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

### 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 + 2M22 = 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 + 1M11 = 1 × d = d

A12 = (– 1)1 + 2M12 = (– 1) × b = – b

A21 = (– 1)2 + 1M21 = (– 1) × c = – c

A22 = (– 1)2 + 2M22 = 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 + 1M11 = 1 × 1 = 1,

A12 = (– 1)1 + 2M12 = – 1 × 0 = 0

A13 = (– 1)1 + 3M13 = 1 × 0 = 0\

Cofactors of elements of second row are

A21 = (– 1)2 + 1M21 = – 1 × 0 = 0,

A22 = (– 1)2 + 2M22 = 1 × 1 = 1

A23 = (– 1)2 + 3M23 = – 1 × 0 = 0

Cofactors of elements of third row are

A31 = (– 1)3 + 1M31 = 1 × 0 = 0,

A32 = (– 1)3 + 2M32 = – 1 × 0 = 0

A33 = (– 1)3 + 3M33 = 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 + 1M11 = 1 × 11 = 11,

A12 = (– 1)1 + 2M12 = –1 × 6 = –6

A13 = (– 1)1 + 3M13 = 1 × 3 = 3

Cofactors of elements of second row are

A21 = (– 1)2 + 1M21 = –1 × –4 = 4

A22 = (– 1)2 + 2M22 = 1 × 2 = 2

A23 = (– 1)2 + 3M23 = – 1 × 1 = – 1

Cofactors of elements of third row are

A31 = (– 1)3 + 1M31 = 1 × – 20 = – 20

A32 = (– 1)3 + 2M32 = – 1 × – 13 = 13

A33 = (– 1)3 + 3M33 = 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 Aij is cofactors of aij, 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 + a32A32 + a33A33

or a11A11 + a21A21 + a31A31

or a12A12 + a22A22 + a32A32

or a13A13 + a23A23 + a33A33

Sum of the products of the elements of first column with their corresponding cofactors is

Δ = a11A11 + a21A21 + a31A31.