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If 2[adbc]+3[1−102]=3[3546]2 \begin{bmatrix} a & d \\ b & c \end{bmatrix} + 3 \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 3 \begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}, then the value of ∣a+b−c−d∣|a + b - c - d| is

Solution

✅ Correct Option: 3

2[adbc]+3[1−102]=[9151218]2\begin{bmatrix} a & d \\ b & c \end{bmatrix} + 3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}

So [2a+32d−32b2c+6]=[9151218]\begin{bmatrix} 2a+3 & 2d-3 \\ 2b & 2c+6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}.

This gives a=3, d=9, b=6, c=6a=3,\ d=9,\ b=6,\ c=6. Then ∣a+b−c−d∣=∣3+6−6−9∣=6|a+b-c-d| = |3+6-6-9| = 6.

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