Question

Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

Answer 1

The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

$\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$· X·$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =I.

First, we have to identify the matrix I. As it was said, the matrix is the identiy matrix, which means

I = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

So, $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$· X·$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =  $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Isolating the X, we have

X·$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$= $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ -  $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Resolving:

X·$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$= $\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]$

X·$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$=$\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X=$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$⁻¹·$\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$·$\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$=$\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a=$\frac{2}{11}$; b=$\frac{7}{33}$; c=$\frac{-1}{11}$ and d=$\frac{-3}{11}$. Substituting:

X= $\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right]$·$\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Multiplying the matrices, we have

X=$\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11} \end{array}\right]$