12th NCERT Matrices Exercise 4.1 Questions 8

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Evaluate the determinants in Exercises 1 and 2

Question (1)

\[\left| {\begin{array}{*{20}{c}}2&4\\{ - 5}&{ - 1}\end{array}} \right|\]

Solution

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

Question (2)

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

Solution

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

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

\[A = \left[ {\begin{array}{*{20}{c}}1&2\\4&2\end{array}} \right]\] \[\left| A \right| = 2 - 8 = - 6\] \[RHS = 4\left| A \right| = 4\left( { - 6} \right) = - 24\] \[2A = 2\left[ {\begin{array}{*{20}{c}}1&2\\4&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&4\\8&4\end{array}} \right]\] \[LHS = \left| {2A} \right| = \left| {\begin{array}{*{20}{c}}2&4\\8&4\end{array}} \right| = 8 - 32 = - 24 = RHS\]

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

LHS
\[A = \left[ {\begin{array}{*{20}{c}}1&0&1\\0&1&2\\0&0&4\end{array}} \right]\] \[3A = 3\left[ {\begin{array}{*{20}{c}}1&0&1\\0&1&2\\0&0&4\end{array}} \right]\] \[3A = \left[ {\begin{array}{*{20}{c}}3&0&3\\0&3&6\\0&0&{12}\end{array}} \right]\] \[ = 3\left( {36 - 0} \right) + 3\left( {0 - 0} \right) = 108\] RHS
\[\left| A \right| = \left[ {\begin{array}{*{20}{c}}1&0&1\\0&1&2\\0&0&1\end{array}} \right]\] \[\left| A \right| = 1\left( {4 - 0} \right) - 0 + 1\left( 0 \right) = 4\] RHS = 27|A| = 27(4) = 108
∴ LHS = RHS

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

\[(i)\left| {\begin{array}{*{20}{c}}3&{ - 1}&{ - 2}\\0&0&{ - 1}\\3&{ - 5}&0\end{array}} \right|\] \[ = 3\left( {0 - 5} \right) + 1\left( {0 + 3} \right) - 2\left( {0 - 0} \right)\] \[ = - 15 + 3 = - 12\]
\[(ii)\left| {\begin{array}{*{20}{c}}3&{ - 4}&5\\1&1&{ - 2}\\2&3&1\end{array}} \right|\] \[ = 3\left( {1 + 6} \right) + 4\left( {1 + 4} \right) + 5\left( {3 - 2} \right)\] \[ = 21 + 4\left( 5 \right) + 5\left( 1 \right) = 46\]
\[(iii)\left| {\begin{array}{*{20}{c}}0&1&2\\{ - 1}&0&{ - 3}\\{ - 2}&3&0\end{array}} \right|\] \[ = 0 - 1\left( {0 - 6} \right) + 2\left( { - 3 - 0} \right) = 0\]
\[(iv)\left| {\begin{array}{*{20}{c}}2&{ - 1}&{ - 2}\\0&2&{ - 1}\\3&{ - 5}&0\end{array}} \right|\] \[ = 2\left( {0 - 5} \right) + 1\left( {0 + 3} \right) - 2\left( {0 - 6} \right)\] \[ = - 10 + 3 + 12 = 5\]

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

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

Question (7)

Find value of x, if
\[\left( i \right)\left| {\begin{array}{*{20}{c}}2&4\\5&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{2x}&4\\6&x\end{array}} \right|\] \[\left( {ii} \right)\left| {\begin{array}{*{20}{c}}2&3\\4&5\end{array}} \right| = \left| {\begin{array}{*{20}{c}}x&3\\{2x}&5\end{array}} \right|\]

Solution

\[\left( i \right)\left| {\begin{array}{*{20}{c}}2&4\\5&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{2x}&4\\6&x\end{array}} \right|\] \[ \Rightarrow 2 - 20 = 2{x^2} - 24\] \[ - 18 + 24 = 2{x^2}\] \[2{x^2} = 6\] \[x = \pm \sqrt 3 \]
\[\left( {ii} \right)\left| {\begin{array}{*{20}{c}}2&3\\4&5\end{array}} \right| = \left| {\begin{array}{*{20}{c}}x&3\\{2x}&5\end{array}} \right|\] \[10 - 12 = 5x - 6x\] \[x = 2\]

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) ± 6     (C) -6     (D) 0

Solution

\[\left| {\begin{array}{*{20}{c}}x&2\\{18}&x\end{array}} \right| = \left| {\begin{array}{*{20}{c}}6&2\\{18}&6\end{array}} \right|\] \[{x^2} - 36 = 36 - 36\] \[{x^2} - 36 = 0\] \[x = \pm 6\] ∴ B is correct answer