Lesson Video: Two-by-Two Determinants Mathematics • 10th Grade

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.