Question

find matrix A such that (after this reger the photo to see the full question)
plzz answer this
​

Answer 1

Answer:

$A\begin{bmatrix}2&3\\ 4&5\end{bmatrix}=\begin{bmatrix}0&-4\\ 10&3\end{bmatrix}\quad :\quad A=\begin{pmatrix}-8&4\\ -19&12\end{pmatrix}$

Step-by-step explanation:

$A\begin{bmatrix}2&3\\ 4&5\end{bmatrix}=\begin{bmatrix}0&-4\\ 10&3\end{bmatrix}$

$\mathrm{Multiply\:both\:sides\:of\:the\:equation\:by}\:\begin{bmatrix}2&3\\ 4&5\end{bmatrix}^{-1}\:\mathrm{from\:the\:right}$

$XA=B\quad \Rightarrow \quad \:X=BA^{-1}$

$A=\begin{bmatrix}0&-4\\ 10&3\end{bmatrix}\begin{bmatrix}2&3\\ 4&5\end{bmatrix}^{-1}$

$\begin{bmatrix}0&-4\\ 10&3\end{bmatrix}\begin{bmatrix}2&3\\ 4&5\end{bmatrix}^{-1}$

$\begin{bmatrix}2&3\\ 4&5\end{bmatrix}^{-1}$

$\\mathrm{Find\:2x2\:matrix\:inverse\:according\:to\:the\:formula}:\quad \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}^{-1}=\dfrac{1}{\det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}}$

$\begin{pmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{pmatrix}$

$=\dfrac{1}{\det \begin{bmatrix}2&3\\ 4&5\end{bmatrix}}\begin{pmatrix}5&-3\\ -4&2\end{pmatrix}$

$\det \begin{bmatrix}2&3\\ 4&5\end{bmatrix}$

$\mathrm{Find\:the\:matrix\:determinant\:according\:to\:formula}:\quad \det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}\:=\:ad-bc$

$=2\cdot \:5-3\cdot \:4$

$=10-12$

$=-2$

$=\dfrac{1}{-2}\begin{pmatrix}5&-3\\ -4&2\end{pmatrix}$

$\dfrac{1}{-2}\begin{pmatrix}5&-3\\ -4&2\end{pmatrix}$

$\mathrm{Scalar\:multiplication:\:Multiply\:each\:of\:the\:matrix\:elements\:by\:a\:scalar}$

$=\begin{pmatrix}\dfrac{1}{-2}\cdot \:5&\dfrac{1}{-2}\left(-3\right)\\ \dfrac{1}{-2}\left(-4\right)&\dfrac{1}{-2}\cdot \:2\end{pmatrix}$

$\mathrm{Simplify\:each\:element}$

$=\begin{pmatrix}-\dfrac{5}{2}&\dfrac{3}{2}\\ 2&-1\end{pmatrix}$

$=\begin{bmatrix}0&-4\\ 10&3\end{bmatrix}\begin{pmatrix}-\dfrac{5}{2}&\dfrac{3}{2}\\ 2&-1\end{pmatrix}$

$\begin{bmatrix}0&-4\\ 10&3\end{bmatrix}\begin{pmatrix}-\dfrac{5}{2}&\dfrac{3}{2}\\ 2&-1\end{pmatrix}$

$\mathrm{Multiply\:the\:rows\:of\:the\:first\:matrix\:by\:the\:columns\:of\:the\:second\:matrix}$

$\begin{pmatrix}0&-4\end{pmatrix}\begin{pmatrix}-\dfrac{5}{2}\\ 2\end{pmatrix}=0\cdot \left(-\dfrac{5}{2}\right)+\left(-4\right)\cdot \:2$

$\begin{pmatrix}0&-4\end{pmatrix}\begin{pmatrix}\dfrac{3}{2}\\ -1\end{pmatrix}=0\cdot \dfrac{3}{2}+\left(-4\right)\left(-1\right)$

$\begin{pmatrix}10&3\end{pmatrix}\begin{pmatrix}-\dfrac{5}{2}\\ 2\end{pmatrix}=10\left(-\dfrac{5}{2}\right)+3\cdot \:2$

$\begin{pmatrix}10&3\end{pmatrix}\begin{pmatrix}\dfrac{3}{2}\\ -1\end{pmatrix}=10\cdot \dfrac{3}{2}+3\left(-1\right)$

$=\begin{pmatrix}0\cdot \left(-\dfrac{5}{2}\right)+\left(-4\right)\cdot \:2&0\cdot \dfrac{3}{2}+\left(-4\right)\left(-1\right)\\ 10\left(-\dfrac{5}{2}\right)+3\cdot \:2&10\cdot \dfrac{3}{2}+3\left(-1\right)\end{pmatrix}$

$\mathrm{Simplify\:each\:element}$

$=\begin{pmatrix}-8&4\\ -19&12\end{pmatrix}$

$\therefore A=\begin{pmatrix}-8&4\\ -19&12\end{pmatrix}$