12th NCERT Matrices Exercise 4.2 Questions 16
Do or do not
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Using the property of determinants and without expanding in Exercise 1 to 5 prove that:

Question (1)

$\left| {\begin{array}{*{20}{c}}x&a&{x + a}\\y&b&{y + b}\\z&c&{z + c}\end{array}} \right| = 0$

Solution

$LHS = \left| {\begin{array}{*{20}{c}}x&a&{x + a}\\y&b&{y + b}\\z&c&{z + c}\end{array}} \right|$ ${C_1} \to {C_1} + {C_2}$ $LHS = \left| {\begin{array}{*{20}{c}}{x + a}&a&{x + a}\\{y + b}&b&{y + b}\\{z + c}&c&{z + c}\end{array}} \right|$ $\text {As} \quad {C_1} = {C_3}$ $LHS = 0$

Question (2)

$\left| {\begin{array}{*{20}{c}}{a - b}&{b - c}&{c - a}\\{b - c}&{c - a}&{a - b}\\{c - a}&{a - b}&{b - c}\end{array}} \right| = 0$

Solution

$D = \left| {\begin{array}{*{20}{c}}{a - b}&{b - c}&{c - a}\\{b - c}&{c - a}&{a - b}\\{c - a}&{a - b}&{b - c}\end{array}} \right|$ ${R_1} \to {R_1} + {R_2} + {R_3}$ $D = \left| {\begin{array}{*{20}{c}}0&0&0\\{b - c}&{c - a}&{a - b}\\{c - a}&{a - b}&{b - c}\end{array}} \right|$ ${\rm{As}}\;{\rm{all}}\;{\rm{elements}}\;{\rm{of}}\;{{\rm{R}}_{\rm{1}}}{\rm{ = 0}}$ $D = 0$

Question (3)

$\left| {\begin{array}{*{20}{c}}2&7&{65}\\3&8&{75}\\5&9&{86}\end{array}} \right| = 0$

Solution

$D = \left| {\begin{array}{*{20}{c}}2&7&{65}\\3&8&{72}\\5&9&{86}\end{array}} \right|$ ${C_3} \to {C_3} - {C_1}$ $D = \left| {\begin{array}{*{20}{c}}2&7&{63}\\3&8&{72}\\5&9&{81}\end{array}} \right|$ ${C_3}\left( {\frac{1}{9}} \right)$ $D = 9\left| {\begin{array}{*{20}{c}}2&7&7\\3&8&9\\5&9&9\end{array}} \right|$ $\text{As} \quad {C_2} = {C_3}$ $D = 9\left( 0 \right)$ $D = 0$

Question (4)

$\left| {\begin{array}{*{20}{c}}1&{bc}&{a\left( {b + c} \right)}\\1&{ca}&{b\left( {c + a} \right)}\\1&{ab}&{c\left( {a + b} \right)}\end{array}} \right| = 0$

Solution

$\left| {\begin{array}{*{20}{c}}1&{bc}&{a\left( {b + c} \right)}\\1&{ca}&{b\left( {c + a} \right)}\\1&{ab}&{c\left( {a + b} \right)}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}1&{bc}&{ab + ac}\\1&{ca}&{bc + ba}\\1&{ab}&{ca + cb}\end{array}} \right|$ ${C_3} \to {C_3} + {C_2}$ $= \left| {\begin{array}{*{20}{c}}1&{bc}&{ab + ac + ba}\\1&{ca}&{bc + ba + ca}\\1&{ab}&{ca + cb + ab}\end{array}} \right|$ ${C_3} \to {C_3}\left( {\frac{1}{{ab + bc + ca}}} \right)$ $= \left( {ab + bc + ca} \right)\left| {\begin{array}{*{20}{c}}1&{bc}&1\\1&{ca}&1\\1&{ab}&1\end{array}} \right|$ $\text {As}\;{C_1} = {C_3}$ $= 0$

Question (5)

$\left| {\begin{array}{*{20}{c}}{b + c}&{q + r}&{y + z}\\{c + a}&{r + p}&{z + x}\\{a + b}&{p + q}&{x + y}\end{array}} \right| = 2\left| {\begin{array}{*{20}{c}}a&p&x\\b&q&y\\c&r&z\end{array}} \right|$

Solution

$LHS = \left| {\begin{array}{*{20}{c}}{b + c}&{q + r}&{y + z}\\{c + a}&{r + p}&{z + x}\\{a + b}&{p + q}&{x + y}\end{array}} \right|$ ${R_1} \to {R_1} + {R_2} + {R_3}$ $LHS = \left| {\begin{array}{*{20}{c}}{2\left( {a + b + c} \right)}&{2\left( {p + q + r} \right)}&{2\left( {x + y + z} \right)}\\{c + a}&{r + p}&{z + x}\\{a + b}&{p + q}&{x + y}\end{array}} \right|$ ${R_1} \to \frac{{{R_1}}}{2}$ $LHS = 2\left| {\begin{array}{*{20}{c}}{a + b + c}&{p + q + r}&{x + y + z}\\{c + a}&{r + p}&{z + x}\\{a + b}&{p + q}&{x + y}\end{array}} \right|$ ${R_2} \to {R_2} - {R_1};{R_3} \to {R_3} - {R_1}$ $LHS = 2\left| {\begin{array}{*{20}{c}}{a + b + c}&{p + q + r}&{x + y + z}\\{ - b}&{ - q}&{ - y}\\{ - c}&{ - r}&{ - z}\end{array}} \right|$ ${R_1} \to {R_1} + {R_2} + {R_3}$ $LHS = 2\left| {\begin{array}{*{20}{c}}a&p&x\\{ - b}&{ - q}&{ - y}\\{ - c}&{ - r}&{ - z}\end{array}} \right|$ ${R_2} \to \frac{{{R_2}}}{{ - 1}};{R_3} \to \frac{{{R_3}}}{{ - 1}}$ $LHS = 2\left| {\begin{array}{*{20}{c}}a&p&x\\b&q&y\\c&r&z\end{array}} \right| = RHS$

By using proprties of determinanta, is Exercises 6 to 14, show that:

Question (6)

$\left| {\begin{array}{*{20}{c}}0&a&{ - b}\\{ - a}&0&{ - c}\\b&c&0\end{array}} \right| = 0$

Solution

$D = \left| {\begin{array}{*{20}{c}}0&a&{ - b}\\{ - a}&0&{ - c}\\b&c&0\end{array}} \right|$ ${R_{ij}} \to {C_{ij}}$ $D = \left| {\begin{array}{*{20}{c}}0&{ - a}&b\\a&0&c\\{ - b}&{ - c}&0\end{array}} \right|$ ${R_1} \to \frac{{{R_1}}}{{ - 1}};{R_2} \to \frac{{{R_2}}}{{ - 1}};{R_3} \to \frac{{{R_3}}}{{ - 1}}$ $D = - \left| {\begin{array}{*{20}{c}}0&a&{ - b}\\{ - a}&0&{ - c}\\b&c&0\end{array}} \right|$ ${D} = - D$ $2D = 0$ $D = 0$

Question (7)

$\left| {\begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}\\{ba}&{ - {b^2}}&{bc}\\{ca}&{cb}&{ - {c^2}}\end{array}} \right| = 4{a^2}{b^2}{c^2}$

Solution

$D = \left| {\begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}\\{ba}&{ - {b^2}}&{bc}\\{ca}&{cb}&{ - {c^2}}\end{array}} \right|$ ${R_1} \to \frac{{{R_1}}}{a};{R_2} \to \frac{{{R_2}}}{b};{R_3} \to \frac{{{R_3}}}{c}$ $D = abc\left| {\begin{array}{*{20}{c}}{ - a}&b&c\\a&{ - b}&c\\a&b&{ - c}\end{array}} \right|$ ${C_1} \to \frac{{{C_1}}}{a};{C_2} \to \frac{{{C_2}}}{b};{C_3} \to \frac{{{C_3}}}{c}$ $D = {a^2}{b^2}{c^2}\left| {\begin{array}{*{20}{c}}{ - 1}&1&1\\1&{ - 1}&1\\1&1&{ - 1}\end{array}} \right|$ ${R_1} \to {R_1} + {R_{3.}};{R_2} \to {R_2} - {R_3}$ $D = {a^2}{b^2}{c^2}\left| {\begin{array}{*{20}{c}}0&2&0\\0&{ - 2}&2\\1&1&{ - 1}\end{array}} \right|$ $D = {a^2}{b^2}{c^2}\left[ {0 - 2\left( {0 - 2} \right) + 0} \right]$ $D = 4{a^2}{b^2}{c^2} = RHS$

Question (8)

$\left( i \right)\left| {\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}} \right| = \left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)$ $\left( {ii} \right)\left| {\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}} \right| = \left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)\left( {a + b + c} \right)$

Solution

$\left( i \right)\left| {\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}} \right|$ ${R_2} \to {R_2} - {R_{1.}};{R_3} \to {R_3} - {R_1}$ $LHS = \left| {\begin{array}{*{20}{c}}1&a&{{a^2}}\\0&{b - a}&{{b^2} - {a^2}}\\0&{c - a}&{{c^2} - {a^2}}\end{array}} \right|$ $LHS = \left| {\begin{array}{*{20}{c}}1&a&{{a^2}}\\0&{b - a}&{\left( {b - a} \right)\left( {b + a} \right)}\\0&{c - a}&{\left( {c - a} \right)\left( {c + a} \right)}\end{array}} \right|$ ${R_2}\left( {\frac{1}{{b - a}}} \right);{R_3}\left( {\frac{1}{{c - a}}} \right)$ $LHS = \left( {b - a} \right)\left( {c - a} \right)\left| {\begin{array}{*{20}{c}}1&a&{{a^2}}\\0&1&{b + a}\\0&1&{c + a}\end{array}} \right|$ ${R_3} \to {R_3} - {R_2}$ $LHS = \left( {b - a} \right)\left( {c - a} \right)\left| {\begin{array}{*{20}{c}}1&a&{{a^2}}\\0&1&{b + a}\\0&0&{c - b}\end{array}} \right|$ $LHS = \left( {b - a} \right)\left( {c - a} \right)\left( {c - b} \right)$ $LHS = - \left( {a - b} \right)\left( {c - a} \right)\left[ { - \left( {b - c} \right)} \right]$ $LHS = \left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right) = RHS$
$\left( {ii} \right)D = \left| {\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}} \right|$ ${C_2} \to {C_2} - {C_1};{C_3} \to {C_3} - {C_1}$ $D = \left| {\begin{array}{*{20}{c}}1&0&0\\a&{b - a}&{c - a}\\{{a^3}}&{{b^3} - {a^3}}&{{c^3} - {a^3}}\end{array}} \right|$ $\text {Form identity} \quad {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$ $D = \left| {\begin{array}{*{20}{c}}1&0&0\\a&{b - a}&{c - a}\\{{a^3}}&{\left( {b - a} \right)\left( {{b^2} + ba + {a^2}} \right)}&{\left( {c - a} \right)\left( {{c^2} + ca + {a^2}} \right)}\end{array}} \right|$ ${C_2}\left( {\frac{1}{{b - a}}} \right);{C_3}\left( {\frac{1}{{c - a}}} \right)$ $D = \left( {b - a} \right)\left( {c - a} \right)\left| {\begin{array}{*{20}{c}}1&0&0\\a&1&1\\{{a^3}}&{{b^2} + ba + {a^2}}&{{c^2} + ca + {a^2}}\end{array}} \right|$ ${C_3} \to {C_3} - {C_2}$ $D = \left( {b - a} \right)\left( {c - a} \right)\left| {\begin{array}{*{20}{c}}1&0&0\\a&1&0\\{{a^3}}&{{b^2} + ab + {a^2}}&{ac + {c^2} - {b^2} - ab}\end{array}} \right|$ $D = \left( {b - a} \right)\left( {c - a} \right)\left[ {ac - ab + {c^2} - {b^2}} \right]$ $D = \left( {b - a} \right)\left( {c - a} \right)\left[ {\left( {c + b} \right)\left( {c - b} \right) + a\left( {c - b} \right)} \right]$ $D = \left( {b - a} \right)\left( {c - a} \right)\left( {c - b} \right)\left( {c + b + a} \right)$ $D = - \left( {a - b} \right)\left( {c - a} \right)\left( {c - b} \right)\left[ { - \left( {b - c} \right)} \right]\left( {a + b + c} \right)$ $D = \left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)\left( {a + b + c} \right)$ $= RHS$

Question (9)

$\left| {\begin{array}{*{20}{c}}x&{{x^2}}&{yz}\\y&{{y^2}}&{zx}\\z&{{z^2}}&{xy}\end{array}} \right| = \left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)\left( {xy + yz + zx} \right)$

Solution

$D = \left| {\begin{array}{*{20}{c}}x&{{x^2}}&{yz}\\y&{{y^2}}&{zx}\\z&{{z^2}}&{xy}\end{array}} \right|$ ${R_2} \to {R_2} - {R_1};{R_3} \to {R_3} - {R_1}$ $D = \left| {\begin{array}{*{20}{c}}x&{{x^2}}&{yz}\\{y - x}&{\left( {y - x} \right)\left( {y + x} \right)}&{ - z\left( {y - x} \right)}\\{z - x}&{\left( {z - x} \right)\left( {z + x} \right)}&{ - y\left( {z - x} \right)}\end{array}} \right|$ ${R_2}\left( {\frac{1}{{y - x}}} \right);{R_3}\left( {\frac{1}{{z - x}}} \right)$ $D = \left( {y - x} \right)\left( {z - x} \right)\left| {\begin{array}{*{20}{c}}x&{{x^2}}&{yz}\\1&{y + x}&{ - z}\\1&{z + x}&{ - y}\end{array}} \right|$ ${R_3} \to {R_3} - {R_2}$ $D = \left( {y - x} \right)\left( {z - x} \right)\left| {\begin{array}{*{20}{c}}x&{{x^2}}&{yz}\\1&{y + x}&{ - z}\\0&{z - y}&{z - y}\end{array}} \right|$ ${R_3} \to {R_3}\left( {\frac{1}{{z - y}}} \right)$ $D = \left( {y - x} \right)\left( {z - x} \right)\left( {z - y} \right)\left| {\begin{array}{*{20}{c}}x&{{x^2}}&{yz}\\1&{y + x}&{ - z}\\0&1&1\end{array}} \right|$ ${C_2} \to {C_2} - {C_3}$ $D = \left( {y - x} \right)\left( {z - x} \right)\left( {z - y} \right)\left| {\begin{array}{*{20}{c}}x&{{x^2} - yz}&{yz}\\1&{y + x + z}&{ - z}\\0&0&1\end{array}} \right|$ $= - \left( {x - y} \right)\left( {z - x} \right)\left[ { - \left( {y - z} \right)} \right]\left[ {x\left( {x + y + z} \right) - 0 - \left( {{x^2} - yz} \right)\left( {1 - 0} \right) + yz\left( 0 \right)} \right]$ $= \left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)\left[ { \require{cancel} \cancel{{x^2}} + xy + xz - \cancel{{x^2}} + yz} \right]$ $= \left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)\left( {xy + yz + zx} \right)$

Question (10)

$\left( i \right)\left| {\begin{array}{*{20}{c}}{x + 4}&{2x}&{2x}\\{2x}&{x + 4}&{2x}\\{2x}&{2x}&{x + 4}\end{array}} \right| = \left( {5x + 4} \right){\left( {4 - x} \right)^2}$ $\left( {ii} \right)\left| {\begin{array}{*{20}{c}}{y + k}&y&y\\y&{y + k}&y\\y&y&{y + k}\end{array}} \right| = {k^2}\left( {3y + k} \right)$

Solution

$\left( i \right)\left| {\begin{array}{*{20}{c}}{x + 4}&{2x}&{2x}\\{2x}&{x + 4}&{2x}\\{2x}&{2x}&{x + 4}\end{array}} \right|$ ${R_1} \to {R_1} + {R_2} + {R_3}$ $\left| {\begin{array}{*{20}{c}}{5x + 4}&{5x + 4}&{5x + 4}\\{2x}&{x + 4}&{2x}\\{2x}&{2x}&{x + 4}\end{array}} \right|$ ${R_1}\left( {\frac{1}{{5x + 4}}} \right)$ $\left( {5x + 4} \right)\left| {\begin{array}{*{20}{c}}1&1&1\\{2x}&{x + 4}&{2x}\\{2x}&{2x}&{x + 4}\end{array}} \right|$ ${C_2} \to {C_2} - {C_1};{C_3} \to {C_3} - {C_1}$ $\left( {5x + 4} \right)\left| {\begin{array}{*{20}{c}}1&0&0\\{2x}&{4 - x}&0\\{2x}&0&{4 - x}\end{array}} \right|$ $= \left( {5x + 4} \right){\left( {4 - x} \right)^2}$ $= RHS$

Question (11)

$\left( i \right)\left| {\begin{array}{*{20}{c}}{a - b - c}&{2a}&{2a}\\{2b}&{b - c - a}&{2b}\\{2c}&{2c}&{c - a - b}\end{array}} \right| = {\left( {a + b + c} \right)^3}$ $\left( {ii} \right)\left| {\begin{array}{*{20}{c}}{x + y + 2z}&x&y\\z&{y + z + 2x}&y\\z&x&{z + x + 2y}\end{array}} \right| = 2{\left( {x + y + z} \right)^3}$

Solution

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

Question (12)

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

Solution

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

Question (13)

$\left| {\begin{array}{*{20}{c}}{1 + {a^2} - {b^2}}&{2ab}&{ - 2b}\\{2ab}&{1 - {a^2} + {b^2}}&{2a}\\{2b}&{ - 2a}&{1 - {a^2} - {b^2}}\end{array}} \right| = {\left( {1 + {a^2} + {b^2}} \right)^3}$

Solution

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

Question (14)

$\left| {\begin{array}{*{20}{c}}{{a^2} + 1}&{ab}&{ac}\\{ab}&{{b^2} + 1}&{bc}\\{ca}&{cb}&{{c^2} + 1}\end{array}} \right| = 1 + {a^2} + {b^2} + {c^2}$

Solution

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

Choose the correct answer in Exercise 15 and 16

Question (15)

Let A be a square matrix or order 3×3, the |kA| is equal to
(A) k|A|     (B) k2|A|
(C) k3|A|     (D) 3k|A|

Solution

Baisc property

Question (16)

Which of the following is correct
(A) Determinant is a square matrix
(B) Determinant is a number associate to matrix
(C) Determinant is a number associated to a square matrix
(D) None of these

Solution

Baisc property