NCERT Class 12 Maths Determinants

NCERT Class 12 Maths Determinants

Chapter 4 Ex.4.1 Question 1

Evaluate the determinant \(\left| {\begin{array}{*{20}{c}}2&4\\{ - 5}&{ - 1}\end{array}} \right|\)

Solution

Let \(\left| A \right| = \left| {\begin{array}{*{20}{c}}2&4\\{ - 5}&{ - 1}\end{array}} \right|\)

Hence,

\[\begin{align}\left| A \right| &= \left| {\begin{array}{*{20}{c}}2&4\\{ - 5}&{ - 1}\end{array}} \right|\\ &= 2\left( { - 1} \right) - 4\left( { - 5} \right)\\ &= - 2 + 20\\& = 18\end{align}\]

Chapter 4 Ex.4.1 Question 2

Evaluate the determinants:

(i) \(\left| {\begin{array}{*{20}{c}}{\cos \theta }&{ - \sin \theta }\\{\sin \theta }&{\cos \theta }\end{array}} \right|\)

(ii) \(\left| {\begin{array}{*{20}{c}}{{x^2} - x + 1}&{x - 1}\\{x + 1}&{x + 1}\end{array}} \right|\)

Solution

(i) \[\begin{align}\left| {\begin{array}{*{20}{c}}{\cos \theta }&{ - \sin \theta }\\{\sin \theta }&{\cos \theta }\end{array}} \right| &= \left( {\cos \theta } \right)\left( {\cos \theta } \right) - \left( { - \sin \theta } \right)\left( {\sin \theta } \right)\\ &= {\cos ^2}\theta + {\sin ^2}\theta \\& = 1\end{align}\]

(ii) \[\begin{align}\left| {\begin{array}{*{20}{c}}{{x^2} - x + 1}&{x - 1}\\{x + 1}&{x + 1}\end{array}} \right| &= \left( {{x^2} - x + 1} \right)\left( {x + 1} \right) - \left( {x - 1} \right)\left( {x + 1} \right)\\ &= {x^3} - {x^2} + x + {x^2} - x + 1 - \left( {{x^2} - 1} \right)\\ &= {x^3} + 1 - {x^2} + 1\\& = {x^3} - {x^2} + 2\end{align}\]

Chapter 4 Ex.4.1 Question 3

If \(A = \left[ {\begin{array}{*{20}{c}}1&2\\4&2\end{array}} \right]\), then show that \(\left| {2A} \right| = 4\left| A \right|\)

Solution

The given matrix is \(A = \left[ {\begin{array}{*{20}{c}}1&2\\4&2\end{array}} \right]\)

Therefore,

\[\begin{align}2A &= 2\left( {\begin{array}{*{20}{c}}1&2\\4&2\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}2&4\\8&4\end{array}} \right)\end{align}\]

Hence,

\[\begin{align}LHS &= \left| {2A} \right|\\ &= \left| {\begin{array}{*{20}{c}}2&4\\8&4\end{array}} \right|\\ &= 2 \times 4 - 4 \times 8\\ &= 8 - 32\\ &= - 24\end{align}\]

Now,

\[\begin{align}\left| A \right|& = \left| {\begin{array}{*{20}{c}}1&2\\4&2\end{array}} \right| = 1 \times 2 - 2 \times 4\\ &= 2 - 8\\& = - 6\end{align}\]

Therefore,

\[\begin{align}RHS &= 4\left| A \right|\\& = 4\left( { - 6} \right)\\& = - 24\end{align}\]

Thus, \(\left| {2A} \right| = 4\left| A \right|\) proved.

Chapter 4 Ex.4.1 Question 4

If \(A = \left( {\begin{array}{*{20}{c}}1&0&1\\0&1&2\\0&0&4\end{array}} \right)\), then show that \(\left| {3A} \right| = 27\left| A \right|\)

Solution

The given matrix is \(A = \left( {\begin{array}{*{20}{c}}1&0&1\\0&1&2\\0&0&4\end{array}} \right)\)

It can be observed that in the first column, two entries are zero. Thus, we expand along the first column \(\left( {{C_1}} \right)\) for easier calculation.

\[\begin{align}\left| A \right| &= 1\left| {\begin{array}{*{20}{c}}1&2\\0&4\end{array}} \right| - 0\left| {\begin{array}{*{20}{c}}0&1\\1&4\end{array}} \right| + 0\left| {\begin{array}{*{20}{c}}0&1\\1&2\end{array}} \right|\\ &= 1\left( {4 - 0} \right) - 0 + 0\\& = 4\end{align}\]

Therefore,

\[\begin{align}27\left| A \right| &= 27\left| 4 \right|\\& = 108 \qquad \qquad \ldots \left( 1 \right)\end{align}\]

Now,

\[3A = 3\left( {\begin{array}{*{20}{c}}1&0&1\\0&1&2\\0&0&4\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&0&3\\0&3&6\\0&0&{12}\end{array}} \right)\]

Therefore,

\[\begin{align}\left| {3A} \right| &= 3\left| {\begin{array}{*{20}{c}}3&6\\0&{12}\end{array}} \right| - 0\left| {\begin{array}{*{20}{c}}0&3\\0&{12}\end{array}} \right| + 0\left| {\begin{array}{*{20}{c}}0&3\\3&6\end{array}} \right|\\ &= 3\left( {36 - 0} \right)\\ &= 36\left( {36} \right)\\ &= 108 \qquad \qquad \ldots \left( 2 \right)\end{align}\]

From equations \(\left( 1 \right)\) and \(\left( 2 \right)\),

\(\left| {3A} \right| = 27\left| A \right|\)

Thus, \(\left| {3A} \right| = 27\left| A \right|\) proved.

Chapter 4 Ex.4.1 Question 5

Evaluate the determinants

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

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

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

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

Solution

(i) Let \(A = \left| {\begin{array}{*{20}{c}}3&{ - 1}&{ - 2}\\0&0&{ - 1}\\3&{ - 5}&0\end{array}} \right|\)

It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.

Hence,

\[\begin{align}\left| A \right| &= - 0\left| {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\\{ - 5}&0\end{array}} \right| + 0\left| {\begin{array}{*{20}{c}}3&{ - 2}\\3&0\end{array}} \right| - \left( { - 1} \right)\left| {\begin{array}{*{20}{c}}3&{ - 1}\\3&{ - 5}\end{array}} \right|\\ &= \left( { - 15 + 3} \right)\\& = - 12\end{align}\]

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

Hence,

\[\begin{align}\left| A \right|& = 3\left| {\begin{array}{*{20}{c}}1&{ - 2}\\3&1\end{array}} \right| + 4\left| {\begin{array}{*{20}{c}}1&{ - 2}\\2&1\end{array}} \right| + 5\left| {\begin{array}{*{20}{c}}1&1\\2&3\end{array}} \right|\\& = 3\left( {1 + 6} \right) + 4\left( {1 + 4} \right) + 5\left( {3 - 2} \right)\\& = 3\left( 7 \right) + 4\left( 5 \right) + 5\left( 1 \right)\\& = 21 + 20 + 5\\ &= 46\end{align}\]

(iii) Let \(A = \left| {\begin{array}{*{20}{c}}0&1&2\\{ - 1}&0&{ - 3}\\{ - 2}&3&0\end{array}} \right|\)

Hence,

\[\begin{align}\left| A \right| &= 0\left| {\begin{array}{*{20}{c}}0&{ - 3}\\3&0\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}{ - 1}&{ - 3}\\{ - 2}&0\end{array}} \right| + 2\left| {\begin{array}{*{20}{c}}{ - 1}&0\\{ - 2}&3\end{array}} \right|\\& = 0 - 1\left( {0 - 6} \right) + 2\left( { - 3 - 0} \right)\\& = - 1\left( { - 6} \right) + 2\left( { - 3} \right)\\& = 6 - 6 = 0\end{align}\]

(iv) Let \(A = \left| {\begin{array}{*{20}{c}}2&{ - 1}&{ - 2}\\0&2&{ - 1}\\3&{ - 5}&0\end{array}} \right|\)

Hence,

\[\begin{align}\left| A \right| &= 2\left| {\begin{array}{*{20}{c}}2&{ - 1}\\{ - 5}&0\end{array}} \right| - 0\left| {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\\{ - 5}&0\end{array}} \right| + 3\left| {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\\2&{ - 1}\end{array}} \right|\\ &= 2\left( {0 - 5} \right) - 0 + 3\left( {1 + 4} \right)\\ &= - 10 + 15 = 5\end{align}\]

Chapter 4 Ex.4.1 Question 6

If \(A = \left( {\begin{array}{*{20}{c}}1&1&{ - 2}\\2&1&{ - 3}\\5&4&{ - 9}\end{array}} \right)\), find \(\left| A \right|\)

Solution

Let \(A = \left( {\begin{array}{*{20}{c}}1&1&{ - 2}\\2&1&{ - 3}\\5&4&{ - 9}\end{array}} \right)\)

Hence,

\[\begin{align}\left| A \right| &= 1\left| {\begin{array}{*{20}{c}}1&{ - 3}\\4&{ - 9}\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}2&{ - 3}\\5&{ - 9}\end{array}} \right| - 2\left| {\begin{array}{*{20}{c}}2&1\\5&4\end{array}} \right|\\ &= 1\left( { - 9 + 12} \right) - 1\left( { - 18 + 15} \right) - 2\left( {8 - 5} \right)\\ &= 1\left( 3 \right) - 1\left( { - 3} \right) - 2\left( 3 \right)\\ &= 3 + 3 - 6\\& = 0\end{align}\]

Chapter 4 Ex.4.1 Question 7

Find the values of \(x\), if

(i) \(\left| {\begin{array}{*{20}{c}}2&4\\5&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{2x}&4\\6&x\end{array}} \right|\)

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

Solution

(i) \(\left| {\begin{array}{*{20}{c}}2&4\\5&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{2x}&4\\6&x\end{array}} \right|\)

Therefore,

\[\begin{align} &\Rightarrow \;2 \times 1 - 5 \times 4 = 2x \times x - 6 \times 4\\& \Rightarrow \; 2 - 20 = 2{x^2} - 24\\ &\Rightarrow 2{x^2} = 6\\& \Rightarrow {x^2} = 3\\ &\Rightarrow \; x = \pm \sqrt 3 \end{align}\]

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

Therefore,

\[\begin{align} &\Rightarrow \;2 \times 5 - 3 \times 4 = x \times 5 - 3 \times 2x\\& \Rightarrow \;10 - 12 = 5x - 6x\\& \Rightarrow \; - 2 = - x\\ &\Rightarrow \; x = 2\end{align}\]

Chapter 4 Ex.4.1 Question 8

If \(\left| {\begin{array}{*{20}{c}}x&2\\{18}&x\end{array}} \right| = \left| {\begin{array}{*{20}{c}}6&2\\{18}&6\end{array}} \right|\), then \(x\) is equal to

(A) \(6\)              (B) \( \pm 6\)              (C) \( - 6\)              (D) \(0\)

Solution

\(\left| {\begin{array}{*{20}{c}}x&2\\{18}&x\end{array}} \right| = \left| {\begin{array}{*{20}{c}}6&2\\{18}&6\end{array}} \right|\)

Therefore,

\[\begin{align} &\Rightarrow \;{x^2} - 36 = 36 - 36\\& \Rightarrow \;{x^2} - 36 = 0\\& \Rightarrow \; {x^2} = 36\\& \Rightarrow \; x = \pm 6\end{align}\]

Thus, the correct option is \(B\).

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