10. Consider the following matrix B = [ 1 0 -k -1 3 1 0 4k...

10. Consider the following matrix B = [ 1 0 -k -1 3 1 0 4k -8 ] where k is a constant. (a) Use a suitable cofactor expansion to calculate det(B) in terms of k. (b) For which values of k does B not have an inverse? 6. Decide if these statements are true or false. Think carefully about why! (a) The determinant det A is defined for any matrix A. (b) The determinant det A is defined for each square matrix A. (c) The determinant of a square matrix is a scalar. (d) If an n x n matrix A is multiplied by a scalar c, the determinant of the resulting matrix is c det A. (e) If an n x n matrix A is multiplied by a scalar c, the determinant of the resulting matrix is c^n det A. (f) If det A = 2 and det B = 3, then det(A + B) = 5. (g) If det A = 2 and det B = 3, then det(AB) = 6.

Transcript

00:01 In this question we are given a vector d is equal to negative of 1, 0, negative of k, negative of 1, 3, 1, sorry, this is positive.

00:18 0, 4k, negative of 8.

00:23 For the first subpart, we need to obtain the 2 factor expansion of the determinant of the given.

00:39 Matrix.

00:41 Now the determinant of p is given by 1 times the co -factor 3 -1, 4k, negative of 8, that is the 4 -factor of 1, then negative of negative of 1 times co -factor of this negative of 1, that is 0, negative of k, 4k and negative of 8, plus 0 times co -factor of this 0 ,000, and negative of 8, plus 0 times, co -factor of this 0, that is 0 negative of k 3 and 1.

01:16 We have expanded along first column.

01:20 Therefore, solving this we get the determinant as 1 times negative of 24 minus 4k plus 1 times 0 plus 4k square.

01:36 That is, the determinant of v is given by 4k square minus 4k squared minus 4k.

01:44 24.

01:45 This is the expression of determinant of metrics b.

01:50 And this is the solution for the first subpart.

01:54 Further, for the second subpart, we need to obtain the value of k for which the vector b is, matrix b is not invertible.

02:09 Now we know that any matrix is not invertible if it is singular.

02:16 That is if its determinant is 0.

02:19 Now, substituting the value of determinant, we get 4k squared minus 4k minus 24 is equal to 0.

02:30 That is k square minus k minus 6 is equal to 0.

02:41 Now, we're using middle ton supplation.

02:46 The sectors of these quadratic equation are k minus 3.

02:51 And k plus two is equal to 0.

02:54 Equating both the sectors with 0, we obtain the value of k as 3 and negative of 2.

03:00 Therefore, for these values of k, the determinant would be equal to 0 and the metrics will not be invertible.

03:07 Hence, this is the solution for the second subpart.

03:11 Further, for the last subpart, we have some true or false.

03:18 We need to check whether the given statements are true or false.

03:22 The first statement is that the determinant of a is defined for any matrix a...

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