Question Video: Determinants and Invertibility | Nagwa Question Video: Determinants and Invertibility | Nagwa

# Question Video: Determinants and Invertibility Mathematics • First Year of Secondary School

## Join Nagwa Classes

Is the matrix 3, 1 and β3, 1 invertible?

01:47

### Video Transcript

Is the matrix three, one, negative three, one invertible?

For a two-by-two matrix π΄, where π΄ is equal to π, π, π, π, its inverse is given by the formula one over the determinant of π multiplied by π, negative π, negative π, π, where the determinant of π΄ is given by ππ minus ππ.

Notice that this means that if the determinant of the matrix π΄ is zero, the inverse does not exist, since one over the determinant of π΄ would be one over zero, which is undefined.

What we need to decide then is whether the determinant of the matrix three, one, negative three, one is equal to zero. That will tell us whether it has an inverse or not.

To find the determinant, we multiply the first element in the first row by the second element in the second row. Itβs three multiplied by one. Then we subtract the product of the second element in the first row and the first element in the second row. Thatβs one multiplied by negative three. That gives us three minus negative three, which is six.

Since the determinant of our matrix is not zero, it does have an inverse; itβs invertible. In fact, we can substitute the values we now know into our formula for the inverse of a matrix. It becomes one-sixth multiplied by one, negative one, three, three. Multiplying everything by one over the determinant gives us one-sixth, negative one-sixth, one-half, and one-half.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions