Solve for $$ X $$ in the equation, where $$ \begin{array}{l} A=\left[\begin{array}{lll} -5 & 4 & 6 \ -2 & 0 & 2 \end{array}\right] \text { and } B=\left[\begin{array}{rrr} 0 & 1 & -3 \ 4 & 0 & 2 \end{array}\right] \ 4 B=-2 X-2 A \end{array} $$ Step 1 The matrix equation is given as $$ 4 B=-2 X-2 A $$. The first step is to solve the equation for $$ X $$ just as in linear equations. Add $$ 2 A $$ to both sides of the equation and then divide both sides of the equation by -2 to solve for $$ X $$. $$ (-2 \vee-2) B-A=X $$ Step 2 $$ \left.x=(-1)\left[\begin{array}{lllll} -5 & -5 & 4 & 4 & 6 \ -2 & -2 & 0 & 0 & 2 \end{array}\right]+\left(\begin{array}{ll} -2 & 0 \end{array}\right] \begin{array}{ll} -2 \end{array}\right]\left[\begin{array}{lll} 0 & 1 & -3 \ 4 & 0 & 2 \end{array}\right] $$ Step 3 Perform the scalar multiplication before the addition of matrices. Remember you can multiply a matrix by a scalar $$ c $$ by multiplying each entry in the matrix by $$ c $$. Multiply each entry in $$ A $$ by $$ (-1) $$ and each entry in $$ B $$ by $$ (-2) $$. $$ x=\left[\begin{array}{rrr} 5 & -4 & -6 \ 2 & 0 & -2 \end{array}\right]+\left[\begin{array}{lll} \square & \square & \square \ 0 & 0 & \square \ \square & \square \end{array}\right] $$

Question

UpStudy AI · Accepted Answer

**Step 1: Isolate $$ X $$ in the Equation**

We start with the given matrix equation:
$$
4B = -2X - 2A.
$$
Add $$ 2A $$ to both sides:
$$
4B + 2A = -2X.
$$
Now, divide both sides of the equation by $$-2$$:
$$
X = -\frac{1}{2}(4B + 2A).
$$
We can factor the expression inside the parentheses:
$$
X = -\frac{1}{2}\cdot 2(2B + A) = - (2B + A).
$$
Thus,
$$
X = -2B - A.
$$

**Step 2: Compute the Matrices $$-2B$$ and $$-A$$**

The given matrices are:
$$
A=\begin{pmatrix} -5 & 4 & 6 \ -2 & 0 & 2 \end{pmatrix} \quad \text{and} \quad B=\begin{pmatrix} 0 & 1 & -3 \ 4 & 0 & 2 \end{pmatrix}.
$$

First, compute $$-2B$$:
$$
-2B = -2 \begin{pmatrix} 0 & 1 & -3 \ 4 & 0 & 2 \end{pmatrix}
= \begin{pmatrix} -2\cdot0 & -2\cdot1 & -2\cdot(-3) \ -2\cdot4 & -2\cdot0 & -2\cdot2 \end{pmatrix}
= \begin{pmatrix} 0 & -2 & 6 \ -8 & 0 & -4 \end{pmatrix}.
$$

Next, compute $$-A$$:
$$
-A = -1 \cdot \begin{pmatrix} -5 & 4 & 6 \ -2 & 0 & 2 \end{pmatrix}
=\begin{pmatrix} 5 & -4 & -6 \ 2 & 0 & -2 \end{pmatrix}.
$$

**Step 3: Add $$-2B$$ and $$-A$$ to Find $$ X $$**

Now, add the matrices:
$$
X = (-2B) + (-A) = \begin{pmatrix} 0 & -2 & 6 \ -8 & 0 & -4 \end{pmatrix} + \begin{pmatrix} 5 & -4 & -6 \ 2 & 0 & -2 \end{pmatrix}.
$$

Add corresponding entries element-wise:

- For the first row:
  - $$ x_{11}: 0 + 5 = 5 $$
  - $$ x_{12}: -2 + (-4) = -6 $$
  - $$ x_{13}: 6 + (-6) = 0 $$

- For the second row:
  - $$ x_{21}: -8 + 2 = -6 $$
  - $$ x_{22}: 0 + 0 = 0 $$
  - $$ x_{23}: -4 + (-2) = -6 $$

So, the matrix $$ X $$ is:
$$
X = \begin{pmatrix} 5 & -6 & 0 \ -6 & 0 & -6 \end{pmatrix}.
$$
$$ X = \begin{pmatrix} 5 & -6 & 0 \ -6 & 0 & -6 \end{pmatrix} $$

Solución de inteligencia artificial de Upstudy

Responder

Solución

Beyond the Answer

preguntas relacionadas

Latest Algebra Questions