# Miscellaneous Exercise Determinants Solution - NCERT Class 12

## Chapter 4 Ex.4.ME Question 1

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

### Solution

\begin{align}\Delta &= \left| {\begin{array}{*{20}{c}}x&{\sin \theta }&{\cos \theta }\\{ - \sin \theta }&{ - x}&1\\{\cos \theta }&1&x\end{array}} \right|\\ &= 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 + \sin \theta \cos \theta - \sin \theta \cos \theta + x{\cos ^2}\theta \\ &= - {x^3} - x + x\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)\\& = - {x^3} - x + x\\& = - {x^3}\end{align}

Hence, $$\Delta$$ is independent of $$\theta$$.

## Chapter 4 Ex.4.ME Question 2

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

### Solution

\begin{align}LHS& = \left| {\begin{array}{*{20}{c}}a&{{a^2}}&{bc}\\b&{{b^2}}&{ca}\\c&{{c^2}}&{ab}\end{array}} \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| \qquad \qquad \left[ {{R_1} \to a{R_1},{R_2} \to b{R_2},{R_3} \to c{R_3}} \right]\\ &= \frac{1}{{abc}} \cdot abc\left| {\begin{array}{*{20}{c}}{{a^2}}&{{a^3}}&1\\{{b^2}}&{{b^3}}&1\\{{c^2}}&{{c^3}}&1\end{array}} \right| \qquad \quad \left[ {{\text{Taking out factor }}abc{\text{ from }}{C_3}} \right]\\ &= \left| {\begin{array}{*{20}{c}}{{a^2}}&{{a^3}}&1\\{{b^2}}&{{b^3}}&1\\{{c^2}}&{{c^3}}&1\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| \qquad \qquad\;\;\;\left[ {{C_1} \leftrightarrow {C_3}{\text{ and }}{C_2} \leftrightarrow {C_3}} \right]\\& = RHS\end{align}

Hence, proved.

## Chapter 4 Ex.4.ME 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

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

Expanding along $${C_3}$$,

\begin{align}\Delta &= - \sin \alpha \left( { - \sin \alpha {{\sin }^2}\beta - {{\cos }^2}\beta \sin \alpha } \right)\\&\qquad + \cos \alpha \left( {\cos \alpha {{\cos }^2}\beta + \cos \alpha {{\sin }^2}\beta } \right)\\ &= {\sin ^2}\alpha \left( {{{\sin }^2}\beta + {{\cos }^2}\beta } \right) + {\cos ^2}\alpha \left( {{{\cos }^2}\beta + {{\sin }^2}\beta } \right)\\ &= {\sin ^2}\alpha \left( 1 \right) + {\cos ^2}\alpha \left( 1 \right)\\ &= 1\end{align}

## Chapter 4 Ex.4.ME Question 4

If $$a,b,c$$are real numbers and $$\Delta = \left| {\begin{array}{*{20}{c}}{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\\{a + b}&{b + c}&{c + a}\end{array}} \right| = 0$$, show that either $$a + b + c = 0$$ or $$a = b = c$$.

### Solution

\begin{align}\Delta &= \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|\\& = \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| \qquad \qquad \left[ {{R_1} \to {R_1} + {R_2} + {R_3}} \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|\\& = 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| \qquad \qquad \left[ {{C_2} \to {C_2} - {C_1}{\text{ and }}{C_3} \to {C_3} - {C_1}} \right]\end{align}

Expanding $${R_1}$$,

\begin{align}\Delta &= 2\left( {a + b + c} \right)\left( 1 \right)\left[ {\left( {b - c} \right)\left( {c - b} \right) - \left( {b - a} \right)\left( {c - a} \right)} \right]\\& = 2\left( {a + b + c} \right)\left[ { - {b^2} - {c^2} + 2bc - bc + ba + ac - {a^2}} \right]\\ &= 2\left( {a + b + c} \right)\left[ {ab + bc + ca - {a^2} - {b^2} - {c^2}} \right]\end{align}

It is given that $$\Delta = 0$$.

Hence,

$$2\left( {a + b + c} \right)\left[ {ab + bc + ca - {a^2} - {b^2} - {c^2}} \right] = 0$$

Either $$\left( {a + b + c} \right) = 0$$ or $$\left[ {ab + bc + ca - {a^2} - {b^2} - {c^2}} \right] = 0$$

Now,

\begin{align}& \Rightarrow ab + bc + ca - {a^2} - {b^2} - {c^2} = 0\\ &\Rightarrow - 2ab - 2ac - 2ca + 2{a^2} + 2{b^2} + 2{c^2} = 0\\& \Rightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = 0\\ &\Rightarrow {\left( {a - b} \right)^2} = {\left( {b - c} \right)^2} = {\left( {c - a} \right)^2} = 0\\& \qquad \left[ {{{\left( {a - b} \right)}^2},{{\left( {b - c} \right)}^2},{{\left( {c - a} \right)}^2}{\text{are non - negative}}} \right]\\ &\Rightarrow \left( {a - b} \right) = \left( {b - c} \right) = \left( {c - a} \right) = 0\\ &\Rightarrow a = b = c\end{align}

Hence, if $$\Delta = 0$$, then either $$\left( {a + b + c} \right) = 0$$ or $$a = b = c$$.

## Chapter 4 Ex.4.ME Question 5

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

### Solution

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

Expanding along $${R_1}$$,

\begin{align}& \Rightarrow \left( {3x + a} \right)\left[ {1 \times {a^2}} \right] = 0\\& \Rightarrow {a^2}\left( {3x + a} \right) = 0\end{align}

Since $$a \ne 0$$

Therefore,

\begin{align}& \Rightarrow 3x + a = 0\\& \Rightarrow x = - \frac{a}{3}\end{align}

## Chapter 4 Ex.4.ME 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

\begin{align}\Delta & = \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|\\ &= abc\left| {\begin{array}{*{20}{c}}a&c&{a + c}\\{a + b}&b&a\\b&{b + c}&c\end{array}} \right|\\& \qquad \qquad \qquad \left[ {{\text{Taking out common factors }}a,b{\text{ and }}c{\text{ from }}{C_1},{C_2}{\text{ and }}{C_3}} \right]\\ &= abc\left| {\begin{array}{*{20}{c}}a&c&{a + c}\\b&{b - c}&{ - c}\\{b - a}&b&{ - a}\end{array}} \right|\\& \qquad \qquad \qquad\left[ {{R_2} \to {R_2} - {R_1}{\text{ and }}{R_3} \to {R_3} - {R_1}} \right]\\ &= abc\left| {\begin{array}{*{20}{c}}a&c&{a + c}\\{a + b}&b&a\\{b - a}&b&{ - a}\end{array}} \right|\\& \qquad \qquad \qquad\left[ {{R_2} \to {R_2} + {R_1}} \right]\\ &= abc\left| {\begin{array}{*{20}{c}}a&c&{a + c}\\{a + b}&b&a\\{2b}&{2b}&0\end{array}} \right|\\& \qquad \qquad \qquad\left[ {{R_3} \to {R_3} + {R_2}} \right]\\ &= 2a{b^2}c\left| {\begin{array}{*{20}{c}}a&c&{a + c}\\{a + b}&b&a\\1&1&0\end{array}} \right|\\\Delta &= 2a{b^2}c\left| {\begin{array}{*{20}{c}}a&{c - a}&{a + c}\\{a + b}&{ - a}&a\\1&0&0\end{array}} \right|\\& \qquad \qquad \qquad\left[ {{C_2} \to {C_2} - {C_1}} \right]\end{align}

Expanding along $${R_3}$$,

\begin{align}\Delta &= 2a{b^2}c\left[ {a\left( {c - a} \right) + a\left( {a + c} \right)} \right]\\ &= 2a{b^2}c\left[ {ac - {a^2} + {a^2} + ac} \right]\\ &= 2a{b^2}c\left( {2ac} \right)\\& = 4{a^2}{b^2}{c^2}\end{align}

Hence, proved.

## Chapter 4 Ex.4.ME 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 $${\left( {AB} \right)^{ - 1}}$$.

### Solution

We know that $${\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$$.

It is given that $$B = \left| {\begin{array}{*{20}{c}}1&2&{ - 2}\\{ - 1}&3&0\\0&{ - 2}&1\end{array}} \right|$$

Therefore,

\begin{align}\left| B \right| &= 1\left( 3 \right) - 2\left( { - 1} \right) - 2\left( { - 2} \right)\\ &= 3 + 2 - 4\\&= 5 - 4\\ &= 1\end{align}

Now,

\begin{align}&{B_{11}} = 3\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{B_{12}} = 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{B_{13}} = 2\\&{B_{21}} = 2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{B_{22}} = 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{B_{23}} = 2\\&{B_{31}} = 6\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{B_{32}} = 2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{B_{33}} = 5\end{align}

Hence,

$$adjB = \left( {\begin{array}{*{20}{c}}3&2&6\\1&1&2\\2&2&5\end{array}} \right)$$

Now,

\begin{align}{B^{ - 1}} &= \frac{1}{{\left| B \right|}}adjB\\ &= \left( {\begin{array}{*{20}{c}}3&2&6\\ 1&1&2\\2&2&5\end{array}} \right)\end{align}

Therefore,

\begin{align}{\left( {AB} \right)^{ - 1}} &= {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)\\& = \left( {\begin{array}{*{20}{c}}{9 - 30 + 30}&{ - 3 + 12 - 12}&{3 - 10 + 12}\\{3 - 15 + 10}&{ - 1 + 6 - 4}&{1 - 5 + 4}\\{6 - 30 + 25}&{ - 2 + 12 - 10}&{2 - 10 + 10}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}9&{ - 3}&5\\{ - 2}&1&0\\1&0&2\end{array}} \right)\end{align}

Thus, $${\left( {AB} \right)^{ - 1}} = \left( {\begin{array}{*{20}{c}}9&{ - 3}&5\\{ - 2}&1&0\\1&0&2\end{array}} \right)$$.

## Chapter 4 Ex.4.ME Question 8

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

(i) $${\left[ {adjA} \right]^{ - 1}} = adj{\left( A \right)^{ - 1}}$$

(ii) $${\left( {{A^{ - 1}}} \right)^{ - 1}} = A$$

### Solution

It is given that $$A = \left( {\begin{array}{*{20}{c}}1&{ - 2}&1\\{ - 2}&3&1\\1&1&5\end{array}} \right)$$

Therefore,

\begin{align}\left| A \right| &= 1\left( {15 - 1} \right) + 2\left( { - 10 - 1} \right) + 1\left( { - 2 - 3} \right)\\ &= 14 - 22 - 5\\ &= - 13\end{align}

Now,

\begin{align}&{A_{11}} = 14\;\;\;\;\;\;\;\;\;\;\;{A_{12}} = 11\;\;\;\;\;\;\;\;\;\;\;{A_{13}} = - 5\\&{A_{21}} = 11\;\;\;\;\;\;\;\;\;\;\;{A_{22}} = 4\;\;\;\;\;\;\;\;\;\;\;\;{A_{23}} = - 3\\&{A_{31}} = - 5\;\;\;\;\;\;\;\;\;\;\;{A_{32}} = - 3\;\;\;\;\;\;\;\;\;{A_{33}} = - 1\end{align}

Hence,

$$adjA = \left( {\begin{array}{*{20}{c}}{14}&{11}&{ - 5}\\{11}&4&{ - 3}\\{ - 5}&{ - 3}&{ - 1}\end{array}} \right)$$

Now,

\begin{align}{A^{ - 1}} &= \frac{1}{{\left| A \right|}}\left( {adjA} \right)\\ &= - \frac{1}{{13}}\left( {\begin{array}{*{20}{c}}{14}&{11}&{ - 5}\\{11}&4&{ - 3}\\{ - 5}&{ - 3}&{ - 1}\end{array}} \right)\\ &= \frac{1}{{13}}\left( {\begin{array}{*{20}{c}}{ - 14}&{ - 11}&5\\{ - 11}&{ - 4}&3\\5&3&1\end{array}} \right)\end{align}

(i)

\begin{align}\left| {adjA} \right| &= 14\left( { - 4 - 9} \right) - 11\left( { - 11 - 15} \right) - 5\left( { - 33 + 20} \right)\\ &= 14\left( { - 13} \right) - 11\left( { - 26} \right) - 5\left( { - 13} \right)\\ &= - 182 + 286 + 65\\ &= 169\end{align}

We have,

$adj\left( {adjA} \right) = \left( {\begin{array}{*{20}{c}}{ - 13}&{26}&{ - 13}\\{26}&{ - 39}&{ - 13}\\{ - 13}&{ - 13}&{ - 65}\end{array}} \right)$

Therefore,

\begin{align}{\left[ {adjA} \right]^{ - 1}}& = \frac{1}{{\left| {adjA} \right|}}\left( {adj\left( {adjA} \right)} \right)\\& = \frac{1}{{169}}\left( {\begin{array}{*{20}{c}}{ - 13}&{26}&{ - 13}\\{26}&{ - 39}&{ - 13}\\{ - 13}&{ - 13}&{ - 65}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{\frac{{ - 1}}{{13}}}&{\frac{2}{{13}}}&{\frac{{ - 1}}{{13}}}\\{\frac{2}{{13}}}&{\frac{{ - 3}}{{13}}}&{\frac{{ - 1}}{{13}}}\\{\frac{{ - 1}}{{13}}}&{\frac{{ - 1}}{{13}}}&{\frac{{ - 5}}{{13}}}\end{array}} \right)\end{align}

Now,

${A^{ - 1}} = - \frac{1}{{13}}\left( {\begin{array}{*{20}{c}}{14}&{11}&{ - 5}\\{11}&4&{ - 3}\\{ - 5}&{ - 3}&{ - 1}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{{ - 14}}{{13}}}&{\frac{{ - 11}}{{13}}}&{\frac{5}{{13}}}\\{\frac{{ - 11}}{{13}}}&{\frac{{ - 4}}{{13}}}&{\frac{3}{{13}}}\\{\frac{5}{{13}}}&{\frac{3}{{13}}}&{\frac{1}{{13}}}\end{array}} \right)$

Therefore,

\begin{align}adj{\left( A \right)^{ - 1}}& = \left( {\begin{array}{*{20}{c}}{\frac{{ - 13}}{{169}}}&{\frac{{26}}{{169}}}&{\frac{{ - 13}}{{169}}}\\{\frac{{26}}{{169}}}&{\frac{{ - 39}}{{169}}}&{\frac{{ - 13}}{{169}}}\\{\frac{{ - 13}}{{169}}}&{\frac{{ - 13}}{{169}}}&{\frac{{ - 65}}{{169}}}\end{array}} \right)\\\\& = \left( {\begin{array}{*{20}{c}}{\frac{{ - 1}}{{13}}}&{\frac{2}{{13}}}&{\frac{{ - 1}}{{13}}}\\{\frac{2}{{13}}}&{\frac{{ - 3}}{{13}}}&{\frac{{ - 1}}{{13}}}\\{\frac{{ - 1}}{{13}}}&{\frac{{ - 1}}{{13}}}&{\frac{{ - 5}}{{13}}}\end{array}} \right)\end{align}

Hence, $${\left[ {adjA} \right]^{ - 1}} = adj{\left( A \right)^{ - 1}}$$ proved.

${A^{ - 1}} = \frac{1}{{13}}\left( {\begin{array}{*{20}{c}}{ - 14}&{ - 11}&5\\{ - 11}&{ - 4}&3\\5&3&1\end{array}} \right)$

Hence,

(ii)

$$adj{\left( A \right)^{ - 1}} = \left( {\begin{array}{*{20}{c}}{\frac{{ - 1}}{{13}}}&{\frac{2}{{13}}}&{\frac{{ - 1}}{{13}}}\\{\frac{2}{{13}}}&{\frac{{ - 3}}{{13}}}&{\frac{{ - 1}}{{13}}}\\{\frac{{ - 1}}{{13}}}&{\frac{{ - 1}}{{13}}}&{\frac{{ - 5}}{{13}}}\end{array}} \right)$$

Now,

\begin{align}\left| {{A^{ - 1}}} \right| &= {\left( {\frac{1}{{13}}} \right)^3}\left[ { - 14\left( { - 4 - 9} \right) + 11\left( { - 11 - 26} \right) + 5\left( { - 33 + 20} \right)} \right]\\ &= {\left( {\frac{1}{{13}}} \right)^3}\left[ { - 169} \right]\\& = - \frac{1}{{13}}\end{align}

Therefore,

\begin{align}{\left( {{A^{ - 1}}} \right)^{ - 1}} &= \frac{{adj{A^{ - 1}}}}{{\left| A \right|}} = \frac{1}{{\left( { - \frac{1}{{13}}} \right)}} \times \left( {\begin{array}{*{20}{c}}{\frac{{ - 1}}{{13}}}&{\frac{2}{{13}}}&{\frac{{ - 1}}{{13}}}\\{\frac{2}{{13}}}&{\frac{{ - 3}}{{13}}}&{\frac{{ - 1}}{{13}}}\\{\frac{{ - 1}}{{13}}}&{\frac{{ - 1}}{{13}}}&{\frac{{ - 5}}{{13}}}\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}1&{ - 2}&1\\{ - 2}&3&1\\1&1&5\end{array}} \right) = A\end{align}

Hence, $${\left( {{A^{ - 1}}} \right)^{ - 1}} = A$$ proved.

## Chapter 4 Ex.4.ME Question 9

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

### Solution

\begin{align}\Delta &= \left| {\begin{array}{*{20}{c}}x&y&{x + y}\\y&{x + y}&x\\{x + y}&x&y\end{array}} \right|\\ &= \left| {\begin{array}{*{20}{c}}{2\left( {x + y} \right)}&{2\left( {x + y} \right)}&{2\left( {x + y} \right)}\\y&{x + y}&x\\{x + y}&x&y\end{array}} \right| \qquad \left[ {{R_1} \to {R_1} + {R_2} + {R_3}} \right]\\& = 2\left( {x + y} \right)\left| {\begin{array}{*{20}{c}}1&1&1\\y&{x + y}&x\\{x + y}&x&y\end{array}} \right|\\ &= 2\left( {x + y} \right)\left| {\begin{array}{*{20}{c}}1&0&0\\y&x&{x - y}\\{x + y}&{ - y}&{ - x}\end{array}} \right| \qquad \quad \left[ {{C_2} \to {C_2} - {C_1}{\rm{ and }}{C_3} \to {C_3} - {C_1}} \right]\\ &= 2\left( {x + y} \right)\left[ { - {x^2} + y\left( {x - y} \right)} \right] \qquad \qquad \left[ {{\text{Expanding along }}{R_1}} \right]\\& = - 2\left( {x + y} \right)\left( {{x^2} + {y^2} - yx} \right)\\ &= - 2\left( {{x^3} + {y^3}} \right)\end{align}

## Chapter 4 Ex.4.ME Question 10

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

### Solution

\begin{align}\Delta &= \left| {\begin{array}{*{20}{c}}1&x&y\\1&{x + y}&y\\1&x&{x + y}\end{array}} \right|\\ &= \left| {\begin{array}{*{20}{c}}1&x&y\\0&y&0\\0&0&x\end{array}} \right| \qquad \quad \left[ {{R_2} \to {R_2} - {R_1}{\text{ and }}{R_3} \to {R_3} - {R_1}} \right]\\& = 1\left( {xy - 0} \right)\qquad \quad \left[ {{\text{Expanding along }}{C_1}} \right]\\ &= xy\end{align}

## Chapter 4 Ex.4.ME Question 11

Using properties of determinants prove that:

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

### Solution

\begin{align}\Delta &= \left| {\begin{array}{*{20}{c}}\alpha &{{\alpha ^2}}&{\beta + \gamma }\\\beta &{{\beta ^2}}&{\gamma + \alpha }\\\gamma &{{\gamma ^2}}&{\alpha + \beta }\end{array}} \right|\\& = \left| {\begin{array}{*{20}{c}}\alpha &{{\alpha ^2}}&{\beta + \gamma }\\{\beta - \alpha }&{{\beta ^2} - {\alpha ^2}}&{\alpha - \beta }\\{\gamma - \alpha }&{{\gamma ^2} - {\alpha ^2}}&{\alpha - \gamma }\end{array}} \right|\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{R_2} \to {R_2} - {R_1}{\text{ and }}{R_3} \to {R_3} - {R_1}} \right]\\ &= \left( {\beta - \alpha } \right)\left( {\gamma - \alpha } \right)\left| {\begin{array}{*{20}{c}}\alpha &{{\alpha ^2}}&{\beta + \gamma }\\1&{\beta + \alpha }&{ - 1}\\1&{\gamma + \alpha }&{ - 1}\end{array}} \right|\\ &= \left( {\beta - \alpha } \right)\left( {\gamma - \alpha } \right)\left| {\begin{array}{*{20}{c}}\alpha &{{\alpha ^2}}&{\beta + \gamma }\\1&{\beta + \alpha }&{ - 1}\\0&{\gamma - \beta }&0\end{array}} \right|\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{ }}{R_3} \to {R_3} - {R_2}} \right]\\ &= \left( {\beta - \alpha } \right)\left( {\gamma - \alpha } \right)\left[ { - \left( {\gamma - \beta } \right)\left( { - \alpha - \beta - \gamma } \right)} \right]\;\;\;\;\;\;\;\;\;\;\;\left[ {{\text{Expanding along }}{R_3}} \right]\\ &= \left( {\beta - \alpha } \right)\left( {\gamma - \alpha } \right)\left( {\gamma - \beta } \right)\left( {\alpha + \beta + \gamma } \right)\\& = \left( {\alpha - \beta } \right)\left( {\beta - \gamma } \right)\left( {\gamma - \alpha } \right)\left( {\alpha + \beta + \gamma } \right)\end{align}

Hence, proved.

## Chapter 4 Ex.4.ME Question 12

Using properties of determinants prove that:

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

### Solution

\begin{align}\Delta &= \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|\\\Delta &= \left| {\begin{array}{*{20}{c}}x&{{x^2}}&{1 + p{x^3}}\\{y - x}&{{y^2} - {x^2}}&{p\left( {{y^3} - {x^3}} \right)}\\{z - x}&{{z^2} - {x^2}}&{p\left( {{z^3} - {x^3}} \right)}\end{array}} \right|\\& \qquad \left[ {{R_2} \to {R_2} - {R_1}{\rm{ and }}{R_3} \to {R_3} - {R_1}} \right]\\\\\Delta& = \left( {y - x} \right)\left( {z - x} \right)\left| {\begin{array}{*{20}{c}}x&{{x^2}}&{1 + p{x^3}}\\1&{y + x}&{p\left( {{y^2} + {x^2} + xy} \right)}\\1&{z + x}&{p\left( {{z^2} + {x^2} + xz} \right)}\end{array}} \right|\\\Delta &= \left( {y - x} \right)\left( {z - x} \right)\left| {\begin{array}{*{20}{c}}x&{{x^2}}&{1 + p{x^3}}\\1&{y + x}&{p\left( {{y^2} + {x^2} + xy} \right)}\\0&{z - y}&{p\left( {z - y} \right)\left( {x + y + z} \right)}\end{array}} \right|\\& \qquad \left[ {{\rm{ }}{R_3} \to {R_3} - {R_2}} \right]\\\\\Delta &= \left( {y - x} \right)\left( {z - x} \right)\left( {z - y} \right)\left| {\begin{array}{*{20}{c}}x&{{x^2}}&{1 + p{x^3}}\\1&{y + x}&{p\left( {{y^2} + {x^2} + xy} \right)}\\0&1&{p\left( {x + y + z} \right)}\end{array}} \right|\\\Delta &= \left( {x - y} \right)\left( {z - y} \right)\left( {z - x} \right)\left[ {\left( { - 1} \right)\left( p \right)\left( {x{y^2} + {x^3} + {x^2}y} \right) + 1 + p{x^3} + p\left( {x + y + z} \right)\left( {xy} \right)} \right]\\& \qquad \left[ {{\text{Expanding along }}{R_3}} \right]\\\\&= \left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)\left[ { - px{y^2} - p{x^3} - p{x^2}y + 1 + p{x^3} + p{x^2}y + px{y^2} + pxyz} \right]\\ &= \left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)\left( {1 + pxyz} \right)\end{align}

Hence, proved.

## Chapter 4 Ex.4.ME Question 13

Using properties of determinants prove that:

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

### Solution

\begin{align}\Delta &= \left| {\begin{array}{*{20}{c}}{3a}&{ - a + b}&{ - a + c}\\{ - b + a}&{3b}&{ - b + c}\\{ - c + a}&{ - c + b}&{3c}\end{array}} \right|\\& = \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|\\&\qquad\left[ {{C_1} \to {C_1} + {C_2} + {C_3}} \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|\\ &= \left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}}1&{ - a + b}&{ - a + c}\\0&{2b + a}&{a - b}\\0&{a - c}&{2c + a}\end{array}} \right|\\&\qquad \left[ {{R_2} \to {R_2} - {R_1}{\text{ and }}{R_3} \to {R_3} - {R_1}} \right]\\\\& = \left( {a + b + c} \right)\left[ {\left( {2b + a} \right)\left( {2c + a} \right) - \left( {a - b} \right)\left( {a - c} \right)} \right]\\&\qquad \left[ {{\text{ Expanding along }}{C_1}} \right]\\\\& = \left( {a + b + c} \right)\left[ {4bc + 2ab + 2ac + {a^2} - {a^2} + ac + ba - bc} \right]\\ &= \left( {a + b + c} \right)\left( {3ab + 3bc + 3ac} \right)\\& = 3\left( {a + b + c} \right)\left( {ab + bc + ca} \right)\end{align}

Hence, proved.

## Chapter 4 Ex.4.ME Question 14

Using properties of determinants prove that:

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

### Solution

\begin{align}\Delta &= \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|\\& = \left| {\begin{array}{*{20}{c}}1&{1 + p}&{1 + p + q}\\0&1&{2 + p}\\0&3&{7 + 3p}\end{array}} \right|\\& \qquad \left[ {{R_2} \to {R_2} - 2{R_1}{\rm{ and }}{R_3} \to {R_3} - 3{R_1}} \right]\\\\&= \left| {\begin{array}{*{20}{c}}1&{1 + p}&{1 + p + q}\\0&1&{2 + p}\\0&0&1\end{array}} \right|\\&\qquad \left[ {{\rm{ }}{R_3} \to {R_3} - 3{R_2}} \right]\\\\&= 1\left| {\begin{array}{*{20}{c}}1&{2 + p}\\0&1\end{array}} \right|\\&\qquad\left[ {{\text{Expanding along }}{C_1}} \right]\\\\&= 1\left( {1 - 0} \right) = 1\end{align}

Hence, proved.

## Chapter 4 Ex.4.ME Question 15

Using properties of determinants prove that:

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

### Solution

\begin{align}\Delta &= \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|\\& = \frac{1}{{\sin \delta \cos \delta }}\left| {\begin{array}{*{20}{c}}{\sin \alpha \sin \delta }&{\cos \alpha \cos \delta }&{\cos \alpha \cos \delta - \sin \alpha \sin \delta }\\{\sin \beta \sin \delta }&{\cos \beta \cos \delta }&{\cos \beta \cos \delta - \sin \beta \sin \delta }\\{\sin \gamma \sin \delta }&{\cos \gamma \cos \delta }&{\cos \gamma \cos \delta - \sin \gamma \sin \delta }\end{array}} \right|\\ &= \frac{1}{{\sin \delta \cos \delta }}\left| {\begin{array}{*{20}{c}}{\cos \alpha \cos \delta }&{\cos \alpha \cos \delta }&{\cos \alpha \cos \delta - \sin \alpha \sin \delta }\\{\cos \beta \cos \delta }&{\cos \beta \cos \delta }&{\cos \beta \cos \delta - \sin \beta \sin \delta }\\{\cos \gamma \cos \delta }&{\cos \gamma \cos \delta }&{\cos \gamma \cos \delta - \sin \gamma \sin \delta }\end{array}} \right|\qquad\left[ {{C_1} \to {C_1} + {C_3}} \right]\end{align}

Here, two columns $${C_1}$$ and $${C_2}$$ are identical.

Therefore, $$\Delta = 0$$

Hence, proved.

## Chapter 4 Ex.4.ME Question 16

Solve the system of the following equations:

\begin{align}\frac{2}{x} + \frac{3}{y} + \frac{{10}}{z} = 4\\\frac{4}{x} - \frac Let \(\frac{1}{x} = p,\frac{1}{y} = q and $$\frac{1}{z} = r$$.

Then the given system of equations is as follows:

\begin{align}2p + 3q + 10r &= 4\\4p - 6q + 5r &= 1\\6p + 9q - 20r &= 2\end{align}

This system can be written in the form of $$AX = B$$, where

$$A = \left( {\begin{array}{*{20}{c}}2&3&{10}\\4&{ - 6}&5\\6&9&{ - 20}\end{array}} \right),X = \left[ {\begin{array}{*{20}{c}}p\\q\\r\end{array}} \right]$$ and $$B = \left[ {\begin{array}{*{20}{c}}4\\1\\2\end{array}} \right]$$

Therefore,

\begin{align}\left| A \right|& = 2\left( {120 - 45} \right) - 3\left( { - 80 - 30} \right) + 10\left( {36 + 36} \right)\\& = 150 + 330 + 720\\& = 1200\end{align}

Thus, $$A$$ is non-singular.

Therefore, $${A^{ - 1}}$$ exists.

Now,

\begin{align}&{A_{11}} = 75\;\;\;\;\;\;\;\;\;\;\;{A_{12}} = 110\;\;\;\;\;\;\;\;\;\;\;{A_{13}} = 72\\&{A_{21}} = 150\;\;\;\;\;\;\;\;\;{A_{22}} = - 100\;\;\;\;\;\;\;\;\;{A_{23}} = 0\\&{A_{31}} = 75\;\;\;\;\;\;\;\;\;\;\;{A_{32}} = 30\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{33}} = - 24\end{align}

Hence,

\begin{align}{A^{ - 1}}& = \frac{1}{{\left| A \right|}}\left( {adjA} \right)\\& = \frac{1}{{1200}}\left( {\begin{array}{*{20}{c}}{75}&{150}&{75}\\{110}&{ - 100}&{30}\\{72}&0&{ - 24}\end{array}} \right)\end{align}

Now,

\begin{align}& \Rightarrow \; X = {A^{ - 1}}B\\& \Rightarrow \; \left[ {\begin{array}{*{20}{c}}p\\q\\r\end{array}} \right] = \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]\\ &\Rightarrow \; \left[ {\begin{array}{*{20}{c}}p\\q\\r\end{array}} \right] = \frac{1}{{1200}}\left[ {\begin{array}{*{20}{c}}{300 + 150 + 150}\\{440 - 100 + 60}\\{288 + 0 - 48}\end{array}} \right]\\ &\Rightarrow \; \left[ {\begin{array}{*{20}{c}}p\\q\\r\end{array}} \right] = \frac{1}{{1200}}\left[ {\begin{array}{*{20}{c}}{600}\\{400}\\{240}\end{array}} \right]\\ &\Rightarrow \; \left[ {\begin{array}{*{20}{c}}p\\q\\r\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}\\{\frac{1}{3}}\\{\frac{1}{5}}\end{array}} \right]\end{align}

Therefore,

$$p = \frac{1}{2},q = \frac{1}{3}$$ and $$r = \frac{1}{5}$$

Hence, $$x = 2,y = 3$$ and $$z = 5$$.

{y} + \fracabc{z} = 1\\\frac{6}{x} + \frac{9}{y} - \frac{{20}}{z} = 2\end{align}\)

{6}

## Chapter 4 Ex.4.ME Question 17

If $$a,b,c$$ are in A.P, then the determinant $$\left| {\begin{array}{*{20}{c}}{x + 2}&{x + 3}&{x + 2a}\\{x + 3}&{x + 4}&{x + 2b}\\{x + 4}&{x + 5}&{x + 2c}\end{array}} \right|$$ is

(A) $$0$$

(B) $$1$$

(C) $$x$$

(D) $$2x$$

### Solution

\begin{align}\Delta &= \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|\\& = \left| {\begin{array}{*{20}{c}}{x + 2}&{x + 3}&{x + 2a}\\{x + 3}&{x + 4}&{x + \left( {a + c} \right)}\\{x + 4}&{x + 5}&{x + 2c}\end{array}} \right|\\& \qquad \left( {2b = a + c{\rm{ as }}a,b,c{\rm{ are in A}}{\rm{.P}}} \right)\\& = \left| {\begin{array}{*{20}{c}}{ - 1}&{ - 1}&{a - c}\\{x + 3}&{x + 4}&{x + \left( {a + c} \right)}\\1&1&{c - a}\end{array}} \right|\\& \qquad \left[ {{R_1} \to {R_1} - {R_2}{\rm{ and }}{R_3} \to {R_3} - {R_2}} \right]\\ &= \left| {\begin{array}{*{20}{c}}0&0&0\\{x + 3}&{x + 4}&{x + a + c}\\1&1&{c - a}\end{array}} \right|\\& \qquad \left[ {{R_1} \to {R_1} + {R_3}} \right]\end{align}

Here, all the elements of the first row are zero.

Hence, we have $$\Delta = 0$$

Thus, the correct option is A.

## Chapter 4 Ex.4.ME Question 18

If $$x,y,z$$ are non-zero real numbers, then the inverse of matrix $$A = \left( {\begin{array}{*{20}{c}}x&0&0\\0&y&0\\0&0&z\end{array}} \right)$$ is

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

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

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

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

### Solution

It is given that $$A = \left( {\begin{array}{*{20}{c}}x&0&0\\0&y&0\\0&0&z\end{array}} \right)$$

Hence,

\begin{align}\left| A \right| &= x\left( {yz - 0} \right)\\ &= xyz\\& \ne 0\end{align}

Now,

\begin{align}&{A_{11}} = yz\;\;\;\;\;\;\;\;\;\;\;\;{A_{12}} = 0\;\;\;\;\;\;\;\;\;\;\;\;{A_{13}} = 0\\&{A_{21}} = 0\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{22}} = xz\;\;\;\;\;\;\;\;\;\;\;{A_{23}} = 0\\&{A_{31}} = 0\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{32}} = 0\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{33}} = xy\end{align}

Therefore,

\begin{align}{A^{ - 1}} &= \frac{1}{{\left| A \right|}}\left( {adjA} \right)\\ &= \frac{1}{{xyz}}\left( {\begin{array}{*{20}{c}}{yz}&0&0\\0&{xz}&0\\0&0&{xy}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{\frac{{yz}}{{xyz}}}&0&0\\0&{\frac{{xz}}{{xyz}}}&0\\0&0&{\frac{{xy}}{{xyz}}}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{\frac{1}{x}}&0&0\\0&{\frac{1}{y}}&0\\0&0&{\frac{1}{z}}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{{x^{ - 1}}}&0&0\\0&{{y^{ - 1}}}&0\\0&0&{{z^{ - 1}}}\end{array}} \right)\end{align}

Thus, the correct option is A.

## Chapter 4 Ex.4.ME Question 19

Let $$A = \left( {\begin{array}{*{20}{c}}1&{\sin \theta }&1\\{ - \sin \theta }&1&{\sin \theta }\\{ - 1}&{ - \sin \theta }&1\end{array}} \right)$$, where $$0 \le \theta \le 2\pi$$, then:

A) $$Det\left( A \right) = 0$$

B) $$Det\left( A \right) \in \left( {2,\infty } \right)$$

C) $$Det\left( A \right) \in \left( {2,4} \right)$$

D) $$Det\left( A \right) \in \left[ {2,4} \right]$$

### Solution

It is given that$$A = \left( {\begin{array}{*{20}{c}}1&{\sin \theta }&1\\{ - \sin \theta }&1&{\sin \theta }\\{ - 1}&{ - \sin \theta }&1\end{array}} \right)$$

Hence,

\begin{align}\left| A \right| &= 1\left( {1 + {{\sin }^2}\theta } \right) - \sin \theta \left( { - \sin \theta + \sin \theta } \right) + 1\left( {{{\sin }^2}\theta + 1} \right)\\ &= 1 + {\sin ^2}\theta + {\sin ^2}\theta + 1\\& = 2 + 2{\sin ^2}\theta \\& = 2\left( {1 + {{\sin }^2}\theta } \right)\end{align}

Now,

\begin{align}&\Rightarrow \; 0 \le \theta \le 2\pi \\ &\Rightarrow\; - 1 \le \sin \theta \le 1\\& \Rightarrow \;0 \le {\sin ^2}\theta \le 1\\& \Rightarrow \;1 \le 1 + {\sin ^2}\theta \le 2\\& \Rightarrow \;2 \le 2\left( {1 + {{\sin }^2}\theta } \right) \le 4\end{align}

Therefore, $$Det\left( A \right) \in \left[ {2,4} \right]$$

Thus, the correct option is D.

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