Question

Match the LIST-I with LIST-II

LIST-I Matrix \(A\)	LIST-II \(|\text{adj}\,A|\)
A. \(A=\begin{bmatrix}2&4\\1&3\end{bmatrix}\)	I. 1
B. \(A=\begin{bmatrix}5&2\\7&4\end{bmatrix}\)	II. 7
C. \(A=\begin{bmatrix}1&0\\0&1\end{bmatrix}\)	III. 2
D. \(A=\begin{bmatrix}6&1\\5&2\end{bmatrix}\)	IV. 6

Choose the correct answer from the options given below:

• A. A - I, B - II, C - III, D - IV
• B. A - IV, B - II, C - III, D - I
• C. A - III, B - IV, C - I, D - II
• D. A - III, B - IV, C - II, D - I

Hint:

For a 2 by 2 matrix, the determinant of the adjoint equals the determinant of the matrix itself.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For an \(n\times n\) matrix, \(|\text{adj}\,A| = |A|^{n-1}\). For a 2 by 2 matrix, \(n-1=1\), so \(|\text{adj}\,A| = |A|\). We only need the determinant of each matrix.

Step 2: Matrix A.
\[ |A| = 2\times 3 - 4\times 1 = 6-4 = 2 \] So A matches III.

Step 3: Matrix B.
\[ |B| = 5\times 4 - 2\times 7 = 20-14 = 6 \] So B matches IV.

Step 4: Matrix C.
This is the identity matrix, so \[ |C| = 1\times 1 - 0\times 0 = 1 \] So C matches I.

Step 5: Matrix D.
\[ |D| = 6\times 2 - 1\times 5 = 12-5 = 7 \] So D matches II.

Step 6: Choose the option.
The matching is A - III, B - IV, C - I, D - II. Option 1 has A - I, which is wrong. Option 2 has A - IV, which is wrong. Option 4 has C - II and D - I, which is wrong. Option 3 is correct.

Final Answer:
The matching is A - III, B - IV, C - I, D - II, which is option 3. \[ \boxed{\text{A-III, B-IV, C-I, D-II}} \]