### Video Transcript

In this video, we will learn how to
identify determinants and evaluate two-by-two determinants. We will begin by looking at some
definitions and the rule we will use to calculate the determinant of a two-by-two
matrix. The determinant of a matrix is
denoted by absolute value bars instead of square brackets. For example, if we consider the
two-by-two matrix 𝑎, 𝑏, 𝑐, 𝑑, then the determinant of this matrix will be
written as 𝑎, 𝑏, 𝑐, 𝑑 inside absolute value bars.

For any two-by-two matrix, the
determinant is found by subtracting the product of its diagonals. The determinant of the two-by-two
matrix 𝑎, 𝑏, 𝑐, 𝑑 is equal to 𝑎𝑑 minus 𝑏𝑐. We multiply the value in the top
left and bottom right of the matrix and then subtract the product of the values in
the top right and bottom left. We will now look at some specific
questions where we need to calculate the determinant of a two-by-two matrix.

Find the determinant of the
following matrix five, one, negative one, five.

We recall that the determinant of
any matrix is written inside absolute value bars. The determinant of the two-by-two
matrix 𝑎, 𝑏, 𝑐, 𝑑 is equal to 𝑎𝑑 minus 𝑏𝑐. The determinant of the matrix five,
one, negative one, five is equal to five multiplied by five minus one multiplied by
negative one. Five multiplied by five is equal to
25. Multiplying a positive number by a
negative number gives a negative answer. Therefore, one multiplied by
negative one is negative one. We know that subtracting negative
one is the same as adding one. The determinant of the matrix five,
one, negative one, five is therefore equal to 26.

In our next question, we will find
the determinant of a different two-by-two matrix.

Find the value of the determinant
of the matrix negative two, negative nine, seven, seven.

The absolute value bars either side
of our matrix indicate we need to calculate the determinant. We know that in order to calculate
the determinant of a two-by-two matrix 𝑎, 𝑏, 𝑐, 𝑑, we need to subtract 𝑏𝑐 from
𝑎𝑑. In our question, we begin by
multiplying negative two and seven. We then need to subtract negative
nine multiplied by seven. Multiplying a negative and a
positive number gives a negative answer.

So negative two multiplied by seven
is negative 14. We need to subtract negative 63
from this. Subtracting negative 63 is the same
as adding 63. Negative 14 plus 63 is equal to
49. We can see this on the number line
shown as 14 plus 49 is equal to 63. Adding 14 to negative 14 takes us
to zero, and adding another 49 takes us to 49. The determinant of the matrix
negative two, negative nine, seven, seven is 49.

In our next question, we need to
find out the determinant in the form of an algebraic expression.

Find the value of the determinant
of matrix 𝐴: 𝑥, negative 11, 𝑥, negative one.

The determinant of a matrix is
denoted by absolute value bars, and the determinant of any two-by-two matrix of the
form 𝑎, 𝑏, 𝑐, 𝑑 is equal to 𝑎𝑑 minus 𝑏𝑐. This means that the determinant of
the matrix 𝑥, negative 11, 𝑥, negative one is equal to 𝑥 multiplied by negative
one minus negative 11 multiplied by 𝑥. Multiplying the top-left and
bottom-right value gives us negative 𝑥. Multiplying the top-right and
bottom-left value gives us negative 11𝑥. We need to subtract this from
negative 𝑥. This can be simplified to negative
𝑥 plus 11𝑥. The determinant of matrix 𝐴: 𝑥,
negative 11, 𝑥, negative one is therefore equal to 10𝑥.

In our next question, we need to
solve an equation using our knowledge of determinants.

If the determinant of the matrix
one, 𝑥, 𝑥, three is equal to the determinant of the matrix two, one, four, three,
then 𝑥 equals blank.

We recall that the determinant of
any two-by-two matrix of the form 𝑎, 𝑏, 𝑐, 𝑑 is equal to 𝑎𝑑 minus 𝑏𝑐. Let’s begin by looking at the
matrix one, 𝑥, 𝑥, three. Multiplying the top-left and
bottom-right values gives us three. Multiplying the top-right and
bottom-left values gives us 𝑥 squared. This means the determinant of this
matrix is three minus 𝑥 squared. We can repeat this process for our
second matrix. Two multiplied by three is equal to
six. One multiplied by four is equal to
four. As six minus four is equal to two,
the determinant of this matrix is two.

In order to calculate the value of
𝑥, we need to solve the equation three minus 𝑥 squared is equal to two. We can add 𝑥 squared and subtract
two from both sides of this equation. This gives us three minus two is
equal to 𝑥 squared. As 𝑥 squared is equal to one, we
can square root both sides of this equation. The square root of one is equal to
one. This means that 𝑥 could be
positive one or negative one, as after square rooting a number, our answer can be
positive or negative. If the determinant of the two
matrices are equal, then 𝑥 can be equal to one or negative one.

We could check this answer by
substituting these values back into the expression three minus 𝑥 squared. Squaring one or negative one gives
us an answer of one. Therefore, we have three minus
one. As this is equal to two, which was
the determinant of the second matrix, we know that our answers are correct. 𝑥 can be equal to one or negative
one.

In our final question, we will find
the determinant of a matrix using our knowledge of trigonometry.

Evaluate the determinant of a
two-by-two matrix 10 cos 𝑥, negative two sin 𝑥, 10 sin 𝑥, two cos 𝑥.

We know that the determinant of any
two-by-two matrix 𝑎, 𝑏, 𝑐, 𝑑 is equal to 𝑎𝑑 minus 𝑏𝑐. In this question, we begin by
multiplying 10 cos 𝑥 and two cos 𝑥. 10 multiplied by two is equal to
20, and cos 𝑥 multiplied by cos 𝑥 is cos squared 𝑥. 10 cos 𝑥 multiplied by two cos 𝑥
is therefore equal to 20 cos squared 𝑥. We also need to multiply negative
two sin 𝑥 and 10 sin 𝑥. This gives us negative 20 sin
squared 𝑥. The determinant can be rewritten as
20 cos squared 𝑥 plus 20 sin squared 𝑥. At this stage, we can factor out
the highest common factor of 20. This gives us 20 multiplied by cos
squared 𝑥 plus sin squared 𝑥.

One of our trigonometrical
identities states that sin squared 𝜃 plus cos squared 𝜃 is equal to one. This means that cos squared 𝑥 plus
sin squared 𝑥 is also equal to one. We need to multiply 20 by one. The determinant of the two-by-two
matrix 10 cos 𝑥, negative two sin 𝑥, 10 sin 𝑥, two cos 𝑥 is therefore equal to
20.

We will now summarize the key
points of this video. The determinant of a two-by-two
matrix 𝑎, 𝑏, 𝑐, 𝑑 is written as the absolute value of 𝑎, 𝑏, 𝑐, 𝑑. This is equal to 𝑎𝑑 minus
𝑏𝑐. We multiply our values in the top
left and bottom right of the matrix and then subtract the product of the values in
the top right and bottom left of the matrix. It is important to note that whilst
we’re using absolute value symbols, the determinant of a matrix can be negative. It is simply found by subtracting
the products of the diagonals.