NCERT Class 12 Mathematics — Determinants, Exercise 4.1. All 8 questions solved.

Exercise 4.1 is mechanical, and deliberately so: it is where evaluating a determinant becomes automatic before Chapter 4 starts using them for areas, inverses and systems of equations.

For a $2\times2$ matrix the rule is a single cross-multiplication:

$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad – bc$$

For $3\times3$, expand along a row or column, alternating the signs according to the pattern

$$\begin{vmatrix} + & – & + \\ – & + & – \\ + & – & + \end{vmatrix}$$

Key insight. Every row and every column gives the same answer, so you are free to choose — and you should choose the one with the most zeros, because each zero kills an entire $2\times2$ minor. Question 5(i) has two zeros in its second row and collapses to a single term; expanding it along the first row instead means three minors for the same result.

Question 1

Evaluate the determinant $\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$.

Solution. Cross-multiply, main diagonal first.

$$= (2)(-1) – (4)(-5) = -2 + 20 = 18$$

The second minus sign is the one to watch: subtracting $4 \times (-5)$ adds 20.

Question 2

Evaluate the determinants.

Question 2 (i)

$\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$

Solution.

$$= (\cos\theta)(\cos\theta) – (-\sin\theta)(\sin\theta) = \cos^2\theta + \sin^2\theta = 1$$

This matrix rotates the plane by $\theta$, and its determinant being 1 says a rotation preserves area — worth knowing, since determinants and area are the subject of the next section.

Question 2 (ii)

$\begin{vmatrix} x^2 – x + 1 & x – 1 \\ x + 1 & x + 1 \end{vmatrix}$

Solution.

$$= (x^2 – x + 1)(x + 1) – (x – 1)(x + 1)$$

Both products are standard identities, which is quicker than expanding term by term. The first is the sum of cubes $a^3 + b^3 = (a+b)(a^2 – ab + b^2)$ with $b = 1$; the second is the difference of squares.

$$= (x^3 + 1) – (x^2 – 1) = x^3 – x^2 + 2$$

Question 3

If $\mathrm{A} = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$, then show that $|2\mathrm{A}| = 4|\mathrm{A}|$.

Solution. Compute both sides separately.

$$|\mathrm{A}| = (1)(2) – (2)(4) = 2 – 8 = -6$$

Multiplying a matrix by 2 multiplies every entry by 2 — not just one row:

$$2\mathrm{A} = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}, \qquad |2\mathrm{A}| = (2)(4) – (4)(8) = 8 – 32 = -24$$

$$4|\mathrm{A}| = 4(-6) = -24$$

The two agree. The reason is that each of the 2 rows contributes a factor of 2, giving $2^2 = 4$ — an instance of the general rule $|k\mathrm{A}| = k^n|\mathrm{A}|$ for an $n\times n$ matrix.

Question 4

If $\mathrm{A} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}$, then show that $|3\mathrm{A}| = 27|\mathrm{A}|$.

Solution. The first column has two zeros, so expand along it — only the top entry survives.

$$|\mathrm{A}| = 1\begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} = 1(4 – 0) = 4$$

Now $3\mathrm{A}$, again multiplying every entry:

$$3\mathrm{A} = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix}$$

$$|3\mathrm{A}| = 3\begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} = 3(36 – 0) = 108$$

$$27|\mathrm{A}| = 27(4) = 108$$

Here $n = 3$, so the factor is $3^3 = 27$ — one 3 from each of the three rows. That is the whole point of putting questions 3 and 4 side by side.

Question 5

Evaluate the determinants.

Question 5 (i)

$\begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix}$

Solution. The second row is $0, 0, -1$ — two zeros, so expanding along it leaves a single term. The sign for position (2,3) is $(-1)^{2+3} = -1$.

$$= -(-1)\begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix} = +\begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix}$$

$$= (3)(-5) – (-1)(3) = -15 + 3 = -12$$

Question 5 (ii)

$\begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}$

Solution. No zeros anywhere, so expand along the first row and take care with the alternating signs.

$$= 3\begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} – (-4)\begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} + 5\begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix}$$

$$= 3(1 + 6) + 4(1 + 4) + 5(3 – 2) = 21 + 20 + 5 = 46$$

Question 5 (iii)

$\begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix}$

Solution. Expand along the first row; its leading zero removes one minor.

$$= 0 – 1\begin{vmatrix} -1 & -3 \\ -2 & 0 \end{vmatrix} + 2\begin{vmatrix} -1 & 0 \\ -2 & 3 \end{vmatrix}$$

$$= -1\big(0 – 6\big) + 2\big(-3 – 0\big) = 6 – 6 = 0$$

The zero is not a coincidence: this matrix is skew-symmetric (its transpose is its negative), and every skew-symmetric matrix of odd order has determinant zero.

Question 5 (iv)

$\begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix}$

Solution. Expanding along the first row:

$$= 2\begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} – (-1)\begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix} + (-2)\begin{vmatrix} 0 & 2 \\ 3 & -5 \end{vmatrix}$$

$$= 2(0 – 5) + 1(0 + 3) – 2(0 – 6) = -10 + 3 + 12 = 5$$

Question 6

If $\mathrm{A} = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}$, find $|\mathrm{A}|$.

Solution. Expanding along the first row:

$$= 1\begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix} – 1\begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix} + (-2)\begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix}$$

$$= 1(-9 + 12) – 1(-18 + 15) – 2(8 – 5)$$

$$= 3 + 3 – 6 = 0$$

A determinant of zero means the matrix is singular — it has no inverse, and its rows are linearly dependent. (Indeed row 3 is row 1 plus twice row 2.) This matters from Exercise 4.4 onwards, where invertibility decides whether a system of equations has a unique solution.

Question 7

Find values of $x$.

Question 7 (i)

$\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$

Solution. Evaluate both sides and equate.

$$2 – 20 = 2x^2 – 24$$

$$-18 = 2x^2 – 24 \quad\Rightarrow\quad 2x^2 = 6 \quad\Rightarrow\quad x^2 = 3$$

$$x = \pm\sqrt{3}$$

Both roots are valid — nothing in the question rules out negative $x$.

Question 7 (ii)

$\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$

Solution.

$$10 – 12 = 5x – 6x$$

$$-2 = -x \quad\Rightarrow\quad x = 2$$

Here the $x$ terms are linear rather than squared, so there is only one solution.

Question 8

If $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$, then $x$ is equal to

Solution. Evaluate the right-hand side first, since it is fully numerical.

$$\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix} = 36 – 36 = 0$$

$$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = x^2 – 36$$

$$x^2 – 36 = 0 \quad\Rightarrow\quad x^2 = 36 \quad\Rightarrow\quad x = \pm 6$$

Option (A) is the trap: $x = 6$ does work, but so does $x = -6$, and a multiple-choice question offering $\pm 6$ separately is asking whether you stopped at the first root.

Common mistakes

• Losing a sign when the entry being subtracted is negative. In question 1, $-(4)(-5) = +20$. Two negatives appear in almost every part of question 5.
• Forgetting the alternating signs in a $3\times3$ expansion. The middle term is subtracted, so $-(-4)$ becomes $+4$ in question 5(ii). Writing the $+\,-\,+$ pattern above the row before starting costs nothing.
• Scaling only one row when the whole matrix is multiplied. $2\mathrm{A}$ doubles all four entries in question 3, which is why the determinant changes by $2^2$ rather than by 2.
• Applying $|k\mathrm{A}| = k|\mathrm{A}|$. The correct rule is $k^n|\mathrm{A}|$ where $n$ is the order — $4|\mathrm{A}|$ for $2\times2$, $27|\mathrm{A}|$ for $3\times3$. Questions 3 and 4 are a matched pair testing exactly this.
• Expanding along a row with no zeros when a better one exists. Question 5(i) is one $2\times2$ minor if you use the second row, and three if you don’t.
• Taking only the positive square root. Questions 7(i) and 8 both have two solutions, and question 8 puts the incomplete answer in the options on purpose.

Practise next

• Exercise 4.2 — the area of a triangle from a determinant, and the collinearity test that follows from it.
• Exercise 4.4 — minors, cofactors and the adjoint, which build directly on the expansion practised here.