NCERT Solutions For Class 12 Maths Chapter 4 Exercise 4.4

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Chapter 4 Ex.4.4 Question 1

Write Minors and Cofactors of the elements of following determinants:

(i) \(\left| {\begin{array}{*{20}{c}}2&{ - 4}\\0&3\end{array}} \right|\)

(ii) \(\left| {\begin{array}{*{20}{c}}a&c\\b&d\end{array}} \right|\)

Solution

(i) The given determinant is \(\left| {\begin{array}{*{20}{c}}2&{ - 4}\\0&3\end{array}} \right|\)

Minor of element \({a_{ij}}\) is \({M_{ij}}\).

\({M_{11}} = \) minor of element \({a_{11}} = 3\)

\({M_{12}} = \) minor of element \({a_{12}} = 0\)

\({M_{21}} = \) minor of element \({a_{21}} = - 4\)

\({M_{22}} = \) minor of element \({a_{22}} = 2\)

Cofactor of \({a_{ij}}\) is \({A_{ij}} = {\left( { - 1} \right)^{i + j}}{M_{ij}}\)

\(\begin{align}&{A_{11}} = {\left( { - 1} \right)^{1 + 1}}\;{M_{11}} = {\left( { - 1} \right)^2}\left( 3 \right) = 3\\&{A_{12}} = {\left( { - 1} \right)^{1 + 2}}\;{M_{12}} = {\left( { - 1} \right)^3}\left( 0 \right) = 0\\&{A_{21}} = {\left( { - 1} \right)^{2 + 1}}\;{M_{21}} = {\left( { - 1} \right)^3}\left( { - 4} \right) = 4\\&{A_{22}} = {\left( { - 1} \right)^{2 + 2}}\;{M_{22}} = {\left( { - 1} \right)^4}\left( 2 \right) = 2\end{align}\)

(ii) The given determinant is \(\left| {\begin{array}{*{20}{c}}a&c\\b&d\end{array}} \right|\)

Minor of element \({a_{ij}}\) is \({M_{ij}}\).

\({M_{11}} = \)minor of element \({a_{11}} = d\)

\({M_{12}} = \)minor of element \({a_{12}} = b\)

\({M_{21}} = \)minor of element \({a_{21}} = c\)

\({M_{22}} = \)minor of element \({a_{22}} = a\)

Cofactor of \({a_{ij}}\) is \({A_{ij}} = {\left( { - 1} \right)^{i + j}}{M_{ij}}\)

\(\begin{align}&{A_{11}} = {\left( { - 1} \right)^{1 + 1}}\;{M_{11}} = {\left( { - 1} \right)^2}\left( d \right) = d\\&{A_{12}} = {\left( { - 1} \right)^{1 + 2}}\;{M_{12}} = {\left( { - 1} \right)^3}\left( b \right) = - b\\&{A_{21}} = {\left( { - 1} \right)^{2 + 1}}\;{M_{21}} = {\left( { - 1} \right)^3}\left( c \right) = - c\\&{A_{22}} = {\left( { - 1} \right)^{2 + 2}}\;{M_{22}} = {\left( { - 1} \right)^4}\left( a \right) = a\end{align}\)

Chapter 4 Ex.4.4 Question 2

Write Minors and Cofactors of the elements of following determinants:

(i) \(\left| {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right|\)

(ii) \(\left| {\begin{array}{*{20}{c}}1&0&4\\3&5&{ - 1}\\0&1&2\end{array}} \right|\)

Solution

(i) The given determinant is \(\left| {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right|\)

Minor of element \({a_{ij}}\) is \({M_{ij}}\).

\({M_{11}} = \)minor of element \({a_{11}} = \left| {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right| = 1\)

\({M_{12}} = \)minor of element \({a_{12}} = \left| {\begin{array}{*{20}{c}}0&0\\0&1\end{array}} \right| = 0\)

\({M_{13}} = \)minor of element \({a_{13}} = \left| {\begin{array}{*{20}{c}}0&1\\0&0\end{array}} \right| = 0\)

\({M_{21}} = \)minor of element \({a_{21}} = \left| {\begin{array}{*{20}{c}}0&0\\0&1\end{array}} \right| = 0\)

\({M_{22}} = \) minor of element \({a_{22}} = \left| {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right| = 1\)

\({M_{23}} = \)minor of element \({a_{23}} = \left| {\begin{array}{*{20}{c}}1&0\\0&0\end{array}} \right| = 0\)

\({M_{31}} = \)minor of element \({a_{31}} = \left| {\begin{array}{*{20}{c}}0&0\\1&0\end{array}} \right| = 0\)

\({M_{32}} = \)minor of element \({a_{32}} = \left| {\begin{array}{*{20}{c}}1&0\\0&0\end{array}} \right| = 0\)

\({M_{33}} = \)minor of element \({a_{33}} = \left| {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right| = 1\)

Cofactor of \({a_{ij}}\) is \({A_{ij}} = {\left( { - 1} \right)^{i + j}}{M_{ij}}\)

\(\begin{align}&{A_{11}}= {\left( { - 1} \right)^{1 + 1}}\;{M_{11}} = {\left( { - 1} \right)^2}\left( 1 \right) = 1\\&{A_{12}} = {\left( { - 1} \right)^{1 + 2}}\;{M_{12}} = {\left( { - 1} \right)^3}\left( 0 \right) = 0\\&{A_{13}} = {\left( { - 1} \right)^{1 + 3}}\;{M_{13}} = {\left( { - 1} \right)^4}\left( 0 \right) = 0\\&{A_{21}} = {\left( { - 1} \right)^{2 + 1}}\;{M_{21}} = {\left( { - 1} \right)^3}\left( 0 \right) = 0\\&{A_{22}} = {\left( { - 1} \right)^{2 + 2}}\;{M_{22}} = {\left( { - 1} \right)^4}\left( 1 \right) = 1\\&{A_{23}} = {\left( { - 1} \right)^{2 + 3}}\;{M_{23}} = {\left( { - 1} \right)^5}\left( 0 \right) = 0\\&{A_{31}} = {\left( { - 1} \right)^{3 + 1}}\;{M_{31}} = {\left( { - 1} \right)^4}\left( 0 \right) = 0\\&{A_{32}} = {\left( { - 1} \right)^{3 + 2}}\;{M_{32}} = {\left( { - 1} \right)^5}\left( 0 \right) = 0\\&{A_{33}} = {\left( { - 1} \right)^{3 + 3}}\;{M_{33}} = {\left( { - 1} \right)^6}\left( 1 \right) = 1\end{align}\)

The given determinant is \(\left| {\begin{array}{*{20}{c}}1&0&4\\3&5&{ - 1}\\0&1&2\end{array}} \right|\)

Minor of element \({a_{ij}}\) is \({M_{ij}}\).

\({M_{11}} = \)minor of element \({a_{11}} = \left| {\begin{array}{*{20}{c}}5&{ - 1}\\1&2\end{array}} \right| = 11\)

\({M_{12}} = \)minor of element \({a_{12}} = \left| {\begin{array}{*{20}{c}}3&{ - 1}\\0&2\end{array}} \right| = 6\)

\({M_{13}} = \)minor of element \({a_{13}} = \left| {\begin{array}{*{20}{c}}3&5\\0&1\end{array}} \right| = 3\)

\({M_{21}} = \)minor of element \({a_{21}} = \left| {\begin{array}{*{20}{c}}0&4\\1&2\end{array}} \right| = - 4\)

\({M_{22}} = \)minor of element \({a_{22}} = \left| {\begin{array}{*{20}{c}}1&4\\0&2\end{array}} \right| = 2\)

\({M_{23}} = \)minor of element \({a_{23}} = \left| {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right| = 1\)

\({M_{31}} = \)minor of element \({a_{31}} = \left| {\begin{array}{*{20}{c}}0&4\\5&{ - 1}\end{array}} \right| = - 20\)

\({M_{32}} = \)minor of element \({a_{32}} = \left| {\begin{array}{*{20}{c}}1&4\\3&{ - 1}\end{array}} \right| = - 13\)

\({M_{33}} = \)minor of element \({a_{33}} = \left| {\begin{array}{*{20}{c}}1&0\\3&5\end{array}} \right| = 5\)

Cofactor of \({a_{ij}}\) is \({A_{ij}} = {\left( { - 1} \right)^{i + j}}{M_{ij}}\)

\(\begin{align}&{A_{11}} = {\left( { - 1} \right)^{1 + 1}}\;{M_{11}} = {\left( { - 1} \right)^2}\left( {11} \right) = 11\\&{A_{12}} = {\left( { - 1} \right)^{1 + 2}}\;{M_{12}} = {\left( { - 1} \right)^3}\left( 6 \right) = - 6\\&{A_{13}} = {\left( { - 1} \right)^{1 + 3}}\;{M_{13}} = {\left( { - 1} \right)^4}\left( 3 \right) = 3\\&{A_{21}} = {\left( { - 1} \right)^{2 + 1}}\;{M_{21}} = {\left( { - 1} \right)^3}\left( { - 4} \right) = 4\\&{A_{22}} = {\left( { - 1} \right)^{2 + 2}}\;{M_{22}} = {\left( { - 1} \right)^4}\left( 2 \right) = 2\\&{A_{23}} = {\left( { - 1} \right)^{2 + 3}}{M_{23}} = {\left( { - 1} \right)^5}\left( 1 \right) = - 1\\&{A_{31}} = {\left( { - 1} \right)^{3 + 1}}\;{M_{31}} = {\left( { - 1} \right)^4}\left( { - 20} \right) = - 20\\&{A_{32}} = {\left( { - 1} \right)^{3 + 2}}\;{M_{32}} = {\left( { - 1} \right)^5}\left( { - 13} \right) = 13\\&{A_{33}} = {\left( { - 1} \right)^{3 + 3}}\;{M_{33}} = {\left( { - 1} \right)^6}\left( 5 \right) = 5\end{align}\)

Chapter 4 Ex.4.4 Question 3

Using Cofactors of elements of second row, evaluate \(\Delta = \left| {\begin{array}{*{20}{c}}5&3&8\\2&0&1\\1&2&3\end{array}} \right|\)

Solution

The given determinant is \(\left| {\begin{array}{*{20}{c}}5&3&8\\2&0&1\\1&2&3\end{array}} \right|\)

\({M_{21}} = \) minor of element \({a_{21}} = \left| {\begin{array}{*{20}{c}}3&8\\2&3\end{array}} \right| = - 7\)

\({A_{21}} = {\left( { - 1} \right)^{2 + 1}}{M_{21}} = {\left( { - 1} \right)^3}\left( { - 7} \right) = 7\)

\({M_{22}} = \)minor of element \({a_{22}} = \left| {\begin{array}{*{20}{c}}5&8\\1&3\end{array}} \right| = 15 - 8 = 7\)

\({A_{22}} = {\left( { - 1} \right)^{2 + 2}}{M_{22}} = {\left( { - 1} \right)^4}\left( 7 \right) = 7\)

\({M_{23}} = \)minor of element \({a_{23}} = \left| {\begin{array}{*{20}{c}}5&3\\1&2\end{array}} \right| = 7\)

\({A_{23}} = {\left( { - 1} \right)^{2 + 3}}{M_{21}} = {\left( { - 1} \right)^5}\left( 7 \right) = - 7\)

We know that \(\Delta \) is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

Therefore,

\[\begin{align}\Delta &= {a_{21}}{A_{21}} + {a_{22}}{A_{22}} + {a_{23}}{A_{23}}\\&= 2\left( 7 \right) + 0\left( 7 \right) + 1\left( { - 7} \right)\\&= 14 - 7\\&= 7\end{align}\]

Chapter 4 Ex.4.4 Question 4

Using Cofactors of elements of third column, evaluate \(\Delta = \left| {\begin{array}{*{20}{c}}1&x&{yz}\\1&y&{zx}\\1&z&{xy}\end{array}} \right|\)

Solution

The given determinant is \(\left| {\begin{array}{*{20}{c}}1&x&{yz}\\1&y&{zx}\\1&z&{xy}\end{array}} \right|\)

Therefore,

\[\begin{align}&{M_{13}} = \left| {\begin{array}{*{20}{c}}1&y\\1&z\end{array}} \right| = z - y\\&{M_{23}} = \left| {\begin{array}{*{20}{c}}1&x\\1&z\end{array}} \right| = z - x\\&{M_{33}} = \left| {\begin{array}{*{20}{c}}1&x\\1&y\end{array}} \right| = y - x\end{align}\]

\[\begin{align}&{A_{13}} = {\left( { - 1} \right)^{1 + 3}}{M_{13}} = {\left( { - 1} \right)^4}\left( {z - y} \right) = z - y\\&{A_{23}} = {\left( { - 1} \right)^{2 + 3}}{M_{23}} = {\left( { - 1} \right)^5}\left( {z - x} \right) = - \left( {z - x} \right) = x - z\\&{A_{33}} = {\left( { - 1} \right)^{3 + 3}}{M_{33}} = {\left( { - 1} \right)^6}\left( {y - x} \right) = y - x\end{align}\]

We know that \(\Delta \) is equal to the sum of the product of the elements of the third column with their corresponding cofactors.

Therefore,

\[\begin{align}\Delta &= {a_{13}}{A_{13}} + {a_{23}}{A_{23}} + {a_{33}}{A_{33}}\\&= yz\left( {z - y} \right) + zx\left( {x - z} \right) + xy\left( {y - x} \right)\\&= y{z^2} - {y^2}z + {x^2}z - x{z^2} + x{y^2} - {x^2}y\\&= \left( {{x^2}z - {y^2}z} \right) + \left( {y{z^2} - x{z^2}} \right) + \left( {x{y^2} - {x^2}y} \right)\\&= z\left( {{x^2} - {y^2}} \right) + {z^2}\left( {y - x} \right) + xy\left( {y - x} \right)\\&= z\left( {x - y} \right)\left( {x + y} \right) + {z^2}\left( {y - x} \right) + xy\left( {y - x} \right)\\&= \left( {x - y} \right)\left[ {zx + zy - {z^2} - xy} \right]\\&= \left( {x - y} \right)\left[ {z\left( {x - z} \right) + y\left( {z - x} \right)} \right]\\&= \left( {x - y} \right)\left( {z - x} \right)\left[ { - z + y} \right]\\&= \left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)\end{align}\]

Hence, \(\Delta = \left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)\)

Chapter 4 Ex.4.4 Question 5

If \(\Delta = \left| {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right|\) and \({A_{ij}}\) is the cofactor of \({a_{ij}}\), then the value of \(\Delta \) is given by:

(a) \({a_{11}}{A_{31}} + {a_{12}}{A_{32}} + {a_{13}}{A_{33}}\)

(b) \({a_{11}}{A_{11}} + {a_{12}}{A_{21}} + {a_{13}}{A_{31}}\)

(c) \({a_{21}}{A_{11}} + {a_{22}}{A_{12}} + {a_{23}}{A_{13}}\)

(d) \({a_{11}}{A_{11}} + {a_{21}}{A_{21}} + {a_{31}}{A_{31}}\)

Solution

We know that \(\Delta \) is equal to the sum of the product of the elements of a column or row with their corresponding cofactors.

\[\Delta = {a_{11}}{A_{11}} + {a_{21}}{A_{21}} + {a_{31}}{A_{31}}\]

Thus, the correct option is D.

  
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