Cramer’s Rule is one of the easiest ways to solve a given equation. Furthermore, it helps in getting to the solution of any one of the variables. As a result, there is no need to solve the whole given equation. Let’s understand the concepts of Cramer’s rule better.

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## Cramer’s Rule

Given is the solution of equations,

a_{11}x + a_{12}y + a_{13}z = d_{1
}_{b21}x + b_{22}y + b_{23}z = d_{2
}a_{31}x + a_{32}y + a_{33}z = d_{3}

We can show it as:

x = |Δ_{x}| / |Δ|

y = |Δ_{y}| / |Δ|

z = |Δ_{z}|/ |Δ|

where **Δ **is the coefficient matric and the condition for the same is: **Δ ≠ 0**

Coefficient Matrix: \(Δ =\begin{vmatrix}{a_1}{_1} & {a_1}{_2} & {a_1}{_3} \\{a_2}{_1} & {a_2}{_2} & {a_2}{_3} \\{a_3}{_1} & {a_3}{_2} & {a_3}{_3} \end{vmatrix} \)

X Matrix: \(Δ_x =\begin{vmatrix}{d_1} & {a_1}{_2} & {a_1}{_3} \\{d_2} & {a_2}{_2} & {a_2}{_3} \\{d_3} & {a_3}{_2} & {a_3}{_3} \end{vmatrix} \)

Y Matrix: \(Δ_y =\begin{vmatrix}{a_1}{_1} & {d_1} & {a_1}{_3} \\{a_2}{_1} & {d_2} & {a_2}{_3} \\{a_3}{_1} & {d_3} & {a_3}{_3} \end{vmatrix} \)

Z Matrix: \(Δ_z =\begin{vmatrix}{a_1}{_1} & {a_1}{_2} & {d_1} \\{a_2}{_1} & {a_2}{_2} & {d_2} \\{a_3}{_1} & {a_3}{_2} & {d_3} \end{vmatrix} \)

### Note

- To solve the given equations for
*x,*a replacement takes place.*x*with the constant column. - To solve the given equations for y
*,*a replacement takes place. - To solve the given equations for z
*,*a replacement takes place.

### Theorem

Now that we know the basic of Cramer’s rule, let’s know about the theoretical results. Two most significant results that we get from Cramer’s rule about the square system are:

**Theorem 1: **A given square system, say, ** Ax = b **gives a very unique result for every column matrix

**b**. This is possible if and only if the determinant of A is not equal to zero i.e. det

**A ≠ 0.**

**Theorem 2: **A given homogeneous square system, say, * Ax = 0 *is going to have a trivial solution for

**x =0.**This is possible if and only if the determinant of A is not equal to zero i.e. det

**A ≠ 0.**

Cramer’s Rule is more of theoretical importance than practical. This is so because it gives the direct values for unknown variables. Hence, we don’t prefer it a lot as it is not an efficient way to solve a given equation when it comes to a large equation.

Importance of Gaussian elimination is still the primary choice. However, Cramer’s Rule can play its part when we need to get the value of a specific unknown only.

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### Advantages of Cramer’s Rule

One of the biggest advantage that Cramer’s rule offers is that we can easily find the unknown variables without the need to know about the other variables. Another fact is that, if either of x,y, or z is in the fraction form, then there is no need of a fraction to get hold of the other values. We can find each value in an independent manner.

In a nutshell, we can say that Cramer’s Rule helps in reducing a lengthy calculation to a greater extent!

**Point to Remember**

- If the value of Δ = 0, Δ
_{x}= 0, Δ_{y}= 0, Δ_{z}= 0, then in such a case, all the equations will be dependent on one another. Also, the given system of equations will have an infinite number of solutions. - If the value of Δ = 0 and two of the three i.e. Δ
_{x}= 0, Δ_{y}= 0 but Δ_{z}is not equal to zero, then the given system of equations will have solutions. This is where the equations are inconsistent.

## Solved Examples on Cramer’s Rule

**Example 1: **Solve the given system of equations using Cramer’s Rule.

x + 3y + 3z = 5

3x + y – 3z = 4

-3x + 4y + 7z = -7

Solution: So, in order to solve the given equation, we will make four matrices. These matrices will help in getting the values of x, y, and z.

x + 3y + 3z = 5

3x + y – 3z = 4

-3x + 4y + 7z = -7

**Coefficient Matrix**

\(D =\begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & -3 \\-3 & 4 & 7 \end{vmatrix} \)

**X Matrix**

\(D_x =\begin{vmatrix} -5 & 2 & 3 \\ 4 & 1 & -3 \\-7 & 4 & 7 \end{vmatrix} \)

**Y Matrix**

\(D_y =\begin{vmatrix} 1 & -5 & 3 \\ 3 & 4 & -3 \\-3 & -7 & 7 \end{vmatrix} \)

**Z Matrix**

\(D_z =\begin{vmatrix} 1 & 2 & 5 \\ 3 & 1 & 4 \\-3 & 4 & -7 \end{vmatrix} \)

Furthermore, we have to solve each matrix. We can even do this manually. It can be a tidious job but having a good practice is a good skill. On solving the matrices, we get the following values:

|D| = 40

|D_{x}| = -40

|D_{y}| = 40

and, |D_{z}| = -80

We can easily calculate the final answers once we get hold of all the determinants.

x = |D_{x}|/|D| = -40 / 40 = -1

y = |D_{y}|/|D| = 40 / 40 = 1

z = |D_{z}|/|D| = -80 / 40 = 2

Hence, the final answer can be given in point notation: *(x,y,z) = (-1,1,2).*

#### Example 2

Solve the given system of equations using Cramer’s Rule**.**

-y – 2z = -8

x + 3z = 2

7x + y + z = 0

Solution:** **Comparing this question to the last one, it is very easy. This is so because there are zero entries in the equations. The zero entries are in z and y column respectively. This becomes a lot simple. Let’s see how.

**Coefficient Matrix**

\(D =\begin{vmatrix} 0 & 1 & 2 \\ 1 & 0 & 3 \\7 & 1 & 1 \end{vmatrix} \)

**X Matrix**

\(D_x =\begin{vmatrix} -8 & -1 & -2 \\ 2 & 0 & 3 \\0 & 1 & 1 \end{vmatrix} \)

**Y Matrix**

\(D_y =\begin{vmatrix} 0 & -8 & 2 \\ 1 & 2 & 3 \\7 & 0 & 1 \end{vmatrix} \)

**Z Matrix**

\(D_z =\begin{vmatrix} 0 & -1 & 8 \\ 1 & 0 & 2 \\7 & 1 & 0 \end{vmatrix} \)

On solving these, we get the respective values for the determinant x, y, and z.

|D| = -22

|D_{x}| = 22

|D_{y}| = -132

and, |D_{z}| = -22

Now, we can finally get to the answer.

x = |D_{x}|/|D| = 22 / -22 = -1

y = |D_{y}|/|D| = -132 / -22 = 6

z = |D_{z}|/|D| = -22 / -22 = 1

Hence, the answer is: *(x,y,z) = (-1,6,1)*

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