Question

Find the transpose of the matrix A.

Aequals=left bracket Start 2 By 3 Matrix 1st Row 1st Column negative 4 2nd Column negative 5 3rd Column 8 2nd Row 1st Column 8 2nd Column 4 3rd Column negative 8 EndMatrix right bracket

−4 −5 8
8 4 −8
Fill in the matrix with entries of the transpose.

left bracket Start 2 By 3 Matrix 1st Row 1st Column negative 4 2nd Column negative 5 3rd Column 8 2nd Row 1st Column 8 2nd Column 4 3rd Column negative 8 EndMatrix right bracket Superscript Upper T

−4 −5 8
8 4 −8
Tequals=left bracket Start 3 By 2 Matrix 1st Row 1st Column nothing 2nd Column nothing 2nd Row 1st Column nothing 2nd Column nothing 3rd Row 1st Column nothing 2nd Column nothing EndMatrix right bracket

Answer 1

`\therefore A^(T)=(2)/(3) \left[\begin{array}{cc}-4&8\\-5&4\\8&-8\end{array}\right]`

Explanation:

Matrix transpose :The matrix transpose is switches the rows and column elements.

If a constant term multiplies with a matrix , then the constant term will multiply each element of the matrix.

Given

`A=(2)/(3) \left[\begin{array}{ccc}-4&-5&8\\8&4&-8\end{array}\right]`

`=\left[\begin{array}{ccc}(2)/(3) .(-4)&(2)/(3)(-5)&(2)/(3)8\\ \\ (2)/(3).8&(2)/(3).4&(2)/(3).(-8)\end{array}\right]`

`=\left[\begin{array}{ccc}(-8)/(3) &(-10)/(3)&(16)/(3)\\\ \\ (16)/(3)&(8)/(3)&(-16)/(3)\end{array}\right]`

`A^(T)=\left[\begin{array}{cc}(-8)/(3) &(16)/(3)\\\ \\ (-10)/(3)&(8)/(3)\\ \\ (16)/(3)&(-16)/(3)\end{array}\right]`

`=(2)/(3) \left[\begin{array}{cc}-4&8\\-5&4\\8&-8\end{array}\right]`

`\therefore A^(T)=(2)/(3) \left[\begin{array}{cc}-4&8\\-5&4\\8&-8\end{array}\right]`