-d^2=(a-d)(a+d)=-5 \end{align} which gives us $\frac54a=(a-d)\iff-\frac14a=d$ and therefore \begin{align} a(a+d)=-4=a(a-\frac14a)\iff-4=\frac34a^2 \end{align} which leads, if we stay in the field of the real numbers, to a contradiction. However, one might find for example other complex solutions. For a more detailed discussion please check out the comment section.

• What about the solutions with determinant $-7$? – Robert Israel Oct 21 '15 at 15:49
• @RobertIsrael you're completely right, I totally forgot to mention this...the case of $-7$ leads to a contradiction, at least if we operate in field of the reals. I'll add it – user190080 Oct 21 '15 at 16:02
• @RobertIsrael do you happen to know whether there exist other complex solutions in the case of $\det=-7$? I always thought that there can only exist $2$ solutions to this equation. Actually I am sure in this case since the diagonalization gives us two solutions - although this might not be unique... – user190080 Oct 21 '15 at 16:24
• The solutions with determinant $-7$ are $\pm \pmatrix{-4i/\sqrt{3} & 5 i/\sqrt{3}\cr -5 i/\sqrt{3} & i/\sqrt{3}}$. Although they are complex in this case, there are cases where all four solutions are real. – Robert Israel Oct 21 '15 at 16:58
• There can also be infinitely many real solutions. For example, the identity matrix has infinitely many real square roots (e.g. all reflections). – Robert Israel Oct 21 '15 at 17:12

A nonsingular $n \times n$ matrix $M$ will have some square roots that are polynomials in $M$ of degree $\le n-1$. Thus in this case we can look for solutions of the form $X = s M + t I$. By the Cayley-Hamilton theorem, a matrix satisfies its characteristic polynomial: in this case the characteristic polynomial is $p(x) = x^2 - 11 x + 49$, and $M^2 - 11 M + 49 I = 0$. Thus $s M + t I$ will be a square root of $M$ if $(s x + t)^2 - x$ is a multiple of $p(x)$. In this case $$(s x + t)^2 - x - s^2 p(x) = (11 s^2 + 2 s t - 1) x - 49 s^2 + t^2$$ so we want \eqalign{11 s^2 & + 2 s t - 1 = 0\cr -49 s^2 & + t^2 = 0\cr} The solutions are \eqalign{ s &= 1/5, t = 7/5 \cr s &= -1/5, t = -7/5\cr s &= i/\sqrt{3}, t = -7 i/\sqrt{3}\cr s &= -i/\sqrt{3}, t = 7 i/\sqrt{3}\cr} corresponding to $$X = \pmatrix{2 & 1\cr -1 & 3\cr},\ \pmatrix{-2 & -1\cr 1 & -3\cr},\ \pmatrix{-4i/\sqrt{3} & 5i\sqrt{3}\cr -5i/\sqrt{3} & i/\sqrt{3}},\ \pmatrix{4i/\sqrt{3} & -5i\sqrt{3}\cr 5i/\sqrt{3} & -i/\sqrt{3}}$$

hint : you have two lines with bc