Question

Solve the linear system using an inverse matrix. -4x +5y 12 The inverse of the coefficient matrix A, A 1 is (Simplify your answer. Type an integer or simplified fraction for each matrix element) The solution to the system is x-and y Simplify your answers. Type integers or simplified fractions) Enter your answer in each of the answer boxes

Answer 1

apply inverse rule for 2*2 matrix

$\begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}^{-1}=\frac{1}{\det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}}\begin{pmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{pmatrix}$

$=\frac{1}{\det \begin{pmatrix}-1&1\\ -4&5\end{pmatrix}}\begin{pmatrix}5&-1\\ -\left(-4\right)&-1\end{pmatrix}$

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

$=\frac{1}{-5+4}\begin{pmatrix}5&-1\\ -\left(-4\right)&-1\end{pmatrix}$

$=\frac{1}{-1}\begin{pmatrix}5&-1\\ -\left(-4\right)&-1\end{pmatrix}$

$=\begin{pmatrix}-5&1\\ -4&1\end{pmatrix}$ .........inverse matrix

system $Ax=b...........x=A^{-1}b$

$b=\begin{pmatrix}6\\ 12\end{pmatrix}$

$A^{-1}b$

$=\begin{pmatrix}-5&1\\ \:-4&1\end{pmatrix}\begin{pmatrix}6\\ 12\end{pmatrix}$

$=\begin{pmatrix}\left(-5\right)\cdot \:6+1\cdot \:12\\ \left(-4\right)\cdot \:6+1\cdot \:12\end{pmatrix}$

$=\begin{pmatrix}-18\\ -12\end{pmatrix}$

so solution is

${\color{Red} \begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}-18\\ -12\end{pmatrix}}$

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$A=\begin{pmatrix}-3&1\\ -5&2\end{pmatrix}\:\:\:\:\:\:\:\:\:\:\:\:B=\begin{pmatrix}-2&1\\ -5&3\end{pmatrix}$

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

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

${\color{Red} AB=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}}$

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

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

${\color{Red} BA=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}}$

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yes because $AB=BA=I_2$