First slide

Elementary row operations (Transformation)

Question

A matrix is said to be of rank r when it contains at least one non-zero minor of order r and no such minor of order r + 1. The rank of a matrix is denoted by $ρ$ (A).
By means of elementary transformations every non-zero matrix of rank r can be reduced to one of the following forms
$(A) [\begin{matrix} I_{r} & ⋮ & 0 \\ \dots & \dots & \dots \\ 0 & 0 \end{matrix}]$ $(B) [\begin{matrix} I_{r} \\ \dots \\ 0 \end{matrix}]$ $(C) [I_{r} ⋮ 0]$ $(D) [I_{r}]$
Where I_r is a r-rowed unit matrix. These are called the normal forms of the given matrix and the value of r is the rank of the
matrix. In order to find the rank of a given matrix, reduce the matrix to its normal form. This process of reducing a matrix of
rank r to its normal form is known as ‘The sweep-out process’.

Moderate

Question

The rank of the matrix $A = [\begin{array}{llll} 0 & 1 & 2 & 1 \\ 1 & 2 & 3 & 2 \\ 3 & 1 & 1 & 3 \end{array}]$ is

A

1
B

2
C

3
D

4

Solution

we have
$\begin{aligned} [\begin{array}{cccc} 0 & 1 & 2 & 1 \\ 1 & 2 & 3 & 2 \\ 3 & 1 & 1 & 3 \end{array}] ~ [\begin{array}{cccc} 1 & 0 & 2 & 1 \\ 2 & 1 & 3 & 2 \\ 1 & 3 & 1 & 3 \end{array}] C_{1} \leftrightarrow C_{2} \\ ~ [\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 2 & 1 & - 1 & 0 \\ 1 & 3 & - 1 & 2 \end{array}] C_{3} \to C_{3} - 2 C_{1} \\ ~ [\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & - 1 & 0 \\ 0 & 3 & - 1 & 2 \end{array}] (\begin{array}{c} R_{2} \to R_{2} - 2 R_{1} \\ R_{3} \to R_{3} - R_{1} \end{array}) \\ ~ [\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 3 & 2 & 2 \end{array}] (C_{3} \to C_{3} + C_{2}) \\ ~ [\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 2 \end{array}] (R_{3} \to R_{3} - 3 R_{2}) \end{aligned}$
$\begin{aligned} ~ [\begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{array}] (R_{3} \to \frac{1}{2} R_{3}) \\ ~ [\begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}] (C_{4} \to C_{4} - C_{3}) \end{aligned}$
Hence, [I₃ : 0] is the normal form of A and, therefore, the
rank of the matrix A is 3

Question

Rank of the matrix $A = [\begin{array}{rrrr} 1 & - 1 & 2 & - 3 \\ 4 & 1 & 0 & 2 \\ 0 & 3 & 1 & 4 \\ 0 & 1 & 0 & 2 \end{array}]$ is

A

1
B

2
C

3
D

4

Solution

We have,
$\begin{aligned} A & = [\begin{array}{rrrr} 1 & - 1 & 2 & - 3 \\ 4 & 1 & 0 & 2 \\ 0 & 3 & 1 & 4 \\ 0 & 1 & 0 & 2 \end{array}] \\ ~ [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 5 & - 8 & 14 \\ 0 & 3 & 1 & 4 \\ 0 & 1 & 0 & 2 \end{array}] \\ = (\begin{matrix} C_{2} \to C_{2} + C_{1} \\ C_{3} \to C_{3} - 2 C_{1} \\ C_{4} \to C_{4} + 3 C_{1} \end{matrix}) \end{aligned}$
$\begin{aligned} ~ [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 3 & 1 & 4 \\ 0 & 5 & - 8 & 4 \end{array}] (R_{2} \leftrightarrow R_{4}) \\ ~ [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 3 & 1 & - 2 \\ 0 & 5 & - 8 & 4 \end{array}] (C_{4} \to C_{4} - 2 C_{2}) \\ ~ [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & - 2 \\ 0 & 5 & - 8 & 4 \end{array}] (R_{3} \to R_{3} - 3 R_{2}) \\ ~ [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 5 & - 8 & - 12 \end{array}] (C_{4} \to C_{4} + 2 C_{3}) \\ ~ [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & - 12 \end{array}] [\begin{matrix} R_{4} \to R_{4} - 5 R_{1} \\ R_{4} \to R_{4} + 8 R_{3} \end{matrix}] \\ ~ [\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}] = I_{4} [C_{4} \to \frac{- 1}{12} C_{4}] \end{aligned}$
Hence, $ρ$ (A) = 4.

Question

The rank of the matrix $A = [\begin{array}{rrrr} 1 & 3 & 4 & 3 \\ 3 & 9 & 12 & 9 \\ - 1 & - 3 & - 4 & - 3 \end{array}]$ is

A

1
B

2
C

3
D

0

Solution

$\begin{aligned} A = [\begin{array}{rrrr} 1 & 3 & 4 & 3 \\ 3 & 9 & 12 & 9 \\ - 1 & - 3 & - 4 & - 3 \end{array}] \\ ~ [\begin{array}{rrrr} 1 & 3 & 4 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}] R_{2} \to R_{2} - 3 R_{1} \\ R_{3} \to R_{3} + R_{1} \end{aligned}$
The equivalent matrix is in Echelon form. The number of non-zero rows in this matrix is 1. Therefore, the rank of A =1.

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