Algebraic variable elements MCQ Quiz in தமிழ் - Objective Question with Answer for Algebraic variable elements - இலவச PDF ஐப் பதிவிறக்கவும்

Calculation: 

Given $\left|\begin{array}{ccc}\mathrm{x}& 1& 2\\ 2& 0& 3\\ 4& 5& 6\end{array}\right|$ = 0

⇒ x(0 - 15) - 1(12 - 12) + 2(10) = 0

Now, - 15x + 20 = 0

⇒ 15x = 20

⇒ x = 20/15

⇒ x = 4/3

∴ The value of x is 4/3

Calculation:

Here, $\begin{array}{l}\mathrm{p}{\mathrm{x}}^3+\mathrm{q}{\mathrm{x}}^2+\mathrm{r}\mathrm{x}+\mathrm{s}=\left|\begin{array}{ccc}x& 2x& 3x\\ x+3& x& x\\ 1-2x& -x& x+1\end{array}\right|\end{array}$

⇒ $\mathrm{p}{\mathrm{x}}^3+\mathrm{q}{\mathrm{x}}^2+\mathrm{r}\mathrm{x}+\mathrm{s}$ = x[(x + 1)(x) - (x)(-x)] - 2x [(x + 3)(x+1) - x(1 - 2x)] + 3x[-x(x+3)-x(1-2x)]

= $\mathrm{x}({\mathrm{x}}^2+\mathrm{x}+{\mathrm{x}}^2)-2\mathrm{x}({\mathrm{x}}^2+4\mathrm{x}+3-\mathrm{x}+2{\mathrm{x}}^2)+3\mathrm{x}(-{\mathrm{x}}^2-3\mathrm{x}-\mathrm{x}-2{\mathrm{x}}^2)$

= $\mathrm{x}(2{\mathrm{x}}^2+\mathrm{x})-2\mathrm{x}(3{\mathrm{x}}^2+3\mathrm{x}+3)+3\mathrm{x}({\mathrm{x}}^2-4\mathrm{x})$

= $2{\mathrm{x}}^3+{\mathrm{x}}^2-6{\mathrm{x}}^3-6{\mathrm{x}}^2-6\mathrm{x}+3{\mathrm{x}}^3-12{\mathrm{x}}^2$

= $-{\mathrm{x}}^3-17{\mathrm{x}}^2-6\mathrm{x}$

So, p = - 1, q = - 17, r = - 6, s = 0

p + q + r + s = - 1 - 17 - 6 = - 24

Hence, option (3) is correct. 

Calculation:

Let Δ = $\left|\begin{array}{ccc}\mathrm{x}+2& \mathrm{x}+3& \mathrm{x}+5\\ \mathrm{x}+4& \mathrm{x}+6& \mathrm{x}+9\\ \mathrm{x}+8& \mathrm{x}+11& \mathrm{x}+15\end{array}\right|$

$\begin{array}{r}\\ & \text{ By applying }{\mathrm{C}}_2\to {\mathrm{C}}_2-{\mathrm{C}}_1,{\mathrm{C}}_3\to {\mathrm{C}}_3-{\mathrm{C}}_1\\ & =\left|\begin{array}{ccc}\mathrm{x}+2& 1& 3\\ \mathrm{x}+4& 2& 5\\ \mathrm{x}+8& 3& 7\end{array}\right|\\ & \end{array}$

$\text{ By applying }{\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_1,\text{ and }{\mathrm{R}}_3\to {\mathrm{R}}_3-{\mathrm{R}}_2$

$=\left|\begin{array}{ccc}\mathrm{x}+2& 1& 3\\ 2& 1& 2\\ 4& 1& 2\end{array}\right|$

= (x + 2) (0) - 1 (4 - 8) + 3(2 - 4)

= 4 - 6

= -2

Hence, option (2) is correct.

Concept:

Properties of Determinants:

• The determinant evaluated across any row or column is the 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 rows or two columns 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.

Calculation:

Let $\mathrm{\Delta }=\left|\begin{array}{ccc}1-\mathrm{a}& \mathrm{a}-\mathrm{b}-\mathrm{c}& \mathrm{b}+\mathrm{c}\\ 1-\mathrm{b}& \mathrm{b}-\mathrm{c}-\mathrm{a}& \mathrm{c}+\mathrm{a}\\ 1-\mathrm{c}& \mathrm{c}-\mathrm{a}-\mathrm{b}& \mathrm{a}+\mathrm{b}\end{array}\right|$

Applying c₂ → c₂ + c₃, we get

$\mathrm{\Delta }=\left|\begin{array}{ccc}1-\mathrm{a}& \mathrm{a}& \mathrm{b}+\mathrm{c}\\ 1-\mathrm{b}& \mathrm{b}& \mathrm{c}+\mathrm{a}\\ 1-\mathrm{c}& \mathrm{c}& \mathrm{a}+\mathrm{b}\end{array}\right|$

Applying c₁ → c₁ + c₂, c₃ → c₃ + c₂, we get

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

Taking common a + b + c from C₃, we get

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

We know that if two columns of a determinant are identical, the value of the determinant is zero.

∴ Δ = 0

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

Taking common a, b, and c from R₁, R_2, and R₃ respectively,

⇒ $\Delta =abc\left|\begin{array}{ccc}1+\frac{1}{a}& \frac{1}{a}& \frac{1}{a}\\ \frac{1}{b}& 1+\frac{1}{b}& \frac{1}{b}\\ \frac{1}{c}& \frac{1}{c}& 1+\frac{1}{c}\end{array}\right|$

Applying  R1 →  R1 + R2 + R3 

⇒ $\Delta =abc\left|\begin{array}{ccc}1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}& 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}& 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\\ \frac{1}{b}& 1+\frac{1}{b}& \frac{1}{b}\\ \frac{1}{c}& \frac{1}{c}& 1+\frac{1}{c}\end{array}\right|$

Taking common $1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ from R1,

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

Applying C2 →  C2 - C1, C3 →  C3 - C1 

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

⇒ $\Delta =abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$

So, if $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$, then Δ = abc(1 + 0) = abc

⇒ Statement I is correct

And, a-1 + b-1 + c-1 = -1,

⇒  $1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$ then Δ = 0

⇒ Statement II is correct

∴ The correct answer is option (3).

Given:

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

Calculations:

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

Apply R₁ → R₁ + R₂ + R₃

⇒ $\left|\begin{array}{ccc}x+y+z& x+y+z& 2(x+y+z)\\ z& y& x+y\\ x& z& z+x\end{array}\right|=0$

Take (x + y + z) common from R1

⇒ (x + y + z) $\left|\begin{array}{ccc}1& 1& 2\\ z& y& x+y\\ x& z& z+x\end{array}\right|=0$

Apply C₂ → C₁ - C₂ and C₃ → C₃ - 2C1

⇒ $(x+y+z)\left|\begin{array}{ccc}1& 0& 0\\ z& z-y& x+y-2z\\ x& z-x& z-x\end{array}\right|=0$

Now take the determinant, we get

⇒ (x + y + z)[1(z - y)(z - x) - (z - x)(x + y - 2z)] = 0

⇒ (x + y + z) (z - x)(z - y - x - y + 2z) = 0

⇒  x + y + z = 0 and z - x = 0

∴ The value is either x + y = -z or x = z.

Concept:

If we have any matrix with two identical rows or columns then its determinant is equal to zero.

Calculation:

Let Δ = $\left|\begin{array}{ccc}1& a& a^2-bc\\ 1& b& b^2-ac\\ 1& c& c^2-ab\end{array}\right|$

R₁ → R₁ - R₂

R₂ → R₂ - R₃

⇒ Δ = $\left|\begin{array}{ccc}0& a-b& (a-b)(a+b+c)\\ 0& b-c& (b-c)(a+b+c)\\ 1& c& c^2-ab\end{array}\right|$

[(a² - bc) - (b² - ac) = a² - b² + ac - bc = (a + b)(a - b) +c(a - b) = (a - b)(a + b + c)

and (b2 - ac) - (c2 - ab) = b2 - c2 + ab - ac = (b + c)(b - c) +a(b - c) = (b - c)(a + b + c)]

Taking (a - b) and (b - c) common,

⇒ Δ = (a - b)(b - c)$\left|\begin{array}{ccc}0& 1& a+b+c\\ 0& 1& a+b+c\\ 1& c& c^2-ab\end{array}\right|$ = 0   

[If we have any matrix with two identical rows or columns then its determinant is equal to zero.]

∴​ $\left|\begin{array}{ccc}1& a& a^2-bc\\ 1& b& b^2-ac\\ 1& c& c^2-ab\end{array}\right|$ = 0

Concept:

• If two rows or columns of a determinant are identical the value of the determinant is zero.
• In a determinant each element in any row (or column) consists of the sum of two terms, then the determinant can be expressed as the sum of two determinants of the same order. For example,​ $\left|\begin{array}{cc}a+x& b\\ a+y& c\end{array}\right|$$=\left|\begin{array}{cc}a& b\\ a& c\end{array}\right|=\left|\begin{array}{cc}x& b\\ y& c\end{array}\right|$ 

Solution:

Let $\mathrm{\Delta }=\left|\begin{array}{ccc}a& b& c\\ a+2x& b+2y& c+2z\\ x& y& z\end{array}\right|=\left|\begin{array}{ccc}a& b& c\\ a& b& c\\ x& y& z\end{array}\right|+\left|\begin{array}{ccc}a& b& c\\ 2x& 2y& 2z\\ x& y& z\end{array}\right|$

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

$=0+\left|\begin{array}{ccc}a& b& c\\ 2x& 2y& 2z\\ x& y& z\end{array}\right|$$=\left|\begin{array}{ccc}a& b& c\\ 2x& 2y& 2z\\ x& y& z\end{array}\right|$

Apply R₂ → R₂ - R₃ on the above determinant, and we get

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

As we know, If two rows or columns of a determinant are identical the value of the determinant is zero. Therefore, Δ = 0

∴ The correct option is (2)

Formula used:

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

Calculation:

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

C₁ → C1 + C₂ + C₃

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

Take Common (a + b + c - x) from C_1, we get

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

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

⇒ a + b + c = x

⇒ x = 0    [ ∵ a + b + c = 0]

But according to question x ≠ 0

By expanding the determinant we have

⇒ 1[(b - x)(c - x) - a²] - c(c - x - a) + b (a - b + x) = 0

⇒ x² - (a² + b² + c²) + (ab + bc + ca) = 0

⇒ x² = (a2 + b2 + c2) - (ab + bc + ca)     ----(1)

Let (a2 + b2 + c2) = ∑a² and (ab + bc + ca) = ∑ab then,

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

⇒ 0 = ∑ a² + 2∑ ab      [a + b + c = 0]          ----(2)

From (1) and (2), we get

⇒ x = $\pm \sqrt{\frac{3}{2}\sum a^2}$

⇒ x = $\pm \sqrt{\frac{3}{2}(a^2+b^2+c^2})$

Hence, the correct answer is option 3)

Concept:

Properties of determinants:

• If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.
• If we interchange any two rows (columns) of a matrix, then 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:

A = $\left[\begin{array}{ccc}{x}^2& y^2& -2xy\\ x& y& 0\\ 1& 1& 2\end{array}\right]$ 

|A| = $\left|\begin{array}{ccc}{x}^2& y^2& -2xy\\ x& y& 0\\ 1& 1& 2\end{array}\right|$

c₃ = c₃ - c₁ - c₂

|A| = $\left|\begin{array}{ccc}{x}^2& y^2& -(2xy+x^2+y^2)\\ x& y& 0-x-y\\ 1& 1& 2-1-1\end{array}\right|$

|A| = $\left|\begin{array}{ccc}{x}^2& y^2& -(x+y{)}^2\\ x& y& -(x+y)\\ 1& 1& 0\end{array}\right|$

|A| = $\left|\begin{array}{ccc}{x}^2& y^2& 0\\ x& y& 0\\ 1& 1& 0\end{array}\right|$

|A| = 0