Q7. Consider the square matrices A =

Q7. Consider the square matrices $A = \begin{bmatrix} 3 & 5 \\ 2 & 3 \end{bmatrix}$, $O = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$, and $B = \begin{bmatrix} A & A^2 \\ O & A \end{bmatrix}$. Use block matrices to find, a) $B^{-1}$. b) det$(B)$. Note: For $A = \begin{bmatrix} A_1 & A_2 \\ O & A_3 \end{bmatrix}$, $A^{-1} = \begin{bmatrix} A_1^{-1} & -A_1^{-1}A_2A_3^{-1} \\ 0 & A_3^{-1} \end{bmatrix}$ and det$(A) = $det$(A_1)$det$(A_3)$

Transcript

00:02 Let a and b be 4 by 4 matrices, that is, square matrices with 4 rows and 4 columns, with determinant of a equal negative 1 and determinant of b equal 2.

00:16 We will use properties of the determinant to compute in part a determinant of a times b, in part b determinant of b raised to the 5th power, in part c determinant of 2 times a, in part d determinant of a transpose times a, and in part e determinant of b inverse times a times b.

00:40 So let's start with part a and we are going to be talking about the properties of the determinants when we use it.

00:48 In this case, the first property we got to use is the determinant of a part of 2 square matrices a and b of the same order is equal to the same order of matrices is important to be sure we can do the product a times b and that the resulting matrix is a square matrix again.

01:10 So the determinant of a times b is the determinant of a times the determinant of b.

01:20 That's the property of the determinant and so we know these two numbers, determinant of a is negative 1 times determinant of b is 2, so it's negative 2.

01:31 In other words, determinant of a times b is negative 2.

01:40 Part a where we apply property here of determinants, determinant of the product of 2 square matrices of the same order is equal to the product of the determinant of the matrices.

01:53 In part b, determinant of b raised to the 5th power.

01:57 First of all, the 5th power of b can be calculated because b is a square matrix.

02:04 So we can multiply b times b and we get again a 4 by 4 matrix and that result which is b squared can be multiplied again by b and we will get again a 4 by 4 matrix and we can go on that way.

02:18 In other words, this is equal to the determinant of b times b times b times b times b.

02:27 Of course, we can do it by grouping first 4 b's and that times b and we apply the property right here in part a.

02:37 But if we do it, we know we are going to get this is equal to the determinant of b times the determinant of b 5 times this way times continue down here.

03:02 So remember this result comes from the property we talked about in part a because we have determinant of the product of 2 matrices is the product of the determinant of the factor matrices.

03:15 And here we have 5 products of b by itself and we apply recursively by grouping first 4, then 3 and then 2 on one term of this product here and apply successively this property right here we get this that is determinant of the product of b by itself 5 times is equal to the product of the determinant of b by itself 5 times.

03:41 Or in other words, this is determinant of b to the 5th and now determinant of b we know is 2.

03:52 So we get 2 raised to the 5th power and that is 32.

03:59 So determinant of b to the 5th is equal to 32 and that's part b.

04:11 Now in part c, we want to calculate determinant of 2 times a.

04:19 So when we do a scalar times a matrix, the result is equal to the scalar raised to the order of the matrix that is 2 raised to the 4th in this case because a is 4 by 4 times the determinant of the matrix.

04:39 That is we don't have exactly the scalar getting out of the determinant but a power of the scalar and the exponent of that power is just the order of the square matrix a.

04:53 So in this case will be 16 times the determinant of a is negative 1 so we get negative 16...

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