Made easy workbook

Question

Let $A = (a_{ij})_{n\times n}$ such that $a_{ij}=3$ for all $i,j$ then nullity of $A$ is

• A. $n-1$
• B. $n-3$
• C. $n$
• D. $0$

Comments

I didn't get the matrix A's expression properly. Is it that all the elements of the matrix A which is of order 'n' is a_ij (where a_ij=3) ?

---

yes ..all the elements of this matrix = 3..So , rank of this matrix A = 1 ..Since , rank(A) + nullity (A) = no. of columns in A.. So, 1 + nullity(A) = n ...So, nullity(A) = n - 1

---

Yes then it's (n-1).

Answer 1

nullity= number of columns of A- rank of matrix.