Question

(1 point) Solve for X. $\begin{bmatrix} -9 & -8 \ -4 & 4 \end{bmatrix} X + \begin{bmatrix} -7 & 8 \ -9 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \ -1 & 7 \end{bmatrix}$. $\begin{bmatrix} \frac{-10}{9} & \\ -2 & \end{bmatrix}$

          (1 point) Solve for X.
$\begin{bmatrix} -9 & -8 \ -4 & 4 \end{bmatrix} X + \begin{bmatrix} -7 & 8 \ -9 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \ -1 & 7 \end{bmatrix}$.
$\begin{bmatrix} \frac{-10}{9} & \\
-2 & \end{bmatrix}$

(1 point) Solve for X.
< b m a t r i x >
X +
< b m a t r i x >
=
< b m a t r i x >.
< b m a t r i x >

Added by Robert W.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

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Solved by Expert Suman Saurav Thakur

Step 1

Start by simplifying the expression inside the square root: √1 = 1. Show more…

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Solve for X. [-x + [-âˆš1] = -(10âˆš9) X =

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Question

(1 point) Solve for X. $\begin{bmatrix} -9 & -8 \ -4 & 4 \end{bmatrix} X + \begin{bmatrix} -7 & 8 \ -9 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \ -1 & 7 \end{bmatrix}$. $\begin{bmatrix} \frac{-10}{9} & \\ -2 & \end{bmatrix}$

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