Determinants

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XII, Chapter 4.

Two lines, one number

Take the pair of equations $x + 2y = 3$ and $2x + 4y = 7$. Try to solve them and you hit a wall: the left sides are in the ratio $1 : 2$, but the right sides are not, so the lines are parallel and never meet. Now take $x + 2y = 3$ and $2x + 5y = 7$. These meet at exactly one point. What separates the two cases is a single number built from the coefficients: $1 \times 4 - 2 \times 2 = 0$ in the first pair, and $1 \times 5 - 2 \times 2 = 1$ in the second. That number is a determinant. This chapter turns the idea into a tool for areas, inverses and whole systems of equations.

What a determinant is

Every square matrix $A$ has one number attached to it, called its determinant, written $\lvert A \rvert$, $\det A$ or $\Delta$. Only square matrices have determinants. And $\lvert A \rvert$ is read "determinant of $A$", not "modulus of $A$": it can be negative.

For small orders the definition is direct.

For example, $\begin{vmatrix} 3 & -2 \\ 4 & 5 \end{vmatrix} = 3(5) - (-2)(4) = 15 + 8 = 23$. The second product carries a minus sign in front of it and its own negative entry, so the two minus signs combine into a plus.

Minors, cofactors and the sign grid

The minor $M_{ij}$ of the element $a_{ij}$ is the determinant left after deleting row $i$ and column $j$. The cofactor attaches a sign to it:

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

The sign $(-1)^{i+j}$ follows a checkerboard that starts with plus in the top left corner:

$$\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$$

Expansion. The determinant equals the sum of each element of any one row (or column) times its own cofactor. For the first row of a $3 \times 3$ determinant,

$$\lvert A \rvert = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$$

and the value is the same whichever of the three rows or three columns you choose. So choose the row or column with the most zeros: every zero removes a whole $2 \times 2$ calculation.

The other row trap. Multiply the elements of one row by the cofactors of a different row and the total is always $0$, for example $a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0$. Here is why. The cofactors $A_{21}, A_{22}, A_{23}$ do not depend on row 2 at all, so that sum is the expansion along row 2 of the determinant in which row 2 has been replaced by a copy of row 1. A determinant with two identical rows is $0$ (already visible for $2 \times 2$, where $ab - ba = 0$).

Worked example: expanding with a zero. Evaluate $\begin{vmatrix} 2 & 3 & 0 \\ 1 & 4 & 5 \\ 0 & 2 & 1 \end{vmatrix}$.

Expanding along row 1 instead gives $2(4 - 10) - 3(1 - 0) + 0 = -12 - 3 = -15$, the same value.

Common wrong step. Forgetting that the middle term carries a minus sign and adding $+5(4)$, which gives $25$. The checkerboard is the single largest source of lost marks in $3 \times 3$ determinants.

Pause and predict: in the determinant $\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$, what are the minor and the cofactor of the element $6$?

Check your answer: The element $6$ sits in row 2, column 3. Deleting them leaves $\begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix} = 8 - 14 = -6$, so $M_{23} = -6$. Because $2 + 3 = 5$ is odd, the cofactor is $A_{23} = (-1)^5(-6) = 6$.

The scalar rule: $\lvert kA \rvert = k^n \lvert A \rvert$

For an $n \times n$ matrix $A$ and a number $k$,

$$\lvert kA \rvert = k^n \lvert A \rvert$$

The reason is that $kA$ multiplies every row by $k$, and the determinant picks up one factor of $k$ from each of the $n$ rows. In a $2 \times 2$ case, $\begin{vmatrix} ka & kb \\ kc & kd \end{vmatrix} = k^2(ad - bc)$.

Two more results come from the idea that a determinant behaves well under multiplication. For square matrices of the same order, the product rule is

$$\lvert AB \rvert = \lvert A \rvert \, \lvert B \rvert$$

and applying it to $AA^{-1} = I$ gives $\lvert A \rvert \, \lvert A^{-1} \rvert = 1$, so $\lvert A^{-1} \rvert = \dfrac{1}{\lvert A \rvert}$.

Worked example: three rules in one problem. $A$ is a $3 \times 3$ matrix with $\lvert A \rvert = -2$. Find $\lvert 2A \rvert$, $\lvert A^{-1} \rvert$ and, given that $B$ is a $3 \times 3$ matrix with $\lvert B \rvert = 5$, find $\lvert AB \rvert$.

Common wrong step. Using $\lvert 2A \rvert = 2\lvert A \rvert = -4$. The factor $2$ comes out of each of the three rows, so it is cubed.

Area of a triangle and collinear points

The area of the triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is

$$\text{Area} = \dfrac{1}{2} \left\lvert \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right\rvert$$

The outer bars are an absolute value, because area cannot be negative. Three points are collinear exactly when this determinant is $0$, since the triangle then has no area. For the vertices $(3, 8)$, $(-4, 2)$ and $(5, 1)$ the determinant is $3(2 - 1) - 8(-4 - 5) + 1(-4 - 10) = 3 + 72 - 14 = 61$, so the area is $\dfrac{61}{2}$ square units.

Worked example: finding an unknown vertex. The triangle with vertices $(2, -6)$, $(5, 4)$ and $(k, 4)$ has area $35$ square units. Find $k$.

Common wrong step. Dropping the absolute value and solving only $50 - 10k = 70$. A given area allows the determinant to be either $+70$ or $-70$, so there are two positions for the vertex.

Beyond the textbook. For the exam: area of a triangle is half the absolute value of this determinant, and zero means collinear. In reality: the $2 \times 2$ determinant $ad - bc$ measures the area of the parallelogram whose sides are the vectors $(a, c)$ and $(b, d)$. A triangle is half of such a parallelogram, which is where the $\dfrac{1}{2}$ comes from. When the determinant is $0$, the two sides lie along one line and the parallelogram is flat. Why it matters: this "flattening" is exactly why a matrix with determinant $0$ has no inverse. The same idea carries over to volumes in three dimensions.

Adjoint and inverse

The adjoint of a square matrix $A$, written $\operatorname{adj} A$, is the transpose of the matrix of cofactors. Compute every cofactor $A_{ij}$, arrange them as the matrix $[A_{ij}]$, then transpose.

For a $2 \times 2$ matrix there is a shortcut: swap the two diagonal entries and change the sign of the other two.

$$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \quad \Longrightarrow \quad \operatorname{adj} A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$

The key theorem is

$$A \, (\operatorname{adj} A) = (\operatorname{adj} A) \, A = \lvert A \rvert \, I$$

The reason is visible entry by entry. A diagonal entry of $A(\operatorname{adj}A)$ is a row of $A$ times its own cofactors, which is $\lvert A \rvert$. An off-diagonal entry is a row times the cofactors of a different row, which is $0$ by the other row trap above. So the product is $\lvert A \rvert$ times the identity.

If $\lvert A \rvert \ne 0$, divide through by it to get the inverse:

$$A^{-1} = \dfrac{1}{\lvert A \rvert} \operatorname{adj} A$$

Taking determinants of $A(\operatorname{adj}A) = \lvert A \rvert I$ for an $n \times n$ matrix gives $\lvert A \rvert \, \lvert \operatorname{adj}A \rvert = \lvert A \rvert^n$, hence

$$\lvert \operatorname{adj} A \rvert = \lvert A \rvert^{\,n-1}$$

A matrix is singular if $\lvert A \rvert = 0$ and non-singular if $\lvert A \rvert \ne 0$. A square matrix is invertible exactly when it is non-singular.

Worked example: an inverse. Find $A^{-1}$ for $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$.

Common wrong step. Swapping the diagonal but not changing the signs of the off-diagonal entries, or leaving out the final division by $\lvert A \rvert$. Step 4 catches both errors immediately.

Pause and predict: a $2 \times 2$ matrix has $\lvert A \rvert = -3$. What are $\lvert 2A \rvert$ and $\lvert A^{-1} \rvert$?

Check your answer: $\lvert 2A \rvert = 2^2(-3) = -12$ and $\lvert A^{-1} \rvert = -\dfrac{1}{3}$. The scalar rule uses the order $n = 2$ as the power.

Solving a system by the matrix method

A system of three equations $a_ix + b_iy + c_iz = d_i$ for $i = 1, 2, 3$ is written $AX = B$, where $A$ is the matrix of coefficients, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}$. A system is consistent if it has at least one solution and inconsistent if it has none. The outcome depends on $\lvert A \rvert$:

Worked example: a system of three equations. Solve $x - y + 2z = 7$, $\ 3x + 4y - 5z = -5$, $\ 2x - y + 3z = 12$.

Substituting confirms it: $2 - 1 + 6 = 7$, $\ 6 + 4 - 15 = -5$, $\ 4 - 1 + 9 = 12$.

Common wrong step. Using the cofactor matrix directly, without transposing it. The inverse needs the adjoint, which is the transpose of the cofactor matrix. Check step 3: the first row of $\operatorname{adj}A$ is the first column of cofactors.

Worked example: when $\lvert A \rvert = 0$. Decide what happens to $x + 2y = 3$, $\ 2x + 4y = 7$, and then to $x + 2y = 3$, $\ 2x + 4y = 6$.

Common wrong step. Concluding "no solution" as soon as $\lvert A \rvert = 0$. That is only correct when $(\operatorname{adj}A)B \ne O$. When it equals $O$, the system can still be consistent, as the second system shows.

Beyond the textbook. For the exam: find $A^{-1}$ using the adjoint and then compute $X = A^{-1}B$. In reality: for large systems nobody computes cofactors, because the number of operations grows far too fast. Computers solve $AX = B$ by systematically eliminating variables, row by row, which gives the same answer much faster. Why it matters: the adjoint method is a clean way to prove that an inverse exists and to compute small cases by hand, which is exactly what CUET tests.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Matrices introduces invertible matrices and the reversal law; this chapter gives the number $\lvert A \rvert$ that decides whether an inverse exists.
• Next: Continuity and Differentiability starts the calculus half of Class XII.
• Class XI link: Straight Lines has the coordinate formulas for the area of a triangle and for collinear points that the determinant form restates.
• Class XII link: Relations and Functions explains why "invertible" means one-one and onto, the idea behind a non-singular matrix.