Question

If \(2X + \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix}\), then \(X\) is ______.

• A. \(\begin{bmatrix} 2 & 5 \\ 5 & 3 \end{bmatrix}\)
• B. \(\begin{bmatrix} -1 & 3 \\ 2 & -1 \end{bmatrix}\)
• C. \(\begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix}\)
• D. \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

Hint:

Rearrange to 2X = (RHS matrix) - (given matrix), subtract entry-wise, then divide the whole result by 2.

Answer 1

The Correct Option is C

Rather than manipulating the whole matrix in one shot, solve for each of the four unknown entries of $X$ separately, treating them like four independent linear equations. Write $X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. The equation $2X + \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix}$ splits into four scalar equations, one per position: $2a + 1 = 3$, $2b + 2 = 8$, $2c + 3 = 7$, and $2d + 4 = 2$.

Solving the first: $2a = 2$, so $a = 1$. Solving the second: $2b = 6$, so $b = 3$. Solving the third: $2c = 4$, so $c = 2$. Solving the fourth: $2d = -2$, so $d = -1$.

Putting these four numbers back into their positions in $X$ gives the completed matrix.

\[\boxed{X = \begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix}}\]