NCERT Solutions for Class 12 Maths Chapter 4 Determinants - Miscellaneous Exercise

Access Exercises of Class 12 Maths Chapter 4 –Determinants

Exercise 4.1 Solutions: 8 Questions (2 Long, 5 Short Answers, 1 MCQ)

Exercise 4.2 Solutions: 16 Questions(7 Long, 7 Short, 2 MCQs)

Exercise 4.3 Solutions: 5 Questions ( 4 Short Answers, 1 MCQ)

Exercise 4.4 Solutions: 5 Questions (4 Long, 1 MCQ)

Exercise 4.5 Solutions: 18 Questions (11 Long, 5 Short, 2 MCQs)

Exercise 4.6 Solutions: 16 Questions (13 Long, 3 Short)

Miscellaneous Exercise Solutions: 19 Questions (15 Long, 1 Short, 3 MCQs)

Miscellaneous Exercise

1. Prove that the determinant

$$\begin{bmatrix}\textbf{x} &\textbf{sin}\space\theta &\textbf{cos}\space\theta\\\textbf{\normalsize-sin}\space\theta &\textbf{\normalsize-x} &\textbf{1}\\\textbf{cos}\space\theta &\textbf{1} &\textbf{x}\end{bmatrix}$$

is independent of θ.

$$\textbf{Sol.\space}\text{Let A = }\begin{vmatrix}\textbf{x} &\textbf{sin}\space\theta &\textbf{cos}\space\theta\\-\textbf{sin}\space\theta &\textbf{\normalsize-x} &1\\\textbf{cos}\space\theta &\textbf{1} &\textbf{x}\end{vmatrix}$$

Expanding to corresponding first row, we get

$$\text{A} = x\begin{vmatrix}-x &1\\1 &x\end{vmatrix}-\text{sin}\space\theta\begin{vmatrix}-\text{sin}\space\theta &1\\\text{cos}\space\theta &x\end{vmatrix}\\+ \text{cos}\space\theta\begin{vmatrix}-\text{sin}\space\theta &\normalsize-x\\\text{cos}\space\theta &1\end{vmatrix}$$

= x (– x² – 1) – sin θ(– x sin θ – cos θ) + cos θ(– sin θ + x cos θ)

= – x³ – x + x sin²θ + sin θ cos θ – sin θ cos θ + x cos² θ

= – x³ – x + x(sin² θ + cos² θ)

= – x³ – x + x (∵ sin² θ + cos² θ = 1)

= – x³.

Hence, A is independent of θ.

2. Without expanding the determinant, prove that

$$\begin{vmatrix}\textbf{a} &\textbf{a}^{2} &\textbf{bc}\\\textbf{b} &\textbf{b}^{2} &\textbf{ca}\\\textbf{c} &\textbf{c}^{2} &\textbf{ab}\end{vmatrix}\textbf{=}\begin{vmatrix}\textbf{1} &\textbf{a}^{\textbf{2}} &\textbf{a}^{\textbf{3}}\\\textbf{1} &\textbf{b}^{2} &\textbf{b}^{3}\\1 &\textbf{c}^{2} &\textbf{c}^{3}\end{vmatrix}\\\textbf{Sol.\space}\text{LHS = }\begin{vmatrix}a &a^{2} &bc\\b &b^{2} &ca\\c &c^{2} &ab\end{vmatrix}$$

Operating R₁ → aR₁, R₂ → bR₂ and R₃ → cR₃ we obtain

$$=\frac{1}{abc}\begin{vmatrix}a^{2} &a^{3} &abc\\b^{2} &b^{3} &abc\\c^{2} &c^{3} &abc\end{vmatrix}\\=\frac{1}{abc}×abc\begin{vmatrix}a^{2} &a^{3} &1\\b^{2} &b^{3} &1\\c^{2} &c^{3} &1\end{vmatrix}$$

[Taking out factor abc from C₃]

$$=(\normalsize-1)^{2}\begin{vmatrix}1 &a^{2} &a^{3}\\1 &b^{2} &b^{3}\\1 &c^{2} &c^{3}\end{vmatrix}$$

(using C₁ ↔ C₃ and C₂ ↔ C₃)

$$=\begin{vmatrix}1 &a^{2} &a^{3}\\1 &b^{2} &b^{3}\\ 1& c^{2} & c^{3}\end{vmatrix}=\text{RHS}$$

Note : When we multiply any row or column of a determinant by any constant k then it is necessary that determinant divides by k.

$$\textbf{3.\space}\textbf{Evaluate}\\\space\begin{vmatrix}\textbf{cos}\alpha\textbf{cos}\beta &\textbf{cos}\alpha\textbf{sin}\beta&-\textbf{sin\space}\alpha\\\textbf{-sin}\space\beta &\textbf{cos}\space\beta &\textbf{0}\\\textbf{sin}\space\alpha\textbf{cos}\space\beta &\textbf{sin}\space\alpha\textbf{sin}\space\beta &\textbf{cos}\space\alpha\end{vmatrix}$$

Sol. Given, determinant

$$=\begin{vmatrix}\text{cos}\alpha\text{cos}\beta &\text{cos}\alpha\text{sin}\beta &-\text{sin}\space\alpha\\-\text{sin}\space\beta &\text{cos}\space\beta &0\\\text{sin}\alpha\text{cos}\beta &\text{sin}\space\alpha\text{sin}\beta &\text{cos}\space\alpha\end{vmatrix}$$

Expanding corresponding to R₁, we get
= cos α cos β (cos α cos β – 0) – cos α sin β(– cos α sin β – 0) – sin α(– sin² β sin α – cos² β sin α)

= cos² α cos² β + cos² α sin² β + sin² α (sin² β + cos² β)

= cos² α (cos² β + sin² β) + sin² α (sin² β + cos² β)

= cos² α (1) sin² α (1) = cos² α + sin² α = 1.

[∵ sin² θ + cos² θ = 1]

4. If a, b and c are real numbers and

$$\Delta =\begin{vmatrix}\textbf{b+c} &\textbf{c+a} &\textbf{a+b}\\\textbf{c+a} &\textbf{a+b} &\textbf{b+c}\\\textbf{a+b} &\textbf{b+c} &\textbf{c+a}\end{vmatrix} \textbf{= 0.}$$

Show that either

a + b + c = 0 or a = b = c.

$$\textbf{Sol.\space}\text{Given,\space}\Delta =\begin{vmatrix}b+c &c+a &a+b\\\text{c+a} &a+b &b+c\\a+b &b+c &c+a\end{vmatrix}\\=\begin{vmatrix}2(a+b+c) &c+a &a+b\\2(a+b+c) &a+b &b+c\\2(a+b+c) &b+c &c+a\end{vmatrix}$$

(using C₁ → C₁ + C₂ + C₃)

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

(Taking out factor 2(a + b) + c) from C₁)

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

(using R₂ → R₂ – R₁, R₂ → R₃ – R₁)

By expanding along C₁, we get

= 2(a + b + c)[(b – c)(c – b) – (b – a)(c – a)]

= 2(a + b + c) [bc – c² – b² + bc – bc + ac + ab – a²]

= 2(a + b + c)[ab + bc + ca – a² – b² – c²]

It is given that Δ = 0,

∴ (a + b + c) × 2[ab + bc + ca – a² – b² – c²] = 0

⇒ Either (a + b + c) = 0

or 2[ab + bc + ca – a² – b² – c²] = 0

If 2[ab + bc + ca – a² – b² – c²] = 0

⇒ –(2a² + 2b² + 2c² – 2ab – 2bc – 2ca) = 0

[Taking negative sign common]

⇒ 2a² + 2b² + 2c² – 2ab – 2bc – 2ca = 0

⇒ (a² + b² – 2ab) + (b² + c² – 2bc) + (c² + a² – 2ca) = 0

⇒ (a – b)² + (b – c)² + (c – a²)= 0

[∵ (x – y)² = x² + y² – 2xy]

⇒ (a – b)² = (b – c)² = (c – a)² = 0

[∵ (a – b)², (b – c)², (c – a)² are positive]

⇒ (a – b)² = (b – c)² = (c – a)² = 0

⇒ a = b = c

Hence, if Δ = 0, then either a + b + c = 0 or a = b = c.

5. Solve the equation, if

$$\begin{vmatrix}\textbf{x+a} &\textbf{x} &\textbf{x}\\\textbf{x} &\textbf{x+a} &\textbf{x}\\\textbf{x} &\textbf{x} &\textbf{x+a}\end{vmatrix}\\\textbf{= 0, a}\neq\textbf{0,}$$

$$\textbf{Sol.\space}\text{Given,\space}\begin{vmatrix}x+a &x &x\\x &x+a &x\\x &x &x+a\end{vmatrix}\\= 0\\\Rarr\space\begin{vmatrix}3x+a &3x+a &3x+a\\ x &x+a &x\\x &x &x+a\end{vmatrix}=0$$

(using R₁ → R₁ + R₂ + R₃)

$$\Rarr\space(3x+a)\begin{vmatrix}1 &1 &1\\x &x+a &x\\x &x &x+a\end{vmatrix}= 0$$

(Taking out (3x + a) from R₁)

$$\Rarr\space(3x+a)\begin{vmatrix}1 &0 &0\\x &a &0\\x &0 &a\end{vmatrix}=0,$$

(using C₂ → C₂ – C₁ and C₃ → C₃ – C₁)

Expanding corresponding to R₁ (first row), we get

(3x + a)[1(a × a – 0)] = 0

⇒ a²(3x + a) = 0

But a ≠ 0 (given). Therefore, 3x + a = 0

$$\Rarr\space x =-\frac{a}{3}.$$

6. Prove that

$$\begin{vmatrix}\textbf{a}^{\textbf{2}} &\textbf{bc} &\textbf{ac+c}^{\textbf{2}}\\\textbf{a}^{2}\textbf{+ ab} &\textbf{b}^{\textbf{2}} &\textbf{ac}\\\textbf{ab} &\textbf{b}^{\textbf{2}}\textbf{+bc} &\textbf{c}^{\textbf{2}}\end{vmatrix}$$= 4a²b²c².

$$\textbf{Sol.\space}\text{LHS =}\space\begin{vmatrix}a^{2} &bc &ac+c^{2}\\a^{2}+ab &b^{2} &ac\\ab &b^{2}+bc &c^{2}\end{vmatrix}\\= abc\begin{vmatrix}a &c &a+c\\a+b &b &a\\b &b+c &c\end{vmatrix}$$

(Take out a from C₁, b fom C₂ and c from C₃)

$$=abc\begin{vmatrix}0 &c &a+c\\2b &b &a\\2b &b+c &c\end{vmatrix}$$

(using C₁ → C₁ + C₂ – C₃)

$$=\space abc\begin{vmatrix}0 &c &a+c\\0 &\normalsize-c &a-c\\2b &b+c &c\end{vmatrix}$$

$$\lbrack\text{using}\space\text{R}_{2}\xrightarrow{}\text{R}_{2}-\text{R}_{3}\rbrack$$

Expanding along C₁, we get

= (abc) [(2b){c(a – c) + c(a + c)}]

= 2(ab²c)(2ac) = 4a²b²c² = RHS

Hence proved.

$$\textbf{7.\space If A}^{\normalsize-1} =\begin{bmatrix}\textbf{3} &\textbf{\normalsize-1} &\textbf{\normalsize1}\\\textbf{\normalsize-15} &\textbf{6} &\textbf{\normalsize-5}\\\textbf{5} &\textbf{\normalsize-2} &\textbf{2}\end{bmatrix}\\\textbf{and B = }\begin{bmatrix}\textbf{1} &\textbf{2} &\textbf{-2}\\\textbf{-1} &\textbf{3} &\textbf{0}\\\textbf{0} &\textbf{-2} &\textbf{1}\end{bmatrix}\textbf{,}\\\textbf{find\space (AB)}^{\normalsize-1}\textbf{.}$$

Sol. We know that (AB)^–1 = B^–1A^–1 and A^–1 is known, therefore, we proceed to find B^–1.

$$\text{Here,\space |B| =}\begin{vmatrix}1 &2 &\normalsize-2\\\normalsize-1 &3 &0\\0 &\normalsize-2 &1\end{vmatrix}$$

= 1(3 – 0) – 2(– 1 – 0) – 2(2 – 0)

= 3 + 2 – 4 = 1 ≠ 0

∴ B^–1 exists. Cofactors of B are

B₁₁ = (3 – 0) = 3, B₁₂ = – (– 1 – 0) = 1, B₁₃ = (2 – 0) = 2,

B₂₁ = – (2 – 4) = 2,B₂₂ = 1 – 0 = 1, B₂₃ = – (– 2 – 0) = 2,

B₃₁ = (0 + 6) = 6, B₃₂ = – (0 – 2)= 2, B₃₃ = 3 + 2 = 5,

$$\therefore\space\text{adj B} =\begin{bmatrix}3 &1 &2\\2 &1 &2\\6 &2 &5\end{bmatrix}^{\text{T}}=\begin{bmatrix}3 &2 &6\\1 &1 &2\\2 &2 &5\end{bmatrix}\\\therefore\space\text{B}^{\normalsize-1}=\frac{1}{|\text{B}|}\space\text{adj(B)}\\=\frac{1}{1}\begin{bmatrix}3 &2 &6\\1 &1 &2\\2 &2 &5\end{bmatrix}=\begin{bmatrix}3 &2 &6\\1 &1 &2\\2 &2 &5\end{bmatrix}$$

Now, (AB)^–1 = B^–1A^–1

$$=\begin{bmatrix}3 &2 &6\\1 &1 &2\\2 &2 &5\end{bmatrix}\begin{bmatrix}3 &\normalsize-1 &1\\\normalsize-15 &6 &\normalsize-5\\5 &\normalsize-2 &2\end{bmatrix}\\=\\\begin{bmatrix}9-30+30 &-3+12-12 &3-10+12\\3-15+10 &-1+6-4 &1-5+4\\6-30+25 &-2+12-10 &2-10+10\end{bmatrix}\\=\begin{bmatrix}9 &\normalsize-3 &5\\\normalsize-2 &1 &0\\1 &0 &2\end{bmatrix}.$$

$$\textbf{8.\space Let A = }\begin{bmatrix}\textbf{1} &\textbf{2} &\textbf{1}\\\textbf{2} &\textbf{3} &\textbf{1}\\\textbf{1} &\textbf{1} &\textbf{5}\end{bmatrix}\textbf{,}\space\textbf{verify that}$$

(a) [adj A]^–1 = adj (A^–1)

(b) (A^–1)^–1 = A.

$$\textbf{Sol.\space}\text{Given, A = }\begin{vmatrix}1 &2 &1\\2 &3 &1\\1 &1 &5\end{vmatrix}$$

|A| = 1(15 – 1) – 2(10 – 1) + 1(2 – 3)

= 14 – 18 – 1

= – 5 ≠ 0
Hence, its inverse exists.

Cofactors of A

A₁₁ = (– 1)¹⁺¹ (15 – 1) = 14

A₁₂ = (– 1)¹⁺² (10 – 1) = –9

A₁₃ = (– 1)¹⁺³ (2 – 3) = –1

A₂₁ = (– 1)²⁺¹ (10 – 1) = –9

A₂₂ = (– 1)²⁺² (5 – 1) = 4

A23 = (– 1)²⁺³ (1 – 2) = 1

A₃₁ = (– 1)³⁺¹ (2 – 3) = –1

A₃₂ = (– 1)³⁺² (1 – 2) = –1

A₃₃ = (– 1)³⁺³ (3 – 4) = –1

$$\text{adj(A)} = \begin{bmatrix}14 &\normalsize-9 &\normalsize-1\\\normalsize-9 &4 &1\\\normalsize-1 &1 &\normalsize-1\end{bmatrix}\\\therefore\space\text{A}^{\normalsize-1}=\frac{1}{|\text{A}|}\space\text{(adj A)}\\=\frac{1}{\normalsize-5}\begin{bmatrix}14 &\normalsize-9 &\normalsize-1\\\normalsize-9 &4 &1\\\normalsize-1 &1 &\normalsize-1\end{bmatrix}\\\text{(a) Here, adj (A) =}\\\begin{bmatrix}14 &\normalsize-9 &\normalsize-1\\\normalsize-9 &4 &1\\\normalsize-1 &1 &\normalsize-1\end{bmatrix} =\text{B(let)}$$

∴ |B| = 14 (– 4 – 1) + 9(9 + 1) – 1(– 9 + 4)

= –70 + 90 + 5 = 25

$$\therefore\space\text{B}^{\normalsize-1}=\frac{1}{|\text{B}|}\space\text{adj B}$$

C₁₁ = –5, C₁₂ = – 10, C₁₃ = –5, C₂₁ = –10, C₂₂ = –15

C₂₃ = –5, C₃₁ = – 5, C₃₂ = –5, C₃₃ = –25

$$\text{adj B = }\begin{bmatrix}\normalsize-5 &\normalsize-10 &\normalsize-5\\\normalsize-10 &\normalsize-15 &\normalsize-5\\ \normalsize-5 &\normalsize-5 &\normalsize-25\end{bmatrix}\\\text{B}^{\normalsize-1}=\frac{1}{|\text{B}|}\space\text{adj B}\\=\frac{1}{25}\begin{bmatrix}\normalsize-5 &\normalsize-10 &\normalsize-5\\\normalsize-10 &\normalsize-15 &\normalsize-5\\\normalsize-5 &\normalsize-5 &\normalsize-25\end{bmatrix}\\=\begin{bmatrix}\frac{\normalsize-1}{5} &\frac{\normalsize-2}{5} &\frac{\normalsize-1}{5}\\\frac{\normalsize-2}{5} &\frac{\normalsize-3}{5} &\frac{\normalsize-1}{5}\\\frac{\normalsize-1}{5} &\frac{\normalsize-1}{5} &\normalsize-1\end{bmatrix}$$

$$\\\text{LHS : }[\text{adj A}]^{\normalsize-1}=\\\begin{bmatrix}\frac{\normalsize-1}{15} &\frac{\normalsize-2}{5} &\frac{\normalsize-1}{5}\\\frac{\normalsize-2}{5} &\frac{\normalsize-3}{5} &\frac{\normalsize-1}{5}\\\frac{\normalsize-1}{5} &\frac{\normalsize-1}{5} &\normalsize-1\end{bmatrix}$$

RHS : adj (A^–1) = adj C

$$\text{A}^{\normalsize-1} =\text{C} =\begin{bmatrix}\frac{\normalsize-14}{5} &\frac{9}{5} &\frac{1}{5}\\\frac{9}{5} &\frac{\normalsize-4}{5} &\frac{\normalsize-1}{5}\\\frac{1}{5} &\frac{\normalsize-1}{5} &\frac{1}{5}\end{bmatrix}$$

Cofactors of C

B₁₁ = –1/5, B₁₂ = –2/5, B₁₃ = 1/5,

B₂₁ = –2/5, B₂₂ = –3/5, B₂₃ = –1/5,

B₃₁ = –1/5, B₃₂ = –1/5, B₃₃ = –1

RHS = adj C = adj (A^–1) =

$$\begin{bmatrix}\frac{\normalsize-1}{5} &\frac{\normalsize -2}{5} &\frac{\normalsize-1}{5}\\\frac{\normalsize-2}{5} &\frac{\normalsize-3}{5} &\frac{\normalsize-1}{5}\\\frac{\normalsize-1}{5} &\frac{\normalsize-1}{5} &\normalsize-1\end{bmatrix}$$

Hence, LHS = RHS Hence proved.

$$\text{(b)\space (A}^{\normalsize-1})^{\normalsize-1}=\frac{1}{|\text{A}|}\space\text{(adj A}^{\normalsize-1})\\=\frac{1}{\normalsize-5}\begin{bmatrix}\frac{\normalsize-1}{5} &\frac{\normalsize-2}{5} &\frac{\normalsize-1}{5}\\\frac{\normalsize-2}{5} &\frac{\normalsize-3}{5} &\frac{\normalsize-1}{5}\\\frac{\normalsize-1}{5} &\frac{\normalsize-1}{5} &\normalsize-1\end{bmatrix}\\=\begin{bmatrix}1 &2 &1\\ 2 &3 &1\\1 &1 &3\end{bmatrix}=\text{A}\\\textbf{Hence proved.}$$

$$\textbf{9. Evaluate}\space\begin{vmatrix}\textbf{x} &\textbf{y} &\textbf{x+y}\\\textbf{y} &\textbf{x+y} &\textbf{x}\\\textbf{x+y} &\textbf{x} &\textbf{y}\end{vmatrix}\textbf{.}\\\textbf{Sol.\space}\text{Let A =}\begin{vmatrix}x &y &x+y\\y &x+y &x\\x+y &x &y\end{vmatrix}\\=\begin{vmatrix}2(x+y) &y &x+y\\2(x+y) &x+y &x\\2(x+y) &x &y\end{vmatrix}$$

(using C₁ → C₁ + C₂ + C₃)

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

(using R₂ → R₂ – R₁; R₃ → R₃ – R₁)

Expanding along R₁, we get

= 2(x + y) × 1{– x² + y(x – y)}

= 2(x + y)(– x² + xy – y²)

= – 2(x + y)(x² – xy + y²)

= – 2(x² + y²)

10. Evalaute

$$\begin{vmatrix}\textbf{1} &\textbf{x} &\textbf{y}\\\textbf{1} &\textbf{x+y} &\textbf{y}\\\textbf{1} &\textbf{x} &\textbf{x+y}\end{vmatrix}\textbf{.}\\\textbf{Sol.\space}\text{Let A = }\begin{vmatrix}1 &x &y\\1 &x+y &y\\1 &x &x+y\end{vmatrix}\\=\begin{vmatrix}1 &x &y\\0 &y &0\\0 &0 &x\end{vmatrix}$$

(using R₂ → R₂ – R₁ and R₃ → R₃ – R₁)

Expanding along C₁, we get A = 1(yx – 0) = xy

Using properties of determinants in Q. 11 to 15, prove that :

$$\textbf{11.\space}\begin{vmatrix}\alpha &\alpha^{2} &\beta+\gamma\\\beta &\beta^{2} &\gamma + \alpha\\\gamma &\gamma^{2} &\alpha+\beta\end{vmatrix}\\=(\beta - \gamma)(\gamma -\alpha)(\alpha-\beta)(\alpha+\beta+\gamma).$$

$$\textbf{Sol.\space}\text{Let A = }\begin{vmatrix}\alpha &\alpha^{2} &\beta+\gamma\\\beta &\beta^{2} &\gamma +\alpha\\\gamma &\gamma^{2} &\alpha +\beta\end{vmatrix}\\=\begin{vmatrix}\alpha &\alpha^{2} &\alpha+\beta+\gamma\\\beta&\beta^{2} &\alpha +\gamma+\beta\\\gamma &\gamma^{2} &\alpha+\beta+\gamma\end{vmatrix}$$

(using C₃ → C₃ + C₁)

$$=(\alpha +\beta+\gamma)\begin{vmatrix}\alpha &\alpha^{2} &1\\\beta &\beta^{2} &1\\\gamma &\gamma^{2} &1\end{vmatrix}$$

(Taking out (α + β + γ) common from C₁)

$$=\space(\alpha +\beta +\gamma)\begin{vmatrix}\alpha &\alpha^{2} &1\\\beta-\alpha &\beta^{2} -\alpha^{2} &0\\\gamma-\alpha &\gamma^{2}-\alpha^{2} &0\end{vmatrix}$$

(using R₂ → R₂ – R₁ and R₃ → R₃ – R₁)

Expanding along C₃, we get

= (α + β + γ) [(β – α)(γ² – α^{2) – (γ – α)(β² – α²)]}

= (α + β + γ) [(β – α)(γ – α)(γ + α) – (γ – α)(β – α)(β + α)]

= (α + β + γ)(β – α)(γ – α)[γ + α – β – α]

= (α + β + γ)(β – α)(γ – α)(γ – β)

= (α + β + γ)(α – β)(β – γ)(γ – α) = RHS.

Hence proved.

$$\textbf{12.}\space\begin{vmatrix}\textbf{x} &\textbf{x}^{2} &\textbf{1+px}^{\textbf{3}}\\\textbf{y} &\textbf{y}^{\textbf{2}} &\textbf{1+ py}^{3}\\\textbf{z} &\textbf{z}^{2} &\textbf{1 + pz}^{\textbf{3}}\end{vmatrix}$$

= (1 + pxyz)(x – y)(y – z)(z – x),

where p is any scalar.

Sol. Firstly, we write the given determinant as sum of two determinants, since all elements of C₃ are expressed as sum of two terms and then apply operations in both determinant separately.

$$\text{LHS =}\begin{vmatrix}x &x^{2} &1+px^{3}\\y &y^{2} &1+py^{3}\\z &z^{2} &1+pz^{3}\end{vmatrix}\\=\begin{vmatrix}x &x^{2} &1\\y &y^{2} &1\\z &z^{2} &1\end{vmatrix} +\begin{vmatrix}x &x^{2} &px^{3}\\y &y^{2} &py^{3}\\z &z^{2} &pz^{3}\end{vmatrix}$$

In the first derterminant, interchange C₁ and C₃; C₂ and C₃ and in the second determinant take out p from C₃, we get

$$(\normalsize-1)^{2}\begin{vmatrix}1 &x &x^{2}\\1 &y &y^{2}\\1 &z &z^{2}\end{vmatrix}+p\begin{vmatrix}x &x^{2} &x^{3}\\y &y^{2} &y^{3}\\z &z^{2} &z^{3}\end{vmatrix}\\=\begin{vmatrix}1 &x &x^{2}\\1 &y &y^{2}\\1 &z &z^{2}\end{vmatrix} + \text{pxyz}\begin{vmatrix}1 &x &x^{2}\\1 &y &y^{2}\\1 &z &z^{2}\end{vmatrix}$$

[Taking out x from R₁, y from R₂; and z from R₃ in the second determinant]

$$= (1 + pxyz)\begin{vmatrix}1 &x &x^{2}\\ 1 &y &y^{2}\\1 &z &z^{2}\end{vmatrix}\\=\text{(1 + pxyz)}\begin{vmatrix}1 &x &x^{2}\\0 &y-x &y^{2}-x^{2}\\0 &z-x &z^{2}-x^{2}\end{vmatrix}$$

(using R₂ → R₂ – R₁ and R₃ → R₃ – R₁)

Expanding along C₁, we get

= (1 + pxyz)[(y – x)(z² – x²) – (z – x)(y² – x²)

= (1 + pxyz)[(y – x)(z – x)(z + x) – (z – x)(y – x)(y + x)]

= (1 + pxyz)(y – x)(z – x)[(z + x) – (y + x)]

= (1 + pxyz)(y – x)(z – x)[(z + x – y – x)]

= (1 + pxyz)[(y – x)(z – x)(z – y)

= (1 + pxyz)(x – y)(y – z)(z – x)

Hence proved.

$$\textbf{13.\space}\begin{vmatrix}\textbf{3a} &\textbf{-a+b} &\textbf{-a+c}\\\textbf{-b+a} &\textbf{3b} &\textbf{-b+c}\\\textbf{-c+a} &\textbf{-c+b} &\textbf{3c}\end{vmatrix}$$

= 3(a + b + c)(ab + bc + ca).

$$\textbf{Sol.\space}\text{Let A} =\\\begin{vmatrix}3a &-a+b &-a+c\\-b+a &3b &-b+c\\-c+a &-c+b &3c\end{vmatrix}\\=\begin{vmatrix}a+b+c &-a+b &-a+c\\a+b+c &3b &-b+c\\a+b+c &-c+b &3c\end{vmatrix}$$

(using C₁ → C₁ + C₂ + C₃)

$$=(a+b+c)\begin{vmatrix}1 &-a+b &-a+c\\1 &3b &-b+c\\1 &-c+b &3c\end{vmatrix}$$

(Taking out (a + b + c) common from C₁)

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

(using R₂ → R₂ – R₁ and R₃ → R₃ – R₁)

Expanding along C₁, we get

= (a + b + c)[(2b + a)(2c + a) – (a – b)(a – c)]

= (a + b + c)[4bc + 2ab + 2ac + a2 – a2 + ac + ba – bc]

= (a + b + c)(3ab + 3bc + 3ac)

= 3(a + b + c)(ab + bc + ca)

= RHS Hence proved.

$$\textbf{14.\space}\begin{vmatrix}\textbf{1} &\textbf{1+p} &\textbf{1+p+q}\\\textbf{2} &\textbf{3+2p} &\textbf{4+3p+2q}\\\textbf{3} &\textbf{6+3}\textbf{p} &\textbf{10+6p+3q}\end{vmatrix}\\\textbf{= 1.}\\\textbf{Sol.}\space \text{LHS} = \begin{vmatrix}1 &1+p &1+p+q\\2 &3+2p &4+3p+2q\\3&6+3p &10+6p+3q\end{vmatrix}\\=\begin{vmatrix}1 &1+p &1+p+q\\0 &1 &2+p\\0 &3 &7+3p\end{vmatrix}$$

(using R₂ → R₂ – 2R₁ and R₃ → R₃ – 3R₁)

$$=\begin{vmatrix}1 &1+p &1+p+q\\0 &1 &2+p\\0 &0 &1\end{vmatrix}$$

(using R₃ → R₃ – 3R₂)

Expanding along C₁, we get

= 1(1 × 1 – 0) = 1 = RHS.

$$\textbf{15.\space}\begin{vmatrix}\textbf{sin}\space\alpha &\textbf{cos}\space\alpha &\textbf{cos}(\alpha+\delta)\\\textbf{sin}\space\beta &\textbf{cos}\space\beta &\textbf{cos}(\beta +\delta)\\\textbf{sin}\space\gamma &\textbf{cos}\space\gamma &\textbf{cos}\space(\gamma +\delta)\end{vmatrix}\\\textbf{= 0.}\\\textbf{Sol.\space}\text{Let A = }\\\begin{vmatrix}\text{sin}\space\alpha &\text{cos}\space\alpha &\text{cos}(\alpha +\delta)\\\text{sin}\space\beta &\text{cos}\space\beta &\text{cos}(\gamma +\delta)\\\text{sin}\space\gamma &\text{cos}\space\gamma &\text{cos}(\gamma +\delta)\end{vmatrix}\\=\begin{vmatrix}\text{sin}\space\alpha &\text{cos}\space\alpha &\text{cos}\space\alpha\text{cos}\space\delta -\text{sin}\alpha\text{sin}\space\delta\\\text{sin}\space\beta &\text{cos}\space\beta &\text{cos}\beta\text{cos}\delta - \text{sin}\beta\text{sin}\space\delta\\\text{sin}\space\gamma &\text{cos}\space\gamma &\text{cos}\gamma\text{cos}\delta -\text{sin}\gamma\text{sin}\delta\end{vmatrix}\\\lbrack\because\space\text{cos}(\text{A+B}) =\text{cos A cos B - sin A sinB}\rbrack$$

$$=\begin{vmatrix}\text{sin}\space\alpha &\text{cos}\space\alpha &\text{cos}\alpha\text{cos}\delta\\\text{sin}\space\beta &\text{cos}\space\beta &\text{cos}\beta\text{cos}\delta\\\text{sin}\gamma &\text{cos}\space\gamma &\text{cos}\gamma\text{cos}\delta\end{vmatrix} + \\\begin{vmatrix}\text{sin}\space\alpha &\text{cos}\space\alpha &-\text{sin}\space\alpha\text{sin}\delta\\\text{sin}\space\beta &\text{cos}\space\beta &-\text{sin}\beta\text{sin}\delta\\\text{sin}\space\gamma &\text{cos}\space\gamma &-\text{sin}\gamma\text{sin}\space\delta\end{vmatrix}$$

Taking common cos δ from C₃ in first determinant and taking common – sin δ from C₃ in second determinant.

$$=\space\text{cos}\space\delta\begin{vmatrix}\text{sin}\space\alpha &\text{cos}\space\beta &\text{cos}\space\alpha\\\text{sin}\space\beta &\text{cos}\space\beta &\text{cos}\space\beta\\\text{sin}\space\gamma &\text{cos}\space\gamma &\text{cos}\space\gamma\end{vmatrix}-\\\text{sin}\space\delta\begin{vmatrix}\text{sin}\space\alpha &\text{cos}\space\alpha & sin\space\alpha\\\text{sin}\space\beta &\text{cos}\space\beta &\text{sin}\space\beta\\\text{sin}\space\gamma &\text{cos}\space\gamma &\text{sin}\space\gamma\end{vmatrix}$$

=cos δ.0−sin δ.0 = 0

[Since, C₂ and C₃ in first determinant C₁ and C₃ in second determinant are identical so values of determinants are zero.]

Hence proved.

16. Solve the system of equations

$$\frac{\textbf{2}}{\textbf{x}}+\frac{\textbf{3}}{\textbf{y}}+\frac{\textbf{10}}{\textbf{z}} \textbf{= 4;}\\\frac{\textbf{4}}{\textbf{x}}-\frac{\textbf{6}}{\textbf{y}}+\frac{\textbf{5}}{\textbf{z}}\textbf{= 1}\\\textbf{and}\space\frac{\textbf{6}}{\textbf{x}} \textbf{+}\frac{\textbf{9}}{\textbf{y}}\textbf{-}\frac{\textbf{20}}{\textbf{x}} \textbf{= 2.}$$

$$\textbf{consider}\frac{\textbf{1}}{\textbf{x}} = \textbf{p},\frac{\textbf{1}}{\textbf{y}} = \textbf{q},\frac{\textbf{1}}{\textbf{z}} = \textbf{r}$$

to make a standard form of equations and use the matrix method to solve the three equations in p, q and r.

$$\textbf{Sol.\space}\text{Let}\space\frac{1}{x}=p,\frac{1}{y}= q\space\text{and}\frac{1}{z} = r,$$

then the given equations become

2p + 3q + 10r = 4, 4p – 6q + 5r = 1, 6p + 9q – 20r = 2

This system can be written as AX = B, where

$$\text{A =} \begin{bmatrix}2 &3 &10\\4 &\normalsize-6 &5\\6 &9 &\normalsize-20\end{bmatrix}\textbf{,}\\\text{X} =\begin{bmatrix}p\\q\\r\end{bmatrix},\text{B} =\begin{bmatrix}4 \\1\\2\end{bmatrix}\\\text{Here, |A| =}\begin{vmatrix}2 &3 &10\\4 &\normalsize-6 &5\\6 &9 &\normalsize-20\end{vmatrix}$$

= 2(120 – 45) – 3(– 80 – 30) + 10(36 + 36)

= 150 + 330 + 720 = 1200

Thus, A is non-singular. Therefore, its inverse exists.

Therefore, the above system is consistent and has a unique solution given by X = A^–1B

Cofactors of A are

A₁₁ = 120 – 45 = 75, A₁₂ = – (– 80 – 30) = 110, A₁₃ = (36 + 36) = 72

A₂₁ = – (– 60 – 90) = 150, A₂₂ = (– 40 – 60) = – 100, A₂₃ = – (18 – 18) = 0,

A₃₁ = 15 + 60 = 75, A₃₂ = – (10 – 40) = 30, A₃₃ = – 12 – 12 = – 24

$$\therefore\space\text{adj (A)} = \begin{vmatrix}75 &110 &72\\150 &-100 &0\\75 &30 &\normalsize-24\end{vmatrix}^{\text{T}}\\=\begin{bmatrix}75 &150 &75\\110 &-100 &30\\72 &0 &-24\end{bmatrix}\\\text{Now,\space}\text{A}^{\normalsize-1}\frac{1}{|\text{A}|}\space\text{(adj A)}\\=\frac{1}{1200}\begin{bmatrix}75 &150 &75\\110 &-100 &30\\72 &0 &\normalsize-24\end{bmatrix}\\\because\space\text{X} =\text{A}^{\normalsize-1}\text{B}$$

$$\Rarr\space\begin{bmatrix}p\\q\\r\end{bmatrix} =\frac{1}{1200}\\\begin{bmatrix}75 &150 &75\\110 &-100 &30\\72 &0 &-24\end{bmatrix}\begin{bmatrix}4 \\1 \\2\end{bmatrix}\\=\frac{1}{1200}\space\begin{bmatrix}300 + 150+150\\440-100+60\\288+0-48\end{bmatrix}\\=\frac{1}{1200}\begin{bmatrix}600\\400\\240\end{bmatrix}=\begin{bmatrix}\frac{1}{2}\\\frac{1}{3}\\\frac{1}{5}\end{bmatrix}\\\Rarr\space\text{p} =\frac{1}{2}, q =\frac{1}{3}, r =\frac{1}{5}$$

$$\Rarr\space\frac{1}{x} = \frac{1}{2}, \frac{1}{y}=\frac{1}{3}, \frac{1}{z}= \frac{1}{5}$$

⇒ x = 2, y = 3 and z = 5.

Choose the correct answer in Q. 17 to 19.

17. If a, b, c are in AP, then determinant

$$\begin{vmatrix}\textbf{x+2} &\textbf{x+3} &\textbf{x+2a}\\\textbf{x+3} &\textbf{x+4} &\textbf{x+2b}\\\textbf{x+4} &\textbf{x+5} &\textbf{x+2c}\end{vmatrix}\space\textbf{is :}$$

(a) 0

(b) 1

(c) x

(d) 2x

Sol. (a) 0

$$\text{Let A} = \begin{vmatrix}\text{x+2} &\text{x+3} &\text{x+2a}\\\text{x+3} &\text{x+4} &\text{x+2b}\\\text{x+4} &\text{x+5} &\text{x+2c}\end{vmatrix}\\=\frac{1}{2}\begin{vmatrix}x+2 &x+3 &x+2a\\0 &0 &2(2b-a-c)\\\text{x+4} &\text{x+5} &\text{x+2c}\end{vmatrix}$$

(using R₂ → 2R₂ – R₁ – R₃)

But a, b, c are in AP. Using 2b = a + c, we get

$$\text{A}=\frac{1}{2}\begin{vmatrix}x+2 &x+3 &x+2a\\0 &0 &0\\x+4 &x+5 &x+2c\end{vmatrix} = 0$$

[Since, all elements of R₂ are zero]

18. If x, y, z are non-zero real numbers, then the inverse of matrix A =

$$\begin{bmatrix}\textbf{x} &\textbf{0} &\textbf{0}\\\textbf{0} &\textbf{y} &\textbf{0}\\\textbf{0} &\textbf{0} &\textbf{z}\end{bmatrix}\space\textbf{is :}\\\textbf{(a)\space}\begin{bmatrix}\textbf{x}^{\textbf{-1}} &\textbf{0} &\textbf{0}\\\textbf{0} &\textbf{y}^{\normalsize-1} &\textbf{0}\\\textbf{0} &\textbf{0} &\textbf{z}^{\normalsize-1}\end{bmatrix}\\\textbf{(b)\space}\textbf{xyz}\begin{bmatrix}\textbf{x}^{\normalsize-1} &\textbf{0} &\textbf{0}\\\textbf{0} &\textbf{y}^{\normalsize-1} &\textbf{0}\\\textbf{0} &\textbf{0} &\textbf{z}^{\normalsize-1}\end{bmatrix}\\\textbf{(c)\space}\frac{\textbf{1}}{\textbf{xyz}}\begin{bmatrix}\textbf{x} &\textbf{0} &\textbf{0}\\\textbf{0} &\textbf{y} &\textbf{0}\\\textbf{0} &\textbf{0} &\textbf{z}\end{bmatrix}$$

$$\textbf{(d)\space}\frac{\textbf{1}}{\textbf{xyz}}\begin{bmatrix}\textbf{1} &\textbf{0} &\textbf{0}\\\textbf{0} &\textbf{1} &\textbf{0}\\\textbf{0} &\textbf{0} &\textbf{1}\end{bmatrix}\\\textbf{Sol.\space}\text{(a)\space}\begin{bmatrix}x^{\normalsize-1}&0 &0\\0 &y^{\normalsize-1} &0\\0 &0 &z^{\normalsize-1}\end{bmatrix}\\\text{Given A =}\begin{bmatrix}x &0 &0\\0 &y &0\\0 &0 &z\end{bmatrix}\\\text{|A|} =\begin{vmatrix}x &0 &0\\0 &y &0\\0 &0 &z\end{vmatrix}$$

= x(yz - 0) = xyz ≠ 0

(∵ x, y and z are non-zero)

Cofactors of A are

A₁₁ = (yz – 0) = yz, A₁₂ = – (0 – 0) = 0, A₁₃ = 0 – 0 = 0,

A₂₁ = – (0 – 0) = 0, A₂₂ = xz – 0 = xz, A₂₃ = – (0 – 0) = 0,

A₃₁ = 0 – 0 = 0, A₃₂ = – (0 – 0) = 0, A₃₃ = (xy – 0) = xy.

(∵ x, y and z are non-zero)

Cofactors of A are

A₁₁ = (yz – 0) = yz, A₁₂ = – (0 – 0) = 0, A₁₃ = 0 – 0 = 0,

A₂₁ = – (0 – 0) = 0, A₂₂ = xz – 0 = xz, A₂₃ = – (0 – 0) = 0,

A₃₁ = 0 – 0 = 0, A₃₂ = – (0 – 0) = 0, A₃₃ = (xy – 0) = xy.

$$\therefore\space\text{adj (A)}= \begin{bmatrix}yz &0 &0\\0 &xz &0\\0 &0 &xy\end{bmatrix}^{\text{T}}\\=\begin{bmatrix}yz &0 &0\\0 &xz &0\\0 &0 &xy\end{bmatrix}\\\text{Now,\space A}^{\normalsize-1} =\frac{1}{|\text{A}|}\space\text{(adj A)}\\=\frac{1}{xyz}\begin{bmatrix}yz &0 &0\\0 &xz &0\\0 &0 &xy\end{bmatrix}$$

$$=\begin{bmatrix}\frac{1}{x} &0 &0\\0 &\frac{1}{y} &0\\0 &0 &\frac{1}{z}\end{bmatrix}\\=\begin{bmatrix}x^{\normalsize-1} &0 &0\\0 &y^{\normalsize-1} &0\\0 &0 &z^{\normalsize-1}\end{bmatrix}$$

$$\textbf{19.\space Let A =}\begin{bmatrix}\textbf{1} &\textbf{sin}\space\theta &\textbf{1}\\\textbf{-}\textbf{sin}\space\theta &\textbf{1} &\textbf{sin}\space\theta\\\textbf{-1} &-\textbf{sin}\space\theta &\textbf{1}\end{bmatrix}\textbf{,}$$

where 0 ≤ θ ≤ 2π, then

(a) det A = 0

(b) det A ∈ (2, ∞)

(c) det A ∈ (2, 4)

(d) det A ∈ (2, 4).

Sol. (d) det A ∈ (2, 4)

$$\text{Given, A = }\begin{bmatrix}1 &\text{sin}\space\theta &1\\-\text{sin}\space\theta &1 &\text{sin}\space\theta\\\normalsize-1 &-\text{sin}\space\theta &1\end{bmatrix}\\|\text{A}| =\begin{bmatrix}1 &\text{sin}\space\theta &1\\-\text{sin}\space\theta &1 &\text{sin}\space\theta\\\normalsize-1 &-\text{sin}\space\theta &1\end{bmatrix}$$

= 1( 1 + sin²θ) – sin θ(– sin θ + sin θ) + 1(sin²θ + 1)

= 2 + 2 sin² θ

For 0 ≤ θ ≤ 2π,

– 1 ≤ sin θ ≤ 1 ⇒ 0 ≤ sin²θ ≤ 1

⇒ 1 ≤ 1 + sin²θ ≤ 2 ⇒ 2 ≤ 2(1 + sin²θ) ≤ 4

∴ det(A) ∈ [2, 4]

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