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If [abcda+cb+d]=[2−341]\begin{bmatrix} ab & cd \\ a+c & b+d \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ 4 & 1 \end{bmatrix} where aa, bb, cc, dd are integers, then which of the following are true?

(A) a+d=0a + d = 0

(B) b+d=3b + d = 3

(C) b+d=1b + d = 1

(D) c+d=2c + d = 2

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

Given: [abcda+cb+d]=[2−341]\begin{bmatrix} ab & cd \\ a+c & b+d \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ 4 & 1 \end{bmatrix}

By comparing corresponding elements:

ab=2ab = 2

cd=−3cd = -3

a+c=4a + c = 4

b+d=1b + d = 1


Since a,b,c,da, b, c, d are integers, the possible factor pairs are:

For ab=2ab = 2: (a,b)(a, b) can be (1,2)(1, 2), (2,1)(2, 1), (−1,−2)(-1, -2), or (−2,−1)(-2, -1)

For cd=−3cd = -3: (c,d)(c, d) can be (1,−3)(1, -3), (−1,3)(-1, 3), (3,−1)(3, -1), or (−3,1)(-3, 1)


Testing a=1,b=2a = 1, b = 2:

From a+c=4a + c = 4: c=3c = 3

From b+d=1b + d = 1: d=−1d = -1

Check: cd=3×(−1)=−3cd = 3 \times (-1) = -3 ✓

Therefore: a=1,b=2,c=3,d=−1a = 1, b = 2, c = 3, d = -1


Evaluating each statement:

(A) a+d=1+(−1)=0a + d = 1 + (-1) = 0 ✓

(B) b+d=2+(−1)=1≠3b + d = 2 + (-1) = 1 \neq 3 ✗

(C) b+d=2+(−1)=1b + d = 2 + (-1) = 1 ✓

(D) c+d=3+(−1)=2c + d = 3 + (-1) = 2 ✓


Statements (A), (C), and (D) are true.

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