Question

solve the system of equations 3x+y-z=3,-x+y+z=1,x+y+z=3​

Answer 1

$\large\underline{\sf{Solution-}}$

Given system of equations are

$\rm :\longmapsto\:3x + y - z = 3$

$\rm :\longmapsto\: - x + y + z = 1$

$\rm :\longmapsto\: x + y + z = 3$

We use Cramer's Rule to solve this system of equations.

The matrix form of the above system of equations is

$\rm :\longmapsto\:A \: = \: \begin{gathered}\sf \left[\begin{array}{ccc}3&1& - 1\\ - 1&1&1\\1&1&1\end{array}\right]\end{gathered}$

$\begin{gathered}\sf\rm :\longmapsto\: B=\left[\begin{array}{c}3\\1\\3\end{array}\right]\end{gathered}$

$\begin{gathered}\sf\rm :\longmapsto\: X=\left[\begin{array}{c}x\\y\\z\end{array}\right]\end{gathered}$

Now,

Consider,

$\rm :\longmapsto\:A \: = \: \begin{gathered}\sf \left[\begin{array}{ccc}3&1& - 1\\ - 1&1&1\\1&1&1\end{array}\right]\end{gathered}$

So,

$\rm :\longmapsto\: |A| = \begin{gathered}\sf \left | \begin{array}{ccc}3&1& - 1\\ - 1& 1&1\\1&1& 1\end{array}\right | \end{gathered}$

$\rm \: = \: \: 3(1 - 1) - 1( - 1 - 1) - 1( - 1 - 1)$

$\rm \: = \: \: 0 + 2 + 2$

$\rm \: = \: \: 4$

Now, |A| is non - zero, it means system of equations have unique solution.

So,

$\rm :\longmapsto\:D_1 = \begin{gathered}\sf \left | \begin{array}{ccc}3&1& - 1\\ 1& 1&1\\3&1& 1\end{array}\right | \end{gathered}$

$\rm \: = \: \: 3(1 - 1) - 1(1 - 3) - 1(1 - 3)$

$\rm \: = \: \: 0 + 2 + 2$

$\rm \: = \: \: 4$

Now,

$\rm :\longmapsto\:D_2 = \begin{gathered}\sf \left | \begin{array}{ccc}3&3& - 1\\ - 1& 1&1\\1&3& 1\end{array}\right | \end{gathered}$

$\rm \: = \: \: 3(1 - 3) - 3( - 1 - 1) - 1( - 3 - 1)$

$\rm \: = \: \: - 6 + 6 + 4$

$\rm \: = \: \: 4$

Now,

$\rm :\longmapsto\:D_3 = \begin{gathered}\sf \left | \begin{array}{ccc}3&1&3\\ - 1& 1&1\\1&1& 3\end{array}\right | \end{gathered}$

$\rm \: = \: \: 3(3 - 1) - 1( - 3 - 1) + 3( - 1 - 1)$

$\rm \: = \: \: 6 + 4 - 6$

$\rm \: = \: \: 4$

So, we have now

$\red{\rm :\longmapsto\: |A| = 4}$

$\red{\rm :\longmapsto\:D_1 = 4}$

$\red{\rm :\longmapsto\:D_2 = 4}$

$\red{\rm :\longmapsto\:D_3 = 4}$

Now, using Cramer's rule,

$\blue{\bf :\longmapsto\:x = \dfrac{D_1}{ |A| } = \dfrac{4}{4} = 1}$

$\blue{\bf :\longmapsto\:y = \dfrac{D_2}{ |A| } = \dfrac{4}{4} = 1}$

$\blue{\bf :\longmapsto\:z = \dfrac{D_3}{ |A| } = \dfrac{4}{4} = 1}$

Hence,

The solution of system of equations is

• x = 1

• y = 1

• z = 1

Answer 2

Step-by-step explanation:

x = 1

y = 1

z = 1 this is the answer