# Exercise 4.2 Solutions

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Exercise 4.2 Solutions
Using the property of determinants and without expanding in Exercises 1 to 7, prove that:
{1. \left|\begin{array}{ccc} x & a & x + a \\[5pt] y & b & y + b \\[5pt] z & c & z + c \end{array}\right| = 0}
L.H.S. =
{\left|\begin{array}{ccc} x & a & x + a \\[5pt] y & b & y + b \\[5pt] z & c & z + c \end{array}\right|}
=
{\left|\begin{array}{ccc} x & a & x \\[5pt] y & b & y \\[5pt] z & c & z \end{array}\right| + \left|\begin{array}{ccc} x & a & a \\[5pt] y & b & b \\[5pt] z & c & c \end{array}\right|}
=
0 + 0 {\left(∵ \text{C}_1 = \text{C}_3\right.} in Δ_1 and {\text{C}_2 = \text{C}_3} in {\left.Δ_2\right)}
=
0
=
R.H.S.
∴ It is proved that {\left|\begin{array}{ccc} x & a & x + a \\[5pt] y & b & y + b \\[5pt] z & c & z + c \end{array}\right| = 0}
Using the property of determinants and without expanding in Exercises 1 to 7, prove that:
{2. \left|\begin{array}{ccc} a - b & b - c & c - a \\[5pt] b - c & c - a & a - b \\[5pt] c - a & a - b & b - c \end{array}\right| = 0}
We have,
{\text{L.H.S. = } \left|\begin{array}{ccc} a - b & b - c & c - a \\[5pt] b - c & c - a & a - b \\[5pt] c - a & a - b & b - c \end{array}\right|}
Applying {\text{R}_1 → \text{R}_1 + \text{R}_2} we have,
L.H.S.
{= \left|\begin{array}{ccc} a - b + b - c & b - c + c - a & c - a + a - b \\[5pt] b - c & c - a & a - b \\[5pt] c - a & a - b & b - c \end{array}\right|}
{= \left|\begin{array}{ccc} a - c & b - a & c - b \\[5pt] b - c & c - a & a - b \\[5pt] c - a & a - b & b - c \end{array}\right|}
{= \left|\begin{array}{ccc} -(c - a) & -(a - b) & -(b - c) \\[5pt] b - c & c - a & a - b \\[5pt] c - a & a - b & b - c \end{array}\right|}
= 0 (∵ {\text{R}_1} and {\text{R}_3} are proportional)
= R.H.S.
∴ It is proved that {\left|\begin{array}{ccc} a - b & b - c & c - a \\[5pt] b - c & c - a & a - b \\[5pt] c - a & a - b & b - c \end{array}\right| = 0}
Using the property of determinants and without expanding in Exercises 1 to 7, prove that:
{3. \left|\begin{array}{ccc} 2 & 7 & 65 \\[5pt] 3 & 8 & 75 \\[5pt] 5 & 9 & 86 \end{array}\right| = 0}
We have,
{\text{L.H.S. = } \left|\begin{array}{ccc} 2 & 7 & 65 \\[5pt] 3 & 8 & 75 \\[5pt] 5 & 9 & 86 \end{array}\right|}
Applying {\text{C}_3 → \text{C}_3 - \text{C}_1,} we have,
L.H.S.
=
{\left|\begin{array}{ccc} 2 & 7 & 65 - 2 \\[5pt] 3 & 8 & 75 - 3 \\[5pt] 5 & 9 & 86 - 5 \end{array}\right|}
=
{\left|\begin{array}{ccc} 2 & 7 & 63 \\[5pt] 3 & 8 & 72 \\[5pt] 5 & 9 & 81 \end{array}\right|}
=
{\left|\begin{array}{ccc} 2 & 7 & 9 × 7 \\[5pt] 3 & 8 & 9 × 8 \\[5pt] 5 & 9 & 9 × 9 \end{array}\right|}
=
0 (∵ {\text{C}_2} and {\text{C}_3} are proportional)
=
R.H.S.
∴ It is proved that {\left|\begin{array}{ccc} 2 & 7 & 65 \\[5pt] 3 & 8 & 75 \\[5pt] 5 & 9 & 86 \end{array}\right| = 0}
Using the property of determinants and without expanding in Exercises 1 to 7, prove that:
{4. \left|\begin{array}{ccc} 1 & bc & a(b + c) \\[5pt] 1 & ca & b(c + a) \\[5pt] 1 & ab & c(a + b) \end{array}\right| = 0}
We have,
{\text{L.H.S. = } \left|\begin{array}{ccc} 1 & bc & a(b + c) \\[5pt] 1 & ca & b(c + a) \\[5pt] 1 & ab & c(a + b) \end{array}\right|}
Applying {\text{C}_3 = \text{C}_3 + \text{C}_2,}
L.H.S.
{= \left|\begin{array}{ccc} 1 & bc & bc + a(b + c) \\[5pt] 1 & ca & ca + b(c + a) \\[5pt] 1 & ab & ab + c(a + b) \end{array}\right|}
{= \left|\begin{array}{ccc} 1 & bc & ab + bc + ca \\[5pt] 1 & ca & ab + bc + ca \\[5pt] 1 & ab & ab + bc + ca \end{array}\right|}
Taking common factor {ab + bc + ca} from {\text{C}_3}
L.H.S.
{= (ab + bc + ca) \left|\begin{array}{ccc} 1 & bc & 1 \\[5pt] 1 & ca & 1 \\[5pt] 1 & ab & 1 \end{array}\right|}
= 0 (∵ {\text{C}_1} and {\text{C}_3} are identical)
= R.H.S.
∴ It is proved that {\left|\begin{array}{ccc} 1 & bc & a(b + c) \\[5pt] 1 & ca & b(c + a) \\[5pt] 1 & ab & c(a + b) \end{array}\right| = 0}
Using the property of determinants and without expanding in Exercises 1 to 7, prove that:
{5. \left|\begin{array}{ccc} b + c & q + r & y + z \\[5pt] c + a & r + p & z + x \\[5pt] a + b & p + q & x + y \end{array}\right| = 2 \left|\begin{array}{ccc} a & p & x \\[5pt] b & q & y \\[5pt] c & r & z \end{array}\right|}
We have,
{\text{L.H.S. = }\left|\begin{array}{ccc} b + c & q + r & y + z \\[5pt] c + a & r + p & z + x \\[5pt] a + b & p + q & x + y \end{array}\right|}
Applying {\text{R}_1 → \text{R}_1 + \text{R}_2 + \text{R}_3,} we have,
L.H.S.
{= \left|\begin{array}{ccc} b + c + c + a + a + b & q + r + r + p + p + q & y + z + z + x + x + y \\[5pt] c + a & r + p & z + x \\[5pt] a + b & p + q & x + y \end{array}\right|}
{= \left|\begin{array}{ccc} 2(a + b + c) & 2(p + q + r) & 2(x + y + z) \\[5pt] c + a & r + p & z + x \\[5pt] a + b & p + q & x + y \end{array}\right|}
Taking common factor 2 from {\text{R}_1}, have
{\text{L.H.S. = }2\left|\begin{array}{ccc} (a + b + c) & (p + q + r) & (x + y + z) \\[5pt] c + a & r + p & z + x \\[5pt] a + b & p + q & x + y \end{array}\right|}
Applying {\text{R}_1 → \text{R}_1 - \text{R}_2,} we have,
L.H.S.
{= 2\left|\begin{array}{ccc} (a + b + c) - (c + a) & (p + q + r) - (r + p) & (x + y + z) - (z + x) \\[5pt] c + a & r + p & z + x \\[5pt] a + b & p + q & x + y \end{array}\right|}
{= 2\left|\begin{array}{ccc} b & q & y \\[5pt] c + a & r + p & z + x \\[5pt] a + b & p + q & x + y \end{array}\right|}
Applying {\text{R}_3 → \text{R}_3 - \text{R}_1,} we have,
L.H.S.
{= 2\left|\begin{array}{ccc} b & q & y \\[5pt] c + a & r + p & z + x \\[5pt] a + b - b & p + q - q & x + y - y \end{array}\right|}
{= 2\left|\begin{array}{ccc} b & q & y \\[5pt] c + a & r + p & z + x \\[5pt] a & p & x \end{array}\right|}
Applying {\text{R}_2 → \text{R}_2 - \text{R}_3}, we have,
L.H.S.
{= 2\left|\begin{array}{ccc} b & q & y \\[5pt] c + a - a & r + p - p & z + x - x \\[5pt] a & p & x \end{array}\right|}
{= 2\left|\begin{array}{ccc} b & q & y \\[5pt] c & r & z \\[5pt] a & p & x \end{array}\right|}
Applying {\text{R}_1 ↔ \text{R}_3,} we have,
L.H.S. =
{-2\left|\begin{array}{ccc} a & p & x \\[5pt] c & r & z \\[5pt] b & q & y \end{array}\right|}
(∵ If any two rows (or columns) of a determinant are interchanged, then sign of determinant changes)
Applying {\text{R}_2 ↔ \text{R}_3,} we have,
L.H.S. =
{2\left|\begin{array}{ccc} a & p & x \\[5pt] b & q & y \\[5pt] c & r & z \end{array}\right|}
(∵ If any two rows (or columns) of a determinant are interchanged, then sign of determinant changes)
=
R.H.S.
∴ It is proved that {\left|\begin{array}{ccc} b + c & q + r & y + z \\[5pt] c + a & r + p & z + x \\[5pt] a + b & p + q & x + y \end{array}\right| = 2 \left|\begin{array}{ccc} a & p & x \\[5pt] b & q & y \\[5pt] c & r & z \end{array}\right|}
Using the property of determinants and without expanding in Exercises 1 to 7, prove that:
{6. \left|\begin{array}{ccc} 0 & a & -b \\[5pt] -a & 0 & -c \\[5pt] b & c & 0 \end{array}\right| = 0}
This can be solved in two ways.
Method 1:
By adding/subtracting to each row equimultiples of other rows (can also be done by adding/subtracting to each column equimultiples of other columns)
By using the property of the determinants that The value of the determinant remains unchanged if its rows and columns are interchanged.
Method 1:
We have,
{\text{L.H.S. = } \left|\begin{array}{ccc} 0 & a & -b \\[5pt] -a & 0 & -c \\[5pt] b & c & 0 \end{array}\right|}
Taking out the factor \dfrac1c from {\text{R}_1} and \dfrac1b from {\text{R}_2}, we have
{\text{L.H.S. = } \dfrac{1}{bc}\left|\begin{array}{ccc} 0 & ac & -bc \\[5pt] -ab & 0 & -bc \\[5pt] b & c & 0 \end{array}\right|}
Applying {\text{R}_1 → \text{R}_1 - \text{R}_2,} we have
L.H.S.
{= \dfrac{1}{bc}\left|\begin{array}{ccc} 0 - (-ab) & ac - 0 & -bc - (-bc) \\[5pt] -ab & 0 & -bc \\[5pt] b & c & 0 \end{array}\right|}
{= \dfrac{1}{bc}\left|\begin{array}{ccc} ab & ac & 0 \\[5pt] -ab & 0 & -bc \\[5pt] b & c & 0 \end{array}\right|}
Taking out the factor a from {\text{R}_1}
L.H.S.
{= \dfrac{a}{bc}\left|\begin{array}{ccc} b & c & 0 \\[5pt] -ab & 0 & -bc \\[5pt] b & c & 0 \end{array}\right|}
{= \dfrac{a}{bc}\left|\begin{array}{ccc} b & c & 0 \\[5pt] -ab & 0 & -bc \\[5pt] b & c & 0 \end{array}\right|}
= 0 (∵ {\text{R}_1} and {\text{R}_3} are identical.)
= R.H.S.
∴ It is proved that {\left|\begin{array}{ccc} 0 & a & -b \\[5pt] -a & 0 & -c \\[5pt] b & c & 0 \end{array}\right| = 0}
Method 2:
Let the given determinant be {Δ = \left|\begin{array}{ccc} 0 & a & -b \\[5pt] -a & 0 & -c \\[5pt] b & c & 0 \end{array}\right|}
Interchanging the rows and columns, we get,
{Δ = \left|\begin{array}{ccc} 0 & -a & b \\[5pt] a & 0 & c \\[5pt] -b & -c & 0 \end{array}\right|}
Taking out factor (-1) from each of the three rows, we have,
Δ
{= (-1)(-1)(-1)\left|\begin{array}{ccc} 0 & a & -b \\[5pt] -a & 0 & -c \\[5pt] b & c & 0 \end{array}\right|}
{= (-1)^3Δ}
= -Δ
⇒ Δ = -Δ
⇒ Δ + Δ = 0
⇒ 2Δ = 0
⇒ Δ = 0
∴ It is proved that {\left|\begin{array}{ccc} 0 & a & -b \\[5pt] -a & 0 & -c \\[5pt] b & c & 0 \end{array}\right| = 0}
Using the property of determinants and without expanding in Exercises 1 to 7, prove that:
{7. \left|\begin{array}{ccc} -a^2 & ab & ac \\[5pt] ba & -b^2 & bc \\[5pt] ca & cb & -c^2 \end{array}\right| = 4a^2b^2c^2}
We have,
{\text{L.H.S. = } \left|\begin{array}{ccc} -a^2 & ab & ac \\[5pt] ba & -b^2 & bc \\[5pt] ca & cb & -c^2 \end{array}\right|}
Taking out factors a from {\text{R}_1,} b from {\text{R}_2,} and c from {\text{R}_3,} we have,
{\text{L.H.S. = } abc\left|\begin{array}{ccc} -a & b & c \\[5pt] a & -b & c \\[5pt] a & b & -c \end{array}\right|}
Taking out factors a from {\text{C}_1,} b from {\text{C}_2} and c from {\text{C}_3,} we have,
L.H.S.
{= abc.a.b.c\left|\begin{array}{ccc} -1 & 1 & 1 \\[5pt] 1 & -1 & 1 \\[5pt] 1 & 1 & -1 \end{array}\right|}
{= a^2b^2c^2\left|\begin{array}{ccc} -1 & 1 & 1 \\[5pt] 1 & -1 & 1 \\[5pt] 1 & 1 & -1 \end{array}\right|}
Applying {\text{C}_1 → \text{C}_1 + \text{C}_2} and {\text{C}_2 → \text{C}_2 + \text{C}_3,} we have,
L.H.S.
{= a^2b^2c^2\left|\begin{array}{ccc} -1 + 1 & 1 + (-1) & 1 + 1 \\[5pt] 1 + 1 & -1 + 1 & 1 + (-1) \\[5pt] 1 & 1 & -1 \end{array}\right|}
{= a^2b^2c^2\left|\begin{array}{ccc} 0 & 0 & 2 \\[5pt] 2 & 0 & 0 \\[5pt] 1 & 1 & -1 \end{array}\right|}
Taking common factor 2 from {\text{R}_1} and {\text{R}_2,} we have,
L.H.S.
{= a^2b^2c^2 × 2 × 2\left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] 1 & 0 & 0 \\[5pt] 1 & 1 & -1 \end{array}\right|}
{= 4a^2b^2c^2\left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] 1 & 0 & 0 \\[5pt] 1 & 1 & -1 \end{array}\right|}
Expanding along {\text{R}_1,} we have,
L.H.S.
{= 4a^2b^2c^2 × \left(0 \left|\begin{array}{cc} 0 & 0 \\[5pt] 1 & -1 \end{array}\right| - 0 \left|\begin{array}{cc} 1 & 0 \\[5pt] 1 & -1 \end{array}\right| + 1 \left|\begin{array}{cc} 1 & 0 \\[5pt] 1 & 1 \end{array}\right|\right)}
{= 4a^2b^2c^2 × \big[0 - 0 + 1\big(1(1) - 0(0)\big)\big]}
{= 4a^2b^2c^2 × [1(1 - 0)]}
{= 4a^2b^2c^2 × (1 × 1)}
{= 4a^2b^2c^2}
= R.H.S.
∴ It is proved that {\left|\begin{array}{ccc} -a^2 & ab & ac \\[5pt] ba & -b^2 & bc \\[5pt] ca & cb & -c^2 \end{array}\right| = 4a^2b^2c^2}
By using properties of determinants, in Exercises 8 to 14, show that:
8. (i)
{\left|\begin{array}{ccc} 1 & a & a^2 \\[5pt] 1 & b & b^2 \\[5pt] 1 & c & c^2 \end{array}\right| = (a - b)(b - c)(c - a)}
(ii)
{\left|\begin{array}{ccc} 1 & 1 & 1 \\[5pt] a & b & c \\[5pt] a^3 & b^3 & c^3 \end{array}\right| = (a - b)(b -c)(c - a)(a + b + c)}
(i) To show by using the properties of determinants that {\left|\begin{array}{ccc} 1 & a & a^2 \\[5pt] 1 & b & b^2 \\[5pt] 1 & c & c^2 \end{array}\right| = (a - b)(b - c)(c - a)}
We have,
{\text{L.H.S. = } \left|\begin{array}{ccc} 1 & a & a^2 \\[5pt] 1 & b & b^2 \\[5pt] 1 & c & c^2 \end{array}\right|}
Applying {\text{R}_1 → \text{R}_1 - \text{R}_2} and {\text{R}_2 → \text{R}_2 - \text{R}_3,} we have,
L.H.S.
{= \left|\begin{array}{ccc} 1 - 1 & a - b & a^2 - b^2 \\[5pt] 1 - 1 & b - c & b^2 - c^2 \\[5pt] 1 & c & c^2 \end{array}\right|}
{= \left|\begin{array}{ccc} 0 & a - b & (a + b)(a - b) \\[5pt] 0 & b - c & (b + c)(b - c) \\[5pt] 1 & c & c^2 \end{array}\right|}
Taking out the factor {(a - b)} from {\text{R}_1} and {(b - c)} from {\text{R}_2,} we have
{\text{L.H.S.} = (a - b)(b - c)\left|\begin{array}{ccc} 0 & 1 & a + b \\[5pt] 0 & 1 & b + c \\[5pt] 1 & c & c^2 \end{array}\right|}
Expanding along {\text{C}_1,} we have
L.H.S.
{= (a - b)(b - c)\left( 0 \left|\begin{array}{cc} 1 & b + c \\[5pt] c & c^2 \end{array}\right| - 0 \left|\begin{array}{cc} 1 & a + b \\[5pt] c & c^2 \end{array}\right| + 1 \left|\begin{array}{cc} 1 & a + b \\[5pt] 1 & b + c \end{array}\right|\right)}
{= (a - b)(b - c)[0 - 0 + 1 × \left((b + c) - (a + b)\right)]}
{= (a - b)(b - c)(c - a)}
= R.H.S.
∴ It is proved that {\left|\begin{array}{ccc} 1 & a & a^2 \\[5pt] 1 & b & b^2 \\[5pt] 1 & c & c^2 \end{array}\right| = (a - b)(b - c)(c - a)}
(ii) To show by using the properties of determinants that {\left|\begin{array}{ccc} 1 & 1 & 1 \\[5pt] a & b & c \\[5pt] a^3 & b^3 & c^3 \end{array}\right| = (a - b)(b -c)(c - a)(a + b + c)}
We have,
{\text{L.H.S.} = \left|\begin{array}{ccc} 1 & 1 & 1 \\[5pt] a & b & c \\[5pt] a^3 & b^3 & c^3 \end{array}\right|}
Applying {\text{C}_1 → \text{C}_1 - \text{C}_2,} we have,
L.H.S. =
{\left|\begin{array}{ccc} 1 - 1 & 1 & 1 \\[5pt] a - b & b & c \\[5pt] a^3 - b^3 & b^3 & c^3 \end{array}\right|}
=
{\left|\begin{array}{ccc} 0 & 1 & 1 \\[5pt] a - b & b & c \\[5pt] (a - b)(a^2 + ab + b^2) & b^3 & c^3 \end{array}\right|}
{\left[∵ a^3 - b^3 = \left(a - b\right)\left(a^2 + ab + b^2\right)\right]}
Applying {\text{C}_2 → \text{C}_2 - \text{C}_3,} we have,
L.H.S. =
{\left|\begin{array}{ccc} 0 & 1 - 1 & 1 \\[5pt] a - b & b - c & c \\[5pt] (a - b)(a^2 + ab + b^2) & b^3 - c^3 & c^3 \end{array}\right|}
=
{\left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] a - b & b - c & c \\[5pt] (a - b)(a^2 + ab + b^2) & (b - c)(b^2 + bc + c^2) & c^3 \end{array}\right|}
{\left[∵ a^3 - b^3 = \left(a - b\right)\left(a^2 + ab + b^2\right)\right]}
Taking common factors (a - b) from {\text{C}_1} and (b - c) from {\text{C}_2,} we have,
{\text{L.H.S.} = (a - b)(b - c)\left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] 1 & 1 & c \\[5pt] a^2 + ab + b^2 & b^2 + bc + c^2 & c^3 \end{array}\right|}
Again applying {\text{C}_1 → \text{C}_1 - \text{C}_2,} we have,
L.H.S.
{= (a - b)(b - c)\left|\begin{array}{ccc} 0 - 0 & 0 & 1 \\[5pt] 1 - 1 & 1 & c \\[5pt] (a^2 + ab + b^2) - (b^2 + bc + c^2) & b^2 + bc + c^2 & c^3 \end{array}\right|}
{= (a - b)(b - c)\left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] 0 & 1 & c \\[5pt] a^2 + ab + b^2 - b^2 - bc - c^2 & b^2 + bc + c^2 & c^3 \end{array}\right|}
{= (a - b)(b - c)\left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] 0 & 1 & c \\[5pt] a^2 - c^2 + ab + - bc & b^2 + bc + c^2 & c^3 \end{array}\right|}
{= (a - b)(b - c)\left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] 0 & 1 & c \\[5pt] -(c^2 - a^2) - (bc - ab) & b^2 + bc + c^2 & c^3 \end{array}\right|}
{= (a - b)(b - c)\left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] 0 & 1 & c \\[5pt] -(c + a)(c - a) - b(c - a) & b^2 + bc + c^2 & c^3 \end{array}\right|}
Taking common factor (c - a) from {\text{C}_1,} we have,
L.H.S.
{= (a - b)(b - c)(c - a)\left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] 0 & 1 & c \\[5pt] -(c + a) - b & b^2 + bc + c^2 & c^3 \end{array}\right|}
{= (a - b)(b - c)(c - a)\left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] 0 & 1 & c \\[5pt] -(a + b + c) & b^2 + bc + c^2 & c^3 \end{array}\right|}
Expanding along {\text{R}_1,} we have,
L.H.S.
{= (a - b)(b - c)(c - a)\left(0 \left|\begin{array}{cc} 1 & c \\[5pt] b^2 + bc + c^2 & c^3 \end{array}\right| - 0 \left|\begin{array}{cc} 0 & c \\[5pt] -(a + b + c) & c^3 \end{array}\right| + 1 \left|\begin{array}{cc} 0 & 1 \\[5pt] -(a + b + c) & b^2 + bc + c^2 \end{array}\right|\right)}
{= (a - b)(b - c)(c - a)\left[0 + 0 + 1\Big(0(b^2 + bc + c^2) - 1\big(-(a + b + c)\big)\Big)\right]}
{= (a - b)(b - c)(c - a)(a + b + c)}
= R.H.S.
∴ It is proved that {\left|\begin{array}{ccc} 1 & 1 & 1 \\[5pt] a & b & c \\[5pt] a^3 & b^3 & c^3 \end{array}\right| = (a - b)(b -c)(c - a)(a + b + c)}
By using properties of determinants, in Exercises 8 to 14, show that:
{9. \left|\begin{array}{ccc} x & x^2 & yz \\[5pt] y & y^2 & zx \\[5pt] z & z^2 & xy \end{array}\right| = (x - y)(y - z)(z - x)(xy + yz + zx)}
We have,
{\text{L.H.S. = } \left|\begin{array}{ccc} x & x^2 & yz \\[5pt] y & y^2 & zx \\[5pt] z & z^2 & xy \end{array}\right|}
Applying {\text{R}_1 → \text{R}_1 - \text{R}_2,} we have,
L.H.S.
{= \left|\begin{array}{ccc} x - y & x^2 - y^2 & yz - zx \\[5pt] y & y^2 & zx \\[5pt] z & z^2 & xy \end{array}\right|}
{= \left|\begin{array}{ccc} x - y & (x + y)(x - y) & -z(x - y) \\[5pt] y & y^2 & zx \\[5pt] z & z^2 & xy \end{array}\right|}
Applying {\text{R}_2 → \text{R}_2 - \text{R}_3,} we have,
L.H.S.
{= \left|\begin{array}{ccc} x - y & (x + y)(x - y) & -z(x - y) \\[5pt] y - z & y^2 - z^2 & zx - xy \\[5pt] z & z^2 & xy \end{array}\right|}
{= \left|\begin{array}{ccc} x - y & (x + y)(x - y) & -z(x - y) \\[5pt] y - z & (y + z)(y - z) & -x(y - z) \\[5pt] z & z^2 & xy \end{array}\right|}
Taking out the common factor {(x - y)} from {\text{R}_1} and {y - z} from {\text{R}_2,} we have,
{\text{L.H.S. = } (x - y)(y - z)\left|\begin{array}{ccc} 1 & x + y & -z \\[5pt] 1 & y + z & -x \\[5pt] z & z^2 & xy \end{array}\right|}
Applying {\text{R}_1 → \text{R}_1 - \text{R}_2,} we have,
L.H.S.
{= (x - y)(y - z)\left|\begin{array}{ccc} 1 - 1 & (x + y) - (y + z) & -z - (-x) \\[5pt] 1 & y + z & -x \\[5pt] z & z^2 & xy \end{array}\right|}
{= (x - y)(y - z)\left|\begin{array}{ccc} 0 & x + y - y - z & -z + x \\[5pt] 1 & y + z & -x \\[5pt] z & z^2 & xy \end{array}\right|}
{= (x - y)(y - z)\left|\begin{array}{ccc} 0 & x - z & x - z \\[5pt] 1 & y + z & -x \\[5pt] z & z^2 & xy \end{array}\right|}
{= (x - y)(y - z)\left|\begin{array}{ccc} 0 & -(z - x) & -(z - x) \\[5pt] 1 & y + z & -x \\[5pt] z & z^2 & xy \end{array}\right|}
Taking out common factor {(z - x)} from {\text{R}_1,} we have,
{\text{L.H.S. = } (x - y)(y - z)(z - x)\left|\begin{array}{ccc} 0 & -1 & -1 \\[5pt] 1 & y + z & -x \\[5pt] z & z^2 & xy \end{array}\right|}
Applying {\text{C}_2 → \text{C}_2 - \text{C}_3,} we have,
L.H.S.
{= (x - y)(y - z)(z - x)\left|\begin{array}{ccc} 0 & -1 - (-1) & -1 \\[5pt] 1 & y + z - (-x) & -x \\[5pt] z & z^2 - xy & xy \end{array}\right|}
{= (x - y)(y - z)(z - x)\left|\begin{array}{ccc} 0 & 0 & -1 \\[5pt] 1 & x + y + z & -x \\[5pt] z & z^2 - xy & xy \end{array}\right|} ———-❶
Expanding {\left|\begin{array}{ccc} 0 & 0 & -1 \\[5pt] 1 & x + y + z & -x \\[5pt] z & z^2 - xy & xy \end{array}\right|} along {\text{R}_1,} we have,
{\left|\begin{array}{ccc} 0 & 0 & -1 \\[5pt] 1 & x + y + z & -x \\[5pt] z & z^2 - xy & xy \end{array}\right|}
{= 0 \left|\begin{array}{cc} x + y + z & -x \\[5pt] z^2 - xy & xy \end{array}\right| - 0 \left|\begin{array}{cc} 1 & -x \\[5pt] z & xy \end{array}\right| + (-1) \left|\begin{array}{cc} 1 & x + y + z \\[5pt] z & z^2 - xy \end{array}\right|}
{= 0 - 0 - 1 \left[\big(1(z^2 - xy)\big) - \big((x + y + z)z\big)\right]}
{= - 1(z^2 - xy - zx - yz - z^2)}
{= xy + yz + zx} ———-❷
Substituting ❷ in ❶, we have,
L.H.S.
{= (x - y)(y - z)(z - x)(x + y + z)}
= R.H.S.
∴ It is showed that {\left|\begin{array}{ccc} x & x^2 & yz \\[5pt] y & y^2 & zx \\[5pt] z & z^2 & xy \end{array}\right| = (x - y)(y - z)(z - x)(xy + yz + zx)}
By using properties of determinants, in Exercises 8 to 14, show that:
10. (i)
{\left|\begin{array}{ccc} x + 4 & 2x & 2x \\[5pt] 2x & x + 4 & 2x \\[5pt] 2x & 2x & x + 4 \end{array}\right| = (5x + 4)(4 - x)^2}
(ii)
{\left|\begin{array}{ccc} y + k & y & y \\[5pt] y & y + k & y \\[5pt] y & y & y + k \end{array}\right| = k^2(3y + k)}
(i) To show that {\left|\begin{array}{ccc} x + 4 & 2x & 2x \\[5pt] 2x & x + 4 & 2x \\[5pt] 2x & 2x & x + 4 \end{array}\right|}
We have,
{\text{L.H.S. = } \left|\begin{array}{ccc} x + 4 & 2x & 2x \\[5pt] 2x & x + 4 & 2x \\[5pt] 2x & 2x & x + 4 \end{array}\right|}
Applying {\text{R}_1 → \text{R}_1 + \text{R}_2 + \text{R}_3,} we have,
L.H.S.
{= \left|\begin{array}{ccc} x + 4 + 2x + 2x & 2x + (x + 4) + 2x & 2x + 2x + (x + 4) \\[5pt] 2x & x + 4 & 2x \\[5pt] 2x & 2x & x + 4 \end{array}\right|}
{= \left|\begin{array}{ccc} 5x + 4 & 5x + 4 & 5x + 4 \\[5pt] 2x & x + 4 & 2x \\[5pt] 2x & 2x & x + 4 \end{array}\right|}
Taking out the common factor {(5x + 4)} from {\text{R}_1,} we have,
{\text{L.H.S. = } (5x + 4) \left|\begin{array}{ccc} 1 & 1 & 1 \\[5pt] 2x & x + 4 & 2x \\[5pt] 2x & 2x & x + 4 \end{array}\right|}
Applying {\text{C}_3 → \text{C}_3 - \text{C}_2,} we have,
L.H.S.
{= (5x + 4)\left|\begin{array}{ccc} 1 & 1 & 1 - 1 \\[5pt] 2x & x + 4 & 2x - (x + 4) \\[5pt] 2x & 2x & x + 4 - 2x \end{array}\right|}
{= (5x + 4)\left|\begin{array}{ccc} 1 & 1 & 0 \\[5pt] 2x & x + 4 & x - 4 \\[5pt] 2x & 2x & 4 - x \end{array}\right|}
Applying {\text{C}_2 → \text{C}_2 - \text{C}_1,} we have,
L.H.S.
{= (5x + 4)\left|\begin{array}{ccc} 1 & 1 - 1 & 0 \\[5pt] 2x & x + 4 - 2x & x - 4 \\[5pt] 2x & 2x - 2x & x - x \end{array}\right|}
{= (5x + 4)\left|\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 2x & 4 - x & x - 4 \\[5pt] 2x & 0 & 4 - x \end{array}\right|} ———-❶
Expanding {\left|\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 2x & 4 - x & x - 4 \\[5pt] 2x & 0 & 4 - x \end{array}\right|} along {\text{R}_1,} we have,
{\left|\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 2x & 4 - x & x - 4 \\[5pt] 2x & 0 & 4 - x \end{array}\right|}
{= 1 \left|\begin{array}{cc} 4 - x & x - 4 \\[5pt] 0 & 4 - x \end{array}\right| - 0 \left|\begin{array}{cc} 2x & x - 4 \\[5pt] 2x & 4 - x \end{array}\right| + 0 \left|\begin{array}{cc} 2x & 4 - x \\[5pt] 2x & 0 \end{array}\right|}
{= 1[(4 -x)(4 - x) - (x - 4)0] - 0 + 0}
{= (4 -x)^2} ———-❷
Substituting ❷ in ❶, we have,
L.H.S.
{= (5x + 4)(4 - x)^2}
= R.H.S.
∴ It is showed that {\left|\begin{array}{ccc} x + 4 & 2x & 2x \\[5pt] 2x & x + 4 & 2x \\[5pt] 2x & 2x & x + 4 \end{array}\right| = (5x + 4)(4 - x)^2}
(ii) To show that {\left|\begin{array}{ccc} y + x & y & y \\[5pt] y & y + k & y \\[5pt] y & y & y + k \end{array}\right| = k^2(3y + k)}
We have,
{\text{L.H.S. = } \left|\begin{array}{ccc} y + x & y & y \\[5pt] y & y + k & y \\[5pt] y & y & y + k \end{array}\right|}
(ii) To show that {\left|\begin{array}{ccc} y + k & y & y \\[5pt] y & y + k & y \\[5pt] y & y & y + k \end{array}\right| = k^2(3y + k)}
We have,
{\text{L.H.S. = } \left|\begin{array}{ccc} y + k & y & y \\[5pt] y & y + k & y \\[5pt] y & y & y + k \end{array}\right|}
Applying {\text{R}_1 → \text{R}_1 - \text{R}_2,} we have,
L.H.S
{= \left|\begin{array}{ccc} y + k - y & y - (y + k) & y - y \\[5pt] y & y + k & y \\[5pt] y & y & y + k \end{array}\right|}
{= \left|\begin{array}{ccc} k & -k & 0 \\[5pt] y & y + k & y \\[5pt] y & y & y + k \end{array}\right|}
Applying {\text{R}_2 → \text{R}_2 - \text{R}_3,} we have,
L.H.S
{= \left|\begin{array}{ccc} k & -k & 0 \\[5pt] y - y & y + k - y & y - (y + k) \\[5pt] y & y & y + k \end{array}\right|}
{= \left|\begin{array}{ccc} k & -k & 0 \\[5pt] 0 & k & -k \\[5pt] y & y & y + k \end{array}\right|}
Taking out common factor from {\text{R}_1} and {\text{R}_2,} we have,
L.H.S.
{= k.k.\left|\begin{array}{ccc} 1 & -1 & 0 \\[5pt] 0 & 1 & -1 \\[5pt] y & y & y + k \end{array}\right|}
{= k^2 \left|\begin{array}{ccc} 1 & -1 & 0 \\[5pt] 0 & 1 & -1 \\[5pt] y & y & y + k \end{array}\right|} ———-❶
Now expanding {\left|\begin{array}{ccc} 1 & -1 & 0 \\[5pt] 0 & 1 & -1 \\[5pt] y & y & y + k \end{array}\right|} along {\text{C}_1,} we have,
{\left|\begin{array}{ccc} 1 & -1 & 0 \\[5pt] 0 & 1 & -1 \\[5pt] y & y & y + k \end{array}\right|}
{= 1 \left|\begin{array}{cc} 1 & -1 \\[5pt] y & y + k \end{array}\right| - 0 \left|\begin{array}{cc} -1 & 0 \\[5pt] y & y + k \end{array}\right| + y \left|\begin{array}{cc} -1 & 0 \\[5pt] 1 & -1 \end{array}\right|}
{= 1[1(y + k) - (-1)y]} {- 0} {+ y[(-1)(-1) - 0(1)]}
{= y + k + y + y}
{= 3y + k} ———-❷
Substituting ❷ in ❶, we have,
L.H.S.
{= k^2 (3y + k)}
= R.H.S.
∴ It is showed that {\left|\begin{array}{ccc} y + k & y & y \\[5pt] y & y + k & y \\[5pt] y & y & y + k \end{array}\right| = k^2(3y + k)}
By using properties of determinants, in Exercises 8 to 14, show that:
11. (i)
{\left|\begin{array}{ccc} a - b - c & 2a & 2a \\[5pt] 2b & b - c - a & 2b \\[5pt] 2c & 2c & c - b - a \end{array}\right| = (a + b + c)^3}
(ii)
{\left|\begin{array}{ccc} x + y + 2z & x & y \\[5pt] z & y + z + 2x & y \\[5pt] z & x & z + x + 2y \end{array}\right| = 2(x + y + z)^3}
(i) To show that {\left|\begin{array}{ccc} a - b - c & 2a & 2a \\[5pt] 2b & b - c - a & 2b \\[5pt] 2c & 2c & c - b - a \end{array}\right| = (a + b + c)^3}
We have,
{\text{L.H.S. = } \left|\begin{array}{ccc} a - b - c & 2a & 2a \\[5pt] 2b & b - c - a & 2b \\[5pt] 2c & 2c & c - b - a \end{array}\right|}
Applying {\text{R}_1 → \text{R}_1 + \text{R}_2 + \text{R}_3}
L.H.S.
{= \left|\begin{array}{ccc} a - b - c + 2b + 2c & 2a + (b - c - a) + 2c & 2a + 2b + (c - b - c) \\[5pt] 2b & b - c - a & 2b \\[5pt] 2c & 2c & c - b - a \end{array}\right|}
{= \left|\begin{array}{ccc} a + b + c & a + b + c & a + b + c \\[5pt] 2b & b - c - a & 2b \\[5pt] 2c & 2c & c - b - a \end{array}\right|}
Taking common factor {(a + b + c)} from {\text{R}_1,} we have,
{\text{L.H.S. = } (a + b + c) \left|\begin{array}{ccc} 1 & 1 & 1 \\[5pt] 2b & b - c - a & 2b \\[5pt] 2c & 2c & c - b - a \end{array}\right|}
Applying {\text{C}_1 → \text{C}_1 - \text{C}_2,} we have,
L.H.S.
{= (a + b + c) \left|\begin{array}{ccc} 1 - 1 & 1 & 1 \\[5pt] 2b - (b - c - a) & b - c - a & 2b \\[5pt] 2c - 2c & 2c & c - b - a \end{array}\right|}
{= (a + b + c) \left|\begin{array}{ccc} 0 & 1 & 1 \\[5pt] a + b + c & b - c - a & 2b \\[5pt] 0 & 2c & c - b - a \end{array}\right|}
Applying {\text{C}_2 → \text{C}_2 - \text{C}_3,} we have,
L.H.S.
{= (a + b + c) \left|\begin{array}{ccc} 0 & 1 - 1 & 1 \\[5pt] a + b + c & b - c - a - 2b & 2b \\[5pt] 0 & 2c - (c - b - a) & c - b - a \end{array}\right|}
{= (a + b + c) \left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] a + b + c & -(a + b + c) & 2b \\[5pt] 0 & a + b + c & c - b - a \end{array}\right|}
Taking out common factor {(a + b + c)} from both {\text{C}_1} and {\text{C}_2,} we have,
{\text{L.H.S. = } (a + b + c)^3 \left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] 1 & -1 & 2b \\[5pt] 0 & 1 & c - b - a \end{array}\right|} ———-❶
Now expanding {\left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] 1 & -1 & 2b \\[5pt] 0 & 1 & c - b - a \end{array}\right|} along {\text{R}_1,} we have,
{\left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] 1 & -1 & 2b \\[5pt] 0 & 1 & c - b - a \end{array}\right|}
{= 0 \left|\begin{array}{cc} -1 & 2b \\[5pt] 1 & a - b - c \end{array}\right| - 0 \left|\begin{array}{cc} 1 & 2b \\[5pt] 0 & a - b - c \end{array}\right| + 1 \left|\begin{array}{cc} 1 & -1 \\[5pt] 0 & 1 \end{array}\right|}
= 0 – 0 + 1[1(0) – (-1)1]
= 1(0 + 1)
= 1 ———-❷
Substituting ❷ in ❶, we have,
L.H.S.
{= (a + b + c)^3(1)}
{= (a + b + c)^3}
= R.H.S.
∴ It is showed that {\left|\begin{array}{ccc} a - b - c & 2a & 2a \\[5pt] 2b & b - c - a & 2b \\[5pt] 2c & 2c & c - b - a \end{array}\right| = (a + b + c)^3}
(ii) To show that {\left|\begin{array}{ccc} x + y + 2z & x & y \\[5pt] z & y + z + 2x & y \\[5pt] z & x & z + x + 2y \end{array}\right| = 2(x + y + z)^3}
We have,
{\text{L.H.S. = } \left|\begin{array}{ccc} x + y + 2z & x & y \\[5pt] z & y + z + 2x & y \\[5pt] z & x & z + x + 2y \end{array}\right|}
Applying {\text{C}_1 → \text{C}_1 + \text{C}_2 + \text{C}_3,} we have,
L.H.S.
{= \left|\begin{array}{ccc} x + y + 2z + x + y & x & y \\[5pt] z + (y + z + 2x) + y & y + z + 2x & y \\[5pt] z + x + (z + x + 2y) & x & z + x + 2y \end{array}\right|}
{= \left|\begin{array}{ccc} 2(x + y + z) & x & y \\[5pt] 2(x + y + z) & y + z + 2x & y \\[5pt] 2(x + y + z) & x & z + x + 2y \end{array}\right|}
Taking out common factor {2(x + y + z)} from {\text{C}_1,} we have,
{\text{L.H.S. = } 2(x + y + z) \left|\begin{array}{ccc} 1 & x & y \\[5pt] 1 & y + z + 2x & y \\[5pt] 1 & x & z + x + 2y \end{array}\right|}
Applying {\text{R}_1 → \text{R}_1 - \text{R}_2,} we have,
L.H.S.
{= 2(x + y + z) \left|\begin{array}{ccc} 1 - 1 & x - (y + z + 2x) & y - y \\[5pt] 1 & y + z + 2x & y \\[5pt] 1 & x & z + x + 2y \end{array}\right|}
{= 2(x + y + z) \left|\begin{array}{ccc} 0 & -(x + y + z) & 0 \\[5pt] 1 & y + z + 2x & y \\[5pt] 1 & x & z + x + 2y \end{array}\right|}
Applying {\text{R}_2 → \text{R}_2 - \text{R}_3,} we have,
L.H.S.
{= 2(x + y + z) \left|\begin{array}{ccc} 0 & -(x + y + z) & 0 \\[5pt] 1 - 1 & y + z + 2x - x & y - (z + x + 2y) \\[5pt] 1 & x & z + x + 2y \end{array}\right|}
{= 2(x + y + z) \left|\begin{array}{ccc} 0 & -(x + y + z) & 0 \\[5pt] 0 & x + y + z & -(x + y + z) \\[5pt] 1 & x & z + x + 2y \end{array}\right|}
Taking out common factor {(x + y + z)} from {\text{R}_1} and {\text{R}_2,} we have,
{\text{L.H.S. = } 2(x + y + z)^3 \left|\begin{array}{ccc} 0 & -1 & 0 \\[5pt] 0 & 1 & -1 \\[5pt] 1 & x & z + x + 2y \end{array}\right|} ———-❶
Expanding {\left|\begin{array}{ccc} 0 & -1 & 0 \\[5pt] 0 & 1 & -1 \\[5pt] 1 & x & z + x + 2y \end{array}\right|} along {\text{C}_1,} we have,
{\left|\begin{array}{ccc} 0 & -1 & 0 \\[5pt] 0 & 1 & -1 \\[5pt] 1 & x & z + x + 2y \end{array}\right|}
{= 0 \left|\begin{array}{cc} 1 & -1 \\[5pt] x & z + x + 2y \end{array}\right| - 0 \left|\begin{array}{cc} -1 & 0 \\[5pt] x & z + x + 2y \end{array}\right| + 1 \left|\begin{array}{cc} -1 & 0 \\[5pt] 1 & -1 \end{array}\right|}
= 0 – 0 + 1[(-1)(-1) – 0(1)]
= 1(1 – 0)
= 1(1)
= 1 ———-❷
Substituting ❷ in ❶, we have,
L.H.S.
{= 2(x + y + z)^3 × 1}
{= 2(x + y + z)^3}
= R.H.S.
∴ It is showed that {\left|\begin{array}{ccc} x + y + 2z & x & y \\[5pt] z & y + z + 2x & y \\[5pt] z & x & z + x + 2y \end{array}\right| = 2(x + y + z)^3}
By using properties of determinants, in Exercises 8 to 14, show that:
12. {\left|\begin{array}{ccc} 1 & x & x^2 \\[5pt] x^2 & 1 & x \\[5pt] x & x^2 & 1 \end{array}\right| = (1 - x^3)^2}
We have,
{\text{L.H.S. = } \left|\begin{array}{ccc} 1 & x & x^2 \\[5pt] x^2 & 1 & x \\[5pt] x & x^2 & 1 \end{array}\right|}
Applying {\text{R}_1 → \text{R}_1 + \text{R}_2 + \text{R}_3,} we have,
L.H.S.
{= \left|\begin{array}{ccc} 1 + x + x^2 & x + 1 + x^2 & x^2 + x + 1 \\[5pt] x^2 & 1 & x \\[5pt] x & x^2 & 1 \end{array}\right|}
{= \left|\begin{array}{ccc} 1 + x + x^2 & 1 + x + x^2 & 1 + x + x^2 \\[5pt] x^2 & 1 & x \\[5pt] x & x^2 & 1 \end{array}\right|}
Taking common factor {(1 + x + x^2)} from {\text{R}_1,} we have,
{\text{L.H.S. = } (1 + x + x^2) \left|\begin{array}{ccc} 1 & 1 & 1 \\[5pt] x^2 & 1 & x \\[5pt] x & x^2 & 1 \end{array}\right|}
Applying {\text{C}_1 → \text{C}_1 - \text{C}_2,} we have,
L.H.S.
{= (1 + x + x^2) \left|\begin{array}{ccc} 1 - 1 & 1 & 1 \\[5pt] x^2 - 1 & 1 & x \\[5pt] x - x^2 & x^2 & 1 \end{array}\right|}
{= (1 + x + x^2) \left|\begin{array}{ccc} 0 & 1 & 1 \\[5pt] -(1 - x)(1 + x) & 1 & x \\[5pt] x(1 - x) & x^2 & 1 \end{array}\right|}
Applying {\text{C}_2 → \text{C}_2 - \text{C}_3,} we have,
L.H.S.
{= (1 + x + x^2) \left|\begin{array}{ccc} 0 & 1 - 1 & 1 \\[5pt] -(1 - x)(1 + x) & 1 - x & x \\[5pt] x(1 - x) & x^2 - 1 & 1 \end{array}\right|}
{= (1 + x + x^2) \left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] -(1 - x)(1 + x) & 1 - x & x \\[5pt] x(1 - x) & -(1 - x)(1 + x) & 1 \end{array}\right|}
Taking common factor {(1 - x)} from both {\text{C}_1} as well as {\text{C}_2,} we have,
{\text{L.H.S. = } (1 + x + x^2) (1 - x)^2\left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] -(1 + x) & 1 & x \\[5pt] x & -(1 + x) & 1 \end{array}\right|} ———-❶
Expanding {\left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] -(1 + x) & 1 & x \\[5pt] x & -(1 + x) & 1 \end{array}\right|} along {\text{R}_1,} we have,
{\left|\begin{array}{ccc} 0 & 0 & 1 \\[5pt] -(1 + x) & 1 & x \\[5pt] x & -(1 + x) & 1 \end{array}\right|}
{= 0 \left|\begin{array}{cc} 1 & x \\[5pt] -(1 + x) & 1 \end{array}\right| - 0 \left|\begin{array}{cc} -(1 + x) & x \\[5pt] x & 1 \end{array}\right| + 1 \left|\begin{array}{cc} -(1 + x) & 1 \\[5pt] x & -(1 + x) \end{array}\right|}
{= 0 - 0 + 1\left[-(1 + x) × -(1 + x) - 1(x)\right]}
{= 1 × \left[(1 + x)^2 - x\right]}
{= 1 + 2x + x^2 - x}
{= 1 + x + x^2} ———-❷
Substituting ❷ in ❶, we have,
L.H.S.
{= (1 + x + x^2) (1 - x)^2(1 + x + x^2)}
{= \left[(1 + x + x^2)(1 - x)\right]^2}
{= \left(1 - x^3\right)^2} {\left(∵ (1 + x + x^2)(1 - x) = 1 - x^3\right)}
=R.H.S.
∴ It is showed that {\left|\begin{array}{ccc} 1 & x & x^2 \\[5pt] x^2 & 1 & x \\[5pt] x & x^2 & 1 \end{array}\right| = (1 - x^3)^2}
By using properties of determinants, in Exercises 8 to 14, show that:
13. {\left|\begin{array}{ccc} 1 + a^2 - b^2 & 2ab & -2b \\[5pt] 2ab & 1 - a^2 + b^2 & 2a \\[5pt] 2b & -2a & 1 - a^2 - b^2 \end{array}\right| = \left(1 + a^2 + b^2\right)^3}
We have,
{\text{L.H.S. = } \left|\begin{array}{ccc} 1 + a^2 - b^2 & 2ab & -2b \\[5pt] 2ab & 1 - a^2 + b^2 & 2a \\[5pt] 2b & -2a & 1 - a^2 - b^2 \end{array}\right|}
Applying {\text{R}_1 → \text{R}_1 + b\text{R}_3} and {\text{R}_2 → \text{R}_2 - a\text{R}_3,} we have,
L.H.S.
{= \left|\begin{array}{ccc} 1 + a^2 - b^2 + 2b^2 & 2ab + (-2ab) & -2b + b(1 - a^2 - b^2) \\[5pt] 2ab - 2ab & 1 - a^2 + b^2 - (-2a^2) & 2a - a(1 - a^2 - b^2) \\[5pt] 2b & -2a & 1 - a^2 - b^2 \end{array}\right|}
{= \left|\begin{array}{ccc} 1 + a^2 + b^2 & 0 & -2b + b - a^2b - b^3 \\[5pt] 0 & 1 - a^2 + b^2 + 2a^2 & 2a - a + a^3 + ab^2 \\[5pt] 2b & -2a & 1 - a^2 - b^2 \end{array}\right|}
{= \left|\begin{array}{ccc} 1 + a^2 + b^2 & 0 & -b - a^2b - b^3 \\[5pt] 0 & 1 + a^2 + b^2 & a + a^3 + ab^2 \\[5pt] 2b & -2a & 1 - a^2 - b^2 \end{array}\right|}
{= \left|\begin{array}{ccc} 1 + a^2 + b^2 & 0 & -b(1 + a^2 + b^2) \\[5pt] 0 & 1 + a^2 + b^2 & a(1 + a^2 + b^2) \\[5pt] 2b & -2a & 1 - a^2 - b^2 \end{array}\right|}
Taking common factor {(1 + a^2 + b^2)} from both {\text{R}_1} as well as {\text{R}_2,} we have,
{\text{L.H.S. = } \left|\begin{array}{ccc} 1 & 0 & -b \\[5pt] 0 & 1 & a \\[5pt] 2b & -2a & 1 - a^2 - b^2 \end{array}\right|} ———-❶
Expanding {\left|\begin{array}{ccc} 1 & 0 & -b \\[5pt] 0 & 1 & a \\[5pt] 2b & -2a & 1 - a^2 - b^2 \end{array}\right|} along {\text{R}_1,} we have,
{\left|\begin{array}{ccc} 1 & 0 & -b \\[5pt] 0 & 1 & a \\[5pt] 2b & -2a & 1 - a^2 - b^2 \end{array}\right|}
{= 1 \left|\begin{array}{cc} 1 & a \\[5pt] -2a & 1 - a^2 - b^2 \end{array}\right| - 0 \left|\begin{array}{cc} 0 & a \\[5pt] 2b & 1 - a^2 - b^2 \end{array}\right| + (-b) \left|\begin{array}{cc} 0 & 1 \\[5pt] 2b & -2a \end{array}\right|}
{= 1\left[1(1 - a^2 - b^2) - a(-2a)\right] - 0 - b[0(-2a) - 1(2b)]}
{= 1(1 - a^2 - b^2 + 2a^2) - b(0 - 2b)}
{= 1 + a^2 - b^2 + 2b^2}
{= 1 + a^2 + b^2} ———-❷
Substituting ❷ in ❶, we have,
L.H.S.
{= (1 + a^2 + b^2)^2(1 + a^2 + b^2)}
{= (1 + a^2 + b^2)^3}
= R.H.S.
∴ It is showed that {\left|\begin{array}{ccc} 1 + a^2 - b^2 & 2ab & -2b \\[5pt] 2ab & 1 - a^2 + b^2 & 2a \\[5pt] 2b & -2a & 1 - a^2 - b^2 \end{array}\right| = \left(1 + a^2 + b^2\right)^3}
By using properties of determinants, in Exercises 8 to 14, show that:
14. {\left|\begin{array}{ccc} a^2 + 1 & ab & ac \\[5pt] ab & b^2 + 1 & bc \\[5pt] ca & cb & c^2 + 1 \end{array}\right| = 1 + a^2 + b^2 + c^2}
We have,
{\text{L.H.S. = } \left|\begin{array}{ccc} a^2 + 1 & ab & ac \\[5pt] ab & b^2 + 1 & bc \\[5pt] ca & cb & c^2 + 1 \end{array}\right|}
Applying the following three transformations,
Multiplying and dividing {\text{R}_1} with a
Multiplying and dividing {\text{R}_2} with b
Multiplying and dividing {\text{R}_3} with c
We have,
{\text{L.H.S. = } \left|\begin{array}{ccc} \dfrac1a × a × (a^2 + 1) & \dfrac1a × a × ab & \dfrac1a × a × ac \\[10pt] \dfrac1b × b × ab & \dfrac1b × b × (b^2 + 1) & \dfrac1b × b × bc \\[10pt] \dfrac1c × c × ca & \dfrac1c × c × cb & \dfrac1c × c × (c^2 + 1) \end{array}\right|}
Taking common factor
\dfrac1a from {\text{R}_1}
\dfrac1b from {\text{R}_2}
\dfrac1c from {\text{R}_3}
we have,
{\text{L.H.S. = } \dfrac1a × \dfrac1b × \dfrac1c \left|\begin{array}{ccc} a × (a^2 + 1) & a × ab & a × ac \\[5pt] b × ab & b × (b^2 + 1) & b × bc \\[5pt] c × ca & c × cb & c × (c^2 + 1) \end{array}\right|}
Taking common factor
a from {\text{C}_1}
b from {\text{C}_2}
c from {\text{C}_3}
we have,
{\text{L.H.S. = } \dfrac1{abc} × a × b × c × \left|\begin{array}{ccc} a^2 + 1 & a^2 & a^2 \\[5pt] b^2 & b^2 + 1 & b^2 \\[5pt] c^2 & c^2 & c^2 + 1 \end{array}\right|}
Applying {\text{R}_1 → \text{R}_1 + \text{R}_2 + \text{R}_3,} we have,
L.H.S.
{= \left|\begin{array}{ccc} a^2 + 1 + b^2 + c^2 & a^2 + b^2 + 1 + c^2 & a^2 + b^2 + c^2 + 1 \\[5pt] b^2 & b^2 + 1 & b^2 \\[5pt] c^2 & c^2 & c^2 + 1 \end{array}\right|}
{= \left|\begin{array}{ccc} 1 + a^2 + b^2 + c^2 & 1 + a^2 + b^2 + c^2 & 1 + a^2 + b^2 + c^2 \\[5pt] b^2 & b^2 + 1 & b^2 \\[5pt] c^2 & c^2 & c^2 + 1 \end{array}\right|}
Taking common factor {1 + a^2 + b^2 + c^2} from {\text{R}_1,} we have,
{\text{L.H.S. = } (1 + a^2 + b^2 + c^2) \left|\begin{array}{ccc} 1 & 1 & 1 \\[5pt] b^2 & b^2 + 1 & b^2 \\[5pt] c^2 & c^2 & c^2 + 1 \end{array}\right|}
Applying {\text{C}_2 → \text{C}_2 - \text{C}_1} and {\text{C}_3 → \text{C}_3 - \text{C}_1,} we have,
L.H.S.
{= (1 + a^2 + b^2 + c^2) \left|\begin{array}{ccc} 1 & 1 - 1 & 1 - 1 \\[5pt] b^2 & b^2 + 1 - b^2 & b^2 - b^2 \\[5pt] c^2 & c^2 - c^2 & c^2 + 1 - c^2 \end{array}\right|}
{= (1 + a^2 + b^2 + c^2) \left|\begin{array}{ccc} 1 & 0 & 0 \\[5pt] b^2 & 1 & 0 \\[5pt] c^2 & 0 & 1 \end{array}\right|} ———-❶
Expanding {\left|\begin{array}{ccc} 1 & 0 & 0 \\[5pt] b^2 & 1 & 0 \\[5pt] c^2 & 0 & 1 \end{array}\right|} along {\text{R}_1,} we have,
{\left|\begin{array}{ccc} 1 & 0 & 0 \\[5pt] b^2 & 1 & 0 \\[5pt] c^2 & 0 & 1 \end{array}\right|}
{= 1 \left|\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right| - 0 \left|\begin{array}{cc} b^2 & 0 \\[5pt] c^2 & 1 \end{array}\right| + 0 \left|\begin{array}{cc} b^2 & 1 \\[5pt] c^2 & 0 \end{array}\right|}
= 1[1(1) – 0(0)] – 0 + 0
= 1 ———-❷
Substituting ❷ in ❶, we have
L.H.S.
{= (1 + a^2 + b^2 + c^2) × 1}
{= 1 + a^2 + b^2 + c^2}
= R.H.S.
∴ It is showed that {\left|\begin{array}{ccc} a^2 + 1 & ab & ac \\[5pt] ab & b^2 + 1 & bc \\[5pt] ca & cb & c^2 + 1 \end{array}\right| = 1 + a^2 + b^2 + c^2}
Choose the correct answer in Exercises 15 and 16.
15. Let A be a square matrix of order 3 × 3, then |kA| is equal to
(A) k|A|
(B) k^2|A|
(C) k^3|A| ✔
(D) 3k|A|
If A is a square matrix of order n then we can take common factor k from each rows. As the square matrix A has n rows, we can take out the common factor k from each of the n rows. Thus, we have,
{|k\text{A}|}
{= k × k × k ....... n} times |A|
= k^n × |A|
Thus in a 3 × 3 matrix, we’ll have |kA| = k^3|A|.
So, option C is the correct answer.
Alternatively, consider a 3 × 3 square matrix.
For example, consider a square matrix
{\text{A} = \left[\begin{array}{cc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right]}
kA
{= k\left[\begin{array}{cc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right]}
{= \left[\begin{array}{cc} ka_1 & kb_1 & kc_1 \\ ka_2 & kb_2 & kc_2 \\ ka_3 & kb_3 & kc_3 \end{array}\right]}
Now,
∴ |kA|
{= \left|\begin{array}{cc} ka_1 & kb_1 & kc_1 \\ ka_2 & kb_2 & kc_2 \\ ka_3 & kb_3 & kc_3 \end{array}\right|}
Taking common factor k from {\text{R}_1,} {\text{R}_2} and {\text{R}_3,} we have,
|kA|
{= k × k × k × \left|\begin{array}{cc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|}
{= k^3 × \left|\begin{array}{cc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|}
= {k^3}|A|
Choose the correct answer in Exercises 15 and 16.
16. Which of the following is correct
(A) Determinant is a square matrix.
(B) Determinant is a number associated to a matrix.
(C) Determinant is a number associated to a square matrix. ✔
(D) None of these
The following is the definition of Determinant.
Determinant: To every square matrix {\text{A} = \left[a_{ij}\right]} of order {n,} we can associate a number (real or complex) called determinant of the square matrix A, where {a_{ij} = (i, j)^\text{th}} element of A.
Thus, the concept of determinants is applicable only to square matrices.
So, option C is the correct answer.