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If [2x+3y+12x−z3y]=[7−2−4−9]\begin{bmatrix} 2x + 3 & y + 1 \\ 2x - z & 3y \end{bmatrix} = \begin{bmatrix} 7 & -2 \\ -4 & -9 \end{bmatrix}, then matrix [3z+14−24]\begin{bmatrix} 3z + 1 & 4 \\ -2 & 4 \end{bmatrix} is equal to:

Solution

✅ Correct Option: 2

From matrix equality: 2x+3=72x + 3 = 7, so x=2x = 2. 3y=−93y = -9, so y=−3y = -3. 2x−z=−42x - z = -4, so z=2(2)−(−4)=8z = 2(2) - (-4) = 8. Therefore 3z+1=253z + 1 = 25. The required matrix is [254−24]\begin{bmatrix} 25 & 4 \\ -2 & 4 \end{bmatrix}.

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