(9x3 - 2x2 - 3) - (4x3 - 9x + 6)

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dowlingdestiny

31.12.2021

Mathematics

$\therefore A^{T}=\frac{2}{3} \left[\begin{array}{cc}-4&8\\-5&4\\8&-8\end{array}\right]$

Step-by-step 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

$A=\frac{2}{3} \left[\begin{array}{ccc}-4&-5&8\\8&4&-8\end{array}\right]$

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

$=\left[\begin{array}{ccc}\frac{-8}{3} &\frac{-10}{3}&\frac{16}{3}\\\ \\ \frac{16}{3}&\frac{8}{3}&\frac{-16}{3}\end{array}\right]$

$A^{T}=\left[\begin{array}{cc}\frac{-8}{3} &\frac{16}{3}\\\ \\ \frac{-10}{3}&\frac{8}{3}\\ \\ \frac{16}{3}&\frac{-16}{3}\end{array}\right]$

$=\frac{2}{3} \left[\begin{array}{cc}-4&8\\-5&4\\8&-8\end{array}\right]$

$\therefore A^{T}=\frac{2}{3} \left[\begin{array}{cc}-4&8\\-5&4\\8&-8\end{array}\right]$

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