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Question Number 214713 by efronzo1 last updated on 17/Dec/24

$$\mathrm{Find}\mathrm{matrix}\mathrm{B}\mathrm{if}\mathrm{given}\mathrm{AB}=\mathrm{BA}=\left(\begin{array}{c}00\\ 00\end{array}\right)$$$$\mathrm{where}\mathrm{A}=\left(\begin{array}{c}53\\ 53\end{array}\right)\mathrm{and}\mathrm{B}\ne \left(\begin{array}{c}00\\ 00\end{array}\right)$$$$$$

Answered by golsendro last updated on 18/Dec/24

$$\det \left(\mathrm{A}\right)=15-15=0$$$$\mathrm{so}\mathrm{one}\mathrm{of}\mathrm{possible}\mathrm{matrix}\mathrm{B}\mathrm{is}\mathrm{adj}\left(\mathrm{A}\right)$$$$\mathrm{B}=\mathrm{adj}\left(\mathrm{A}\right)=\left(\begin{array}{c}3-3\\ -55\end{array}\right)$$$$\mathrm{verification}$$$$\left(\begin{array}{c}53\\ 53\end{array}\right)\left(\begin{array}{c}3-3\\ -55\end{array}\right)=\left(\begin{array}{c}00\\ 00\end{array}\right)$$

Answered by didoumaths last updated on 17/Dec/24

$$B=\left(\begin{array}{cc}3& 3\\ -5& -5\end{array}\right)$$

Commented by som(math1967) last updated on 18/Dec/24

$$BA=\left(\begin{array}{cc}3& 3\\ -5& -5\end{array}\right)\left(\begin{array}{cc}5& 3\\ 5& 3\end{array}\right)=\left(\begin{array}{cc}30& 18\\ -50& -30\end{array}\right)$$$$\ne \left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$$

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