Evaluate the determinant of the given matrix by first using elementary row operations to reduce

step 1 Section 3 .2 problem number 11. So we're dealing with a 4 -by -4 determinant, and so we're asked

Evaluate the determinant of the given matrix by first using...

Key Concepts

Effect of Row Operations on Determinants

Understanding the effect of row operations on determinants is crucial because while adding a multiple of one row to another does not change the determinant, swapping rows reverses the sign of the determinant and scaling a row multiplies the determinant by the scaling factor. This knowledge ensures correct computation of the determinant after the matrix has been manipulated.

Upper Triangular Matrix

An upper triangular matrix is one in which all the elements below the main diagonal are zero. When a matrix is transformed into this form, either by row operations or by being inherently upper triangular, the determinant can be easily calculated as the product of the diagonal entries. This property greatly simplifies the process of computing determinants.

Determinant

The determinant is a scalar value associated with a square matrix that can provide important insights about the matrix, such as whether it is invertible (non-zero determinant) and its scaling effect on space (e.g., volume scaling factor). It is computed from the elements of the matrix and encapsulates key properties in linear algebra.

Elementary Row Operations

Elementary row operations, which include swapping two rows, multiplying a row by a non-zero scalar, and adding a scalar multiple of one row to another, are used to simplify matrices and solve systems of linear equations. Each of these operations has a specific impact on the determinant, making them useful tools for systematic calculation while requiring adjustments to account for changes in the determinant.

Transcript

00:01 Section 3 .2 problem number 11.

00:05 So we're dealing with a 4 -by -4 determinant, and so we're asked to use elementary row operations to reduce it to upper triangular form, which means zeros everywhere below the diagonals.

00:17 So let's just methodically just start clearing column by column.

00:21 So if i want to clear that first element, so in order to do that, i can do minus seven halves times row 1 plus row 2 and let that become my new row 2.

00:37 So this gives me 2 minus 1, 3, 4.

00:45 So negative 7 halves times 2 is negative 7, add that the 7 and you get a 0.

00:51 Negative 7 halves times negative 1 is 7 halves.

00:55 So 7 halves plus 1, that is 7 halves plus 2 halves, which is 9 halves.

01:01 Negative seven halves times three negative 21 halves negative 21 halves plus two that's negative 21 halves plus four halves which is negative 17 halves and then negative seven halves times four is negative 14 halves negative 14 halves plus three that's a negative 14 halves plus six halves so negative seven halves times four sorry negative 7 has and 4 is negative 28 halves or negative 14 negative 14 plus 3 is negative 11 okay and so now the other two remote rows remain unchanged negative 2 4 8 and 6 6 minus 6 18 and minus 24 now i'm going to have to continue using you know row 1 to clear row 3 and 4 i'm going to use row 2 to clear rows 3 and 4 so i can get rid of the fractions here so if i were to multiply row 2 by 2 so when you multiply 2 times row 2 you're multiplying the determinant by 2 so to keep this the same this is going to be 1 half the determinant of 2 minus 1 3 4 and then multiply row 2 by 2 and you get 0 9 minus 17 minus 22 so i get rid of all the fractions so i don't carry them with me the rest of the journey.

02:49 So 6 minus 6, 18 minus 24.

02:55 So now let's just continue.

02:56 The next thing i would need to do would be to clear that negative 2 that you see there.

03:01 So what i can do with that is just take row 1 plus row 3 and let that become my new row 3.

03:09 So what this will give me is 1 1⁄2.

03:13 So then i say 2 minus 1 3 4 0 9 minus 17 minus 22 and then when i add 2 in negative 2 i get a 0 at negative 1 and 4 and i get a 3 at 8 and 3 and i get 11 and then add 4 and 6 and i get a 10 row 4 remains unchanged my next target would be to eliminate the 6 here so that can happen with negative 3 times row 1 plus row 4 and let that become the new row 4.

04:01 So this becomes 1 1⁄2 minus 1, 3, 4, 09, minus 17, minus 22, 0 3, 11, 10, minus 3, 11, 10.

04:24 Negative 3 times negative 1 is 3 so 3 plus negative 6 is minus 3 negative 3 is negative 9 negative 9 plus 18 is 9 negative 3 times 4 is negative 12 negative 24 is negative 36 next target will be to eliminate this 3 that you see here and i can do that or let's see what i could in this case, you look at it and say, okay, well, in order to use row two, to clear everything, i'm going to have to take one third of that.

05:10 So one thing that can make this a little bit easier, why don't we just switch rows? why don't i switch row two and row four? so if i were to switch row two, interchange it with row four.

05:23 It's going to make my math a little bit easier...

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