Finding All $2\times 2$ Matrices That Satisfy $A^2=A$

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The discussion focuses on finding all 2x2 matrices \(A\) that satisfy the equation \(A^2 = \begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}\). The matrix \(A\) is expressed as \(A = \begin{pmatrix}a & b\\ c & d\end{pmatrix}\), leading to a system of equations derived from the matrix multiplication. The solutions reveal that \(a\) and \(d\) must be equal, resulting in two sets of solutions: \(a=d=1,~b=\frac{1}{2},~c=0\) and \(a=d=-1,~b=-\frac{1}{2},~c=0\).

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How can i find all matrix $A$ of order $2\times 2$ that satisfy the $A^2 = \begin{pmatrix}1 & 1\\
0 & 1
\end{pmatrix}$
 
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jacks said:
How can i find all matrix $A$ of order $2\times 2$ that satisfy the $A^2 = \begin{pmatrix}1 & 1\\
0 & 1
\end{pmatrix}$

Hi jacks,

Let \(A = \begin{pmatrix}a & b\\ c & d\end{pmatrix}\). Then \(A^2 = \begin{pmatrix}a^2+bc & ab+bd\\ ac+cd & bc+d^2\end{pmatrix}\)

Since, \(A^2 = \begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}\)

\[a^2+bc=1~~~~~~~(1)\]

\[ab+bd=1~~~~~~~(2)\]

\[bc+d^2=1~~~~~~~~(3)\]

\[ac+cd=0~~~~~~~~(4)\]

From (1) and (3), \(a=\pm d\). But (4) implies that, \(a=d\). Then substituting for \(d\) in (4), \(ac=0\). If \(a=0\Rightarrow d=0\) and we get a contradiction from (2). Therefore, \(c=0\). Then by (1) and (3), \(a=d=\pm 1\). Finally by (2), \(b=\pm\frac{1}{2}\). So we have two sets of answers,

\[a=d=1,~b=\frac{1}{2},~c=0\mbox{ or }a=d=-1,~b=-\frac{1}{2},~c=0\]
 

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