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CBSE Class 12 Math 2022 Term I Solved Paper

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Question : 18 of 50
Marks: +1, -0
[3c+6a−da+d2−3b]=[122−8−4]\begin{bmatrix} 3c+6 & a-d \\ a+d & 2-3b \end{bmatrix} = \begin{bmatrix} 12 & 2 \\ -8 & -4 \end{bmatrix} are equal, then value of ab−cdab - cd is
Solution:     👈: Video Solution
Explanation: Given, [3c+6a−da+d2−3b]=[122−8−4]\begin{bmatrix} 3c+6 & a-d \\ a+d & 2-3b \end{bmatrix} = \begin{bmatrix} 12 & 2 \\ -8 & -4 \end{bmatrix}
∴  3c+6=12\therefore \; 3c+6 = 12
a−d=2a-d = 2
a+d=−8a+d = -8
2−3b=−42-3b = -4
From eq. (i), we get c=2c = 2
On solving eqs. (ii) and (iii), we get a=−3a = -3 and d=−5d = -5 from eq. (iv), we get b=2b = 2
 Now,   ab−cd=(−3)2−2(−5)\text{ Now, } \; ab - cd = (-3)2 - 2(-5)
⇒ab−cd=−6+10=4\Rightarrow ab - cd = -6 + 10 = 4
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