NCERT Solutions for Class 12 Maths Chapter 4 Determinants - Exercise 4.4

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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)

Exercise 4.4

$$\textbf{1.\space(i)\space}\begin{vmatrix}\textbf{2} &\textbf{\normalsize-4}\\\textbf{0} &\textbf{\normalsize-3}\end{vmatrix}\\\textbf{(ii)\space}\begin{vmatrix}\textbf{a} &\textbf{c}\\\textbf{b} &\textbf{d}\end{vmatrix}\\\textbf{Sol.\space (i)\space}\text{Here, \space}\begin{vmatrix}2 &-4\\0 &3\end{vmatrix}\\\therefore\space \text{Minors, M}_{11} = 3, \text{M}_{12}=0,\\\text{M}_{21}=-4\space\text{and}\space \text{M}_{22}=2$$

Also cofactors,

A₁₁ = (– 1)^{1 + 1} M₁₁ = 1 × 3 = 3

A₁₂ = (– 1)^{1 + 2} M₁₂ = (– 1) × 0 = 0

A₂₁ = (– 1)^{2 + 1} M₂₁ = (– 1) × (– 4) = 4

and A₂₂ = (– 1)^{2 + 2}M₂₂ = 1 × 2 = 2

$$\textbf{(ii)\space}\text{Here,}\space\begin{vmatrix}a &c\\b &d\end{vmatrix}$$

∴ Minors, M₁₁ = d, M₁₂ = b, M₂₁ = c and M₂₂ = a

Also cofactors,

A₁₁ = (– 1)^{1 + 1}M₁₁ = 1 × d = d

A₁₂ = (– 1)^{1 + 2}M₁₂ = (– 1) × b = – b

A₂₁ = (– 1)^{2 + 1}M₂₁ = (– 1) × c = – c

A₂₂ = (– 1)^{2 + 2}M₂₂ = 1 × a = a

$$\textbf{2.\space(i)\space}\begin{vmatrix}\textbf{1} &\textbf{0} &\textbf{0}\\\textbf{0} &\textbf{1} &\textbf{0}\\\textbf{0} &\textbf{0} &\textbf{1}\end{vmatrix}\\\textbf{(ii)\space}\begin{vmatrix}\textbf{1} &\textbf{0} &\textbf{4}\\\textbf{3} &\textbf{5} &\textbf{\normalsize-1}\\\textbf{0} &\textbf{1} &\textbf{2}\end{vmatrix}\\\textbf{Sol.}\space\text{Here,}\begin{vmatrix}1 &0 &0\\0 &1 &0\\0 &0 &1\end{vmatrix}$$

Minors of elements of first row are

$$\text{M}_{11}=\begin{vmatrix}1 &0\\0 &1\end{vmatrix}\\=1-0=1\\\text{M}_{12} = \begin{vmatrix}0 &0\\0 &1\end{vmatrix}\\=0-0=0\\\text{and}\space \text{M}_{13} = \begin{vmatrix}0 &1\\0 &0\end{vmatrix}\\=0-0=0$$

Minors of elements of second row are

$$\text{M}_{21} =\begin{vmatrix}0 &0\\0 &1\end{vmatrix}\\=0-0=0\\\text{M}_{22} = \begin{vmatrix}1 &0\\0 &1\end{vmatrix}\\=1-0=1\\\text{and}\space\text{M}_{23}=\begin{vmatrix}1 &0\\0 &0\end{vmatrix}\\=0-0=0$$

Minors of elements of third row are

$$\text{M}_{31} =\begin{vmatrix}0 &0\\1 &0\end{vmatrix}\\=0-0 = 0\\\text{M}_{32}=\begin{vmatrix}1 &0\\0 &0\end{vmatrix}\\=0-0=0\\\text{and}\space\text{M}_{33} =\begin{vmatrix}1 &0\\0 &1\end{vmatrix}\\=1-0=1$$

Hence, cofactors of elements of first row are

A₁₁ = (– 1)^{1 + 1}M₁₁ = 1 × 1 = 1,

A₁₂ = (– 1)^{1 + 2}M₁₂ = – 1 × 0 = 0

A₁₃ = (– 1)^{1 + 3}M₁₃ = 1 × 0 = 0\

Cofactors of elements of second row are

A₂₁ = (– 1)^{2 + 1}M₂₁ = – 1 × 0 = 0,

A₂₂ = (– 1)^{2 + 2}M₂₂ = 1 × 1 = 1

A₂₃ = (– 1)^{2 + 3}M₂₃ = – 1 × 0 = 0

Cofactors of elements of third row are

A₃₁ = (– 1)^{3 + 1}M₃₁ = 1 × 0 = 0,

A₃₂ = (– 1)^{3 + 2}M₃₂ = – 1 × 0 = 0

A₃₃ = (– 1)^{3 + 3}M₃₃ = 1 × 1 = 1

$$\textbf{(ii)\space}\text{Here,}\space\begin{vmatrix}1 &0 &4\\3 &5 &\normalsize-1\\0 &1 &2\end{vmatrix}$$

Minors of elements of first row are

$$\text{M}_{11} =\begin{vmatrix}5 &-1\\ 1 &2\end{vmatrix}\\=10+1=11,\\\text{M}_{12}=\begin{vmatrix}3 &\normalsize-1\\0 &2\end{vmatrix}\\\text{and}\space\text{M}_{13} =\begin{vmatrix}3 &5\\0 &1\end{vmatrix}\\=3-0 = 3$$

Minors of elements of second row are

$$\text{M}_{21} = \begin{vmatrix}0 &4\\1 &2\end{vmatrix}\\=0-4=\normalsize-4,\\\text{M}_{22} = \begin{vmatrix}1 &4\\0 &2\end{vmatrix}\\=2-0=2\\\text{and}\space\text{M}_{23} =\begin{vmatrix}1 &0\\0 &1\end{vmatrix}\\=1-0 = 1 $$

Minors of elements of third row are

$$\text{M}_{31} =\begin{vmatrix}0 &4\\5 &\normalsize-1\end{vmatrix}\\\text{M}_{32} =\begin{vmatrix}1 &4\\3 &\normalsize-1\end{vmatrix}\\=-1-12=-13\\\text{and}\space\text{M}_{33}=\begin{vmatrix}1 &0\\3 &5\end{vmatrix}\\=5-0=5$$

Hence, cofactors of elements of first row are

A₁₁ = (– 1)^{1 + 1}M_{11 = 1 × 11 = 11,}

A₁₂ = (– 1)^{1 + 2}M₁₂ = –1 × 6 = –6

A₁₃ = (– 1)^{1 + 3}M₁₃ = 1 × 3 = 3

Cofactors of elements of second row are

A₂₁ = (– 1)^{2 + 1}M₂₁ = –1 × –4 = 4

A₂₂ = (– 1)^{2 + 2}M₂₂ = 1 × 2 = 2

A₂₃ = (– 1)^{2 + 3}M₂₃ = – 1 × 1 = – 1

Cofactors of elements of third row are

A₃₁ = (– 1)^{3 + 1M₃₁ = 1 × – 20 = – 20}

A₃₂ = (– 1)^{3 + 2}M₃₂ = – 1 × – 13 = 13

A₃₃ = (– 1)^{3 + 3}M₃₃ = 1 × 5 = 5

3. Using cofactors of the elements of second row,

$$\textbf{evaluate}\space\Delta = \begin{vmatrix}\textbf{5} &\textbf{3} &\textbf{8}\\\textbf{2} &\textbf{0} &\textbf{1}\\\textbf{1} &\textbf{2} &\textbf{3}\end{vmatrix}\textbf{.}\\\textbf{Sol.\space}\text{Given,}\space\Delta = \begin{vmatrix}5 &3 &8\\2 &0 &1\\1 &2 &3\end{vmatrix}$$

Cofactors of the elements of second row

$$\text{A}_{21} =(\normalsize-1)^{2+1}\begin{vmatrix}3 &8\\2 &3\end{vmatrix}\\=-(9-16)=7\\\lbrack\because\space \text{A}_{ij} = (\normalsize-1)^{i+j}\text{M}_{ij}\rbrack\\\text{A}_{22} =(\normalsize-1)^{2+2}\begin{vmatrix}5 &8\\1 &3\end{vmatrix}\\=15-8 =7\\\text{and}\space \text{A}_{23} = (\normalsize-1)^{2+3}\begin{vmatrix}5 &3\\1 &2\end{vmatrix}\\=-(10-3)=\normalsize-7$$

Now, expansion of Δ using cofactors of elements
of second row is given by

Δ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃

= 2 × 7 + 0 × 7 + 1(– 7) = 14 – 7 = 7

4. Using cofactors of elements of third column,

$$\textbf{evaluate}\space\Delta = \begin{vmatrix}\textbf{1} &\textbf{x} &\textbf{yz}\\1 &\textbf{y} &\textbf{zx}\\ \textbf{1} &\textbf{z} &\textbf{xy}\end{vmatrix}\\\textbf{Sol.\space}\text{Given, \space}\Delta = \begin{vmatrix}1 &x &yz\\1 &y &zx\\1 &z &xy\end{vmatrix}$$

Cofactors of the elements of third column are

$$\text{A}_{13} = (\normalsize-1)^{1+3}\begin{vmatrix}1 &y\\1 &z\end{vmatrix}\\= 1(z-y) = z-y\\\text{A}_{23}=(\normalsize-1)^{2+3}\begin{vmatrix}1 &x\\1 &z \end{vmatrix}\\=-1(z-x)=x-2\\\text{and}\space \text{A}_{33} =(\normalsize-1)^{3+3}\begin{vmatrix}1 &x\\1 &y\end{vmatrix}\\= 1(y-x) = y-x$$

Now, expansion of Δ using cofactors elements of
third column is given by

Δ = a₁₃A₁₃ + a₂₃A₂₃ + a₃₃A₃₃ = yz(z – y)
+ zx(x – z) + xy(y – x)

= yz² – y²z + zx² – z²x + xy² – x²y

= x²(z – y) + x(y² – z²) + yz(z – y)

= (z – y){x² – x(y + z) + yz}

= (z – y)(x² – xy – xz + yz)

= (z – y)[x(x – y) – z(x – y)]

= (y – z)(x – y)(z – x)

= (x – y)(y – z)(z – x)

$$\textbf{5. If}\space\Delta = \begin{vmatrix}\textbf{a}_{\textbf{11}} & \textbf{a}_{\textbf{12}} &\textbf{a}_{\textbf{13}}\\\textbf{a}_{\textbf{21}} &\textbf{a}_{\textbf{22}} &\textbf{a}_{\textbf{23}}\\\textbf{a}_{\textbf{31}} &\textbf{a}_{\textbf{32}} &\textbf{a}_{\textbf{33}}\end{vmatrix}$$

and A_ij is cofactors of a_ij, then value of Δ is given by

(a) a₁₁A₃₁ + a₁₂A₃₂ + a₁₃A₃₃

(b) a₁₁A₁₁ + a₁₂A₂₁ + a₁₃A₃₁

(c) a₂₁A₁₁ + a₂₂A₁₂ + a₂₃A₁₃

(d) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

Sol. (d) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

Δ is equal to the sum of the products of the elements of a row (or a column) with their corresponding cofactors.

Δ = a₁₁A₁₁ + a₁₂A₁₂+ a₁₃A₁₃

or a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃

or a₃₁A₃₁ + a₃₂A_{32 +} a₃₃A₃₃

or a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

or a₁₂A₁₂ + a₂₂A₂₂ + a₃₂A₃₂