Determinants and Systems of Linear Equations notes: ISC Class 12 Mathematics

1. Determinants, minors and cofactors

Order $2$ : $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$ .
Minor $M_{ij}$ : delete row $i$ and column $j$ . Cofactor $A_{ij} = (-1)^{i + j}M_{ij}$ .
Order $3$ : expand along any row or column, for example $|A| = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$ ; choose the line with most zeros.
Elements of one row with the cofactors of another row add up to $0$ .

Cofactor $A_{ij} = (-1)^{i + j}M_{ij}$.
	Column 1	Column 2	Column 3
Row 1	$+$	$-$	$+$
Row 2	$-$	$+$	$-$
Row 3	$+$	$-$	$+$

Same value along any row or column.
Step	Working
Choose row 2	two zeros, so one term
Cofactor $A_{22}$	$+\begin{vmatrix} 2 & -1 \\ 5 & 6 \end{vmatrix} = 12 + 5 = 17$
Value	$4 \times 17 = 68$
Check (row 1)	$2(24) - 7(0) + (-1)(-20) = 68$

Matching rows give $|A|$; different rows give $0$.
Sum	Value for $A = \begin{bmatrix} 2 & 7 & -1 \\ 0 & 4 & 0 \\ 5 & -3 & 6 \end{bmatrix}$
Row 1 elements × row 1 cofactors	$68$
Row 1 elements × row 3 cofactors	$0$
Row 3 elements × row 2 cofactors	$0$

Formulas

Cofactor
$A_{ij} = (-1)^{i + j}M_{ij}$

Key terms

Minor: The determinant left after deleting the row and column of an element.
Cofactor: $A_{ij} = (-1)^{i + j}M_{ij}$ .

2. Properties of determinants

$|A^T| = |A|$ .
Interchanging two rows (columns) changes the sign.
Two identical or proportional rows (columns): value $0$ .
A factor common to one row (column) can be taken out; hence $|kA| = k^n|A|$ for order $n$ .
$R_i \to R_i + kR_j$ (or the column version) leaves the value unchanged.
If a row is a sum of two rows, the determinant splits into two determinants.

Strategy: look at row and column sums; make a common factor; create zeros; expand last.

The same rules hold for columns.
Property	Effect on the value
Rows and columns interchanged ( $A^T$ )	unchanged
Two rows (or columns) interchanged	sign changes
Two rows (or columns) identical or proportional	value is $0$
One row multiplied by $k$	value multiplied by $k$
$R_i \to R_i + kR_j$	unchanged
A row written as a sum of two rows	splits into a sum of two determinants

Each value was computed by full expansion.
Operation on $A_2$ (value $77$ )	New value
$A_2^T$	$77$
$R_1 \leftrightarrow R_3$	$-77$
$R_2 \to 3R_2$	$231$
$R_1 \to R_1 + 2R_3$	$77$

Create a common factor first; expand last.

Formulas

Scalar multiple, order n
$|kA| = k^n|A|$

Key terms

Elementary operation: $R_i \to R_i + kR_j$ : does not change the value.

3. Area of a triangle and collinearity

Area of the triangle with vertices $(x_1, y_1)$ , $(x_2, y_2)$ , $(x_3, y_3)$ :

$\Delta = \dfrac12\left|\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right|$ .

The three points are collinear exactly when the determinant is $0$ .
Line through two points: put $(x, y)$ with the two points and set the determinant to $0$ .
When an area is given, solve both $D = 2\Delta$ and $D = -2\Delta$ .

Area $= \dfrac12|21| = \dfrac{21}{2}$ square units.

$y = 5 - 2x$

Zero area means the three points are collinear.

Formulas

Area
$\Delta = \tfrac12\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right|$

Key terms

Collinear: Lying on one straight line; the area determinant is zero.

4. Adjoint and inverse

$\operatorname{adj} A$ = transpose of the matrix of cofactors. For $2 \times 2$ : $\operatorname{adj}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ .
$A(\operatorname{adj} A) = (\operatorname{adj} A)A = |A|\,I$ .
$A$ is singular if $|A| = 0$ , non-singular otherwise.
$A^{-1} = \dfrac{1}{|A|}\operatorname{adj} A$ exists exactly when $|A| \ne 0$ , and it is unique.
For order $3$ : $|\operatorname{adj} A| = |A|^2$ and $\operatorname{adj}(\operatorname{adj} A) = |A|\,A$ .

Do not forget the transpose.

$A(\operatorname{adj} A) = 23I = |A|\,I$.
Quantity	Value
$A$	$\begin{bmatrix} 3 & -4 \\ 2 & 5 \end{bmatrix}$
$\operatorname{adj} A$	$\begin{bmatrix} 5 & 4 \\ -2 & 3 \end{bmatrix}$
$\|A\|$	$23$
$A(\operatorname{adj} A)$	$\begin{bmatrix} 23 & 0 \\ 0 & 23 \end{bmatrix}$

$|A| = 11$; transpose this table to get $\operatorname{adj} A$.
Row	Cofactors
Row 1	$6$ , $-2$ , $-1$
Row 2	$-1$ , $4$ , $2$
Row 3	$-3$ , $1$ , $6$

Formulas

Inverse
$A^{-1} = \dfrac{1}{|A|}\operatorname{adj} A$

Key terms

Adjoint: Transpose of the matrix of cofactors.
Non-singular matrix: A square matrix with non-zero determinant.

5. Properties of the inverse

$(A^{-1})^{-1} = A$ .
$(AB)^{-1} = B^{-1}A^{-1}$ (order reverses).
$(A^T)^{-1} = (A^{-1})^T$ .
$|A^{-1}| = \dfrac{1}{|A|}$ .
$\operatorname{adj}(AB) = (\operatorname{adj} B)(\operatorname{adj} A)$ .

Each follows from $AA^{-1} = I$.
Property	Note
$(A^{-1})^{-1} = A$	undoing the undo
$(AB)^{-1} = B^{-1}A^{-1}$	order reverses
$(A^T)^{-1} = (A^{-1})^T$	transpose and inverse commute
$\|A^{-1}\| = \dfrac{1}{\|A\|}$	from $\|A\|\,\|A^{-1}\| = \|I\| = 1$

Learn these four.
Result	For order $n$
$A(\operatorname{adj} A) = (\operatorname{adj} A)A = \|A\|\,I$	every square $A$
$\|\operatorname{adj} A\| = \|A\|^{n - 1}$	$n = 3$ : $\|A\|^2$
$\|kA\| = k^n\|A\|$	$n = 3$ : $k^3\|A\|$
$A^{-1} = \dfrac{1}{\|A\|}\operatorname{adj} A$	only when $\|A\| \ne 0$

Formulas

Reversal law
$(AB)^{-1} = B^{-1}A^{-1}$

Key terms

Inverse of a product: $(AB)^{-1} = B^{-1}A^{-1}$ .

6. Systems of linear equations

Write the system as $AX = B$ .

$|A| \ne 0$ : consistent, unique solution $X = A^{-1}B$ .
$|A| = 0$ and $(\operatorname{adj} A)B \ne O$ : inconsistent (no solution).
$|A| = 0$ and $(\operatorname{adj} A)B = O$ : either infinitely many solutions or none; check the equations.

Matrix method: matrix form, $|A|$ , cofactors, $\operatorname{adj} A$ , $A^{-1}$ , then $X = A^{-1}B$ , then substitute back.

The test ISC expects.

Always substitute back.
Equation	Left side at $(1, 2, -1)$	Right side
$x + 2y + z = 4$	$4$	$4$
$3x - y + 2z = -1$	$-1$	$-1$
$2x + y - z = 5$	$5$	$5$

Same left side, different right sides.
System	$\|A\|$	$(\operatorname{adj} A)B$	Conclusion
$x + 2y = 4$ , $2x + 4y = 3$	$0$	$\begin{bmatrix} 10 \\ -5 \end{bmatrix}$	no solution
$x + 2y = 4$ , $2x + 4y = 8$	$0$	$\begin{bmatrix} 0 \\ 0 \end{bmatrix}$	infinitely many
$x + 2y = 4$ , $3x - y = 5$	$-7$	—	unique: $(2, 1)$

$x + 2y = 4$
$2x + 4y = 3$

Parallel lines: $|A| = 0$ and $(\operatorname{adj} A)B \ne O$.

Formulas

Matrix method
$X = A^{-1}B$

Key terms

Consistent system: A system with at least one solution.
Inconsistent system: A system with no solution.

Exam tips

How ISC marks are awarded

Evaluate (2 marks): expansion and value.
Prove using properties (3–5 marks): one mark per named operation; final form.
Inverse of a $3 \times 3$ matrix (3 marks): $|A|$ , adjoint, inverse.
Matrix method (5 marks): matrix form, $|A|$ , adjoint, product, answer with a check.
Area (2–3 marks): determinant, modulus, units.

Common mistakes

Forgetting the chessboard sign of a cofactor.
Forgetting to transpose the cofactor matrix.
Writing $|3A| = 3|A|$ for a $3 \times 3$ matrix (it is $27|A|$ ).
Dropping the modulus in the area formula or one sign when solving for $k$ .
Writing $X = BA^{-1}$ instead of $X = A^{-1}B$ .

1. Determinants, minors and cofactors

Formulas

Key terms

2. Properties of determinants

Formulas

Key terms

3. Area of a triangle and collinearity

Formulas

Key terms

4. Adjoint and inverse

Formulas

Key terms

5. Properties of the inverse

Formulas

Key terms

6. Systems of linear equations

Formulas

Key terms

Exam tips

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