Answered

Matrix [tex]$A=\left[\begin{array}{cc}6 & -12 \\ 4 & 8\end{array}\right]$[/tex]. What is matrix [tex]$X$[/tex] if [tex]$\frac{1}{2} X=A$[/tex]?

A. [tex]\left[\begin{array}{cc}3 & -6 \\ 2 & 4\end{array}\right][/tex]

B. [tex]\left[\begin{array}{cc}5.5 & -12.5 \\ 3.5 & 7.5\end{array}\right][/tex]

C. [tex]\left[\begin{array}{cc}6.5 & -11.5 \\ 4.5 & 8.5\end{array}\right][/tex]

D. [tex]\left[\begin{array}{cc}12 & -24 \\ 8 & 16\end{array}\right][/tex]



Answer :

GinnyAnswer
To solve for matrix [tex]\(X\)[/tex] given that [tex]\(\frac{1}{2} X = A\)[/tex], where [tex]\(A = \left[\begin{array}{cc}6 & -12 \\ 4 & 8\end{array}\right]\)[/tex], we need to find a matrix [tex]\(X\)[/tex] such that when halved, equals matrix [tex]\(A\)[/tex].

Here’s a step-by-step approach:

1. Start with the equation [tex]\(\frac{1}{2} X = A\)[/tex].

2. To clear the fraction, multiply both sides of the equation by 2:
[tex]\[ X = 2A \][/tex]

3. Substitute the given matrix [tex]\(A\)[/tex] into the equation:
[tex]\[ X = 2 \left[\begin{array}{cc}6 & -12 \\ 4 & 8\end{array}\right] \][/tex]

4. Perform scalar multiplication by individually multiplying each element in matrix [tex]\(A\)[/tex] by 2:
[tex]\[ X = \left[\begin{array}{cc} 2 \cdot 6 & 2 \cdot (-12) \\ 2 \cdot 4 & 2 \cdot 8 \end{array}\right] \][/tex]

5. This results in:
[tex]\[ X = \left[\begin{array}{cc}12 & -24 \\ 8 & 16\end{array}\right] \][/tex]

So, matrix [tex]\(X\)[/tex], given [tex]\(\frac{1}{2} X = A\)[/tex], is:
[tex]\[ \left[\begin{array}{cc}12 & -24 \\ 8 & 16\end{array}\right]. \][/tex]

Thus, the correct answer from the options is:
[tex]\[ \left[\begin{array}{cc}12 & -24 \\ 8 & 16\end{array}\right]. \][/tex]
Hi1315

Answer:

D

Step-by-step explanation:

Given the matrix  [tex]A = \left[\begin{array}{cc} 6 & -12 \\ 4 & 8 \end{array}\right][/tex]and the equation [tex]\frac{1}{2} X = A[/tex], we need to find matrix  X .

First, let's isolate  X  in the equation. Multiply both sides of the equation by 2:

X = 2A

Now, let's calculate  2A :

[tex]2 \times \left[\begin{array}{cc} 6 & -12 \\ 4 & 8 \end{array}\right] \\\\ \left[\begin{array}{cc} 2 \times 6 & 2 \times -12 \\ 2 \times 4 & 2 \times 8 \end{array}\right] \\\\\left[\begin{array}{cc} 12 & -24 \\ 8 & 16 \end{array}\right][/tex]

Therefore, the correct matrix  X  is:

D.[tex]\left[\begin{array}{cc} 12 & -24 \\ 8 & 16 \end{array}\right][/tex]