Rivalry2

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Original 2026-01-01

Modified 2026-02-21

1 Properties<a>of</a>Cramer's rule are:

2 <ol><li>Cramer’s rule solves systems of<a></a><a>linear equations</a>using<a>determinants</a>.

3 </li>

4 <li>Values of the<a>variables</a>can be determined by the<a>determinants</a>.

5 </li>

6 <li>One determinant comes from the<a></a><a>coefficients</a>of variables (<a>coefficient</a>matrix).

7 </li>

8 <li>Another determinant is created by replacing one column with the<a>constants</a>(right-hand side).

9 </li>

10 <li>The value of a variable = determinant with replaced column ÷ coefficient determinant.

11 </li>

12 </ol>Let us see how Cramer’s rule works in the following system of equations:

13 \(2x + y + z = 3\) \(x - y - z = 0\) \(x + 2y + z = 0\)

14 Write the given equations with all coefficients:

15 \(2x + 1y + 1z = 3\\ 1x - 1y - 1z = 0\\ 1x + 2y + 1z = 0\)

16 Create the coefficient matrix and identify its determinant (D):

17 \(D= \begin{bmatrix} 2 & 1 & 1\\[0.3em] 1& -1 &-1 \\[0.3em] 1 & 2 &1 \end{bmatrix}\)

18 Here, the matrix only contains the coefficients of x, y and z.

19 The answer column:

20 \(\begin{bmatrix} 3\\[0.3em] 0 \\[0.3em] 0 \end{bmatrix}\)

21 Now, form Dx by replacing the 1st column with the answer column:

22 Dx = \( \begin{bmatrix} 3 & 1 & 1\\[0.3em] 0& -1 &-1 \\[0.3em] 0 & 2 &1 \end{bmatrix}\)

23 Here, the determinant with x-values is replaced.

24 Create Dy and Dz by replacing the 2nd and 3rd columns with the answer column.

25 Dy = \(\begin{bmatrix} 2 & 3 & 1\\[0.3em] 1& 0 &-1 \\[0.3em] 1& 0&1 \end{bmatrix}\)

26 Dz = \(\begin{bmatrix} 2 & 1 & 3\\[0.3em] 1& -1&0 \\[0.3em] 1& 2&0\end{bmatrix}\)

27 We can use cofactor expansion (3 × 3 rule). For a general 3 × 3 matrix, the formula is:

28 \(\begin{bmatrix} a& b& c\\[0.3em] d& e&f \\[0.3em] g& h&i\end{bmatrix} = aei +bfg + cdh - ceg -bdi - afh\)

29 D = \(\begin{bmatrix} 2 & 1 & 1\\[0.3em] 1& -1&-1 \\[0.3em] 1& 2&1\end{bmatrix}\)

30 D = Apply the cofactor expansion\((2) (-1) (1) = -2\\ (1) (-1) (1) = -1\\ (1) (1) (2) = 2\\ (1) (-1) (1) = -1\\ (1) (-1) (1) = -1\\ (2) (1) (2) = 4 \)

31 \(D = (-2) + (-1) + 2 - (-1) - (-1) - 4 = -1 - 4 + 2 + 1 + 1 = -1\)

32

33 Hence, D = 3

34 Next \(D_x\)= \(\begin{bmatrix} 3 & 1 & 1\\[0.3em] 0& -1&-1 \\[0.3em] 0& 2&1\end{bmatrix}\)

35 \(D_x\) = Apply the cofactor expansion\((3) (-1) (1) = -3\\ (1) (-1) (0) = 0\\ (1) (0) (2) = 0\\ (1) (-1) (0) = 0\\ (1) (-1) (0) = 0\\ (3) (2) (1) = 6\)

36 \(D_x = -3 + 0 + 0 - 0 - 0 - 6 = -3 + 6 = 3\)

37 Thus, \(D_x = 3\)

38

39 \(D_y\) = \(\begin{bmatrix} 2 & 3& 1\\[0.3em] 1& 0&-1 \\[0.3em] 1& 0&1\end{bmatrix}\)

40 \(D_y\) = Apply the cofactor expansion\((2) (0) (1) = 0\\ (3) (-1) (1) = -3\\ (1) (1) (0) = 0\\ (1) (0) (1) = 0\\ (1) (-1) (2) = -2\\ (2) (0) (1) = 0 \)

41 \(D_y = 0 - 3 + 0 - 0 + 2 - 0 = -3 + 2 = -6\)

42

43 \(D_y = -6\)

44 \(D_z\) = \(\begin{bmatrix} 2 & 1& 3\\[0.3em] 1& -1&0 \\[0.3em] 1& 2&0\end{bmatrix}\)

45 \(D_z\) = Apply the cofactor expansion\((2) (-1) (0) = 0\\ (1) (0) (1) = 0\\ (3) (1) (2) = 6\\ (3) (-1) (1) = -3\\ (2) (0) (2) = 0\\ (2) (1) (0) = 0\)

46 \(D_z = 0 + 0 + 6 - (-3) = 6 + 3 = 9\)

47 \(D_z = 9\)

48

49 Now, we can apply Cramer’s rule. \( x = \frac{D_x}{D} = \frac{3}{3} = 1 \)

50 \( y = \frac{D_y}{D} = \frac{-6}{3} = -2 \)

51 \( z = \frac{D_z}{D} = \frac{9}{3} = 3 \)

52 So, the values for the variables:

53 \(x = 1\\ y = -2\\ z = 3\)

54