Respuesta :
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.
[tex]\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right][/tex]· X·[tex]\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right][/tex] =I.
First, we have to identify the matrix I. As it was said, the matrix is the identiy matrix, which means
I = [tex]\left[\begin{array}{ccc}1&0\\0&1\end{array}\right][/tex]
So, [tex]\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right][/tex]· X·[tex]\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}1&0\\0&1\end{array}\right][/tex]
Isolating the X, we have
X·[tex]\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right][/tex]= [tex]\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right][/tex] - [tex]\left[\begin{array}{ccc}1&0\\0&1\end{array}\right][/tex]
Resolving:
X·[tex]\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right][/tex]= [tex]\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right][/tex]
X·[tex]\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right][/tex]=[tex]\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right][/tex]
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=[tex]\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right][/tex]⁻¹·[tex]\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right][/tex]
Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.
So,
[tex]\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right][/tex]·[tex]\left[\begin{array}{ccc}a&b\\c&d\end{array}\right][/tex]=[tex]\left[\begin{array}{ccc}1&0\\0&1\end{array}\right][/tex]
9a - 3b = 1
7a - 6b = 0
9c - 3d = 0
7c - 6d = 1
Resolving these equations, we have a=[tex]\frac{2}{11}[/tex]; b=[tex]\frac{7}{33}[/tex]; c=[tex]\frac{-1}{11}[/tex] and d=[tex]\frac{-3}{11}[/tex]. Substituting:
X= [tex]\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right][/tex]·[tex]\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right][/tex]
Multiplying the matrices, we have
X=[tex]\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11} \end{array}\right][/tex]
