Video: Checking Whether a Matrix Is Invertible

Does the matrix 𝐴 = 5, −4, −5, and 0, −9, 0 and 2, 7, −2 have a multiplicative inverse?

03:38

Video Transcript

Does the matrix 𝐴 is equal to the three-by-three matrix five, negative four, negative five, zero, negative nine, zero, two, seven, negative two have a multiplicative inverse?

In this question, we’re given a matrix 𝐴, and we can see that our matrix 𝐴 has three rows and three columns. We’re asked to determine whether this matrix 𝐴 has a multiplicative inverse. To answer this question, it’s first important to realize exactly what the question is asking us. We’re asked to determine whether or not our matrix 𝐴 has a multiplicative inverse or not. We’re not asked to calculate this multiplicative inverse if it exists. So to answer this question, all we need to do is determine whether this multiplicative inverse exists or not. And to do this, we’re going to need to recall some information about matrices.

To start, we’ll recall that our matrix 𝐴 will always have a multiplicative inverse if the determinant of 𝐴 is nonzero. And in fact, if we know the determinant of our matrix is not equal to zero, then we can always calculate its inverse. We could do this by using the adjoint method. Similarly, we also know that a matrix 𝐴 will not have a multiplicative inverse in any other situation. For example, if the determinant of our matrix is equal to zero, we call our matrix singular.

However, there is another case when we can’t calculate the determinant of our matrix. This would be when our matrix is rectangular and not a square matrix. Then our matrix does not have a multiplicative inverse because it might have a left or right multiplicative inverse, but these won’t be equal. This means all we need to do to check whether a matrix has a multiplicative inverse is to determine whether its determinant is equal to zero or not. So let’s start by calculating the determinant of our three-by-three matrix 𝐴. And to do this, we have a few different options. Because 𝐴 is a three-by-three matrix, we’re going to do this by expanding over either a row or a column.

And remember, whenever we want to evaluate the determinant of a matrix by expanding over a row or a column, we should always look for the row or column with the most number of zeros since this will make the calculation the easiest. In our case, we can see this is the second row because it has two zeros, and we know when we expand over this row, the entries are going to be coefficients of the terms. So the first term and the third term will have a coefficient of zero. In other words, they’re just going to be equal to zero. Therefore, our only nonzero term will be the term we get expanding over the second row and second column.

Remember, we need to multiply negative nine by the determinant of the matrix minor we get by removing the second row and second column from matrix 𝐴. And we also need to multiply this by negative one raised to the power of the row number plus the column number. And in this case, our coefficient comes from the second row and second column, so this is negative one to the power of two plus two. Therefore, we’ve shown the determinant of matrix 𝐴 is negative one to the power of two plus two multiplied by negative nine times the determinant of the two-by-two matrix five, negative five, two, negative two.

To evaluate this expression, we’re going to need to recall how we evaluate the determinant of a two-by-two matrix. We recall the determinant of the two-by-two matrix 𝑎, 𝑏, 𝑐, 𝑑 is equal to 𝑎𝑑 minus 𝑏𝑐. We take the difference in the products of the diagonals. Applying this to our equation and simplifying negative one to the power of two plus two to be one, we get one times negative nine multiplied by five times negative two minus negative five times two. And if we calculate this expression, we get negative nine times negative 10 plus 10 which simplifies to give us negative nine multiplied by zero, which is of course just equal to zero.

Therefore, we’ve shown matrix 𝐴 has zero determinant, and we know that this means that our matrix 𝐴 must have no multiplicative inverse. Therefore, we were able to show the three-by-three matrix five, negative four, negative five, zero, negative nine, zero, two, seven, negative two does not have a multiplicative inverse because its determinant is equal to zero.

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