Question

2x+y-z=3,x+y+z=1,x-2y-3z=4 using cramers method​

Answer 1

Answer:

x = 2 y = -1 z= 0

Step-by-step explanation:

First Let us know the concept to solve this problem .

To use cramer's method we should find Δ , Δ₁ , Δ₂, Δ₃

then we can solve by x =  Δ₁/ Δ

                                    y =  Δ₂/ Δ

                                    z =  Δ₃/ Δ

Now let us see what are these Δ , Δ₁ , Δ₂, Δ₃

Δ  = determinant of the coefficient matrix

Δ₁ = determinant of coefficient matrix with the first column replaced by constants that are on the other side of equation

Δ₂ = determinant of coefficient matrix with the second column replaced by constants that are on the other side of equation

Δ₃ = determinant of coefficient matrix with the third column replaced by constants that are on the other side of equation

if the following equations are considered

ax + by + cz = d

ex + fy + gz  = h

ix   + jy + kz  = l

Then ,

Δ  = $\left[\begin{array}{ccc}a&b&c\\e&f&g\\i&j&k\end{array}\right]$

Δ₁ = $\left[\begin{array}{ccc}d&b&c\\h&f&g\\l&j&k\end{array}\right]$

Δ₂ = $\left[\begin{array}{ccc}a&d&c\\e&h&g\\i&l&k\end{array}\right]$

Δ₃ = $\left[\begin{array}{ccc}a&b&d\\e&f&h\\i&j&l\end{array}\right]$

By these we can find x,y,z

Now let us go into the problem

Δ  = $\left[\begin{array}{ccc}2&1&-1\\1&1&1\\1&-2&-3\end{array}\right]$

    = 2(-3 + 2) -1(-3 - 1) -1(-2-1)

    = -2 + 4 + 3

    = 5

Δ₁ = $\left[\begin{array}{ccc}3&1&-1\\1&1&1\\4&-2&-3\end{array}\right]$

    = 3(-3+2) -1(-3-4) -1(-2-4)

    = -3 + 7 + 6

    = 10

Δ₂ = $\left[\begin{array}{ccc}2&3&-1\\1&1&1\\1&4&-3\end{array}\right]$

    = 2(-3-4) -3(-3-1) -1 (4-1)

    = -14+12-3

    = -5

Δ₃ = $\left[\begin{array}{ccc}2&1&3\\1&1&1\\1&-2&4\end{array}\right]$

    = 2(6) -1 (4-1) +3(-2-1)

    = 12 -3 -9

    =  0

So let us find x, y, z

x =  Δ₁/ Δ

   =  10/5 = 2

y =  Δ₂/ Δ

   =  -5/5 = -1

z =   Δ₃/ Δ

   =   0/5 = 0

Therefore , the values of x,y,z respectively are 2,-1,0 .