Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | \( \begin{vmatrix}x+2 & 3\\x+5 & 4\end{vmatrix}=3 \) | I. | \( x=10 \) |
| B. | \( \begin{vmatrix}x+2 & x+5\\3 & 4\end{vmatrix}=13 \) | II. | \( x=60 \) |
| C. | \( \begin{vmatrix}x+4 & x-4\\8 & 7\end{vmatrix}=0 \) | III. | \( x=0 \) |
| D. | \( \begin{vmatrix}x+4 & 8\\x-4 & 7\end{vmatrix}=60 \) | IV. | \( x=20 \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A - I, B - IV, C - II, D - III
A 2 × 2 determinant is \( \begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc \). The determinants in A and B are transposes of each other, and so are those in C and D, so each pair has the same expression in x.
(A) \( \begin{vmatrix}x+2 & 3\\x+5 & 4\end{vmatrix}=3 \)
This matches (I) \( x=10 \).
\( \displaystyle 4(x+2)-3(x+5)=3 \)
\( \displaystyle x-7=3 \)
\( \displaystyle x=10 \)
(B) \( \begin{vmatrix}x+2 & x+5\\3 & 4\end{vmatrix}=13 \)
This matches (IV) \( x=20 \).
\( \displaystyle 4(x+2)-3(x+5)=13 \)
\( \displaystyle x-7=13 \)
\( \displaystyle x=20 \)
(C) \( \begin{vmatrix}x+4 & x-4\\8 & 7\end{vmatrix}=0 \)
This matches (II) \( x=60 \).
\( \displaystyle 7(x+4)-8(x-4)=0 \)
\( \displaystyle 60-x=0 \)
\( \displaystyle x=60 \)
(D) \( \begin{vmatrix}x+4 & 8\\x-4 & 7\end{vmatrix}=60 \)
This matches (III) \( x=0 \).
\( \displaystyle 7(x+4)-8(x-4)=60 \)
\( \displaystyle 60-x=60 \)
\( \displaystyle x=0 \)
So the matching is A - I, B - IV, C - II, D - III.