Question

x+2y=2,2x+3y=3 by inversion method

Answer 1

2x + 3y = 3

Write in matrix form:

\begin{lgathered}\left(\begin{array}{cc} 1 \ \ 2 \\ 2 \ \ 3\end{array}\right)\left(\begin{array}{cc} x\\ y\end{array}\right) = \left(\begin{array}{cc} 2\\ 3\end{array}\right)\end{lgathered}

(

1 2

2 3

)(

x

y

)=(

2

3

)

Find the determinant:

\begin{lgathered}\left|\begin{array}{cc} 1 \ \ 2 \\ 2 \ \ 3\end{array}\right| = (1)(3) - (2)(2) = -1\end{lgathered}

∣

∣

∣

∣

1 2

2 3

∣

∣

∣

∣

=(1)(3)−(2)(2)=−1

Find the inverse matrix:

\begin{lgathered}\text {inverse of} \left(\begin{array}{cc} 1 \ \ 2 \\ 2 \ \ 3\end{array}\right) = (-1)\left(\begin{array}{cc} 3 \ \ -2 \\ -2 \ \ \ \ 1\end{array}\right) = \left(\begin{array}{cc} -3 \ \ \ 2 \\ 2 \ \ -1\end{array}\right)\end{lgathered}

inverse of(

1 2

2 3

)=(−1)(

3 −2

−2 1

)=(

−3 2

2 −1

)

Multiply the inverse to both sides:

\begin{lgathered}\left(\begin{array}{cc} 1 \ \ 2 \\ 2 \ \ 3\end{array}\right) \left(\begin{array}{cc} -3 \ \ \ 2 \\ 2 \ \ -1\end{array}\right) \left(\begin{array}{cc} x\\ y\end{array}\right) = \left(\begin{array}{cc} 2\\ 3\end{array}\right) \left(\begin{array}{cc} -3 \ \ \ 2 \\ 2 \ \ -1\end{array}\right)\end{lgathered}

(

1 2

2 3

)(

−3 2

2 −1

)(

x

y

)=(

2

3

)(

−3 2

2 −1

)

Left Hand Side (LHS):

\begin{lgathered}\left(\begin{array}{cc} 1 \ \ 2 \\ 2 \ \ 3\end{array}\right) \left(\begin{array}{cc} -3 \ \ \ 2 \\ 2 \ \ -1\end{array}\right) \left(\begin{array}{cc} x\\ y\end{array}\right)\end{lgathered}

(

1 2

2 3

)(

−3 2

2 −1

)(

x

y

)

\begin{lgathered}= \left(\begin{array}{cc} (1)(-1) + (2)(2) \ \ (1)(2) + (2)(-1) \\ 2(-3) + 3(2) \ \ 2(2) + 3(-1)\end{array}\right) \ \left(\begin{array}{cc} x\\ y\end{array}\right)\end{lgathered}

=(

(1)(−1)+(2)(2) (1)(2)+(2)(−1)

2(−3)+3(2) 2(2)+3(−1)

) (

x

y

)

\begin{lgathered}= \left(\begin{array}{cc}1 \ \ 0 \\0 \ \ 1 \end{array}\right) \ \left(\begin{array}{cc} x\\ y\end{array}\right)\end{lgathered}

=(

1 0

0 1

) (

x

y

)

\begin{lgathered}= \left(\begin{array}{cc} x\\ y\end{array}\right)\end{lgathered}

=(

x

y

)

Right Hand Side (RHS):

\begin{lgathered}\left(\begin{array}{cc} 2\\ 3\end{array}\right) \left(\begin{array}{cc} -3 \ \ \ 2 \\ 2 \ \ -1\end{array}\right)\end{lgathered}

(

2

3

)(

−3 2

2 −1

)

\begin{lgathered}= \left(\begin{array}{cc} 2(-3) + 3(2)\\ 2(2) + 3(-1) \end{array}\right)\end{lgathered}

=(

2(−3)+3(2)

2(2)+3(−1)

)

\begin{lgathered}= \left(\begin{array}{cc}0\\ 1 \end{array}\right)\end{lgathered}

=(

0

1

)

Find x and y:

LHS = RHS

\begin{lgathered}\left(\begin{array}{cc} x\\ y\end{array}\right) = \left(\begin{array}{cc}0\\ 1 \end{array}\right)\end{lgathered}

(

x

y

)=(

0

1

)

x = 0x=0

y = 1y=1

Answer: x = 0, y = 1

Answer 2

Answer:

x = 0 and y = 1

Step-by-step explanation:

$\displaystyle\left(\begin{array}{cc}1&2\\2&3\end{array}\right)\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}2\\3\end{array}\right)\\\\\Rightarrow\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}1&2\\2&3\end{array}\right)^{-1}\left(\begin{array}{c}2\\3\end{array}\right)\\\\\Rightarrow\left(\begin{array}{c}x\\y\end{array}\right)=\frac1{(1)(3)-(2)(2)}\left(\begin{array}{cc}3&-2\\-2&1\end{array}\right)\left(\begin{array}{c}2\\3\end{array}\right)$

$\displaystyle\Rightarrow\left(\begin{array}{c}x\\y\end{array}\right)=\frac1{-1}\left(\begin{array}{cc}3&-2\\-2&1\end{array}\right)\left(\begin{array}{c}2\\3\end{array}\right)\\\\\Rightarrow\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}-3&2\\2&-1\end{array}\right)\left(\begin{array}{c}2\\3\end{array}\right)\\\\\Rightarrow\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}(-3)(2)+(2)(3)\\(2)(2)+(-1)(3)\end{array}\right)$

$\displaystyle\Rightarrow\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}0\\1\end{array}\right)$

So x = 0 and y = 1