Let A = < b m a t r i x > and B =

Let $A = \begin{bmatrix} 8 & 7 \\ 0 & 9 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 5 \\ 1 & 3 \end{bmatrix}$. \\ Find the matrix $C$ of the linear transformation $T(x) = B(A(x))$. \\ $C = \begin{bmatrix} \\ \\ \end{bmatrix}$

Transcript

00:01 Hi there, in this problem a linear transformation is given and t 80 square plus pt plus c is equal to this matrix.

00:08 Basis b and c are given and we have to find the matrix representation.

00:14 So let us start with the solution to this problem.

00:18 Firstly we will find t of t square.

00:22 Note that here a is equal to 1, b is equal to 0 and c is equal to 0.

00:32 So t of t squared will be given by t of t square is equal to matrix 2 multiplied by 1 minus 3 multiplied by 0.

00:52 4 multiplied by 1 this will be plus and this is minus 5 multiplied by 0.

00:58 And in the third row we have 6 multiplied by 0 plus 6 multiplied by 0.

01:04 So this is equal to 240 and since c is a basis so we can write this is equal to c1 multiplied by 1 00 plus c2 multiplied by 110 plus c3 multiplied by 1 1 1 and 1 so this means that the matrix 240 is equal to the matrix c1 plus c2 plus c3, c2 plus c3 and in the third row we have c3.

02:03 So comparing the two above matrix we can get c3 is equal to 0, c2 plus c3 is equal to 4 and c1 plus c2 plus c3 is equal to 2.

02:22 From c2 plus c3 equals 4, if we substitute the value of c3 we will get c2 plus 0 is equal to 4 and this means that c2 is equal to 4.

02:37 Again, c1 plus c2 plus c3 is equal to 2 gives c1 4 plus 0 is equal to 2 and from here we can get c1 is equal to negative 2.

02:57 So we get that the matrix 240 is equal to negative 2 multiplied by matrix 100 plus 4 multiplied by matrix 110 plus 0 multiplied by matrix 1110 plus 0 multiplied by matrix 111.

03:29 Again we will find t of t squared minus 1.

03:39 Note that here a is equal to 1, b is equal to 0 and c is equal to negative 1.

03:47 So we will get t of t squared minus 1 is equal to the matrix 2 multiplied by 1 minus plus 3 multiplied by here b is equal to 0.

04:11 In the next row we have 4 multiplied by 1 minus 5 multiplied by negative 1 and in the next row we have 6 multiplied by 0 plus 6 multiplied by negative 1.

04:30 So this is equal to the matrix 2, 4 plus 5 is equal to 9 and negative 6.

04:43 This would be equal to c1 multiplied by 100 plus c2 multiplied by 1110 plus c3 multiplied by 111.

05:04 This means that t of t2 minus 1 is equal to matrix 29 minus 6 which is equal to matrix c1 plus c2 plus c3.

05:22 C2 plus c3 and c3...

Question

Let $A = \begin{bmatrix} 8 & 7 \\ 0 & 9 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 5 \\ 1 & 3 \end{bmatrix}$. \\ Find the matrix $C$ of the linear transformation $T(x) = B(A(x))$. \\ $C = \begin{bmatrix} \\ \\ \end{bmatrix}$

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