Determinants Class 12: notes and important questions

Determinants — revision notes

1. Evaluating determinants

• \(\begin{vmatrix}a&b\\c&d\end{vmatrix} = ad - bc\).
• Order 3: expand along any row or column, \(\Delta = \sum a_{ij}A_{ij}\), where the cofactor \(A_{ij} = (-1)^{i+j}M_{ij}\) and the minor \(M_{ij}\) is the determinant left after deleting row \(i\) and column \(j\). Choose the row/column with most zeros.
• Elements of one row times the cofactors of a different row always sum to \(0\).

2. Area of a triangle and collinearity

• Area \(= \dfrac12\left|\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\right|\). The determinant can be negative — take the modulus; if an area is given, use \(\pm\) and keep both cases.
• Three points are collinear iff this determinant is \(0\). Line through \((x_1, y_1), (x_2, y_2)\): \(\begin{vmatrix}x&y&1\\x_1&y_1&1\\x_2&y_2&1\end{vmatrix} = 0\).

3. Adjoint and inverse

• \(\mathrm{adj}\,A\) = transpose of the matrix of cofactors. For order 2, \(\mathrm{adj}\begin{bmatrix}a&b\\c&d\end{bmatrix} = \begin{bmatrix}d&-b\\-c&a\end{bmatrix}\).
• \(A(\mathrm{adj}\,A) = (\mathrm{adj}\,A)A = |A|I\); \(A^{-1} = \dfrac{1}{|A|}\mathrm{adj}\,A\) exists iff \(|A| \ne 0\) (non-singular).
• For order \(n\): \(|kA| = k^n|A|\), \(|AB| = |A||B|\), \(|A^T| = |A|\), \(|A^{-1}| = \dfrac{1}{|A|}\), \(|\mathrm{adj}\,A| = |A|^{n-1}\), \((AB)^{-1} = B^{-1}A^{-1}\), \((A^T)^{-1} = (A^{-1})^T\).

4. Linear systems \(AX = B\)

• \(|A| \ne 0\): unique solution \(X = A^{-1}B\) (consistent).
• \(|A| = 0\): find \((\mathrm{adj}\,A)B\). If it is \(\ne O\), no solution (inconsistent). If it is \(= O\), the system has either infinitely many solutions or none — check the equations directly.

Worked example 1

Area of the triangle with vertices \((0, 0)\), \((4, 1)\), \((1, 3)\): \(\dfrac12\begin{vmatrix}0&0&1\\4&1&1\\1&3&1\end{vmatrix} = \dfrac12(12 - 1) = \dfrac{11}{2}\) sq units.

Worked example 2

\(A = \begin{bmatrix}1&2\\3&5\end{bmatrix}\): \(|A| = -1\), \(A^{-1} = -\begin{bmatrix}5&-2\\-3&1\end{bmatrix} = \begin{bmatrix}-5&2\\3&-1\end{bmatrix}\). So \(x + 2y = 4,\ 3x + 5y = 11\) gives \(X = A^{-1}\begin{bmatrix}4\\11\end{bmatrix} = \begin{bmatrix}2\\1\end{bmatrix}\).

Worked example 3

If \(|A| = 3\) for order 3: \(|2\,\mathrm{adj}\,A| = 2^3|A|^2 = 72\).

Common errors

• Sign pattern of cofactors \(\begin{smallmatrix}+&-&+\\-&+&-\\+&-&+\end{smallmatrix}\) forgotten, especially for \(A_{12}, A_{21}, A_{23}, A_{32}\).
• Writing the matrix of cofactors instead of its transpose as \(\mathrm{adj}\,A\).
• \(|kA| = k|A|\) instead of \(k^n|A|\).
• Forgetting the modulus (or one of the two cases) in area problems.
• Writing \(X = BA^{-1}\) instead of \(X = A^{-1}B\); for \(XA = C\) the answer is \(X = CA^{-1}\).
• Concluding “no solution” from \(|A| = 0\) alone.

Board-exam tips

• In the 5-mark “solve using matrix method” question, marks are given for \(|A|\), adj A (or cofactors), \(A^{-1}\) and the solution separately — show each clearly, and substitute your answer back into one equation.
• Properties of determinants (row/column operations) are not in the current syllabus: expand directly.