Original JEE Main and CUET-style questions on matrices and determinants, with a worked solution and the usual trap for each. Tap an option, or type a numerical answer, to check it.
Learn the chapters first: Class 12, Matrices · Class 12, Determinants.
For a \(3 \times 3\) matrix \(A\) with \(|A| = 2\), the value of \(|\operatorname{adj}(3A)|\) is
\(|3A| = 3^3 \cdot 2 = 54\), and for a \(3 \times 3\) matrix \(|\operatorname{adj} B| = |B|^2\).
So \(|\operatorname{adj}(3A)| = 54^2 = 2916\).
Trap: \(|3A| = 3|A|\) is wrong for a \(3 \times 3\) matrix: each of the three rows is multiplied by 3.
If \(A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}\), then \(A^{-1}\) is
\(|A| = 8 - 3 = 5\). Swap the diagonal entries and change the signs of the others: \(A^{-1} = \dfrac15\begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}\).
Find \(k\) for which the system \(x + y + z = 1\), \(x + 2y + 3z = 2\), \(x + 3y + kz = 3\) has infinitely many solutions.
The determinant of coefficients is \(1(2k - 9) - 1(k - 3) + 1(3 - 2) = k - 5\), which must be 0, so \(k = 5\).
Check consistency at \(k = 5\): the third equation minus the second gives \(y + 2z = 1\), the same as the second minus the first, so there are infinitely many solutions.
Trap: A zero determinant alone could also mean no solution: always check the equations are consistent.
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