90 views
Given that a matrix $[A]_{4\times4},$any one row/column is dependent on the others, and given matrix are singular matrix$(|A|=0)$.

And another matrix $B=adj(A),$then find them,

$1)$Rank of the matrix $B$

$2)$Rank of the marix $adj(B)$
0

let take a 3x3 matrix instead of 4x4 matrix for simplicity

The rank of a matrix is defined as the maximum number of linearly independent row/ col vectors in the matrix

Here it's given that  " one row/column is dependent on the others"

which means out of 3 rows , let R1 , R2 rows are linearly independent and  R3 is linear dependent

therefore , rank of the matrix B  = 2

therefore rank of (A) = 2

Now if A is singular then adj(A) is also a singular

A = $\begin{bmatrix} a & b & c & \\ d & e&f & \\ 0&0 & 0 & \end{bmatrix}$

$\begin{bmatrix} 0 & 0 & 0 & \\ 0 & 0&0 & \\ bf -e c&dc-af & ae -db & \end{bmatrix}$

$\begin{bmatrix} bf -e c & dc-af & ae -db& \\ 0 & 0&0 & \\ 0 & 0& 0 & \end{bmatrix}$ = rank  = 1

Now adj(B) is a null matrix

therefore rank = 0

edited by
0
let me know whether it's right or wrong
0
Rank of $A=3$

$1)$Rank of $B=1$

$2)$Rank of $adj(B)=0$
0
i guess u are doing mistake

if rank of A is 3 then  rank of adj(A) =1

so atleast one minor of it will not be 0 so adj(A) will not be singular
0
$A$ is singular,i don't know $adj(A)$ is singular or not?
0
it will not be
0
but in answer rank of B is 1 and rank of adj(B)=0
0
Now check it !
0
wait it did small mistake :3
0
can you edit your matrix  A is not $4\times4$
0
just for simplicity and for the understanding i take matrix 3x3

procedure is imp

adj of 4x4 take so much time :3
+1
Yes, it is right

I also develop one more concept, such a huge dimension of the matrix(Take small dimension).
0

If we take 4 x 4 also

A rank = 3

the last row is  0 0 0 0 right ??

then also when we take adj of it , it also  gives rank = 1

just analyze it yourself :3

In gate exam you don't have enough to calculate the  adj of 4x4 matrix

0
Yes good brother
0

thanks :p

0
can we take $A_{2\times2}$ matrix for simplicity?

like $A=\begin{bmatrix} a &b \\ 0&0 \end{bmatrix}$

and $B=adj(A)=\begin{bmatrix} 0 &-b \\ 0&a \end{bmatrix}$

and $C=adj(B)=\begin{bmatrix} a &b \\ 0&0 \end{bmatrix}=A$

this is right??
0
No because In this question there's 3 condition

rank of A

for 2x2 you can't analyze it in a proper way

cuz rank A = 1

0
for 2x2 matrix you can't get the null matrix
0
what is wrong to take $2\times2$
0

what is wrong to take 2×2

Rank of A=3

1)Rank of B=1

you couldn't get this 3 conditions

+1 vote
1
2
+1 vote