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If [0x21]+2[31y0]=[6581]\begin{bmatrix} 0 & x \\ 2 & 1 \end{bmatrix} + 2\begin{bmatrix} 3 & 1 \\ y & 0 \end{bmatrix} = \begin{bmatrix} 6 & 5 \\ 8 & 1 \end{bmatrix}, then the value of 2x+3y2x+3y is :

Solution

✅ Correct Option: 1

Adding gives [6x+22+2y1]\begin{bmatrix} 6 & x+2 \\ 2+2y & 1 \end{bmatrix}. Comparing entries, x+2=5x+2 = 5 and 2+2y=82+2y = 8, so x=3x = 3, y=3y = 3 and 2x+3y=6+9=152x+3y = 6+9 = 15.

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