Question

%20x%20%2B%203%20y%20%3D%2047%20%7D%20%5Cend%7Bcases%7D%20%5Cright." id="TexFormula1" title="\left. \begin{cases} { 8 x + 2 y = 46 } \\ { 7 x + 3 y = 47 } \end{cases} \right." alt="\left. \begin{cases} { 8 x + 2 y = 46 } \\ { 7 x + 3 y = 47 } \end{cases} \right." align="absmiddle" class="latex-formula">
Solve for x & y using MATRICES!! Help...​

Answer 1

$\huge \boxed{\mathfrak{Question} \downarrow}$

$\left. \begin{cases} { 8 x + 2 y = 46 } \\ { 7 x + 3 y = 47 } \end{cases} \right.$

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\left. \begin{cases} { 8 x + 2 y = 46 } \\ { 7 x + 3 y = 47 } \end{cases} \right.$

First, write both the equations in its standard form.

$8x+2y=46\\ 7x+3y=47$

Now, write the equations in form of matrix.

$\left(\begin{matrix}8&2\\7&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}46\\47\end{matrix}\right)$

Then, multiply the equation towards the left by using the inverse of matrix $\left(\begin{matrix}8&2\\7&3\end{matrix}\right)$

$\sf \: inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}8&2\\7&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)$

The product of the matrix & its inverse will be the identity matrix.

$\sf\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)$

Now, multiply the matrices that lie on the left-hand side of the equal sign.

$\sf\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)$

For the 2 × 2 matrix $\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$, the inverse matrix is ⇨ $\left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right)$.

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{8\times 3-2\times 7}&-\frac{2}{8\times 3-2\times 7}\\-\frac{7}{8\times 3-2\times 7}&\frac{8}{8\times 3-2\times 7}\end{matrix}\right)\left(\begin{matrix}46\\47\end{matrix}\right)$

Do the calculations.

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}&-\frac{1}{5}\\-\frac{7}{10}&\frac{4}{5}\end{matrix}\right)\left(\begin{matrix}46\\47\end{matrix}\right)$

Multiply the matrices.

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}\times 46-\frac{1}{5}\times 47\\-\frac{7}{10}\times 46+\frac{4}{5}\times 47\end{matrix}\right)$

Do the arithmetics again.

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{22}{5} \\\frac{27}{5}\end{matrix}\right)$

Finally, extract the matrix elements x & y & write them separately.

$\large \boxed{ \boxed{ \bf \: x=\frac{22}{5},y=\frac{27}{5} }}$

Answer 2

Hey!

$\left \{ {8x~+~2y ~=~ 46} \atop {7x~+~3y~=~46}}\huge$

Solve for x, and y using Matrices.

$\underline{EXPLANATION :\;}$

Put the equations in standard form and then use matrices to solve the system of equations.

$8x~+~2y=46, ~7x~+~3y=47$

Write the equations in matrix form.

$\left(\begin{array}{ccc}8&2\\\\7&3\end{array}\right) \left(\begin{array}{ccc}x\\\\y\end{array}\right) = \left(\begin{array}{ccc}46\\\\47\end{array}\right)$

Left multiply the equation by the inverse matrix of : $\left(\begin{array}{ccc}8&2\\\\7&3\end{array}\right).$

$= \left(\begin{array}{ccc}46\\\\47\end{array}\right)\left(\begin{array}{ccc}x\\\\y\endarray\right) = inverse(\left(\begin{array}{ccc}8&2\\\\7&3\end{array}\right))\left(\begin{array}{ccc}46\\\\47\end{array}\right)$

The product of a matrix and its inverse is the identity matrix.

$\left(\begin{array}{ccc}1&&0\\\\0&&1\end{array}\right)$ $\left(\begin{array}{ccc}x\\\\y\end{array}\right)$ $inverse(\left(\begin{array}{ccc}8&2\\\\7&3\end{array}\right)) \left(\begin{array}{ccc}46\\\\47\end{array}\right)$

Multiply the matrices on the left hand side of the equal sign.

$\left(\begin{array}{ccc}x\\\\y\end{array}\right) = inverse(\left(\begin{array}{ccc}8&2\\\\7&3\end{array}\right)) \left(\begin{array}{ccc}46\\\\47\end{array}\right)$

For the $2~x~2$ matrix $\left(\begin{array}{ccc}a&&b\\\\c&&d\end{array}\right)$the inverse matrix is $\left(\begin{array}{ccc}\frac{d}{ad-bc}&&\frac{-b}{ad-bc} \\\\\f\frac{-c}{ad-bc} &&\frac{a}{ad-bc} \end{array}\right)$ so the matrix equation can be rewritten as a matrix multiplication problem.

$\left(\begin{array}{ccc}x\\\\y\end{array}\right) = \left(\begin{array}{ccc}\frac{3}{10} &&-\frac{1}{5} \\-\frac{7}{10} &&\frac{4}{5} \\\end{array}\right) \left(\begin{array}{ccc}46\\\\47\end{array}\right)$

Multiply the matrices.

$\left(\begin{array}{ccc}x\\\\y\end{array}\right) = \left(\begin{array}{ccc}\frac{3}{10}~x~46~-~\frac{1}{5}~~x~47\\\\-\frac{7}{10}~x~46~+~\frac{4}{5}~x~47 \end{array}\right)$

Do the arithmetic.

$\left(\begin{array}{ccc}x\\\\y\end{array}\right) = \left(\begin{array}{ccc}\frac{22}{5}~\\\frac{27}{5} \\\end{array}\right)$

Extract the matrix elements x and y.

$x = \frac{22}{5}, y = \frac{27}{5}$

Hope It Helps...

Sorry, that I answer this question so late...

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