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Exercise 7.5 · Q2

Q.What must be the matrix XX, if 2X+[1234]=[3872]2X + \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix}?

(1) [132−1]\begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix}
(2) [1−32−1]\begin{bmatrix} 1 & -3 \\ 2 & -1 \end{bmatrix}
(3) [264−2]\begin{bmatrix} 2 & 6 \\ 4 & -2 \end{bmatrix}
(4) [2−64−2]\begin{bmatrix} 2 & -6 \\ 4 & -2 \end{bmatrix}
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✓ Free question

Rearrange the matrix equation and divide by 22.

This uses matrix subtraction and scalar multiplication.

Step 1. Move the known matrix across: 2X=[3872]−[1234]2X = \begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}.

Step 2. Subtract entry-wise: 2X=[264−2]2X = \begin{bmatrix} 2 & 6 \\ 4 & -2 \end{bmatrix}.

Step 3. Multiply by 12\tfrac{1}{2}: X=[132−1]X = \begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix}.

✓Final answer

The correct option is (1) [132−1]\begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix}.

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