12th NCERT Matrices Miscellaneous Exercise Questions 19
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Question (1)

Prove that the detrminant $\left| {\begin{array}{*{20}{c}}x&{\sin \theta }&{\cos \theta }\\{ - \sin \theta }&{ - x}&1\\{\cos \theta }&1&x\end{array}} \right|$ is independent of θ \[\]

Solution

\[ = x\left( { - {x^2} - 1} \right) - \sin \theta \left( { - x\sin \theta - \cos \theta } \right) + \cos \theta \left( { - \sin \theta + x\cos \theta } \right)\] \[ = - {x^3} - x + x{\sin ^2}\theta + \require{cancel} \cancel{\sin \theta \cos \theta} -\cancel{ \sin \theta \cos \theta} + x{\cos ^2}\theta \] \[ = - {x^3} - x + x\] \[ = - {x^3}\]

Question (2)

Without expanding the determinant, prove that
\[\left| {\begin{array}{*{20}{c}}a&{{a^2}}&{bc}\\b&b&{ca}\\c&{{c^2}}&{ab}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}1&{{a^2}}&{{a^3}}\\1&{{b^2}}&{{b^3}}\\1&{{c^2}}&{{c^3}}\end{array}} \right|\]

Solution

\[LHS = \left| {\begin{array}{*{20}{c}}a&{{a^2}}&{bc}\\b&{{b^2}}&{ca}\\c&{{c^2}}&{ab}\end{array}} \right|\] \[{R_1}\left( a \right),{R_2}\left( b \right),{R_3}\left( c \right)\] \[\frac{1}{{abc}}\left| {\begin{array}{*{20}{c}}{{a^2}}&{{a^3}}&{abc}\\{{b^2}}&{{b^3}}&{abc}\\{{c^2}}&{{c^3}}&{abc}\end{array}} \right|\] \[{C_3}\left( {\frac{1}{{abc}}} \right)\] \[\frac{{abc}}{{abc}}\left| {\begin{array}{*{20}{c}}{{a^2}}&{{a^3}}&1\\{{b^2}}&{{b^3}}&1\\{{c^2}}&{{c^3}}&1\end{array}} \right|\] \[{C_1} \leftrightarrow {C_3}\] \[ - \left| {\begin{array}{*{20}{c}}1&{{a^3}}&{{a^2}}\\1&{{b^3}}&{{b^2}}\\1&{{c^3}}&{{c^2}}\end{array}} \right|\] \[{C_2} \leftrightarrow {C_3}\] \[ \left| {\begin{array}{*{20}{c}}1&{{a^2}}&{{a^3}}\\1&{{b^2}}&{{b^3}}\\1&{{c^2}}&{{c^3}}\end{array}} \right| = RHS\]

Question (3)

Evaluate $\left| {\begin{array}{*{20}{c}}{\cos \alpha \cos \beta }&{\cos \alpha \sin \beta }&{ - \sin \alpha }\\{ - \sin \beta }&{\cos \beta }&0\\{\sin \alpha \cos \beta }&{\sin \alpha \sin \beta }&{\cos \alpha }\end{array}} \right|$

Solution

\[\left| {\begin{array}{*{20}{c}}{\cos \alpha \cos \beta }&{\cos \alpha \sin \beta }&{ - \sin \alpha }\\{ - \sin \beta }&{\cos \beta }&0\\{\sin \alpha \cos \beta }&{\sin \alpha \sin \beta }&{\cos \alpha }\end{array}} \right|\] \[ = \cos \alpha \cos \beta \left( {\cos \alpha \cos \beta - 0} \right) - \cos \alpha \sin \beta \left( { - \cos \alpha \sin \beta - 0} \right) - \sin \alpha \left( { - \sin \alpha {{\sin }^2}\beta - \sin \alpha {{\cos }^2}\beta } \right)\] \[ = {\cos ^2}\alpha {\cos ^2}\beta + {\cos ^2}\alpha {\sin ^2}\beta + {\sin ^2}\alpha {\sin ^2}\beta + {\sin ^2}\alpha {\cos ^2}\beta \] \[ = {\cos ^2}\alpha \left( {{{\cos }^2}\beta + {{\cos }^2}\alpha } \right) + {\sin ^2}\alpha \left( {{{\sin }^2}\beta + {{\cos }^2}\beta } \right)\] \[ = {\cos ^2}\alpha + {\sin ^2}\beta = 1\]

Question (4)

If a, b, c are real number, and
\[\left| \Delta \right| = \left| {\begin{array}{*{20}{c}}{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\\{a + b}&{b + c}&{c + a}\end{array}} \right| = 0\] Show that either a+b+c = 0 or a = b = c

Solution

\[\left| {\begin{array}{*{20}{c}}{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\\{a + b}&{b + c}&{c + a}\end{array}} \right| = 0\] \[{R_1} \to {R_1} + {R_2} + {R_3}\] \[\left| {\begin{array}{*{20}{c}}{2\left( {a + b + c} \right)}&{2\left( {a + b + c} \right)}&{2\left( {a + b + c} \right)}\\{c + a}&{a + b}&{b + c}\\{a + b}&{b + c}&{c + a}\end{array}} \right| = 0\] \[{R_1}\left( {\frac{1}{{2\left( {a + b + c} \right)}}} \right)\] \[2\left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}}1&1&1\\{c + a}&{a + b}&{b + c}\\{a + b}&{b + c}&{c + a}\end{array}} \right| = 0\] \[{C_2} \to {C_2} - {C_1},{C_3} \to {C_3} - {C_1}\] \[2\left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}}1&0&0\\{c + a}&{b - c}&{b - a}\\{a + b}&{c - a}&{c - b}\end{array}} \right| = 0\] \[2\left( {a + b + c} \right)\left[ {\left( {b - c} \right)\left( {c - b} \right) - \left( {b - a} \right)\left( {c - a} \right)} \right] = 0\] \[2\left( {a + b + c} \right)\left[ {bc - {b^2} - {c^2} + \cancel{bc} - \cancel{bc} + ba + ac - {a^2}} \right] = 0\] \[ - 2\left( {a + b + c} \right)\left[ {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right] = 0\] \[ - \left( {a + b + c} \right)\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] = 0\] \[a + b + c = 0\; \;or \;\; {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = 0\] \[ \Rightarrow a - b = 0;b - c = 0;c - a = 0\] \[a + b + c = 0\; \;or \;\;a = b = c\]

Question (5)

Solve the equation $\left| {\begin{array}{*{20}{c}}{x + a}&x&x\\x&{x + a}&x\\x&x&{x + a}\end{array}} \right| = 0,a \ne 0$

Solution

\[\left| {\begin{array}{*{20}{c}}{x + a}&x&x\\x&{x + a}&x\\x&x&{x + a}\end{array}} \right| = 0\] \[{R_1} \to {R_1} + {R_2} + {R_3}\] \[\left| {\begin{array}{*{20}{c}}{3x + a}&{3x + a}&{3x + a}\\x&{x + a}&x\\x&x&{x + a}\end{array}} \right| = 0\] \[{R_1}\left( {\frac{1}{{3x + a}}} \right)\] \[\left( {3x + a} \right)\left| {\begin{array}{*{20}{c}}1&1&1\\x&{x + a}&x\\x&x&{x + a}\end{array}} \right| = 0\] \[{C_2} \to {C_2} - {C_1};{C_3} \to {C_3} - {C_1}\] \[\left( {3x + a} \right)\left| {\begin{array}{*{20}{c}}1&0&0\\x&a&0\\x&0&a\end{array}} \right| = 0\] \[\left( {3x + a} \right)\left( {{a^2}} \right) = 0\] \[3x + a = 0 \Rightarrow x = \frac{{ - a}}{3}\]

Question (6)

Prove that $\left| {\begin{array}{*{20}{c}}{{a^2}}&{bc}&{ac + {c^2}}\\{{a^2} + ab}&{{b^2}}&{ac}\\{ab}&{{b^2} + bc}&{{c^2}}\end{array}} \right| = 4{a^2}{b^2}{c^2}$ \[\]

Solution

\[\left| {\begin{array}{*{20}{c}}{{a^2}}&{bc}&{ac + {c^2}}\\{{a^2} + ab}&{{b^2}}&{ac}\\{ab}&{{b^2} + bc}&{{c^2}}\end{array}} \right| \] \[{C_1}\left( {\frac{1}{a}} \right);{C_2}\left( {\frac{1}{b}} \right);{C_3}\left( {\frac{1}{c}} \right)\] \[abc\left| {\begin{array}{*{20}{c}}a&c&{a + c}\\{{a} + b}&b&a\\b&{b + c}&c\end{array}} \right|\] \[{R_1} \to {R_1} + {R_2} + {R_3}\] \[abc\left| {\begin{array}{*{20}{c}}{2\left( {a + b} \right)}&{2\left( {b + c} \right)}&{2\left( {a + c} \right)}\\{a + b}&b&a\\b&{b + c}&c\end{array}} \right|\] \[{R_1}\left( {\frac{1}{2}} \right)\] \[2abc\left| {\begin{array}{*{20}{c}}{a + b}&{b + c}&{a + c}\\{a + b}&b&a\\b&{b + c}&c\end{array}} \right|\] \[{R_2} \to {R_2} - {R_1};{R_3} \to {R_3} - {R_1}\] \[2abc\left| {\begin{array}{*{20}{c}}{a + b}&{b + c}&{a + c}\\0&{ - c}&{ - c}\\{ - a}&0&{ - a}\end{array}} \right|\] \[{R_2}\left( {\frac{1}{{ - c}}} \right),{R_3}\left( {\frac{1}{{ - a}}} \right)\] \[2{a^2}b{c^2}\left| {\begin{array}{*{20}{c}}{a + b}&{b + c}&{a + c}\\0&1&1\\1&0&1\end{array}} \right|\] \[{C_3} \to {C_3} - {C_2}\] \[2{a^2}b{c^2}\left| {\begin{array}{*{20}{c}}{a + b}&{b + c}&{a - b}\\0&1&0\\1&0&1\end{array}} \right|\] \[ = 2{a^2}b{c^2}\left[ {\left( {a + b} \right) - \left( {b + c} \right)\left( 0 \right) + \left( {a - b} \right)\left( {0 - 1} \right)} \right]\] \[ = 2{a^2}b{c^2}\left[ {\cancel{a} + b -\cancel{a
} + b} \right]\] \[ = 2{a^2}b{c^2}\left( {2b} \right)\] \[ = 4{a^2}{b^2}{c^2} = RHS\]

Question (7)

If ${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}3&{ - 1}&1\\{ - 15}&6&{ - 5}\\5&{ - 2}&2\end{array}} \right]$ and
$B = \left[ {\begin{array}{*{20}{c}}1&2&{ - 2}\\{ - 1}&3&0\\0&{ - 2}&1\end{array}} \right]$, find (AB)-1

Solution

\[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}3&{ - 1}&1\\{ - 15}&6&{ - 5}\\5&{ - 2}&2\end{array}} \right];B = \left[ {\begin{array}{*{20}{c}}1&2&{ - 2}\\{ - 1}&3&0\\0&{ - 2}&1\end{array}} \right]\] \[{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}\] \[\left| B \right| = 1\left( 3 \right) - 2\left( { - 1} \right) - 2\left( 2 \right) = 1\] \[\text{cofactor}\;B = \left[ {\begin{array}{*{20}{c}}3&1&2\\2&1&2\\6&2&5\end{array}} \right]\] \[adjB = \left[ {\begin{array}{*{20}{c}}3&2&6\\1&1&2\\2&2&5\end{array}} \right]\] \[{B^{ - 1}} = \frac{1}{{\left| B \right|}}adjB\] \[{B^{ - 1}} = \frac{1}{1}\left[ {\begin{array}{*{20}{c}}3&2&6\\1&1&2\\2&2&5\end{array}} \right]\] \[{B^{ - 1}}{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}3&2&6\\1&1&2\\2&2&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3&{ - 1}&1\\{ - 15}&6&{ - 5}\\5&{ - 2}&2\end{array}} \right]\] \[{B^{ - 1}}{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}9&{ - 3}&5\\{ - 2}&1&0\\1&0&2\end{array}} \right]\]

Question (8)

Let $A = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\{ - 2}&3&1\\1&1&5\end{array}} \right]$ , verify that
(i) [Adj A]-1 = Adj (A-1)     (ii) (A-1)-1 = A

Solution

\[A = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\{ - 2}&3&1\\1&1&5\end{array}} \right]\] \[\left( i \right)\quad \left| A \right| = 1\left( {14} \right) + 2\left( { - 11} \right) + 1\left( { - 5} \right)\] \[\left| A \right| = 14 - 22 - 5 = - 13\] \[cofactorofA = \left[ {\begin{array}{*{20}{c}}{14}&{11}&{ - 5}\\{11}&4&{ - 3}\\{ - 5}&{ - 3}&{ - 1}\end{array}} \right]\] AdjA=(cofactor of A)'
\[AdjA = \left[ {\begin{array}{*{20}{c}}{14}&{11}&{ - 5}\\{11}&4&{ - 3}\\{ - 5}&{ - 3}&{ - 1}\end{array}} \right]\] \[LHS = {\left( {AdjA} \right)^{ - 1}}\] \[\left| {AdjA} \right| = \left| {\begin{array}{*{20}{c}}{14}&{11}&{ - 5}\\{11}&4&{ - 3}\\{ - 5}&{ - 3}&{ - 1}\end{array}} \right|\] \[\left| {AdjA} \right| = 14\left( { - 13} \right) - 11\left( { - 26} \right) - 5\left( {-13} \right)\] \[\left| {AdjA} \right| = - 182 + 286 + 65 = 169\] \[ \text{cofactor of(adjA)} = \left[ {\begin{array}{*{20}{c}}{ - 13}&{26}&{ - 13}\\{26}&{ - 39}&{ - 13}\\{ - 13}&{ - 13}&{ - 65}\end{array}} \right]\] \[Adj(adjA) = \left[ {\begin{array}{*{20}{c}}{ - 13}&{26}&{ - 13}\\{26}&{ - 39}&{ - 13}\\{ - 13}&{ - 13}&{ - 65}\end{array}} \right]\] \[{\left( {adjA} \right)^{ - 1}} = \frac{1}{{\left| A \right|}}Adj{\left( {adjA} \right)^{ - 1}}\] \[ = \frac{1}{{169}}\left[ {\begin{array}{*{20}{c}}{ - 13}&{26}&{ - 13}\\{26}&{ - 39}&{ - 13}\\{ - 13}&{ - 13}&{ - 65}\end{array}} \right]\] \[ = \frac{ \cancel{{ - 13}}}{{\cancel{169}_{13}}}\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\{ - 2}&3&1\\1&1&5\end{array}} \right] - - - \left( 1 \right)\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{{ - 1}}{{13}}\left[ {\begin{array}{*{20}{c}}{14}&{11}&{ - 5}\\{11}&4&{ - 3}\\{ - 5}&{ - 3}&{ - 1}\end{array}} \right]\] \[RHS = adj\left( {{A^{ - 1}}} \right) = \left| {{A^{ - 1}}} \right|\] \[{A^{ - 1}} = \frac{{ - 1}}{{{{13}^2}}}\left[ {\begin{array}{*{20}{c}}{14}&{11}&{ - 5}\\{11}&4&{ - 3}\\{ - 5}&{ - 3}&{ - 1}\end{array}} \right]\] \[\left| {{A^{ - 1}}} \right| = \frac{{ - 1}}{{{{13}^2}}}\left( {169} \right) = - 13\] \[\text {cofactor of}\;{A^{ - 1}} = \frac{{ - 1}}{{{{13}^2}}}\left[ {\begin{array}{*{20}{c}}{ - 13}&{26}&{ - 13}\\{26}&{ - 39}&{ - 13}\\{ - 13}&{ - 13}&{ - 65}\end{array}} \right]\] adj of (A-1 = [cofactor of A-1]'
\[\text{adj of}\;{A^{ - 1}} = \frac{{ - 1}}{{{{13}^2}}}\left[ {\begin{array}{*{20}{c}}{ - 13}&{26}&{ - 13}\\{26}&{ - 39}&{ - 13}\\{ - 13}&{ - 13}&{ - 65}\end{array}} \right]\] \[\text{adj of}\;{A^{ - 1}} = \frac{{ - 13}}{{169}}\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\{ - 2}&3&1\\1&1&5\end{array}} \right] - - - (2)\] from (1) and (2)
(adj A)-1 = adj(A-1)

\[\left( {ii} \; \; \right){\left( {{A^{ - 1}}} \right)^{ - 1}} = A\] Let A-1 = B
\[B = \frac{{ - 1}}{{13}}\left[ {\begin{array}{*{20}{c}}{14}&{11}&{ - 5}\\{11}&4&{ - 3}\\{ - 5}&{ - 3}&{ - 1}\end{array}} \right]\] \[ \text{cofactor of} \; B = \frac{{ - 1}}{{{{13}^2}}}\left[ {\begin{array}{*{20}{c}}{ - 13}&{26}&{ - 13}\\{26}&{ - 39}&{ - 13}\\{ - 13}&{ - 13}&{ - 65}\end{array}} \right]\] \[adjB = \frac{1}{{169}}\left[ {\begin{array}{*{20}{c}}{ - 13}&{26}&{ - 13}\\{26}&{ - 39}&{ - 13}\\{ - 13}&{ - 13}&{ - 65}\end{array}} \right]\] \[{B^{ - 1}} = \frac{1}{{\left| B \right|}}adjB\] \[{B^{ - 1}} = \frac{1}{{\frac{1}{{13}}}}\frac{1}{{169}} \times - 13\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\{ - 2}&3&1\\1&1&5\end{array}} \right]\] \[{B^{ - 1}} = \cancel{13} \times \frac{1}{\cancel{{169}}} \times - \cancel{13}\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\{ - 2}&3&1\\1&1&5\end{array}} \right]\] \[{B^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\{ - 2}&3&1\\1&1&5\end{array}} \right]\] \[{B^{ - 1}} = A \Rightarrow {\left( {{A^{ - 1}}} \right)^{ - 1}} = A\]

Question (9)

Evaluate $\left| {\begin{array}{*{20}{c}}x&y&{x + y}\\y&{x + y}&x\\{x + y}&x&y\end{array}} \right|$

Solution

\[\left| {\begin{array}{*{20}{c}}x&y&{x + y}\\y&{x + y}&x\\{x + y}&x&y\end{array}} \right|\] \[{C_1} \to {C_1} + {C_2} + {C_3}\] \[\left| {\begin{array}{*{20}{c}}{2\left( {x + y} \right)}&y&{x + y}\\{2\left( {x + y} \right)}&{x + y}&x\\{2\left( {x + y} \right)}&x&y\end{array}} \right|\] \[{C_1}\left( {\frac{1}{{2\left( {x + y} \right)}}} \right)\] \[ = 2\left( {x + y} \right)\left| {\begin{array}{*{20}{c}}1&y&{x + y}\\1&{x + y}&x\\1&x&y\end{array}} \right|\] \[{R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}\] \[ = 2\left( {x + y} \right)\left| {\begin{array}{*{20}{c}}1&y&{x + y}\\0&x&{ - y}\\0&{x - y}&{ - x}\end{array}} \right|\] \[ = 2\left( {x + y} \right)\left[ { - {x^2} + y\left( {x - y} \right)} \right]\] \[ = 2\left( {x + y} \right)\left[ { - {x^2} + xy - {y^2}} \right]\] \[ = - 2\left( {x + y} \right)\left[ {{x^2} - xy + {y^2}} \right]\] \[ = - 2\left( {{x^3} + {y^3}} \right)\]

Question (10)

Evaluate $\left| {\begin{array}{*{20}{c}}1&x&y\\1&{x + y}&y\\1&x&{x + y}\end{array}} \right|$

Solution

$\left| {\begin{array}{*{20}{c}}1&x&y\\1&{x + y}&y\\1&x&{x + y}\end{array}} \right|$ \[{R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}\] \[\left| {\begin{array}{*{20}{c}}1&x&y\\0&y&0\\0&0&x\end{array}} \right|\] \[ = 1\left( {xy} \right) - x\left( 0 \right) + y\left( 0 \right)\] \[ = xy\]

USing properties of determinants in Exercise 11 to 15, prove that:

Question (11)

\[\left| {\begin{array}{*{20}{c}}\alpha &{{\alpha ^2}}&{\beta + \gamma }\\\beta &{{\beta ^2}}&{\gamma + \alpha }\\\gamma &{{\gamma ^2}}&{\alpha + \beta }\end{array}} \right| = \left( {\beta - \gamma } \right)\left( {\gamma - \alpha } \right)\left( {\alpha - \beta } \right)\left( {\alpha + \beta + \gamma } \right)\]

Solution

\[\left| {\begin{array}{*{20}{c}}\alpha &{{\alpha ^2}}&{\beta + \gamma }\\\beta &{{\beta ^2}}&{\gamma + \alpha }\\\gamma &{{\gamma ^2}}&{\alpha + \beta }\end{array}} \right|\] \[{C_3} \to {C_3} + {C_1}\] \[\left| {\begin{array}{*{20}{c}}\alpha &{{\alpha ^2}}&{\alpha + \beta + \gamma }\\\beta &{{\beta ^2}}&{\alpha + \beta + \gamma }\\\gamma &{{\gamma ^2}}&{\alpha + \beta + \gamma }\end{array}} \right|\] \[{C_3}\left( {\frac{1}{{\alpha + \beta + \gamma }}} \right)\] \[\left( {\alpha + \beta + \gamma } \right)\left| {\begin{array}{*{20}{c}}\alpha &{{\alpha ^2}}&1\\\beta &{{\beta ^2}}&1\\\gamma &{{\gamma ^2}}&1\end{array}} \right|\] \[{R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}\] \[\left( {\alpha + \beta + \gamma } \right)\left| {\begin{array}{*{20}{c}}\alpha &{{\alpha ^2}}&1\\{\beta - \alpha }&{{\beta ^2} - {\alpha ^2}}&0\\{\gamma - \alpha }&{{\gamma ^2} - {\alpha ^2}}&0\end{array}} \right|\] \[\left( {\alpha + \beta + \gamma } \right)\left| {\begin{array}{*{20}{c}}\alpha &{{\alpha ^2}}&1\\{\beta - \alpha }&{\left( {\beta - \alpha } \right)\left( {\beta + \alpha } \right)}&0\\{\gamma - \alpha }&{\left( {\gamma - \alpha } \right)\left( {\gamma + \alpha } \right)}&0\end{array}} \right|\] \[{R_2}\left( {\frac{1}{{\beta - \alpha }}} \right);{R_3}\left( {\frac{1}{{\gamma - \alpha }}} \right)\] \[\left( {\alpha + \beta + \gamma } \right)\left( {\beta - \alpha } \right)\left( {\gamma - \alpha } \right)\left| {\begin{array}{*{20}{c}}\alpha &{{\alpha ^2}}&1\\1&{\beta + \alpha }&0\\1&{\gamma + \alpha }&0\end{array}} \right|\] \[{R_3} \to {R_3} - {R_2}\] \[\left( {\alpha + \beta + \gamma } \right)\left( {\beta - \alpha } \right)\left( {\gamma - \alpha } \right)\left| {\begin{array}{*{20}{c}}\alpha &{{\alpha ^2}}&1\\1&{\beta + \alpha }&0\\0&{\gamma - \beta }&0\end{array}} \right|\] \[{R_3}\left( {\frac{1}{{\gamma - \beta }}} \right)\] \[\left( {\alpha + \beta + \gamma } \right)\left[ { - \left( {\alpha - \beta } \right)} \right]\left[ { - \left( {\beta - \gamma } \right)} \right]\left( {\gamma - \alpha } \right)\left| {\begin{array}{*{20}{c}}\alpha &{{\alpha ^2}}&1\\1&{\beta + \alpha }&0\\0&1&0\end{array}} \right|\] \[ = \left( {\alpha - \beta } \right)\left( {\beta - \gamma } \right)\left( {\gamma - \alpha } \right)\left( {\alpha + \beta + \gamma } \right)\left[ {\alpha \left( 0 \right) + {\alpha ^2}\left( 0 \right) + 1\left( 1 \right)} \right]\] \[ = \left( {\alpha - \beta } \right)\left( {\beta - \gamma } \right)\left( {\gamma - \alpha } \right)\left( {\alpha + \beta + \gamma } \right) = RHS\]

Question (12)

\[\left| {\begin{array}{*{20}{c}}x&{{x^2}}&{1 + p{x^3}}\\y&{{y^2}}&{1 + p{y^3}}\\\gamma &{{\gamma ^2}}&{1 + p{z^3}}\end{array}} \right| = \left( {1 + pxyz} \right)\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)\]

Solution

\[\left| {\begin{array}{*{20}{c}}x&{{x^2}}&{1 + p{x^3}}\\y&{{y^2}}&{1 + p{y^3}}\\z&{{z^2}}&{1 + p{z^3}}\end{array}} \right|\] By Property
\[ = \left| {\begin{array}{*{20}{c}}x&{{x^2}}&1\\y&{{y^2}}&1\\z&{{z^2}}&1\end{array}} \right| + \left| {\begin{array}{*{20}{c}}x&{{x^2}}&{p{x^3}}\\y&{{y^2}}&{p{y^3}}\\z&{{z^2}}&{p{z^3}}\end{array}} \right|\] \[{C_3}\left( {\frac{1}{p}} \right),{R_1}\left( {\frac{1}{x}} \right),{R_2}\left( {\frac{1}{y}} \right),{R_3}\left( {\frac{1}{z}} \right)\] \[ = \left| {\begin{array}{*{20}{c}}x&{{x^2}}&1\\y&{{y^2}}&1\\z&{{z^2}}&1\end{array}} \right| + pxyz\left| {\begin{array}{*{20}{c}}1&x&{{x^2}}\\1&y&{{y^2}}\\1&z&{{z^2}}\end{array}} \right|\] \[{C_2} \leftrightarrow {C_3}\] \[ = \left| {\begin{array}{*{20}{c}}x&{{x^2}}&1\\y&{{y^2}}&1\\z&{{z^2}}&1\end{array}} \right| - pxyz\left| {\begin{array}{*{20}{c}}1&{{x^2}}&x\\1&{{y^2}}&y\\1&{{z^2}}&z\end{array}} \right|\] \[{C_1} \leftrightarrow {C_3}\] \[ = \left| {\begin{array}{*{20}{c}}x&{{x^2}}&1\\y&{{y^2}}&1\\z&{{z^2}}&1\end{array}} \right| + pxyz\left| {\begin{array}{*{20}{c}}x&{{x^2}}&1\\y&{{y^2}}&1\\z&{{z^2}}&1\end{array}} \right|\] \[ = \left( {1 + pxyz} \right)\left| {\begin{array}{*{20}{c}}x&{{x^2}}&1\\y&{{y^2}}&1\\z&{{z^2}}&1\end{array}} \right|\] \[{R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}\] \[ = \left( {1 + pxyz} \right)\left| {\begin{array}{*{20}{c}}x&{{x^2}}&1\\{y - x}&{\left( {y - x} \right)\left( {y + x} \right)}&0\\{z - x}&{\left( {z - x} \right)\left( {z + x} \right)}&0\end{array}} \right|\] \[{R_2}\left( {\frac{1}{{y - x}}} \right),{R_3}\left( {\frac{1}{{z - x}}} \right)\] \[\left( {1 + pxyz} \right)\left( {y - x} \right)\left( {z - x} \right)\left| {\begin{array}{*{20}{c}}x&{{x^2}}&1\\1&{y + x}&0\\1&{z + x}&0\end{array}} \right|\] \[{R_3} \to {R_3} - {R_2}\] \[\left( {1 + pxyz} \right)\left( {y - x} \right)\left( {z - x} \right)\left| {\begin{array}{*{20}{c}}x&{{x^2}}&1\\1&{y + x}&0\\0&{z - y}&0\end{array}} \right|\] \[ = \left( {1 + pxyz} \right)\left( {y - x} \right)\left( {z - x} \right)\left[ {1\left( {z - y} \right)} \right]\] \[ = \left( {1 + pxyz} \right)\left[ { - \left( {x - y} \right)} \right]\left[ { - \left( {y - z} \right)} \right]\left( {z - x} \right)\] \[ = \left( {1 + pxyz} \right)\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)\]

Question (13)

\[\left| {\begin{array}{*{20}{c}}{3a}&{ - a + b}&{ - a + c}\\{ - b + a}&{3b}&{ - b + c}\\{ - c + a}&{ - c + b}&{3c}\end{array}} \right| = 3\left( {a + b + c} \right)\left( {ab + bc + ca} \right)\]

Solution

\[\left| {\begin{array}{*{20}{c}}{3a}&{ - a + b}&{ - a + c}\\{ - b + a}&{3b}&{ - b + c}\\{ - c + a}&{ - c + b}&{3c}\end{array}} \right|\] \[{C_1} \to {C_1} + {C_2} + {C_3}\] \[\left| {\begin{array}{*{20}{c}}{a + b + c}&{ - a + b}&{ - a + c}\\{a + b + c}&{3b}&{ - b + c}\\{a + b + c}&{ - c + b}&{3c}\end{array}} \right|\] \[{R_1}\left( {\frac{1}{{a + b + c}}} \right)\] \[\left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}}1&{ - a + b}&{ - a + c}\\1&{3b}&{ - b + c}\\1&{ - c + b}&{3c}\end{array}} \right|\] \[{R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}\] \[\left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}}1&{ - a + b}&{ - a + c}\\0&{2b + a}&{ - b + a}\\0&{ - c + a}&{2c + a}\end{array}} \right|\] \[=\left( {a + b + c} \right)\left[ {\left( {2b + a} \right)\left( {2c + a} \right) - \left( { - b + a} \right)\left( { - c + a} \right) - \left( { - a + b} \right)0 + \left( { - a + c} \right)0} \right]\] \[\left( {a + b + c} \right)\left[ {4bc + 2ab + 2ac + \cancel{{a^2}} - bc + ba + ac - \cancel{{a^2}}} \right]\] \[ = \left( {a + b + c} \right)\left( {3bc + 3ab + 3ac} \right)\] \[ = 3\left( {a + b + c} \right)\left( {ab + bc + ca} \right) = LHS\]

Question (14)

\[\left| {\begin{array}{*{20}{c}}1&{1 + p}&{1 + p + q}\\2&{3 + 2p}&{4 + 3p + 2q}\\3&{6 + 3p}&{10 + 6p + 3q}\end{array}} \right| = 1\]

Solution

\[LHS = \left| {\begin{array}{*{20}{c}}1&{1 + p}&{1 + p + q}\\2&{3 + 2p}&{4 + 3p + 2q}\\3&{6 + 3p}&{10 + 6p + 3q}\end{array}} \right|\] \[{R_2} \to {R_2} - 2{R_1};{R_3} \to {R_3} - 3{R_1}\] \[LHS = \left| {\begin{array}{*{20}{c}}1&{1 + p}&{1 + p + q}\\0&1&{2 + p}\\0&3&{7 + 3p}\end{array}} \right|\] \[{R_3} \to {R_3} - 3{R_2}\] \[LHS = \left| {\begin{array}{*{20}{c}}1&{1 + p}&{1 + p + q}\\0&1&{2 + p}\\0&0&1\end{array}} \right|\] \[ = 1\left( {1 - 0} \right) - \left( {1 - p} \right)\left( 0 \right) + \left( {1 + p + q} \right)\left( 0 \right)\] \[ = 1 = RHS\]

Question (15)

\[\left| {\begin{array}{*{20}{c}}{\sin \alpha }&{\cos \alpha }&{\cos \left( {\alpha + \delta } \right)}\\{\sin \beta }&{\cos \beta }&{\cos \left( {\beta + \delta } \right)}\\{\sin \gamma }&{\cos \gamma }&{\cos \left( {\gamma + \delta } \right)}\end{array}} \right| = 0\]

Solution

\[LHS = \left| {\begin{array}{*{20}{c}}{\sin \alpha }&{\cos \alpha }&{\cos \left( {\alpha + \delta } \right)}\\{\sin \beta }&{\cos \beta }&{\cos \left( {\beta + \delta } \right)}\\{\sin \gamma }&{\cos \gamma }&{\cos \left( {\gamma + \delta } \right)}\end{array}} \right|\] \[LHS = \left| {\begin{array}{*{20}{c}}{\sin \alpha }&{\cos \alpha }&{\cos \alpha \cos \delta - \sin \alpha \sin \delta }\\{\sin \beta }&{\cos \beta }&{\cos \beta \cos \delta - \sin \beta \sin \delta }\\{\sin \gamma }&{\cos \gamma }&{\cos \gamma \cos \delta - \sin \gamma \sin \delta }\end{array}} \right|\] \[{C_3} \to {C_3} + \sin \delta {C_1}\] \[LHS = \left| {\begin{array}{*{20}{c}}{\sin \alpha }&{\cos \alpha }&{\cos \alpha \cos \delta }\\{\sin \beta }&{\cos \beta }&{\cos \beta \cos \delta }\\{\sin \gamma }&{\cos \gamma }&{\cos \gamma \cos \delta }\end{array}} \right|\] \[{C_3}\left( {\frac{1}{{\cos \delta }}} \right)\] \[LHS = \cos \delta \left| {\begin{array}{*{20}{c}}{\sin \alpha }&{\cos \alpha }&{\cos \alpha }\\{\sin \beta }&{\cos \beta }&{\cos \beta }\\{\sin \gamma }&{\cos \gamma }&{\cos \gamma }\end{array}} \right|\] \[\text{As} \quad {C_2} = {C_3}\] \[ = 0 = RHS\]

Question (16)

Solve the system of the following equations
\[\frac{2}{x} + \frac{3}{y} + \frac{{10}}{z} = 4\] \[\frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1\] \[\frac{6}{x} + \frac{9}{y} - \frac{{20}}{z} = 2\]

Solution

\[\text{Let}\quad \frac{1}{x} = a;\frac{1}{y} = b;\frac{1}{z} = c\] \[2a + 3b + 10c = 4\] \[4a - 6b + 5c = 1\] \[6a + 9b - 20c = 2\] \[\left[ {\begin{array}{*{20}{c}}2&3&{10}\\4&{ - 6}&5\\6&9&{ - 20}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4\\1\\2\end{array}} \right]\] \[AX = B\] \[X = {A^{ - 1}}B\] \[\left| A \right| = \left| {\begin{array}{*{20}{c}}2&3&{10}\\4&{ - 6}&5\\6&9&{ - 20}\end{array}} \right|\] \[ = 2\left( {120 - 45} \right) - 3\left( { - 80 - 30} \right) + 10\left( {36 + 36} \right)\] \[ = 150 + 330 + 720 = 1200\] \[ \text{cofactor of A} = \left[ {\begin{array}{*{20}{c}}{75}&{110}&{72}\\{150}&{ - 100}&0\\{75}&{30}&{ - 24}\end{array}} \right]\] Adj A = (cofactor A)'
\[adjA = \left[ {\begin{array}{*{20}{c}}{75}&{150}&{75}\\{110}&{ - 100}&{30}\\{72}&0&{ - 24}\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{{1200}}\left[ {\begin{array}{*{20}{c}}{75}&{150}&{75}\\{110}&{ - 100}&{30}\\{72}&0&{ - 24}\end{array}} \right]\] \[X = {A^{ - 1}}B\] \[x = \frac{1}{{1200}}\left[ {\begin{array}{*{20}{c}}{75}&{150}&{75}\\{110}&{ - 100}&{30}\\{72}&0&{ - 24}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}4\\1\\2\end{array}} \right]\] \[x = \frac{1}{{1200}}\left[ {\begin{array}{*{20}{c}}{600}\\{400}\\{240}\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}\\{\frac{1}{3}}\\{\frac{1}{5}}\end{array}} \right]\] \[ \Rightarrow a = \frac{1}{2};b = \frac{1}{3};c = \frac{1}{5}\] \[ \Rightarrow \frac{1}{x} = \frac{1}{2};\frac{1}{y} = \frac{1}{3};\frac{1}{z} = \frac{1}{5}\] \[ \Rightarrow x = 2;y = 3;z = 5\]

Choose the correct answer in Exercise 17 to 19

Question (17)

If a, b, c are in A.P., then the determinant is
\[\left| {\begin{array}{*{20}{c}}{x + 2}&{x + 3}&{x + 2a}\\{x + 3}&{x + 4}&{x + 2b}\\{x + 4}&{x + 5}&{x + 2c}\end{array}} \right|\] (A) 0     (B) 1     (C) x     (D) 2x

Solution

a, b, c are in A.P.
∴ a + c = 2b
\[\left| {\begin{array}{*{20}{c}}{x + 2}&{x + 3}&{x + 2a}\\{x + 3}&{x + 4}&{x + 2b}\\{x + 4}&{x + 5}&{x + 2c}\end{array}} \right|\] \[{R_1} \to {R_1} + {R_3}\] \[\left| {\begin{array}{*{20}{c}}{2x + 6}&{2x + 8}&{2x + 2a + 2c}\\{x + 3}&{x + 4}&{x + 2b}\\{x + 4}&{x + 5}&{x + 2c}\end{array}} \right|\] \[{R_1} \to {R_1}\left( {\frac{1}{2}} \right) \quad \text{and relace} a + c = 2b\] \[2\left| {\begin{array}{*{20}{c}}{x + 3}&{x + 4}&{x + 2b}\\{x + 3}&{x + 4}&{x + 2b}\\{x + 4}&{x + 5}&{x + 2c}\end{array}} \right| = 0\]\[ \text{As} \quad {R_1} = {R_2}\]

Question (18)

If x, y, z are nonzero real numbers, then the inverse of matrix
\[A = \left[ {\begin{array}{*{20}{c}}x&0&0\\0&y&0\\0&0&z\end{array}} \right]\]
OPTIONS
$\left( A \right)\left[ {\begin{array}{*{20}{c}}{{x^{ - 1}}}&0&0\\0&{{y^{ - 1}}}&0\\0&0&{{z^{ - 1}}}\end{array}} \right]$

$\left( B \right)xyz\left[ {\begin{array}{*{20}{c}}{{x^{ - 1}}}&0&0\\0&{{y^{ - 1}}}&0\\0&0&{{z^{ - 1}}}\end{array}} \right]$

$\left( C \right)\frac{1}{{xyz}}\left[ {\begin{array}{*{20}{c}}x&0&0\\0&y&0\\0&0&z\end{array}} \right]$

$\left( D \right)\frac{1}{{xyz}}\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$

Solution

\[A = \left[ {\begin{array}{*{20}{c}}x&0&0\\0&y&0\\0&0&z\end{array}} \right]\] \[\left| A \right| = xyz\] \[ \text{cofactor of }A = \left[ {\begin{array}{*{20}{c}}{yz}&0&0\\0&{xz}&0\\0&0&{xy}\end{array}} \right]\] AdjA=(cofactor of A)'
\[AdjA = \left[ {\begin{array}{*{20}{c}}{yz}&0&0\\0&{xz}&0\\0&0&{xy}\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}\left[ {adjA} \right]\] \[{A^{ - 1}} = \frac{1}{{xyz}}\left[ {\begin{array}{*{20}{c}}{yz}&0&0\\0&{xz}&0\\0&0&{xy}\end{array}} \right]\] \[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{{x^{ - 1}}}&0&0\\0&{{y^{ - 1}}}&0\\0&0&{{z^{ - 1}}}\end{array}} \right]\] Option (A) is correct

Question (19)

Let $A = \left[ {\begin{array}{*{20}{c}}1&{\sin \theta }&0\\{ - \sin \theta }&1&{\sin \theta }\\{ - 1}&{ - \sin \theta }&1\end{array}} \right]$, where 0 ≤ θ ≤ 2π Then
(A) Det(A) = 0     (B) Det(A) ∈ (2, ∞)
(C) Det(A) ∈ (2, 4)     (D) Det(A) ∈ [2, 4]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}1&{\sin \theta }&0\\{ - \sin \theta }&1&{\sin \theta }\\{ - 1}&{ - \sin \theta }&1\end{array}} \right]\] \[\left| A \right| = \left( {1 + {{\sin }^2}\theta } \right) - \sin \theta \left( { - \sin \theta + \sin \theta } \right) + 1\left( {{{\sin }^2}\theta + 1} \right)\] \[\left| A \right| = 1 + {\sin ^2}\theta + {\sin ^2}\theta + 1\] \[\left| A \right| = 2 + 2{\sin ^2}\theta \] Range of sinθ is [-1, 1]
Range of sin2θ is [0, 1]
0 ≤ sin2θ ≤ 1
multiply by 2
0 ≤ 2sin2θ ≤ 2
add 2 on both sides
2 ≤ 2+2sin2θ ≤ 4
∴ Det(A) ∈ [2, 4]
Option "D" is correct
Exercise 4.6 ⇐
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