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Reduce the matrix A to normal form and hence find its rank where \[ A=\begin{bmatrix}0 &1 &-3 &-1 \\1 &0 &4 &3 \\3 &1 &0 &2 \\1 &1 &-2 &0 \end{bmatrix} \]
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Step 1:

R1 ⇔ R2

A = $\left[\begin{array}{cccc} 1 & 0 & 4 & 3 \\ 0 & 1 & -3 & -1 \\ 3 & 1 & 0& 2 \\ 1 & 1 & -2 & 0 \end{array}\right]$

 

Step 2:

R3 ⇒ R3 - 3(R1)

R4 ⇒ R4 - R1

A = $\left[\begin{array}{cccc} 1 & 0 & 4 & 3 \\ 0 & 1 & -3 & -1 \\ 0 & 1 & -12 & -7 \\ 0 & 1 & -6 & -3 \end{array}\right]$

 

Step 3:

C3 ⇒ C3 - 4(C1)

C4 ⇒ C4 - 3(C1)

A = $\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & -3 & -1 \\ 0 & 1 & -12 & -7 \\ 0 & 1 & -6 & -3 \end{array}\right]$

 

Step 4:

R3 ⇒ R3 - R2

R4 ⇒ R4 - R2

A = $\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & -3 & -1 \\ 0 & 0 & -9 & -6 \\ 0 & 0 & -3 & -2 \end{array}\right]$

 

Step 5:

C3 ⇒ C3 + 3(C2)

C4 ⇒ C4 + C2

A = $\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -9 & -6 \\ 0 & 0 & -3 & -2 \end{array}\right]$

 

Step 6:

R3 ⇒ R3 + 3(R4)

A = $\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -3 & -2 \end{array}\right]$

 

Step 7:

C3 ⇒ C3 / (-3)

A = $\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & -2 \end{array}\right]$

 

Step 8:

C4 ⇒ C4 + 2(C3)

A = $\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}\right]$

 

Step 9:

R3 ⇔ R4

A = $\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$

Rank of matrix A = number of non zero rows

i.e. Rank of matrix A = 3

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