Question

2x-y+3z=8,-x+2y+z=4,3x+y-4z=0 by crammers rule

Answer 1

.e)⎡⎣⎢11220−3130⎤⎦⎥⎡⎣⎢xyz⎤⎦⎥=⎡⎣⎢7111⎤⎦⎥[1211032−30][xyz]=[7111]

Where A=⎡⎣⎢11220−3130⎤⎦⎥,X=⎡⎣⎢xyz⎤⎦⎥A=[1211032−30],X=[xyz] and B=⎡⎣⎢7111⎤⎦⎥B=[7111]

Let us now find the determinant value of A

|A|=⎡⎣⎢11220−3130⎤⎦⎥[1211032−30]

=1(0+9)−2(0−6)+1(−3−0)=1(0+9)−2(0−6)+1(−3−0)

=9+12−3=18≠0.

Answer 2

Given,

Equations,

2x - y + 3z = 8,

-x + 2y + z = 4,

3x + y - 4z = 0.

To find,

Solution by Cramer's rule.

Solution,

We can solve this problem simply by following the below process.

Here, we have to solve the system of given equations using Cramer's rule.

So, to solve a given system of equations by this rule, first, we arrange them in matrix form,

AX = B

Where,

A = coefficient matrix,

X = column matrix with variables,

B = column matrix with constant numbers which are on the right-hand side of equations.

So the given equations are,

$2x-y+3z=8,\\-x+2y+z=4,\\3x+y-4z=0$

In matrix form,

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

$X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$

$B=\left[\begin{array}{ccc}8\\4\\0\end{array}\right]$

Now, we can find x, y, and z as,

$x=\frac{D_x}{D} ,y=\frac{D_y}{D} ,z=\frac{D_z}{D}$

Where,

D is the determinant of matrix A that is |A|, and

$D_x$ = determinant of matrix obtained by replacing the elements in the first column of matrix A with column matrix B,

$D_y$ = determinant of matrix obtained by replacing the elements in the second column of matrix A with column matrix B,

$D_z$ = determinant of matrix obtained by replacing the elements in the third column of matrix A with column matrix B,

So,

$D=\left|\begin{array}{ccc}2&-1&3\\-1&2&1\\3&1&-4\end{array}\right|$

⇒ $D =$ -38

$D_x=\left|\begin{array}{ccc}8&-1&3\\4&2&1\\0&1&-4\end{array}\right|$

⇒ $D_x =$ -76

$D_y=\left|\begin{array}{ccc}2&8&3\\-1&4&1\\3&0&-4\end{array}\right|$

⇒ $D_y=$ -76

$D_z=\left|\begin{array}{ccc}2&-1&8\\-1&2&4\\3&1&0\end{array}\right|$

⇒ $D_z=$ -76

So now,

$x=\frac{-76}{-38}=2,\\y=\frac{-76}{-38}=2,\\z=\frac{-76}{-38}=2$

So, the values of x, y, and z are

$x =2,\\y=2,\\z=2$

Therefore, the solution for the given system of equations by Cramer's rule will be x = 2, y = 2, and, z = 2.