Find an elementary matrix E such that EA = B.A = 2 1 3-2 4 53 1 4B = 2 1 33 1 4-2 4 5

The Elementary Matrix E is [tex]\left[\begin{array}{ccc}1&0&0\\0&0&1\\0&1&0\end{array}\right][/tex].

We know that the Matrix

[tex]A = \left[\begin{array}{ccc}2&1&3\\-2&4&5\\3&1&4\end{array}\right] , B = \left[\begin{array}{ccc}2&1&3\\3&1&4\\-2&4&5\end{array}\right][/tex]

[tex]EA = B[/tex]

⇒ [tex]E = B A^{-1}[/tex]

[tex]A^{-1} = \frac{1}{|A|} \times \left[\begin{array}{ccc}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{array}\right][/tex]

[tex]|A| = \left|\begin{array}{ccc}2&1&3\\-2&4&5\\3&1&4\end{array}\right|[/tex]

[tex]= 2 \times4\times4-2\times5\times1 - 1\times(-2)\times4+1\times5\times3+3\times(-2)\times1 - 3\times4\times3\\= 32 -10+8+15-6-36 = 3[/tex]

[tex]A^{-1} = \frac{1}{3} \left[\begin{array}{ccc}11&-1&-7\\23&-1&-16\\-14&1&10\end{array}\right] \\A^{-1} = \left[\begin{array}{ccc}\frac{11}{3} &\frac{-1}{3}&\frac{-7}{3}\\\frac{23}{3}&\frac{-1}{3}&\frac{-16}{3}\\\frac{-14}{3}&\frac{1}{3}&\frac{10}{3}\end{array}\right][/tex]

[tex]\therefore E =\left[\begin{array}{ccc}2&1&3\\3&1&4\\-2&4&5\end{array}\right] \times \left[\begin{array}{ccc}\frac{11}{3} &\frac{-1}{3}&\frac{-7}{3}\\\frac{23}{3}&\frac{-1}{3}&\frac{-16}{3}\\\frac{-14}{3}&\frac{1}{3}&\frac{10}{3}\end{array}\right][/tex]

[tex]E = \left[\begin{array}{ccc}\frac{22+23-42}{3}&\frac{2+1-3}{3}&\frac{-14-16+30}{3}\\\frac{33+23-56}{3}&\frac{-3-1+4}{3}&\frac{-21-16+41}{3}\\\frac{-22+92-70}{3}&\frac{2-4+5}{3}&\frac{14-64+50}{3}\end{array}\right][/tex]

[tex]E = \left[\begin{array}{ccc}1&0&0\\0&0&1\\0&1&0\end{array}\right][/tex]

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