# matrix in linear algebra

#### dxzre

##### New member
Let A be a 2 x 2 matrix. The matrix A is said to be the square root of the divide
Matrix B 2 x 2 matrix, if A^2 = B.

the question is:
Prove that there is no square root of:
0 1
0 0

Take the square root of:
1 1
0 0

i need your opinion to solve the question

#### lex

##### Full Member
I would simply write A as $$\displaystyle \left(\begin{matrix}a&b\\c&d\\\end{matrix}\right)$$
Compute $$\displaystyle A^2$$ and then equate it to e.g. $$\displaystyle \left(\begin{matrix}0&1\\0&0\\\end{matrix}\right)$$. This will give equations which will have no solution.
Similarly for $$\displaystyle \left(\begin{matrix}1&1\\0&0\\\end{matrix}\right)$$ but it will yield equations with solutions, which you can find.

#### HenryTorres

##### New member
i tried it 2 times and thanks God i solved it.

Great

#### HallsofIvy

##### Elite Member
Since dxzre has solved this:
Suppose $$\displaystyle \begin{bmatrix}a & b \\ c & d \end{bmatrix}^2= \begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}a & b \\ c & d \end{bmatrix}= \begin{bmatrix}a^2+ bc & ab+ bd \\ ac+ cd & bc+ d^2\end{bmatrix}= \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}$$.

$$\displaystyle a^2+ bc= 0$$, $$\displaystyle ab+ bd= 1$$, $$\displaystyle ac+ cd= 0$$, $$\displaystyle bc+ d^2= 0$$.

From ac+ cd= c(a+ d)= 0, either c= 0 or a+ d= 0.
If c= 0, $$\displaystyle a^2+ bc= a^2= 0$$ so a= 0 and $$\displaystyle bc+ d^2= d^2= 0$$ so d= 0.
ab+bd= 0, not 1.

If c is not 0 then a+ d= 0 so d= -a. ab+ bd= ab- ab= 0, not 1 so there is no such matrix.

Suppose $$\displaystyle \begin{bmatrix}a & b \\ c & d \end{bmatrix}^2= \begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}a & b \\ c & d \end{bmatrix}= \begin{bmatrix}a^2+ bc & ab+ bd \\ ac+ cd & bc+ d^2\end{bmatrix}= \begin{bmatrix}1 & 1 \\ 0 & 0 \end{bmatrix}$$.

$$\displaystyle a^2+ bc= 1$$, $$\displaystyle ab+ bd= 1$$, $$\displaystyle ac+ cd= 0$$, $$\displaystyle bc+ d^2= 0$$.A
As before, ac+ cd= c(a+ d)= 0 so either c= 0 or a+d= 0.
If c= 0, $$\displaystyle a^2+ bd= a^2= 1$$ so a= 1 or -1. $$\displaystyle bc+ d^2= d^2= 0$$, d= 0.
If a= 1, ab+ bd= b= 1
a= b= 1, c= d= 0 so $$\displaystyle \begin{bmatrix}1 & 1 \\ 0 & 0 \end{bmatrix}$$.

Then $$\displaystyle \begin{bmatrix}1 & 1 \\ 0 & 0 \end{bmatrix}^2= \begin{bmatrix}1 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix}1 & 1 \\ 0 & 0 \end{bmatrix}= \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$$.

If c= 0 and a= -1 then ab+ bd= -b= 1 so b= -1.
a= b= -1, c= d= 0 so $$\displaystyle \begin{bmatrix}-1 & -1 \\ 0 & 0 \end{bmatrix}$$.

Then $$\displaystyle \begin{bmatrix}-1 & -1 \\ 0 & 0 \end{bmatrix}^2= \begin{bmatrix}=-1 & -1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix}-1 & -1 \\ 0 & 0 \end{bmatrix}= \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$$.

#### lex

##### Full Member
Since dxzre has solved this:
He probably has by now!

And for the second part, if $$\displaystyle a=-d$$, then the final equation $$\displaystyle bc+d^2=0 \rightarrow bc+a^2=0$$, contradicting the first equation $$\displaystyle a^2+bc=1$$