Question Video: Evaluating Determinants | Nagwa Question Video: Evaluating Determinants | Nagwa

Question Video: Evaluating Determinants Mathematics • First Year of Secondary School

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Evaluate the shown determinant |𝑖², 0, 1 + 𝑖 and 1 + 𝑖, 𝑖², 0 and 1 βˆ’ 𝑖, βˆ’π‘–, 𝑖²|, where 𝑖² = βˆ’1.

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Video Transcript

Evaluate the shown determinant where 𝑖 squared is equal to negative one.

Let’s start by using the instruction that 𝑖 squared is equal to negative one and replacing every instance of 𝑖 squared in our matrix with negative one. To find the determinant of the three-by-three matrix shown, we multiply each of the elements of the top row by the determinant of the two-by-two matrix that is not in their row or column. This is called their cofactor.

Let’s start by evaluating the first cofactor. To find the determinant of a two-by- two matrix, we multiply the top left and bottom right elements and subtract the product of the top right and bottom left. Negative one multiplied by negative one minus zero multiplied by negative 𝑖 is one. The first part of our three-by-three determinant is negative one multiplied by one.

We now need to work out the determinant of the second two- by-two matrix. One plus 𝑖 multiplied by negative one minus zero multiplied by one minus 𝑖 is negative one minus 𝑖. The second part in finding the determinant for the three-by-three matrix is therefore zero multiplied by negative one minus 𝑖. Finally, we need to find the determinant of the third two-by-two matrix.

Once again we find the product of the top left and bottom right and subtract the product of the top right and bottom left. That gives us negative 𝑖 minus 𝑖 squared plus one minus 𝑖, which becomes negative two 𝑖 minus negative one plus one, which simplifies to negative two 𝑖 plus two. Substituting this back into our formula for the determinant gives us one plus 𝑖 multiplied by negative two 𝑖 plus two.

The first expression simplifies to negative one, and the second part of the expression simplifies to zero. Expanding these two brackets gives us negative one plus negative two 𝑖 plus two minus two 𝑖 squared plus two 𝑖. And then we can simplify further to give us one minus two 𝑖 squared. Remember, 𝑖 squared is negative one. So our expression becomes one minus two times negative one, which is three. The shown determinant has a value of three.

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