Question

Express the inverse of the following matrix (assuming it exists) as a matrix containing expressions in terms of k.

Answer 1

Answer:

the Inverse of the matrix is; A⁻¹

$= \left[\begin{array}{ccc}\frac{-3}{(2k)} &\frac{1}{k}&\frac{15}{(2k)}\\\frac{-3k+15}{(10k)}&\frac{-1}{k}&\frac{-75+13k}{(10k)}\\\frac{1}{2}&0&\frac{-5}{2}\end{array}\right]$

Step-by-step explanation:

Given the data in the question;

Matrix A = $\left[\begin{array}{ccc}-25&-25&-13\\k&0&3\\-5&-5&-3\end{array}\right]$

To find the Inverse of matrix A

A⁻¹ = Adj.A / det.A

so

Determinant of the matrix A will be

|A| = -25( 0 + 15) + 25( -3K + 15 ) - 13( -15K + 0 )

= -375 - 75K + 375 + 65K

= -10K

Now, the adjoin of matrix A will be

Adj.A = $\left[\begin{array}{ccc}15&-10&-75\\3k-15&10&75-13k\\-5k&0&25k\end{array}\right]$

The Inverse of the matrix is;

A⁻¹ = Adj.A / det.A

$= \frac{1}{-10k} \left[\begin{array}{ccc}15&-10&-75\\3k-15&10&75-13k\\-5k&0&25k\end{array}\right]$

$= \left[\begin{array}{ccc}\frac{15}{-10k} &\frac{-10}{-10k}&\frac{-75}{-10k}\\\frac{3k-15}{-10k}&\frac{10}{-10k}&\frac{75-13k}{-10k}\\\frac{-5k}{-10k}&\frac{0}{-10k}&\frac{25k}{-10k}\end{array}\right]$

$= \left[\begin{array}{ccc}\frac{-3}{2k} &\frac{1}{k}&\frac{15}{2k}\\\frac{-3k+15}{10k}&\frac{-1}{k}&\frac{-75+13k}{10k}\\\frac{1}{2}&0&\frac{-5}{2}\end{array}\right]$

$= \left[\begin{array}{ccc}\frac{-3}{(2k)} &\frac{1}{k}&\frac{15}{(2k)}\\\frac{-3k+15}{(10k)}&\frac{-1}{k}&\frac{-75+13k}{(10k)}\\\frac{1}{2}&0&\frac{-5}{2}\end{array}\right]$

Therefore the Inverse of the matrix is; A⁻¹

$= \left[\begin{array}{ccc}\frac{-3}{(2k)} &\frac{1}{k}&\frac{15}{(2k)}\\\frac{-3k+15}{(10k)}&\frac{-1}{k}&\frac{-75+13k}{(10k)}\\\frac{1}{2}&0&\frac{-5}{2}\end{array}\right]$