NCERT Solutions For Class 12 Maths Chapter 4 Exercise 4.2

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Chapter 4 Ex.4.2 Question 1

Using the property of determinants and without expanding, prove that:

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

Solution

\[\begin{align}\Delta &= \left| {\begin{array}{*{20}{c}}x&a&{x + a}\\y&b&{y + b}\\z&c&{z + c}\end{array}} \right|\\&= \left| {\begin{array}{*{20}{c}}x&a&x\\y&b&y\\z&c&z\end{array}} \right| + \left| {\begin{array}{*{20}{c}}x&a&a\\y&b&b\\z&c&c\end{array}} \right|\end{align}\]

Here, two columns of each determinant are identical.

Hence,

\[\begin{align}\Delta &= 0 + 0\\ &= 0\end{align}\]

Chapter 4 Ex.4.2 Question 2

Using the property of determinants and without expanding, prove that:

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

Solution

\[\begin{align}\Delta &= \left| {\begin{array}{*{20}{c}}{a - b}&{b - c}&{c - a}\\{b - c}&{c - a}&{a - b}\\{c - a}&{a - b}&{b - c}\end{array}} \right|\\ & = \left| {\begin{array}{*{20}{c}}{a - c}&{b - a}&{c - b}\\{b - c}&{c - a}&{a - b}\\{ - \left( {a - c} \right)}&{ - \left( {b - a} \right)}&{ - \left( {c - b} \right)}\end{array}} \right| \qquad \left[ {{R_1} \to {R_1} + {R_2}} \right]\\ & = \left| {\begin{array}{*{20}{c}}{a - c}&{b - a}&{c - b}\\{b - c}&{c - a}&{a - b}\\{a - c}&{b - a}&{c - b}\end{array}} \right|\end{align}\]

Here, the two rows \({R_1}\) and \({R_3}\) are identical.

Hence, \(\Delta = 0\)

Chapter 4 Ex.4.2 Question 3

Using the property of determinants and without expanding, prove that:

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

Solution

\[\begin{align} \Delta &= \left| {\begin{array}{*{20}{c}} 2&7&{65} \\ 3&8&{75} \\ 5&9&{86} \end{array}} \right| \\&= \left| {\begin{array}{*{20}{c}} 2&7&{63 + 2} \\ 3&8&{72 + 3} \\ 5&9&{81 + 5} \end{array}} \right| \\ & = \left| {\begin{array}{*{20}{c}} 2&7&{63} \\ 3&8&{72} \\ 5&9&{81} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} 2&7&2 \\ 3&8&3 \\ 5&9&5 \end{array}} \right| \\& = \left| {\begin{array}{*{20}{c}} 2&7&{9\left( 7 \right)} \\ 3&8&{9\left( 8 \right)} \\ 5&9&{9\left( 9 \right)} \end{array}} \right| + 0 \qquad \qquad \left[ {\because {\text{Two columns are identical}}} \right] \\ & = 9\left| {\begin{array}{*{20}{c}} 2&7&7 \\ 3&8&8 \\ 5&9&9 \end{array}} \right| \\ & = 0 \qquad \qquad \left[ {\because {\text{Two columns are identical}}} \right] \\ \end{align} \]

Chapter 4 Ex.4.2 Question 4

Using the property of determinants and without expanding, prove that:

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

Solution

\[\begin{align}\Delta &= \left| {\begin{array}{*{20}{c}}1&{bc}&{a\left( {b + c} \right)}\\1&{ca}&{b\left( {c + a} \right)}\\1&{ab}&{c\left( {a + b} \right)}\end{array}} \right|\\ &= \left| {\begin{array}{*{20}{c}}1&{bc}&{ab + bc + ca}\\1&{ca}&{ab + bc + ca}\\1&{ab}&{ab + bc + ca}\end{array}} \right| \qquad \left[ {{C_3} \to {C_3} + {C_2}} \right]\end{align}\]

Here, the two columns \({C_1}\) and \({C_3}\) are proportional.

Hence, \(\Delta = 0\)

Chapter 4 Ex.4.2 Question 5

Using the property of determinants and without expanding, prove that:

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

Solution

\[\begin{align}\Delta &= \left| {\begin{array}{*{20}{c}}{b + c}&{q + r}&{y + z}\\{c + a}&{r + p}&{z + x}\\{a + b}&{p + q}&{x + y}\end{array}} \right|\\ &= \left| {\begin{array}{*{20}{c}}{b + c}&{q + r}&{y + z}\\{c + a}&{r + p}&{z + x}\\a&p&x\end{array}} \right| + \left| {\begin{array}{*{20}{c}}{b + c}&{q + r}&{y + z}\\{c + a}&{r + p}&{z + x}\\b&q&y\end{array}} \right|\\& = {\Delta _1} + {\Delta _2} \dots\dots\dots\dots\dots\dots \left( 1 \right)\end{align}\]

Now,

\[\begin{align}{\Delta _1} &= \left| {\begin{array}{*{20}{c}}{b + c}&{q + r}&{y + z}\\{c + a}&{r + p}&{z + x}\\a&p&x\end{array}} \right|\\ &= \left| {\begin{array}{*{20}{c}}{b + c}&{q + r}&{y + z}\\c&r&z\\a&p&x\end{array}} \right| \qquad \left[ {{R_2} \to {R_2} - {R_3}} \right]\\& = \left| {\begin{array}{*{20}{c}}b&q&y\\c&r&z\\a&p&x\end{array}} \right|\qquad \left[ {{R_1} \to {R_1} - {R_2}} \right]\\& = {\left( { - 1} \right)^2}\left| {\begin{array}{*{20}{c}}a&p&x\\b&q&y\\c&r&z\end{array}} \right|\\{\Delta _1} &= \left| {\begin{array}{*{20}{c}}a&p&x\\b&q&y\\c&r&z\end{array}} \right| \qquad \;\left[ {{R_1} \leftrightarrow {R_3}{\text{ and }}{R_2} \leftrightarrow {R_3}} \right] \qquad \qquad \ldots \left( 2 \right)\end{align}\]

\[\begin{align}{\Delta _2} &= \left| {\begin{array}{*{20}{c}}{b + c}&{q + r}&{y + z}\\{c + a}&{r + p}&{z + x}\\b&q&y\end{array}} \right|\\{\Delta _2}&= \left| {\begin{array}{*{20}{c}}c&r&z\\{c + a}&{r + p}&{z + x}\\b&q&y\end{array}} \right| \qquad\quad \left[ {{R_1} \to {R_1} - {R_3}} \right]\\{\Delta _2}& = \left| {\begin{array}{*{20}{c}}c&r&z\\a&p&x\\b&q&y\end{array}} \right| \qquad \qquad \qquad \qquad\left[ {{R_2} \to {R_2} - {R_1}} \right]\\{\Delta _2} &= {\left( { - 1} \right)^2}\left| {\begin{array}{*{20}{c}}a&p&x\\b&q&y\\c&r&z\end{array}} \right|\\& = \left| {\begin{array}{*{20}{c}}a&p&x\\b&q&y\\c&r&z\end{array}} \right| \qquad \qquad \left[ {{R_1} \leftrightarrow {R_2}{\text{ and }}{\rm{ }}{R_2} \leftrightarrow {R_3}} \right] \qquad \ldots \left( 3 \right)\end{align}\]

From \(\left( 1 \right),\left( 2 \right)\)and \(\left( 3 \right)\), we have

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

Hence, \(\left| {\begin{array}{*{20}{c}}{b + c}&{q + r}&{y + z}\\{c + a}&{r + p}&{z + x}\\{a + b}&{p + q}&{x + y}\end{array}} \right| = 2\left| {\begin{array}{*{20}{c}}a&p&x\\b&q&y\\c&r&z\end{array}} \right|\)proved.

Chapter 4 Ex.4.2 Question 6

Using the property of determinants and without expanding, prove that:

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

Solution

\[\begin{align}\Delta & = \left| {\begin{array}{*{20}{c}}0&a&{ - b}\\{ - a}&0&{ - c}\\b&c&0\end{array}} \right|\\ &= \frac{1}{c}\left| {\begin{array}{*{20}{c}}0&{ac}&{ - bc}\\{ - a}&0&{ - c}\\b&c&0\end{array}} \right| \qquad \left[ {{R_1} \to c{R_1}} \right]\\ &= \frac{1}{c}\left| {\begin{array}{*{20}{c}}{ab}&{ac}&0\\{ - a}&0&{ - c}\\b&c&0\end{array}} \right| \qquad \left[ {{R_1} \to {R_1} - b{R_2}} \right]\\ &= \frac{a}{c}\left| {\begin{array}{*{20}{c}}b&c&0\\{ - a}&0&{ - c}\\b&c&0\end{array}} \right|\end{align}\]

Here, the two rows \({R_1}\) and \({R_3}\) are identical.

Hence, \(\Delta = 0\)

Chapter 4 Ex.4.2 Question 7

Using the property of determinants and without expanding, prove that:

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

Solution

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

Chapter 4 Ex.4.2 Question 8

By using properties of determinants show that:

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

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

Solution

(i) Let \(\Delta = \left| {\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}} \right|\)

\[\begin{align}\Delta &= \left| {\begin{array}{*{20}{c}}0&{a - c}&{{a^2} - {c^2}}\\0&{b - c}&{{b^2} - {c^2}}\\1&c&{{c^2}}\end{array}} \right| \qquad \quad \left[ {{R_1} \to {R_1} - {R_3}{\text{ and }}{R_2} \to {R_2} - {R_3}} \right]\\ &= \left( {c - a} \right)\left( {b - c} \right)\left| {\begin{array}{*{20}{c}}0&{ - 1}&{ - a - c}\\0&1&{b + c}\\1&c&{{c^2}}\end{array}} \right|\\ &= \left( {b - c} \right)\left( {c - a} \right)\left| {\begin{array}{*{20}{c}}0&0&{ - a + b}\\0&1&{b + c}\\1&c&{{c^2}}\end{array}} \right| \qquad \quad \left[ {{R_1} \to {R_1} + {R_2}} \right]\\ &= \left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)\left| {\begin{array}{*{20}{c}}0&0&{ - 1}\\0&1&{b + c}\\1&c&{{c^2}}\end{array}} \right|\\& = \left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)\left| {\begin{array}{*{20}{c}}0&{ - 1}\\1&{b + c}\end{array}} \right|\\ &= \left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)\end{align}\]

Hence, \(\left| {\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}} \right| = \left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)\) proved.

(ii) Let \(\Delta = \left| {\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}} \right|\)

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

Applying \({C_1} \to {C_1} + {C_2}\),

\[\begin{align}\Delta &= \left( {c - a} \right)\left( {b - c} \right)\left| {\begin{array}{*{20}{c}}0&0&1\\0&1&c\\{\left( {{b^2} - {a^2}} \right) + \left( {bc - ac} \right)}&{\left( {{b^2} + bc + {c^2}} \right)}&{{c^3}}\end{array}} \right|\\\\&= \left( {b - c} \right)\left( {c - a} \right)\left( {a - b} \right)\left| {\begin{array}{*{20}{c}}0&0&1\\0&1&c\\{ - \left( {a + b + c} \right)}&{\left( {{b^2} + bc + {c^2}} \right)}&{{c^3}}\end{array}} \right|\\ &= \left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)\left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}}0&0&1\\0&1&c\\{ - 1}&{\left( {{b^2} + bc + {c^2}} \right)}&{{c^3}}\end{array}} \right|\\\\& = \left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)\left( {a + b + c} \right)\left( { - 1} \right)\left| {\begin{array}{*{20}{c}}0&1\\1&c\end{array}} \right|\\\\& = \left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)\left( {a + b + c} \right)\end{align}\]

Hence, \(\left| {\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}} \right| = \left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)\left( {a + b + c} \right)\) proved.

Chapter 4 Ex.4.2 Question 9

By using properties of determinants show that:

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

Solution

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

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

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

Chapter 4 Ex.4.2 Question 10

By using properties of determinants show that:

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

(ii) \(\left| {\begin{array}{*{20}{c}}{y + k}&y&y\\y&{y + k}&y\\y&y&{y + k}\end{array}} \right| = {k^2}\left( {3y + k} \right)\)

Solution

\(\Delta = \left| {\begin{array}{*{20}{c}}{x + 4}&{2x}&{2x}\\{2x}&{x + 4}&{2x}\\{2x}&{2x}&{x + 4}\end{array}} \right|\)

\[\begin{align}\Delta &= \left| {\begin{array}{*{20}{c}}{5x + 4}&{5x + 4}&{5x + 4}\\{2x}&{x + 4}&{2x}\\{2x}&{2x}&{x + 4}\end{array}} \right|\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{R_1} \to {R_1} + {R_2} + {R_3}} \right]\\ &= \left( {5x + 4} \right)\left| {\begin{array}{*{20}{c}}1&1&1\\{2x}&{x + 4}&{2x}\\{2x}&{2x}&{x + 4}\end{array}} \right|\\ &= \left( {5x + 4} \right)\left| {\begin{array}{*{20}{c}}1&0&0\\{2x}&{ - x + 4}&0\\{2x}&0&{ - x + 4}\end{array}} \right|\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{C_2} \to {C_2} - {C_1}{\text{ and }}{C_3} \to {C_3} - {C_1}} \right]\\ &= \left( {5x + 4} \right)\left( {4 - x} \right)\left( {4 - x} \right)\left| {\begin{array}{*{20}{c}}1&0&0\\{2x}&1&0\\{2x}&0&1\end{array}} \right|\\&= \left( {5x + 4} \right){\left( {4 - x} \right)^2}\left| {\begin{array}{*{20}{c}}1&0\\{2x}&1\end{array}} \right|\\& = \left( {5x + 4} \right){\left( {4 - x} \right)^2}\end{align}\]

Hence, \(\left| {\begin{array}{*{20}{c}}{x + 4}&{2x}&{2x}\\{2x}&{x + 4}&{2x}\\{2x}&{2x}&{x + 4}\end{array}} \right| = \left( {5x + 4} \right){\left( {4 - x} \right)^2}\) proved.

\(\Delta = \left| {\begin{array}{*{20}{c}}{y + k}&y&y\\y&{y + k}&y\\y&y&{y + k}\end{array}} \right|\)

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

Expanding along \({C_3}\)

\[\begin{align}\Delta &= {k^2}\left( {3y + k} \right)\left| {\begin{array}{*{20}{c}}1&0\\y&1\end{array}} \right|\\& = {k^2}\left( {3y + k} \right)\end{align}\]

Hence, \(\left| {\begin{array}{*{20}{c}}{y + k}&y&y\\y&{y + k}&y\\y&y&{y + k}\end{array}} \right| = {k^2}\left( {3y + k} \right)\) proved.

Chapter 4 Ex.4.2 Question 11

By using properties of determinants show that:

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

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

Solution

(i) \(\Delta = \left| {\begin{array}{*{20}{c}}{a - b - c}&{2a}&{2a}\\{2b}&{b - c - a}&{2b}\\{2c}&{2c}&{c - a - b}\end{array}} \right|\)

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

Hence, proved.

(ii) \(\Delta = \left| {\begin{array}{*{20}{c}}{x + y + 2z}&x&y\\z&{y + z + 2x}&y\\z&x&{z + x + 2y}\end{array}} \right|\)

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

Hence, proved.

Chapter 4 Ex.4.2 Question 12

By using properties of determinants show that:

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

Solution

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

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

Expanding along \({R_1}\)

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

Hence, proved.

Chapter 4 Ex.4.2 Question 13

By using properties of determinants show that:

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

Solution

\[\begin{align}\Delta &= \left| {\begin{array}{*{20}{c}}{1 + {a^2} - {b^2}}&{2ab}&{ - 2b}\\{2ab}&{1 - {a^2} + {b^2}}&{2a}\\{2b}&{ - 2a}&{1 - {a^2} - {b^2}}\end{array}} \right|\\ &= \left| {\begin{array}{*{20}{c}}{1 + {a^2} + {b^2}}&0&{ - b\left( {1 + {a^2} + {b^2}} \right)}\\0&{1 + {a^2} + {b^2}}&{a\left( {1 + {a^2} + {b^2}} \right)}\\\\{2b}&{ - 2a}&{1 - {a^2} - {b^2}}\end{array}} \right|\\\\&\qquad \left[ {{R_1} \to {R_1} + b{R_3}{\text{ and }}{R_2} \to {R_2} - a{R_3}} \right]\\\\ &= {\left( {1 + {a^2} + {b^2}} \right)^2}\left| {\begin{array}{*{20}{c}}1&0&{ - b}\\0&1&a\\{2b}&{ - 2a}&{1 - {a^2} - {b^2}}\end{array}} \right|\\ &= {\left( {1 + {a^2} + {b^2}} \right)^2}\left[ {\left( 1 \right)\left| {\begin{array}{*{20}{c}}1&a\\{ - 2a}&{1 - {a^2} - {b^2}}\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}0&1\\\\{2b}&{ - 2a}\end{array}} \right|} \right]\\\\& = {\left( {1 + {a^2} + {b^2}} \right)^2}\left[ {1 - {a^2} - {b^2} + 2{a^2} - b\left( { - 2b} \right)} \right]\\\\& = {\left( {1 + {a^2} + {b^2}} \right)^2}\left( {1 + {a^2} + {b^2}} \right)\\\\& = {\left( {1 + {a^2} + {b^2}} \right)^3}\end{align}\]

Hence, proved.

Chapter 4 Ex.4.2 Question 14

By using properties of determinants show that:

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

Solution

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

Taking out common factors \(a,b,c\) from \({R_1},{R_2},{R_3}\) respectively,

\[\begin{align}\Delta &= abc\left| {\begin{array}{*{20}{c}}{a + \frac{1}{a}}&b&c\\a&{b + \frac{1}{b}}&c\\a&b&{c + \frac{1}{c}}\end{array}} \right|\\& = abc\left| {\begin{array}{*{20}{c}}{a + \frac{1}{a}}&b&c\\{ - \frac{1}{a}}&{\frac{1}{b}}&0\\{ - \frac{1}{a}}&0&{\frac{1}{c}}\end{array}} \right| \qquad \quad \left[ {{R_2} \to {R_2} - {R_1}{\text{ and }}{R_3} \to {R_3} - {R_1}} \right]\\ &= abc \times \frac{1}{{abc}}\left| {\begin{array}{*{20}{c}}{{a^2} + 1}&{{b^2}}&{{c^2}}\\{ - 1}&1&0\\{ - 1}&0&1\end{array}} \right|\;\;\;\;\;\;\;\left[ {{C_1} \to a{C_1},{C_2} \to b{C_2}{\text{ and }}{C_3} \to c{C_3}} \right]\\ &= \left| {\begin{array}{*{20}{c}}{{a^2} + 1}&{{b^2}}&{{c^2}}\\{ - 1}&1&0\\{ - 1}&0&1\end{array}} \right|\\ &= - 1\left| {\begin{array}{*{20}{c}}{{b^2}}&{{c^2}}\\1&0\end{array}} \right| + 1\left| {\begin{array}{*{20}{c}}{{a^2} + 1}&{{b^2}}\\{ - 1}&1\end{array}} \right|\\ &= - 1\left( { - {c^2}} \right) + \left( {{a^2} + 1 + {b^2}} \right)\\& = 1 + {a^2} + {b^2} + {c^2}\end{align}\]

Hence, proved.

Chapter 4 Ex.4.2 Question 15

Let \(A\) be a square matrix of order \(3 \times 3\), then \(\left| {kA} \right|\) is equal to:

(A) \(k\left| A \right|\)

(B) \({k^2}\left| A \right|\)

(C) \({k^3}\left| A \right|\)

(D) 3\(k\left| A \right|\)

Solution

Let \(A = \left( {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}} \right)\)

Then,

\[\begin{align}kA &= \left( {\begin{array}{*{20}{c}}{k{a_1}}&{k{b_1}}&{k{c_1}}\\{k{a_2}}&{k{b_2}}&{k{c_2}}\\{k{a_3}}&{k{b_3}}&{k{c_3}}\end{array}} \right)\\\\\left| {kA} \right|& = \left| {\begin{array}{*{20}{c}}{k{a_1}}&{k{b_1}}&{k{c_1}}\\{k{a_2}}&{k{b_2}}&{k{c_2}}\\{k{a_3}}&{k{b_3}}&{k{c_3}}\end{array}} \right|\end{align}\]

Taking out common factors \(k\) from each row

\[\begin{align}\left| {kA} \right| &= {k^3}\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}} \right|\\\\& = {k^3}\left| A \right|\end{align}\]

The correct option is \(C\).

Chapter 4 Ex.4.2 Question 16

Which of the following is correct?

Determinant is a square matrix.

Determinant is a number associated to a matrix.

Determinant is a number associated to a square matrix.

None of the above.

Solution

We know that to every square matrix, \(A = \left[ {{a_{ij}}} \right]\) of order \(n\), we can associate a number called the determinant of square matrix \(A\), where \({a_{ij}} = {\left( {i,j} \right)^{th}}\) element of \(A\).

Thus, the determinant is a number associated to a square matrix.

Hence, the correct option is \(C\).

  
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