Question

Solve the equations x + y = 0, y + z = 1 and x + z = 3 for y by Cramer's rule.
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Answer 1

Answer:

Final answers will be as follows:

x =2

y = -2

z = -1

Answer 2

Given:

$x+y=0$

$y+z=1$

$x+z=3$

To find:

Value of $y$ by Cramer's rule.

Solution:

If we have a system of linear equations in 3 variables,

$a_{1}x+b_{1}y+c_{1}z=d_{1}$

$a_{2}x+b_{2}y+c_{2}z=d_{2}$

$a_{3}x+b_{3}y+c_{3}z=d_{3}$

By Cramer's rule to solve for $x,y,z$, we have $x=\frac{D_{x}}{D}, y=\frac{D_{y}}{D} ,z=\frac{D_{z}}{D}$

where $D$ is a 3×3 determinant of matrix of all coefficients of $x,y,z$.

$D=\left[\begin{array}{ccc}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{array}\right]$

The first row and first column is eliminated and the remaining is maintained as such to calculate determinant of first element, first row and middle column is eliminated and the remaining is maintained as such to calculate determinant of second element and last row and last column is eliminated and the remaining is maintained as such to calculate determinant of third element.

$D=a_{1}\left[\begin{array}{cc}b_{2}&c_{2}\\b_{3}&c_{3}\\\end{array}\right] -b_{1}\left[\begin{array}{cc}a_{2}&c_{2}\\a_{3}&c_{3}\\\end{array}\right] +c_{1}\left[\begin{array}{cc}a_{2}&b_{2}\\a_{3}&b_{3}\\\end{array}\right]$

$D=a_{1}(b_{2}c_{3}-b_{3}c_{2})-b_{1}(a_{2}c_{3}-a_{3}c_{2})+c_{1}(a_{2}b_{3}-a_{3}b_{2})$

Here, we are given three equations,

$x+y=0$

$y+z=1$

$x+z=3$

The coefficients of $x$ will be written in first column, coefficients of $y$ in second column and coefficients of $z$ in last column of the matrix.

$D=\left[\begin{array}{ccc}1&1&0\\0&1&1\\1&0&1\end{array}\right]$

Calculate the determinant by following as shown below:

$D=1\left[\begin{array}{cc}1&1\\0&1\\\end{array}\right] -1\left[\begin{array}{cc}0&1\\1&1\\\end{array}\right] +0\left[\begin{array}{cc}0&1\\1&0\\\end{array}\right]$

$D=1(1*1-0*1)-1(0*1-1*1)+0(0*0-1*1)$

$D=1(1-0)-1(0-1)+0(0-1)$

$D=1(1)-1(-1)+0(-1)$

$D=1+1+0=2$

To solve for $y$,

$D_{y}=\left[\begin{array}{ccc}a_{1}&d_{1}&c_{1}\\a_{2}&d_{2}&c_{2}\\a_{3}&d_{3}&c_{3}\end{array}\right]$

Find its determinant as shown below:

$D=a_{1}\left[\begin{array}{cc}d_{2}&c_{2}\\d_{3}&c_{3}\\\end{array}\right] -d_{1}\left[\begin{array}{cc}a_{2}&c_{2}\\a_{3}&c_{3}\\\end{array}\right] +c_{1}\left[\begin{array}{cc}a_{2}&d_{2}\\a_{3}&d_{3}\\\end{array}\right]$

$D=a_{1}(d_{2}c_{3}-d_{3}c_{2})-d_{1}(a_{2}c_{3}-a_{3}c_{2})+c_{1}(a_{2}d_{3}-a_{3}d_{2})$

Substituting the values in the above formula to solve for $y$,

$D_{y}=1\left[\begin{array}{cc}1&1\\3&1\\\end{array}\right] -0\left[\begin{array}{cc}0&1\\1&1\\\end{array}\right] +0\left[\begin{array}{cc}0&1\\1&3\\\end{array}\right]$

$D_{y}=1(1*1-1*3)-0(0*1-1*1)+0(0*3-1*1)$

$D_{y}=1(1-3)-0(0-1)+0(0-1)$

$D_{y}=1(-2)-0(-1)+0(-1)$

$D_{y}=-2-0+0=-2$

Now,

$y=\frac{D_{y}}{D}$

$y=\frac{-2}{2}=-1$

∴ Value of $y$ by Cramer's rule is $-1$.

The value of $y$ by Cramer's rule is -1.