NCERT Exemplar Solutions Class 12 Maths Determinants — Fill in the Blanks and True or False Q38–58

NCERT Exemplar Problems, Class 12 Mathematics — Determinants, Fill in the Blanks questions 38 to 47 and True or False questions 48 to 58. All 21 questions solved.

Twenty-one short questions. About half are facts about how the determinant behaves under scaling, inversion and the adjoint — $|kA| = k^n|A|$, $|A^{-1}| = |A|^{-1}$, $|\operatorname{adj}A| = |A|^{n-1}$ — which are worth memorising together (38, 39, 42, 48 to 52, 54). The other half are determinants that collapse to $0$ or a single factor after one column operation (40, 45 to 47, 53, 55 to 58).

A note on the syllabus. The rationalised NCERT textbook no longer has a section on properties of determinants, though its adjoint results (and $|kA| = k^n|A|$, in Exercise 4.1) remain. JEE still tests the properties, and the solutions below use them.

Key insight. For an $n \times n$ matrix, every scalar that comes out of a determinant comes out once per row: $|kA| = k^n|A|$. The same counting gives $|\operatorname{adj}A| = |A|^{n-1}$, from $A\,\operatorname{adj}A = |A|\,I$. Most True or False statements here are a check of the exponent.

Question 38

If $A$ is a matrix of order $3 \times 3$, then $|3A| = $ ______.

Solution. Multiplying $A$ by $3$ multiplies every one of its three rows by $3$, and each row contributes one factor: $|3A| = 3^3|A|$.

Question 39

If $A$ is an invertible matrix of order $3 \times 3$, then $|A^{-1}| = $ ______.

Solution. $AA^{-1} = I$, so $|A|\,|A^{-1}| = |I| = 1$.

Question 40

If $x, y, z \in \mathbb{R}$, then the value of the determinant $\begin{vmatrix} (2^x+2^{-x})^2 & (2^x-2^{-x})^2 & 1 \\ (3^x+3^{-x})^2 & (3^x-3^{-x})^2 & 1 \\ (4^x+4^{-x})^2 & (4^x-4^{-x})^2 & 1 \end{vmatrix}$ is equal to ______.

Solution. $(p + q)^2 – (p – q)^2 = 4pq$, and here $pq = a^x \cdot a^{-x} = 1$. So $C_1 \to C_1 – C_2$ turns the first column into $(4, 4, 4)$, which is $4$ times the third column. Two proportional columns make the determinant $0$.

Question 41

If $\cos2\theta = 0$, then $\begin{vmatrix} 0 & \cos\theta & \sin\theta \\ \cos\theta & \sin\theta & 0 \\ \sin\theta & 0 & \cos\theta \end{vmatrix}^2 = $ ______.

Solution. Expand along the first row:

$$\Delta = 0 – \cos\theta(\cos^2\theta – 0) + \sin\theta(0 – \sin^2\theta) = -(\cos^3\theta + \sin^3\theta).$$

$\cos2\theta = 0$ means $\cos^2\theta = \sin^2\theta = \frac12$, so $|\cos\theta| = |\sin\theta| = \frac{1}{\sqrt2}$. There are two cases.

• If $\sin\theta$ and $\cos\theta$ have the same sign ($\theta = \frac\pi4$, $\frac{5\pi}{4}$, …), $\cos^3\theta + \sin^3\theta = \pm\frac{2}{2\sqrt2} = \pm\frac{1}{\sqrt2}$ and $\Delta^2 = \frac12$.
• If they have opposite signs ($\theta = \frac{3\pi}{4}$, $\frac{7\pi}{4}$, …), the cubes cancel and $\Delta^2 = 0$.

The printed answer, $\frac12$, is the first case — the one you get by taking $\theta = \frac\pi4$.

Question 42

If $A$ is a matrix of order $3 \times 3$, then $(A^2)^{-1} = $ ______.

Solution. For an invertible $A$ (the inverse needs $|A| \ne 0$), $A^2 \cdot (A^{-1})^2 = A(AA^{-1})A^{-1} = I$, so the inverse of $A^2$ is $(A^{-1})^2$.

Question 43

If $A$ is a matrix of order $3 \times 3$, then the number of minors in the determinant of $A$ is ______.

Solution. There is one minor for each element, and a $3 \times 3$ matrix has $9$ elements.

Question 44

The sum of the products of the elements of any row with the cofactors of the corresponding elements is equal to ______.

Solution. This is exactly the expansion of the determinant along that row: $a_{i1}A_{i1} + a_{i2}A_{i2} + a_{i3}A_{i3} = |A|$. (With the cofactors of a different row, the same sum is $0$.)

Question 45

If $x = -9$ is a root of $\begin{vmatrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{vmatrix} = 0$, then the other two roots are ______.

Solution. Expand along the first row:

$$x(x^2 – 12) – 3(2x – 14) + 7(12 – 7x) = x^3 – 67x + 126.$$

Since $x = -9$ is a root, divide by $x + 9$: $x^3 – 67x + 126 = (x+9)(x^2 – 9x + 14) = (x+9)(x-2)(x-7)$.

Question 46

$\begin{vmatrix} 0 & xyz & x-z \\ y-x & 0 & y-z \\ z-x & z-y & 0 \end{vmatrix} = $ ______.

Solution. This looks skew-symmetric, but it is not: the $(1, 2)$ entry is $xyz$, whereas skew-symmetry would need $x – y$ there. So expand along the first row:

$$\Delta = -xyz\big[(y-x)\cdot 0 – (y-z)(z-x)\big] + (x-z)\big[(y-x)(z-y) – 0\big]$$

$$= xyz(y-z)(z-x) + (x-z)(y-x)(z-y).$$

The second term equals $(y-z)(z-x)(y-x)$ (two sign changes). Taking $(y-z)(z-x)$ out:

Question 47

If $f(x) = \begin{vmatrix} (1+x)^{17} & (1+x)^{19} & (1+x)^{23} \\ (1+x)^{23} & (1+x)^{29} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47} \end{vmatrix} = A + Bx + Cx^2 + \ldots$, then $A = $ ______.

Solution. The constant term of a polynomial is its value at $x = 0$. Every entry becomes $1^n = 1$, so $f(0) = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} = 0$ (identical rows).

Question 48

$(A^3)^{-1} = (A^{-1})^3$, where $A$ is a square matrix and $|A| \ne 0$.

Solution. $A^3(A^{-1})^3 = AA(AA^{-1})A^{-1}A^{-1} = \ldots = I$, cancelling one pair at a time from the middle. So $(A^{-1})^3$ is the inverse of $A^3$.

Question 49

$(aA)^{-1} = \dfrac1a A^{-1}$, where $a$ is any real number and $A$ is a square matrix.

Solution. The identity holds when $a \ne 0$ and $A$ is invertible, since $(aA)\left(\frac1a A^{-1}\right) = I$. But the statement says any real $a$ and any square $A$: for $a = 0$, $\frac1a$ does not exist and $0 \cdot A$ has no inverse; and a singular $A$ has no $A^{-1}$ at all.

Question 50

$|A^{-1}| \ne |A|^{-1}$, where $A$ is a non-singular matrix.

Solution. From $|A|\,|A^{-1}| = |I| = 1$, the two are always equal (Question 39).

Question 51

If $A$ and $B$ are matrices of order 3 and $|A| = 5$, $|B| = 3$, then $|3AB| = 27 \times 5 \times 3 = 405$.

Solution. $|3AB| = 3^3|AB| = 27\,|A|\,|B| = 27 \times 15 = 405$.

Question 52

If the value of a third-order determinant is 12, then the value of the determinant formed by replacing each element by its cofactor will be 144.

Solution. The determinant of cofactors is $|\operatorname{adj}A|$ (the adjoint is its transpose, which has the same determinant). From $A\,\operatorname{adj}A = |A|\,I_3$, $|A|\,|\operatorname{adj}A| = |A|^3$, so $|\operatorname{adj}A| = |A|^2 = 144$.

Question 53

$\begin{vmatrix} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{vmatrix} = 0$, where $a, b, c$ are in A.P.

Solution. $R_1 \to R_1 + R_3 – 2R_2$ gives $(0,\ 0,\ a + c – 2b)$. Since $a, b, c$ are in A.P., $2b = a + c$, so the whole row is zero and the determinant is $0$.

Question 54

$|\operatorname{adj}A| = |A|^2$, where $A$ is a square matrix of order two.

Solution. For order $n$, $|\operatorname{adj}A| = |A|^{n-1}$; for $n = 2$ that is $|A|$, not $|A|^2$. For example, $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ has $\operatorname{adj}A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$, whose determinant is $ad – bc = |A|$. With $A = 2I_2$: $|A| = 4$, $|\operatorname{adj}A| = 4$, not $16$.

Question 55

The determinant $\begin{vmatrix} \sin A & \cos A & \sin A+\cos B \\ \sin B & \cos A & \sin B+\cos B \\ \sin C & \cos A & \sin C+\cos B \end{vmatrix}$ is equal to zero.

Solution. $C_3 \to C_3 – C_1$ leaves the column $(\cos B, \cos B, \cos B)$. The second column is $(\cos A, \cos A, \cos A)$. Two columns with all entries equal are proportional (and if either is the zero column, the determinant is $0$ anyway).

Question 56

If the determinant $\begin{vmatrix} x+a & p+u & l+f \\ y+b & q+v & m+g \\ z+c & r+w & n+h \end{vmatrix}$ splits into exactly $K$ determinants of order 3, each element of which contains only one term, then the value of $K$ is 8.

Solution. A determinant whose column is a sum of two columns splits into two determinants. Each of the three columns here is a sum of two, so splitting all of them gives $2 \times 2 \times 2 = 8$ determinants.

Question 57

Let $\Delta = \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} = 16$. Then $\Delta_1 = \begin{vmatrix} p+x & a+x & a+p \\ q+y & b+y & b+q \\ r+z & c+z & c+r \end{vmatrix} = 32$.

Solution. $C_1 \to C_1 + C_2 + C_3$ makes the first column $2(a + p + x)$, $2(b + q + y)$, $2(c + r + z)$; take out the $2$. Then $C_2 \to C_2 – C_1$ and $C_3 \to C_3 – C_1$ leave $-p$ and $-x$:

$$\Delta_1 = 2\begin{vmatrix} a+p+x & -p & -x \\ b+q+y & -q & -y \\ c+r+z & -r & -z \end{vmatrix}.$$

$C_1 \to C_1 + C_2 + C_3$ reduces the first column to $(a, b, c)$, and taking $-1$ out of each of the other two columns gives $(-1)^2 = 1$. So $\Delta_1 = 2\Delta = 32$.

Question 58

The maximum value of $\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+\sin\theta & 1 \\ 1 & 1 & 1+\cos\theta \end{vmatrix}$ is $\dfrac12$.

Solution. $R_2 \to R_2 – R_1$, $R_3 \to R_3 – R_1$ give a triangular determinant:

$$\begin{vmatrix} 1 & 1 & 1 \\ 0 & \sin\theta & 0 \\ 0 & 0 & \cos\theta \end{vmatrix} = \sin\theta\cos\theta = \frac12\sin2\theta,$$

whose maximum is $\frac12$, at $\theta = \frac\pi4$. (Compare Question 31, where the $\cos\theta$ sits in a different place and the sign flips, but the maximum is the same.)

Common mistakes

• One factor of $k$ instead of $k^n$. $|3A| = 27|A|$ for a $3 \times 3$ matrix, not $3|A|$ (Questions 38 and 51).
• Using $|A|^{n-1}$ with the wrong $n$. $|\operatorname{adj}A|$ is $|A|^2$ for order 3 but $|A|$ for order 2 (Questions 52 and 54).
• Calling a determinant skew-symmetric by its look. Question 46 has zeros on the diagonal and $y – x$, $z – x$ below it, but the $xyz$ entry breaks the pattern, so the value is not $0$.
• Taking one solution of a trigonometric condition. $\cos2\theta = 0$ allows $\theta = \frac{3\pi}{4}$ as well as $\frac\pi4$, and the two give different answers in Question 41.

Practise next

• Exemplar Short and Long Answer Q1–23, where Question 22 uses the result of Question 52 to finish a proof.
• Exemplar Objective Type Q24–37, the multiple-choice questions from this unit.
• NCERT textbook Exercise 4.4 for the adjoint and inverse, the source of the $|\operatorname{adj}A|$ results here.
• NCERT textbook Miscellaneous Exercise on Determinants for more identities of the kind in Questions 53 and 57.