Question

the matrix Find the the normal form and rank of matrix a =[8 1 3 6 - 0 3 2 2 - -8 -1 -3 4]
​

Answer 1

Rank of matrix A is 3

Step-by-step explanation:

$A=\left[\begin{array}{cccc}8&1&3&6\\0&3&2& 2\\-8&-1&-3&4\end{array}\right]$

The order of matrix A is 3 × 4

∴ The rank of the matrix  ≤ 3

Now consider any third order minor of A.

$=\left[\begin{array}{ccc}8&1&3\\0&3&2\\-8&-1&3\end{array}\right]$

By Expanding the first row

$=8[9-(-2)]-1[0-(-16)]+3[0-(-24)]$

[tex]= 8(9+2) -1(0+16)+3(0+24)\\ = 8(11)-1(16)+3(24)\\ =88-16+72\\ =144\neq 0[/tex]

Therefore there is a minor or 3 which is not zero

Therefore the rank of the matrix A is 3.

Answer 2

Step-by-step explanation:

tep-by-step explanation:

\begin{gathered}A=\left[\begin{array}{cccc}8&1&3&6\\0&3&2& 2\\-8&-1&-3&4\end{array}\right] \end{gathered}

A=

⎣

⎢

⎡

8

0

−8

1

3

−1

3

2

−3

6

2

4

⎦

⎥

⎤

The order of matrix A is 3 × 4

∴ The rank of the matrix ≤ 3

Now consider any third order minor of A.

\begin{gathered}=\left[\begin{array}{ccc}8&1&3\\0&3&2\\-8&-1&3\end{array}\right] \end{gathered}

=

⎣

⎢

⎡

8

0

−8

1

3

−1

3

2

3

⎦

⎥

⎤

By Expanding the first row

=8[9-(-2)]-1[0-(-16)]+3[0-(-24)]=8[9−(−2)]−1[0−(−16)]+3[0−(−24)]

\begin{gathered}= 8(9+2) -1(0+16)+3(0+24)\\ = 8(11)-1(16)+3(24)\\ =88-16+72\\ =144\neq 0\end{gathered}

=8(9+2)−1(0+16)+3(0+24)

=8(11)−1(16)+3(24)

=88−16+72

=144



=0

Therefore there is a minor or 3 which is not zero

Therefore the rank of the matrix A is 3.