Algebraic variable elements MCQ Quiz - Objective Question with Answer for Algebraic variable elements - Download Free PDF

Calculation:

Given:

x² + y² + z² = 1

then, $\left|\begin{array}{ccc}1& \mathrm{z}& -\mathrm{y}\\ -\mathrm{z}& 1& \mathrm{x}\\ \mathrm{y}& -\mathrm{x}& 1\end{array}\right|$

Expanding along R₁, we get-

1(1 + x²) + z(xy + z) − y(xz − y)

1(1 + x2) - z(-z - xy) − y(xz − y)

= 1 + x² + xyz + z² − xyz + y²

= 1 + x² + y² + z²

= 1 + 1   (∵ x2 + y2 + z2 = 1 (given))

= 2

The correct answer is option "3"

Explanation:

$\left|\begin{array}{ccc}{\log }_{x}x& {\log }_{x}y& {\log }_{x}z\\ {\log }_{y}x& {\log }_{y}y& {\log }_{y}z\\ {\log }_{z}x& {\log }_{z}y& {\log }_{z}z\end{array}\right|$

= $\left|\begin{array}{ccc}1& \frac{{\log }_{e}y}{{\log }_{e}x}& \frac{{\log }_{e}z}{{\log }_{e}x}\\ \frac{{\log }_{e}x}{{\log }_{e}y}& 1& \frac{{\log }_{e}z}{{\log }_{e}y}\\ \frac{{\log }_{e}x}{{\log }_{e}z}& \frac{{\log }_{e}y}{{\log }_{e}z}& 1\end{array}\right|$

= 1(1 - 1) + $\frac{{\log }_{e}y}{{\log }_{e}x}(\frac{{\log }_{e}x}{{\log }_{e}y}-\frac{{\log }_{e}x}{{\log }_{e}y})$ + $\frac{{\log }_{e}z}{{\log }_{e}x}(\frac{{\log }_{e}x}{{\log }_{e}z}-\frac{{\log }_{e}x}{{\log }_{e}z})$

= 0 + 0 + 0 

= 0

Option (4) is true.

Calculation:

Given:

x² + y² + z² = 1

then, $\left|\begin{array}{ccc}1& \mathrm{z}& -\mathrm{y}\\ -\mathrm{z}& 1& \mathrm{x}\\ \mathrm{y}& -\mathrm{x}& 1\end{array}\right|$

Expanding along R₁, we get-

1(1 + x²) + z(xy + z) − y(xz − y)

1(1 + x2) - z(-z - xy) − y(xz − y)

= 1 + x² + xyz + z² − xyz + y²

= 1 + x² + y² + z²

= 1 + 1   (∵ x2 + y2 + z2 = 1 (given))

= 2

The correct answer is option "3"

Calculation:

We have, Δ = $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$

= $\left|\begin{array}{ccc}a+b+c& a+b+c& a+b+c\\ b& c& a\\ c& a& b\end{array}\right|$[Applying R₁→R₁ + R₂ + R₃]

= (a + b + c)$\left|\begin{array}{ccc}1& 1& 1\\ b& c& a\\ c& a& b\end{array}\right|$

Applying C₁ → C₁ - C₂ and C₂ → C₂ - C₃

= (a + b + c) $\left|\begin{array}{ccc}0& 0& 1\\ b-c& c-a& a\\ c-a& a-b& b\end{array}\right|$

= (a + b + c)[(b - c)(a - b) - (c - a)²]

= (a + b + c) [ab - b² - ac + bc - (c² + a² - 2ac)]

= (a + b + c) [ ab - b² - ac + bc - c² - a² + 2ac ]

= (a + b + c) [- a² - b² - c² + ab + bc + ca]

= - (a+ b+ c)[a² + b² + c² - ab - bc - ca]

= $-\frac{1}{2}$ (a + b + c) [2a² + 2b² + 2c² - 2ab -2bc - 2ca]

= $-\frac{1}{2}$ (a + b + c)[(a² +b² - 2ab) + (b² + c² - 2bc) + (c² + a² - 2ac)]

⇒ Δ = $-\frac{1}{2}$ (a + b + c)[(a - b)² + (b - c)² + (c - a)²]

According to the question a ≠ b ≠ c and all are positive.

⇒ (a + b + c) > 0 and (a - b)2 > 0, (b - c)2 > 0,  (c - a)² > 0

∴ Δ < 0

Calculation:

 $\left|\begin{array}{ccc}0& x-a& x-b\\ x+a& 0& x-c\\ x+b& x+c& 0\end{array}\right|=0$

Solving determinants we get 

⇒ 0[0 -(x + c)(x - c)] - (x - a)[(0 - (x + b)(x - c)] + (x - b)[(x + a)(x +c)   - 0] = 0

⇒ (x - a) (x + b) (x - c) + (x - b) x + a) (x + c) = 0 

The only value that satisfies the above equation is x = 0

∴ The value of x is 0. 

Concept:

Properties of Determinant of a Matrix:

• If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.
• For any square matrix say A, |A| = |AT|.
• If we interchange any two rows (columns) of a matrix, the determinant is multiplied by -1.
• If any two rows (columns) of a matrix are same then the value of the determinant is zero.

Calculation:

$\left|\begin{array}{ccc}\mathrm{x}-\mathrm{a}& \mathrm{y}-\mathrm{b}& \mathrm{z}-\mathrm{c}\\ \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{x}& \mathrm{y}& \mathrm{z}\end{array}\right|$

Apply R3 → R3 - R2

= $\left|\begin{array}{ccc}\mathrm{x}-\mathrm{a}& \mathrm{y}-\mathrm{b}& \mathrm{z}-\mathrm{c}\\ \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{x}-\mathrm{a}& \mathrm{y}-\mathrm{b}& \mathrm{z}-\mathrm{c}\end{array}\right|$

As we can see that the first and the third row of the given matrix are equal. 

We know that, if any two rows (columns) of a matrix are same then the value of the determinant is zero.

$\left|\begin{array}{ccc}\mathrm{x}-\mathrm{a}& \mathrm{y}-\mathrm{b}& \mathrm{z}-\mathrm{c}\\ \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{x}& \mathrm{y}& \mathrm{z}\end{array}\right|$ = 0

Concept:

Elementary row or column transformations do not change the value of the determinant of a matrix.

Calculation:

$\left|\begin{array}{ccc}1& 1& 1\\ 1& 1+\mathrm{x}& 1\\ 1& 1& 1+\mathrm{y}\end{array}\right|$

Applying R₂ → R₂ – R₁, R₃ → R₃ – R₁, we get

$=\left|\begin{array}{ccc}1& 1& 1\\ 0& \mathrm{x}& 0\\ 0& 0& \mathrm{y}\end{array}\right|$

Now, Expanding along C₁

= 1 (xy – 0) – 0 + 0 = xy

Calculation:

Given $\left|\begin{array}{ccc}\mathrm{x}& 2& 3\\ 1& \mathrm{x}& 1\\ 3& 2& \mathrm{x}\end{array}\right|$ = 0

⇒ x(x² - 2) - 2(x - 3) + 3(2 - 3x) = 0

Now, x³ - 2x - 2x + 6 + 6 - 9x = 0

⇒ x³ - 13x + 12 = 0

∵ x = 3 is a root of the equation, ∴ (x - 3) = 0

If we divide (x3 - 13x + 12) from (x - 3) we will get (x2 + 3x - 4)

⇒ (x - 3)(x² + 3x - 4) = 0

⇒ x2 + 3x - 4 = 0

⇒ (x + 4)(x - 1) = 0

⇒ x = 1, -4

Concept:

Properties of Determinant of a Matrix:

• If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.
• For any square matrix say A, |A| = |AT|.
• If we interchange any two rows (columns) of a matrix, the determinant is multiplied by -1.
• If any two rows (columns) of a matrix are same then the value of the determinant is zero.

Calculation:

$\left|\begin{array}{ccc}\mathrm{z}& \mathrm{x}& \mathrm{y}\\ 1& 1& 1\\ \mathrm{x}+\mathrm{y}& \mathrm{y}+\mathrm{z}& \mathrm{z}+\mathrm{x}\end{array}\right|$

Apply R1 → R1 + R₃

= $\left|\begin{array}{ccc}\mathrm{x}+\mathrm{y}+\mathrm{z}& \mathrm{x}+\mathrm{y}+\mathrm{z}& \mathrm{x}+\mathrm{y}+\mathrm{z}\\ 1& 1& 1\\ \mathrm{x}+\mathrm{y}& \mathrm{y}+\mathrm{z}& \mathrm{z}+\mathrm{x}\end{array}\right|$

Taking common (x + y + z) from Row 1, we get

= $(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ \mathrm{x}+\mathrm{y}& \mathrm{y}+\mathrm{z}& \mathrm{z}+\mathrm{x}\end{array}\right|$

As we can see that the first and the second row of the given matrix are equal. 

We know that, if any two rows (columns) of a matrix are same then the value of the determinant is zero.

∴$\left|\begin{array}{ccc}\mathrm{z}& \mathrm{x}& \mathrm{y}\\ 1& 1& 1\\ \mathrm{x}+\mathrm{y}& \mathrm{y}+\mathrm{z}& \mathrm{z}+\mathrm{x}\end{array}\right|$ = 0

Calculation:

$\left|\begin{array}{ccc}1+x& 1& 1\\ 1& 1+y& 1\\ 1& 1& 1+z\end{array}\right|=0$

R₁ = R₁ - R₂

$\left|\begin{array}{ccc}x& -y& 0\\ 1& 1+y& 1\\ 1& 1& 1+z\end{array}\right|=0$

R₂ = R₂ - R₃

$\left|\begin{array}{ccc}x& -y& 0\\ 0& y& -z\\ 1& 1& 1+z\end{array}\right|=0$

Now, Expanding along R₁, we get

⇒ x [y(1 + z) - (-z)] - (-y) [0 - (-z)] + 0 = 0

⇒ x [y + yz + z] + y [z] = 0

⇒ xy + xyz + xz + yz = 0

Dividing by xyz

⇒ $\frac{1}{\mathrm{z}}+1+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{x}}=0$

⇒ $\text{boldsymbol}{\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}+{\mathrm{z}}^{-1}=-1$

Formula used:

$\mathrm{A}=\left|\begin{array}{ccc}{a}_1& a_2& a_1\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$

det(A) = ${\mathrm{a}}_1\left|\begin{array}{cc}{b}_2& b_3\\ c_2& c_3\end{array}\right|-{\mathrm{a}}_2\left|\begin{array}{cc}{b}_1& b_3\\ c_1& c_3\end{array}\right|+{\mathrm{a}}_3\left|\begin{array}{cc}{b}_1& b_2\\ c_1& c_2\end{array}\right|$

= a1(b2 c3 - b3 c2) - a2(b1c3 - b3c1) + a3 (b1c2 - b2c1)

Calculation:

$\left|\begin{array}{ccc}1& -x& x\\ 1& 1& -y\\ 1& z& 1\end{array}\right|$

⇒  $\left|\begin{array}{ccc}1& -\frac{a}{b-c}& \frac{a}{b-c}\\ 1& 1& -\frac{b}{c-a}\\ 1& \frac{c}{a-b}& 1\end{array}\right|$ = Δ 

⇒ Δ = 1(1 + $\frac{\mathrm{b}\mathrm{c}}{(\mathrm{a}-\mathrm{b})(\mathrm{c}-\mathrm{a})}$) + $\frac{\mathrm{a}}{\mathrm{b}-\mathrm{c}}$($1+\frac{\mathrm{b}}{\mathrm{c}-\mathrm{a}}$) + $\frac{\mathrm{a}}{\mathrm{b}-\mathrm{c}}$ $(\frac{\mathrm{c}}{\mathrm{a}-\mathrm{b}}-1)$

⇒ Δ = 1 + $\frac{\mathrm{b}\mathrm{c}}{(\mathrm{a}-\mathrm{b})(\mathrm{c}-\mathrm{a})}+\frac{\mathrm{a}\mathrm{b}}{(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})}+\frac{\mathrm{a}\mathrm{c}}{(\mathrm{b}-\mathrm{c})(\mathrm{a}-\mathrm{b})}$

⇒ Δ = 1 + $\frac{a}{(b-c)}\times \frac{(b-c)(a-b-c)}{(a-b)(c-a)}+\frac{bc}{(a-b)(c-a)}$

⇒ Δ = $\frac{ac-a^2-bc+ab+a^2-ab-ac+bc}{(a-b)(c-a)}$

∴ Δ =  0

Shortcut TrickPut x = 0, b = 1 and c = 2

⇒ $x=\frac{a}{b-c}$ = 0, $y=\frac{b}{c-a}$ = 1/2, $z=\frac{c}{a-b}$ = -2

Let, $\left|\begin{array}{ccc}1& -x& x\\ 1& 1& -y\\ 1& z& 1\end{array}\right|$ = Δ 

⇒  Δ = $\left|\begin{array}{ccc}1& 0& 0\\ 1& 1& -1/2\\ 1& -2& 1\end{array}\right|$

⇒  Δ = 1[1 - (-1/2)(-2)] 

∴   Δ = 0

Concept:

a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca) 

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

Calculation:

Let Δ  = $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$ 

= a (cb - a²) - b (b² - ca) + c (ba - c²)

= abc - a³ - b³ + abc + abc - c³

= -[a³ + b³ + c³ - 3abc]

= -(a + b + c)(a² + b² + c² - ab - bc - ca)      ....(i)

As we know,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

⇒ 4² = a2 + b2 + c2 + 2 × 0 

⇒ 16 = a2 + b2 + c2

From equ (i)

Δ = -[4 × (16 - 0)]

= - 64

Concept:

If $\mathrm{A}=\left[\begin{array}{ccc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}& {\mathrm{a}}_{13}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}& {\mathrm{a}}_{23}\\ {\mathrm{a}}_{31}& {\mathrm{a}}_{32}& {\mathrm{a}}_{33}\end{array}\right]$ then determinant of A is given by:

• |A| = a₁₁ × {(a₂₂ × a₃₃) – (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) – (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) – (a₂₂ × a₃₁)}
• Elementary row or column transformations do not change the value of the determinant of a matrix.

Calculation:

Given:

$\mathrm{L}\mathrm{e}\mathrm{t}\mathrm{\Delta }=\left|\begin{array}{ccc}1& 1& 1\\ 1& 1+\mathrm{x}\mathrm{y}\mathrm{z}& 1\\ 1& 1& 1+\mathrm{x}\mathrm{y}\mathrm{z}\end{array}\right|$

Apply R₂ → R₂ – R₁ and R₃ → R₃ – R₁, we get

$=\left|\begin{array}{ccc}1& 1& 1\\ 0& \mathrm{x}\mathrm{y}\mathrm{z}& 0\\ 0& 0& \mathrm{x}\mathrm{y}\mathrm{z}\end{array}\right|$

Expanding along R₁, we get

Δ = 1 (x²y²z² – 0) – 0 + 0

∴ Δ = x²y²z²

Concept:

Properties of Determinant of a Matrix:

• If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.
• For any square matrix say A, |A| = |AT|.
• If we interchange any two rows (columns) of a matrix, the determinant is multiplied by -1.
• If any two rows (columns) of a matrix are same then the value of the determinant is zero.

Calculation:

$\left|\begin{array}{ccc}\mathrm{x}-\mathrm{y}& \mathrm{y}-\mathrm{z}& \mathrm{z}-\mathrm{x}\\ \mathrm{y}-\mathrm{z}& \mathrm{z}-\mathrm{x}& \mathrm{x}-\mathrm{y}\\ \mathrm{z}-\mathrm{x}& \mathrm{x}-\mathrm{y}& \mathrm{y}-\mathrm{z}\end{array}\right|$

Apply R1 → R1 + R₂ + R₃, We get

= $\left|\begin{array}{ccc}0& 0& 0\\ \mathrm{y}-\mathrm{z}& \mathrm{z}-\mathrm{x}& \mathrm{x}-\mathrm{y}\\ \mathrm{z}-\mathrm{x}& \mathrm{x}-\mathrm{y}& \mathrm{y}-\mathrm{z}\end{array}\right|$

As we can see that the entry of the first row is zero. 

We know that,

If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.

∴ $\left|\begin{array}{ccc}\mathrm{x}-\mathrm{y}& \mathrm{y}-\mathrm{z}& \mathrm{z}-\mathrm{x}\\ \mathrm{y}-\mathrm{z}& \mathrm{z}-\mathrm{x}& \mathrm{x}-\mathrm{y}\\ \mathrm{z}-\mathrm{x}& \mathrm{x}-\mathrm{y}& \mathrm{y}-\mathrm{z}\end{array}\right|$ = 0

Concept:

If two rows or columns of a determinant are identical the value of the determinant is zero.

Calculation:

Let Δ = $\left|\begin{array}{ccc}1& \mathrm{x}& \mathrm{y}+\mathrm{z}\\ 1& \mathrm{y}& \mathrm{z}+\mathrm{x}\\ 1& \mathrm{z}& \mathrm{x}+\mathrm{y}\end{array}\right|$

Apply C3 ↔ C2 + C3 on the above determinant, we get

$=\left|\begin{array}{ccc}1& \mathrm{x}& \mathrm{x}+\mathrm{y}+\mathrm{z}\\ 1& \mathrm{y}& \mathrm{x}+\mathrm{y}+\mathrm{z}\\ 1& \mathrm{z}& \mathrm{x}+\mathrm{y}+\mathrm{z}\end{array}\right|$

Taking common x + y + z from column 3rd, we get

$=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc}1& \mathrm{x}& 1\\ 1& \mathrm{y}& 1\\ 1& \mathrm{z}& 1\end{array}\right|$

As we know that, if any two rows (columns) of a matrix are same then the value of the determinant is zero.

$\left|\begin{array}{ccc}1& \mathrm{x}& \mathrm{y}+\mathrm{z}\\ 1& \mathrm{y}& \mathrm{z}+\mathrm{x}\\ 1& \mathrm{z}& \mathrm{x}+\mathrm{y}\end{array}\right|=0$

As we know, If two rows or columns of a determinant are identical the value of the determinant is zero.

Therefore, Δ = 0

Properties of Determinants:

• Determinant evaluated across any row or column is same.
• If all the elements of a row or column are zeroes, then the value of the determinant is zero.
• The determinant of an Identity matrix is 1.
• If rows and columns are interchanged then the value of determinant remains the same (value does not change).
• If any two-row or two-column of a determinant are interchanged the value of the determinant is multiplied by -1.
• If two rows or columns of a determinant are identical the value of the determinant is zero.