Rivalry2

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Original 2026-01-01

Modified 2026-02-28

1 Let’s learn how to find the rank of a matrix using the minor method. In the minor method, we focus on the<a>determinants</a>of the minors. Follow these steps to find the rank of a matrix:

2 <ul><li>First, find the determinant of the matrix. Let’s consider A as a<a>square</a>matrix If A ≠ 0, then the rank and<a>order of matrix</a>A are the same If A = 0, then the rank is determined by the largest order of any minor that is non-zero. </li>

3 </ul><ul><li>If all the minors of the order are zero, we repeat the steps above for the non-zero minor of one order smaller than before. Follow these till you find a non-zero minor. </li>

4 </ul>The minor method involves the following steps:

5 <ul><li>If A is a square matrix, find the determinant of A. When det(A) ≠ 0, the order of the matrix is equal to the rank of A. When det(A) = 0, the highest order of any non-zero minor that can be found inside the matrix is equal to the rank of A. </li>

6 </ul><ul><li>If every minor of the order is zero, repeat the steps. Then, find a non-zero minor of order that is 1<a>less than</a>the order from the previous step.</li>

7 </ul>Finding det(A), using the 3 × 3 determinant<a>formula</a>:

8 \(det(A) = a (ei - fh) - b (di - fg) + c (dh -eg)\)

9 Substituting the values:

10 \(det(A) = 1 (4 × 6 - 6 × 5) - 2 (2 × 6 - 6 × 4) + 3 (2 × 5 - 4 × 4)\)

11 Now, we can calculate:

12 \( = 1 (4 × 6 - 6 × 5) = 1 (24 - 30) = 1 (-6) = -6\)

13 \( = - 2 (2 × 6 - 6 × 4) = -2 (-12) = 24\)

14 \( = 3 (2 × 5 - 4 × 4) = 3 (-6) = -18\)

15 \(det(A) = -6 + 24 - 18 = 0 \)

16 Hence, \(det(A) = 0\), the rank of the matrix is less than 3.

17 Check for non-zero minors of order 2.

18 Determinant \(= det(A) = ad - bc \) \( = (1 × 5) - (2 × 4) = 5 - 8 \) \( = -3 ≠ 0 \) As the result is non-zero, the rank of the matrix is:

19 The rank of \(A (ρ(A)) = 2\).

20 Using Echelon Form:Identifying a non-zero determinant for finding the rank of a matrix using minors is less efficient for large matrices. We can find the rank of a matrix more easily by using a technique known as the Echelon form. The echelon form is used when the matrix is in the form of an upper or lower<a>triangular matrix</a>. By using the<a>elementary row operations</a>, we can convert a matrix to its Echelon form:

21 <ul><li>Interchanging two rows. </li>

22 <li>Multiplying a scalar by a row. </li>

23 <li>Adding a multiple of one row to another row. </li>

24 </ul>To calculate the rank of a matrix using the Echelon form, we have to follow several steps:

25 <ul><li>Using the row/column transformations, convert the matrix into Echelon form. </li>

26 <li>Then, the number of non-zero rows in the resulting matrix equals the rank of the matrix. </li>

27 </ul>A row in a matrix where at least one element is non-zero is called a non-zero row. For example, find the rank of the matrix A \(A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 4 & 5 & 6 \end{bmatrix} \)

28 Now, we convert the matrix to its Echelon form using the elementary row operations.

29 For that, apply the row transformation formula:

30 \(Ri → Ri - k·Rj \)

31 Where Ri = the row to be changed

32 Rj = the pivot row.

33 k= the scalar multiple used to eliminate the entry.

34 Applying \(R2 → R2 -4R1\) and \(R3 → R3 - 7R1\), to eliminate 4 and 7 in row 1.

35 Next, we will apply \(R3 → R3 - 2R2\), we will get:

36 A row containing at least one non-zero element is called a non-zero row. In the final matrix, there are 2 non-zero rows. Therefore, the rank of A = ρ(A) = 2

37 Using Normal Form:The structure of a matrix in normal form is:

38 Where Ir = the identity matrix of order “r”, and the other values in the matrix will be zero. For a rectangular matrix, A is converted into the standard form using the elementary row transformations and column operations. This method is used to calculate the rank of both rectangular matrices and square matrices. For example, find the rank of the matrix A \(A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \)

39 Step 1: We aim to eliminate elements below the first pivot (1) in the first column:

40 \(R2 → R2 - R1\)

41 \(R3 → R3 - 2R1\)

42 \(R4 → R4 - 3R1\)

43 Hence, the matrix becomes: \(\begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -12 \end{bmatrix} \)

44 Step 2:Next, eliminate elements below and above the pivot in column 2:

45 \(R1 → R1 - 2R2\)

46 \(R4 →R4 - R2\)

47 Then, the matrix becomes: \(\begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & 0 & 0 \end{bmatrix} \)

48 Step 3:Next, eliminate above the pivot in column 3:

49 \(R1 → R1 + R3\)

50 \(R2 → R2 - R3\)

51 Step 4:Now, eliminate the 2 in column 4 to obtain the normal form of the matrix, \(C4 → C4 - 2C1\) Now, the matrix becomes:

52 This is in the form:

53 The identity matrix I3 appears on the top-left side of the matrix, and all the other rows are zero. Therefore, the rank of matrix A is:

54 \(ρ(A) = 3\)