NCERT Exemplar Solutions Class 12 Maths Determinants — Short and Long Answer Q1–23

NCERT Exemplar Problems, Class 12 Mathematics — Determinants, Short Answer questions 1 to 17 and Long Answer questions 18 to 23. All 23 questions solved.

Twenty-three questions in two kinds. Most of the Short Answer questions are “evaluate or prove using properties”: add all the columns (or rows) together, take out the common factor, subtract one row from the others to make zeros, and expand along the row that is now nearly empty (Questions 1 to 9, 13, 14, 21, 22). The rest apply determinants to geometry (11, 15, 16) and, in the Long Answer questions, to inverses and systems of linear equations (17 to 20).

A note on the syllabus. The rationalised NCERT textbook dropped its section on properties of determinants; its exercises now evaluate determinants only by expansion. Nearly every question on this page is built for those properties. JEE Main and Advanced still test them, so the solutions below use them, stating each property the first time it is used. Questions 17 to 20 (inverse and matrix method) are fully inside the current textbook.

Key insight. A determinant is unchanged when a multiple of one row is added to another, and a factor common to a whole row or column can be taken outside. So look for an operation that makes every entry of one row or column the same; take that common value out, and the 1s left behind let you create zeros and expand in one line.

Question 1

Using the properties of determinants, evaluate $\begin{vmatrix} x^2-x+1 & x-1 \\ x+1 & x+1 \end{vmatrix}$.

Solution. The second row has the common factor $x + 1$. A factor common to every entry of a row can be taken outside the determinant:

$$\Delta = (x+1)\begin{vmatrix} x^2-x+1 & x-1 \\ 1 & 1 \end{vmatrix} = (x+1)\left[(x^2-x+1)-(x-1)\right] = (x+1)(x^2-2x+2).$$

Multiplying out, $x^3 – 2x^2 + 2x + x^2 – 2x + 2 = x^3 – x^2 + 2$.

Question 2

Evaluate $\begin{vmatrix} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{vmatrix}$.

Solution. Every row adds up to $a + x + y + z$, so $C_1 \to C_1 + C_2 + C_3$ fills the first column with that sum. Adding columns does not change the value. Take the sum out:

$$\Delta = (a+x+y+z)\begin{vmatrix} 1 & y & z \\ 1 & a+y & z \\ 1 & y & a+z \end{vmatrix}.$$

Now $R_2 \to R_2 – R_1$ and $R_3 \to R_3 – R_1$:

$$\Delta = (a+x+y+z)\begin{vmatrix} 1 & y & z \\ 0 & a & 0 \\ 0 & 0 & a \end{vmatrix} = (a+x+y+z)\,a^2.$$

Question 3

Evaluate $\begin{vmatrix} 0 & xy^2 & xz^2 \\ x^2y & 0 & yz^2 \\ x^2z & zy^2 & 0 \end{vmatrix}$.

Solution. Take $x^2$ out of the first column, $y^2$ out of the second and $z^2$ out of the third:

$$\Delta = x^2y^2z^2\begin{vmatrix} 0 & x & x \\ y & 0 & y \\ z & z & 0 \end{vmatrix}.$$

Expand along the first row: $0 – x(0 – yz) + x(yz – 0) = 2xyz$. So $\Delta = x^2y^2z^2 \cdot 2xyz$.

Question 4

Evaluate $\begin{vmatrix} 3x & -x+y & -x+z \\ x-y & 3y & z-y \\ x-z & y-z & 3z \end{vmatrix}$.

Solution. Each row sums to $x + y + z$ (for the first, $3x – x + y – x + z$). Apply $C_1 \to C_1 + C_2 + C_3$ and take the sum out:

$$\Delta = (x+y+z)\begin{vmatrix} 1 & -x+y & -x+z \\ 1 & 3y & z-y \\ 1 & y-z & 3z \end{vmatrix}.$$

$R_2 \to R_2 – R_1$, $R_3 \to R_3 – R_1$:

$$\Delta = (x+y+z)\begin{vmatrix} 1 & -x+y & -x+z \\ 0 & x+2y & x-y \\ 0 & x-z & x+2z \end{vmatrix}.$$

Expand along the first column:

$$(x+2y)(x+2z) – (x-y)(x-z) = (x^2 + 2xy + 2xz + 4yz) – (x^2 – xy – xz + yz) = 3xy + 3yz + 3zx.$$

Question 5

Evaluate $\begin{vmatrix} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{vmatrix}$.

Solution. Each row sums to $3x + 4$. $C_1 \to C_1 + C_2 + C_3$, take $3x + 4$ out, then $R_2 \to R_2 – R_1$, $R_3 \to R_3 – R_1$:

$$\Delta = (3x+4)\begin{vmatrix} 1 & x & x \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{vmatrix} = 16(3x+4).$$

Question 6

Evaluate $\begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$.

Solution. This time the columns have a common sum: each column adds to $a + b + c$ (the first is $a – b – c + 2b + 2c$). So $R_1 \to R_1 + R_2 + R_3$:

$$\Delta = (a+b+c)\begin{vmatrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}.$$

$C_2 \to C_2 – C_1$, $C_3 \to C_3 – C_1$:

$$\Delta = (a+b+c)\begin{vmatrix} 1 & 0 & 0 \\ 2b & -(a+b+c) & 0 \\ 2c & 0 & -(a+b+c) \end{vmatrix} = (a+b+c)\cdot(a+b+c)^2.$$

Question 7

Prove that $\begin{vmatrix} y^2z^2 & yz & y+z \\ z^2x^2 & zx & z+x \\ x^2y^2 & xy & x+y \end{vmatrix} = 0$.

Solution. The rows are “missing” $x$, $y$ and $z$ respectively, so multiply $R_1$ by $x$, $R_2$ by $y$, $R_3$ by $z$ and divide by $xyz$ to compensate (for $xyz \ne 0$):

$$\Delta = \frac{1}{xyz}\begin{vmatrix} xy^2z^2 & xyz & xy+xz \\ x^2yz^2 & xyz & yz+xy \\ x^2y^2z & xyz & xz+yz \end{vmatrix}.$$

Now $xyz$ is a factor of the first column and of the second:

$$\Delta = \frac{(xyz)^2}{xyz}\begin{vmatrix} yz & 1 & xy+xz \\ zx & 1 & yz+xy \\ xy & 1 & xz+yz \end{vmatrix}.$$

Apply $C_1 \to C_1 + C_3$: every entry of the first column becomes $xy + yz + zx$, so the first column is $(xy+yz+zx)$ times the second. A determinant with two proportional columns is $0$.

If one of $x, y, z$ is $0$, the result still holds: both sides are polynomials that agree whenever $xyz \ne 0$, so they agree everywhere.

Question 8

Prove that $\begin{vmatrix} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{vmatrix} = 4xyz$.

Solution. $R_1 \to R_1 – R_2 – R_3$ clears the $y$ and $z$ terms from the first row:

$$R_1 = \big(y+z-z-y,\; z-(z+x)-x,\; y-x-(x+y)\big) = (0,\,-2x,\,-2x).$$

Take $-2x$ out of the first row:

$$\Delta = -2x\begin{vmatrix} 0 & 1 & 1 \\ z & z+x & x \\ y & x & x+y \end{vmatrix}.$$

$C_2 \to C_2 – C_3$:

$$\Delta = -2x\begin{vmatrix} 0 & 0 & 1 \\ z & z & x \\ y & -y & x+y \end{vmatrix} = -2x\,\big(z(-y) – z \cdot y\big) = -2x(-2yz) = 4xyz.$$

Question 9

Prove that $\begin{vmatrix} a^2+2a & 2a+1 & 1 \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{vmatrix} = (a-1)^3$.

Solution. The third column is all 1s, so subtracting rows makes two zeros in it. $R_1 \to R_1 – R_2$ and $R_2 \to R_2 – R_3$:

$$\Delta = \begin{vmatrix} a^2-1 & a-1 & 0 \\ 2a-2 & a-1 & 0 \\ 3 & 3 & 1 \end{vmatrix}.$$

Expand along the third column and take out $(a – 1)$ from each of the first two rows:

$$\Delta = (a-1)^2\begin{vmatrix} a+1 & 1 \\ 2 & 1 \end{vmatrix} = (a-1)^2\,(a+1-2) = (a-1)^3.$$

Question 10

If $A + B + C = 0$, prove that $\begin{vmatrix} 1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix} = 0$.

Solution. Expand along the first row:

$$\Delta = (1 – \cos^2 A) – \cos C(\cos C – \cos A\cos B) + \cos B(\cos A\cos C – \cos B)$$

$$= 1 – \cos^2 A – \cos^2 B – \cos^2 C + 2\cos A\cos B\cos C.$$

Since $C = -(A+B)$, $\cos C = \cos(A+B)$. Group the two terms containing $\cos C$:

$$-\cos^2 C + 2\cos A\cos B\cos C = -\cos(A+B)\big[\cos(A+B) – 2\cos A\cos B\big] = \cos(A+B)\cos(A-B),$$

because $\cos(A+B) – 2\cos A\cos B = -\cos(A-B)$. And $\cos(A+B)\cos(A-B) = \cos^2 A – \sin^2 B$. So

$$\Delta = 1 – \cos^2 A – \cos^2 B + \cos^2 A – \sin^2 B = 1 – (\cos^2 B + \sin^2 B) = 0.$$

Question 11

If the coordinates of the vertices of an equilateral triangle with sides of length $a$ are $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$, then $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}^2 = \dfrac{3a^4}{4}$.

Solution. The area of a triangle is half the absolute value of this determinant, so $\Delta = \pm 2 \times \text{area}$. An equilateral triangle of side $a$ has area $\frac{\sqrt3}{4}a^2$. Therefore

$$\Delta^2 = 4 \times \left(\frac{\sqrt3}{4}a^2\right)^2 = 4 \times \frac{3a^4}{16} = \frac{3a^4}{4}.$$

Question 12

Find the value of $\theta$ satisfying $\begin{vmatrix} 1 & 1 & \sin 3\theta \\ -4 & 3 & \cos 2\theta \\ 7 & -7 & -2 \end{vmatrix} = 0$.

(The question prints square brackets; a matrix cannot equal the number $0$, so the determinant is meant.)

Solution. Expand along the third column:

$$\sin3\theta\,(28 – 21) – \cos2\theta\,(-7 – 7) + (-2)(3 + 4) = 7\sin3\theta + 14\cos2\theta – 14.$$

Use $\cos2\theta = 1 – 2\sin^2\theta$ and $\sin3\theta = 3\sin\theta – 4\sin^3\theta$:

$$7(3\sin\theta – 4\sin^3\theta) – 28\sin^2\theta = -7\sin\theta\,(4\sin^2\theta + 4\sin\theta – 3) = -7\sin\theta\,(2\sin\theta + 3)(2\sin\theta – 1).$$

So $\sin\theta = 0$, or $\sin\theta = \frac12$; $\sin\theta = -\frac32$ is impossible.

Question 13

If $\begin{vmatrix} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{vmatrix} = 0$, find the values of $x$.

Solution. Each row sums to $12 + x$. $C_1 \to C_1 + C_2 + C_3$, take it out, then $R_2 \to R_2 – R_1$, $R_3 \to R_3 – R_1$:

$$(12+x)\begin{vmatrix} 1 & 4+x & 4+x \\ 0 & -2x & 0 \\ 0 & 0 & -2x \end{vmatrix} = 4x^2(12+x) = 0.$$

Question 14

If $a_1, a_2, a_3, \ldots, a_r$ are in G.P., prove that $\begin{vmatrix} a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21} \end{vmatrix}$ is independent of $r$.

Solution. Write the G.P. as $a_n = AR^{\,n-1}$. Then the rows are

$$AR^{r}(1,\ R^4,\ R^8),\qquad AR^{r+6}(1,\ R^4,\ R^8),\qquad AR^{r+10}(1,\ R^6,\ R^{10}).$$

Take $AR^{r}$, $AR^{r+6}$ and $AR^{r+10}$ out of the three rows:

$$\Delta = A^3R^{3r+16}\begin{vmatrix} 1 & R^4 & R^8 \\ 1 & R^4 & R^8 \\ 1 & R^6 & R^{10} \end{vmatrix} = 0,$$

since the first two rows are identical. The determinant is $0$ for every $r$, so it is independent of $r$.

Question 15

Show that the points $(a+5, a-4)$, $(a-2, a+3)$ and $(a, a)$ do not lie on a straight line for any value of $a$.

Solution. Three points are collinear exactly when the area determinant is $0$. Apply $R_1 \to R_1 – R_3$, $R_2 \to R_2 – R_3$:

$$\begin{vmatrix} a+5 & a-4 & 1 \\ a-2 & a+3 & 1 \\ a & a & 1 \end{vmatrix} = \begin{vmatrix} 5 & -4 & 0 \\ -2 & 3 & 0 \\ a & a & 1 \end{vmatrix} = 15 – 8 = 7.$$

The value does not depend on $a$ and is never $0$; the triangle always has area $\frac72$.

Question 16

Show that $\triangle ABC$ is an isosceles triangle if the determinant

$$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1+\cos A & 1+\cos B & 1+\cos C \\ \cos^2 A+\cos A & \cos^2 B+\cos B & \cos^2 C+\cos C \end{vmatrix} = 0.$$

(The question prints square brackets; the determinant is meant.)

Solution. Write $p = \cos A$, $q = \cos B$, $r = \cos C$. Apply $R_3 \to R_3 – R_2$, which leaves $p^2 – 1$ and so on, then $R_3 \to R_3 + R_1$ and $R_2 \to R_2 – R_1$:

$$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ p & q & r \\ p^2 & q^2 & r^2 \end{vmatrix} = (p-q)(q-r)(r-p),$$

the standard Vandermonde form (subtract columns and take out the factors to check it). So $\Delta = 0$ forces $\cos A = \cos B$, or $\cos B = \cos C$, or $\cos C = \cos A$. The cosine is one-one on $(0, \pi)$, where every angle of a triangle lies, so two angles are equal and the triangle is isosceles.

Question 17

Find $A^{-1}$ if $A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}$, and show that $A^{-1} = \dfrac{A^2 – 3I}{2}$.

Solution. $|A| = 0(0 – 1) – 1(0 – 1) + 1(1 – 0) = 2$. Each cofactor is a $2 \times 2$ determinant: the diagonal ones are $0 – 1 = -1$ and the off-diagonal ones are $1$. The matrix is symmetric, so

$$\operatorname{adj}A = \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix},\qquad A^{-1} = \frac12\begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}.$$

Now $A^2 = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix}$ (each diagonal entry is $1 + 1$, each off-diagonal entry is $1$), so $A^2 – 3I = \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}$, and half of it is exactly $A^{-1}$.

Question 18

If $A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix}$, find $A^{-1}$. Using $A^{-1}$, solve the system $x – 2y = 10$, $2x – y – z = 8$, $-2y + z = 7$.

Solution. Expand along the first row: $|A| = 1(-1 – 2) – 2(-2 – 0) + 0 = 1$. The cofactors are

$$A_{11} = -3,\ A_{12} = 2,\ A_{13} = 2;\quad A_{21} = -2,\ A_{22} = 1,\ A_{23} = 1;\quad A_{31} = -4,\ A_{32} = 2,\ A_{33} = 3.$$

The adjoint is the transpose of the cofactor matrix, and $|A| = 1$, so

$$A^{-1} = \begin{bmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{bmatrix}.$$

The system’s coefficient matrix is $M = \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{bmatrix}$. This is not $A$ but its transpose: the rows of $M$ are the columns of $A$. That is why the question says “using $A^{-1}$”: $M^{-1} = (A^T)^{-1} = (A^{-1})^T$. So

$$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{bmatrix}\begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} = \begin{bmatrix} -30+16+14 \\ -20+8+7 \\ -40+16+21 \end{bmatrix} = \begin{bmatrix} 0 \\ -5 \\ -3 \end{bmatrix}.$$

Check: $0 + 10 = 10$, $0 + 5 + 3 = 8$, $10 – 3 = 7$.

Question 19

Using the matrix method, solve $3x + 2y – 2z = 3$, $x + 2y + 3z = 6$, $2x – y + z = 2$.

Solution. With $A = \begin{bmatrix} 3 & 2 & -2 \\ 1 & 2 & 3 \\ 2 & -1 & 1 \end{bmatrix}$, $|A| = 3(2 + 3) – 2(1 – 6) – 2(-1 – 4) = 15 + 10 + 10 = 35 \ne 0$, so there is a unique solution $X = A^{-1}B$. The cofactors give

$$\operatorname{adj}A = \begin{bmatrix} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{bmatrix},$$

$$X = \frac{1}{35}\begin{bmatrix} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{bmatrix}\begin{bmatrix} 3 \\ 6 \\ 2 \end{bmatrix} = \frac{1}{35}\begin{bmatrix} 35 \\ 35 \\ 35 \end{bmatrix}.$$

Question 20

Given $A = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}$, find $BA$ and use it to solve $y + 2z = 7$, $x – y = 3$, $2x + 3y + 4z = 17$.

Solution. Multiplying row by column,

$$BA = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = 6I,$$

so $B\left(\frac16 A\right) = I$ and $B^{-1} = \frac16 A$. Rearrange the equations to match the rows of $B$: $x – y = 3$, $2x + 3y + 4z = 17$, $y + 2z = 7$. That is $BX = \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix}$, so

$$X = \frac16\begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}\begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} = \frac16\begin{bmatrix} 12 \\ -6 \\ 24 \end{bmatrix}.$$

Question 21

If $a + b + c \ne 0$ and $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$, prove that $a = b = c$.

Solution. $C_1 \to C_1 + C_2 + C_3$, take out $a + b + c$, then $R_2 \to R_2 – R_1$, $R_3 \to R_3 – R_1$:

$$\Delta = (a+b+c)\begin{vmatrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \end{vmatrix} = (a+b+c)\big[-(b-c)^2 – (a-c)(a-b)\big]$$

$$= -(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = -\tfrac12(a+b+c)\big[(a-b)^2+(b-c)^2+(c-a)^2\big].$$

Since $a + b + c \ne 0$, $\Delta = 0$ forces the sum of three squares to be $0$, so each square is $0$: $a = b = c$.

Question 22

Prove that $\begin{vmatrix} bc-a^2 & ca-b^2 & ab-c^2 \\ ca-b^2 & ab-c^2 & bc-a^2 \\ ab-c^2 & bc-a^2 & ca-b^2 \end{vmatrix}$ is divisible by $a + b + c$, and find the quotient.

Solution. These entries are exactly the cofactors of the determinant in Question 21, $D = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$. For instance the cofactor of $a$ is $cb – a^2$, of the first-row $b$ is $-(b^2 – ca) = ca – b^2$, and of the first-row $c$ is $ab – c^2$; the other rows follow the same cycle.

For a $3 \times 3$ matrix, $A\,(\operatorname{adj}A) = |A|\,I$ gives $|A|\,|\operatorname{adj}A| = |A|^3$, so the determinant of the cofactors is $|A|^2$ (transposing does not change it). Hence

$$\Delta = D^2 = (a+b+c)^2(a^2+b^2+c^2-ab-bc-ca)^2,$$

using $D$ from Question 21. It contains the factor $a + b + c$ (twice, in fact).

Question 23

If $x + y + z = 0$, prove that $\begin{vmatrix} xa & yb & zc \\ yc & za & xb \\ zb & xc & ya \end{vmatrix} = xyz\begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix}$.

Solution. Expand the left side along the first row:

$$xa(yz\,a^2 – x^2bc) – yb(y^2ac – xz\,b^2) + zc(xy\,c^2 – z^2ab)$$

$$= xyz(a^3 + b^3 + c^3) – abc(x^3 + y^3 + z^3).$$

When $x + y + z = 0$, $x^3 + y^3 + z^3 = 3xyz$ (from $x^3+y^3+z^3-3xyz = (x+y+z)(\ldots)$). So the left side is $xyz(a^3 + b^3 + c^3 – 3abc)$.

The right-hand determinant expands to $a(a^2 – bc) – b(ca – b^2) + c(c^2 – ab) = a^3 + b^3 + c^3 – 3abc$. The two sides agree.

Common mistakes

• Taking a factor out of a row twice. A factor common to a row comes out once; taking $k$ out of every row of a $3 \times 3$ determinant gives $k^3$, not $k$ (Question 3 takes a different factor from each column).
• Forgetting to compensate. In Question 7, multiplying rows by $x$, $y$, $z$ multiplies the determinant by $xyz$, so you must divide by it.
• Using $A^{-1}$ when the system’s matrix is $A^T$. In Question 18 the coefficient matrix is the transpose of $A$; multiplying by $A^{-1}$ itself gives $(-74, 42, 49)$, which fails the first equation.
• Reading a bracketed determinant as a matrix. Questions 12, 13 and 16 print square brackets but set the expression equal to $0$, which only makes sense for a determinant.
• Stopping at $\sin\theta = \frac12$. Question 12 also has the root $\sin\theta = 0$, which comes from the factor you might divide out.

Practise next

• Exemplar Objective Type Q24–37, where the same properties settle a multiple-choice question in two lines.
• Exemplar Fill in the Blanks and True or False Q38–58, the rest of the unit, including $|\operatorname{adj}A|$ and $|kA|$.
• NCERT textbook Exercise 4.2 for area and collinearity, as in Questions 11 and 15.
• NCERT textbook Exercise 4.5 for solving systems by the matrix method, as in Questions 18 to 20.