True or false? If the row space of A is R^n then det(A) does not equal to 0.

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rakesh05's profile pic

rakesh05 | High School Teacher | (Level 1) Assistant Educator

Posted on

True.

If we take the matrix A to be n x n.

Because the row space of A is R^n. Means the rows of the matrix A forms a basis of R^n. i.e. the row vectors are linearly independent. So by the definition of linear independence the determinant of the matrix corresponding to these vectors will be non zero.

i.e.  if  `v_1=(v_11,v_12,v_13, .....,v_(1n)),`

         `v_2=(v_21,v_22,v_23, ....v_(2n)),`

             .......................................

         `v_n=(v_(n1),v_(n2),v_(n3),......,v_(n n))`  are vectors in R^n.

These vectors will be linearly independent if and only if

`det [[v_11,v_12, ....., v_(1n)],[v_21,v_22,......,v_(2n)],......,[v_(n1),v_(n2),......,v_(n n)]]`

`!=0` .

pramodpandey's profile pic

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

True.

Since

Row space of A is  `R^n` .

`=> ` rows of A are linearly independent.

`det(A)!=0`

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