Question Video: Evaluating Determinants with Complex Numbers | Nagwa Question Video: Evaluating Determinants with Complex Numbers | Nagwa

# Question Video: Evaluating Determinants with Complex Numbers Mathematics • First Year of Secondary School

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Evaluate the determinant of the matrix 7 0 π + 1, 0 1 9π, βπ + 1 β4π β10.

03:05

### Video Transcript

Evaluate the determinant of the matrix given by seven, zero, π plus one, zero, one, nine π, negative π plus one, negative four π, and negative 10.

Finding the determinant of a three-by-three matrix is a little more complicated than finding the determinant for a two-by-two. So we need to be really careful carrying out each step. The determinant for a three-by-three matrix is found using determinants of two-by-two matrices as shown. This may look a little complicated.

However, it is simply the elements of the top row multiplied by the determinant of their minors, that is, the two-by-two matrix thatβs left when we eliminate the row and the column weβre looking at.

Letβs substitute the values from our matrix into this formula. Next, we need to find the determinants of the two-by-two matrices. Remember, we work this out by finding the product of the top-left and bottom-right elements and subtracting the product of the top-right and bottom-left elements.

The determinant of our first two-by-two matrix is, therefore, one multiplied by negative 10 minus 9π multiplied by negative four π. Remember, π is the square root of negative one. So π squared is just negative one. Our expression becomes negative 10 minus 36, which is negative 46. We can substitute this back into our original formula.

Now at this stage, we could work out the determinant for the second two-by-two matrix. However, ultimately, weβre just going to multiply it by zero, which will give a value of zero. Instead, we can go straight on to finding the determinant of our final two-by-two matrix.

The determinant here is zero multiplied by negative four π minus one multiplied by negative π plus one. This simplifies to π minus one. Again, letβs substitute this back into our formula. And our determinant is going to be calculated by working out the product of seven and negative 46, subtracting zero, then adding the product of π plus one and π minus one.

Seven multiplied by negative 46 is negative 322. And π plus one multiplied by π minus one can be calculated pretty quickly using the rule of the difference of two squares to get π squared minus one. Once again, π squared gives us a value of negative one. So our expression becomes negative 322 minus one minus one. This gives us negative 324. The determinant of this three-by-three matrix is negative 324.

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