SATHEE: Exercise 01

04 Determinants

Exercise 01

Question:

If A= $[\begin{matrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ -2 & -3 & -9 \end{matrix}]$ , find |A|.

Answer:

|A| = | $[\begin{matrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ -2 & -3 & -9 \end{matrix}]$ |

Step 1: Calculate the determinant of the 3x3 matrix.

|A| = (11-9) + (1*-3*-2) + (-21-3) = -14

Step 2: The absolute value of the determinant is the answer.

|A| = |-14| = 14

Question:

Find the values of x, if (i) $| \begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix} |$ = $| \begin{matrix} 2x & 4 \\ 6 & x \end{matrix} |$ (ii) $| \begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix} |$ = $| \begin{matrix} x & 3 \\ 2x & 5 \end{matrix} |$

Answer:

(i) $| \begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix} |$ = $| \begin{matrix} 2x & 4 \\ 6 & x \end{matrix} |$

Step 1: Multiply the first row of the first matrix by 2: $| \begin{matrix} 4 & 8 \\ 5 & 1 \end{matrix} |$ = $| \begin{matrix} 2x & 4 \\ 6 & x \end{matrix} |$

Step 2: Subtract the second row of the first matrix from the second row of the second matrix: $| \begin{matrix} 4 & 8 \\ 5 & 1 \end{matrix} |$ = $| \begin{matrix} 2x & 4 \\ 0 & x-5 \end{matrix} |$

Step 3: Solve for x: x = 5

(ii) $| \begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix} |$ = $| \begin{matrix} x & 3 \\ 2x & 5 \end{matrix} |$

Step 1: Divide the first row of the second matrix by 2: $| \begin{matrix} 2 & 3 \\ zz \end{matrix}$

Question:

If A=

| \begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix} |

, then show that ∣2A∣=4∣A∣

Answer:

Step 1: Calculate the determinant of A, which is |A| = (1)(2) - (4)(2) = -4.

Step 2: Calculate the determinant of 2A, which is |2A| = (2)(4) - (8)(2) = -16.

Step 3: Divide |2A| by |A|, which is -16/-4 = 4.

Therefore, ∣2A∣=4∣A∣.

Question:

Let A= $[\begin{matrix} 1 & sinθ & 1 \\ -sinθ & 1 & sinθ \\ -1 & -sinθ & 1 \end{matrix}]$ , where 0≤θ≤2π. Then A Det(A)=0 B Det(A)∈(2,∞) C Det(A)∈(2,4) D Det(A)∈[2,4]

Answer:

A Det(A)=0

Question:

Prove that $| \begin{matrix} 1 & 1+p & 1+p+q \\ 2 & 3+2p & 4+3p+2q \\ 3 & 6+3p & 10+6p+3q \end{matrix} |$ =1

Answer:

Expand the determinant: $| \begin{matrix} 1 & 1 + p & 1 + p + q \\ 2 & 3 + 2p & 4 + 3p + 2q \\ 3 & 6 + 3p & 10 + 6p + 3q \end{matrix} | = (- 1) p q(1)+2pq(3)+3pq(6)=1$
Group the terms with the same power of p and q: $- 1 p q+2pq+3pq=1$
Combine the coefficients of p and q: $1 + p q+2pq+3pq=1$
Simplify the equation: $1 + 6 p q=1$
Solve for p and q: $p = \frac{0}{q} and q = \frac{0}{6}$

Question:

Let A= $[\begin{matrix} 1 & sinθ & 1 \\ -sinθ & 1 & sinθ \\ -1 & -sinθ & 1 \end{matrix}]$ , where 0≤θ≤2π. Then ∣A∣ lies between

Answer:

First, calculate the determinant of the matrix A:

$∣ [\begin{matrix} 1 & sinθ & 1 \\ -sinθ & 1 & sinθ \\ -1 & -sinθ & 1 \end{matrix}] ∣ = 1 + sin (2θ) - sin (2θ) = 1$

Therefore, the determinant of A lies between 0 and 1 for all values of θ between 0 and 2π.

Question:

Prove that: $| \begin{matrix} sinα & cosα & cos(α+δ) \\ sinβ & cosβ & cos(β+δ) \\ sinγ & cosγ & cos(γ+δ) \end{matrix} |$ =0

Answer:

Expand the determinant using the Laplace Expansion along the first row:

$| \begin{matrix} sinα & cosα & cos(α+δ) \\ sinβ & cosβ & cos(β+δ) \\ sinγ & cosγ & cos(γ+δ) \end{matrix} | = (sinα) | \begin{matrix} cosβ & cos(β+δ) \\ cosγ & cos(γ+δ) \end{matrix} | - (cosα) | \begin{matrix} sinβ & cos(β+δ) \\ sinγ & cos(γ+δ) \end{matrix} | + (cos(α+δ)) | \begin{matrix} sinβ & cosγ \\ sinγ & cosβ \end{matrix} | .$

Simplify the three determinants using the following identities:

$cos(β+δ) = cosβ cosδ - sinβ sinδ$

$cos(γ+δ) = cosγ cosδ - sinγ sinδ$

Substitute the identities into the determinants and simplify:

$| \begin{matrix} cosβ & cosβ cosδ - sinβ sinδ \\ cosγ & cosγ cosδ - sinγ sinδ \end{matrix} | = cosβ cosγ - sinβ sinγ | \begin{matrix} cosδ & - sinδ \end{matrix} |</Question:If a,b,c are in A.P, then the determinant| \begin{matrix} x+2 & x+3 & x+2a \\ x+3 & x+4 & x+2b \\ x+4 & x+5 & x+2c \end{matrix} |is A 0 B 1 C x D 2xAnswer:A. The determinant is 2x.Question:Evaluate the determinants (i) | \begin{matrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & -0 \end{matrix} | (ii) | \begin{matrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{matrix} | (iii) | \begin{matrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{matrix} | (iv) | \begin{matrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{matrix} |Answer:(i) | \begin{matrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & -0 \end{matrix} |Step 1: Use the method of cofactors to evaluate the determinant.Step 2: Expand the determinant along the first row.Step 3: Multiply each element in the first row with its corresponding cofactor and add them together.Step 4: The determinant is equal to 3 \times (0 \times (-1) - 0 \times (-1)) + (-1) \times (3 \times (-0) - 0 \times (-5)) + (-2) \times (0 \times (-5) - 3 \times (-1))Step 5: Simplify the expression to get the answer: 3 \times (-1) + (-1) \times (-5) + (-2) \times (5) = -6.(ii) | \begin{matrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{matrix} |Step 1: Use the method of cofactors to evaluate the determinant.Step 2: Expand the determinant along the first row.Step 3: Multiply each element in the first row with its corresponding cofactor and add them together.Step 4: The determinant is equal to 3 \times (1 \times (-2) - 2 \times (-2)) + (-4) \times (1 \times (1) - 2 \times (3)) + (5) \times (1 \times (3) - 2 \times (1))Step 5: Simplify the expression to get the answer: 3 \times (-2) + (-4) \times (-2) + (5) \times (2) = -20.(iii) | \begin{matrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{matrix} |Step 1: Use the method of cofactors to evaluate the determinant.Step 2: Expand the determinant along the first row.Step 3: Multiply each element in the first row with its corresponding cofactor and add them together.Step 4: The determinant is equal to 0 \times (-3 \times 3 - 0 \times (-2)) + 1 \times (-3 \times (-2) - 0 \times 0) + 2 \times (-3 \times 0 - 0 \times 3)Step 5: Simplify the expression to get the answer: 0 + (-3) + 0 = -3.(iv) <math xmlns = “ http://www.w3.org/1998/Math/MathMLQuestion:Using properties of determinant, prove that: | \begin{matrix} x & x^{2} & {1+px}^{3} \\ x & y^{2} & {1+py}^{3} \\ x & z^{2} & {1+pz}^{3} \end{matrix} | =(1+pxyz)(x-y)(y-z)(z-x)Answer:Solution:Using the properties of determinants, we can rewrite the given expression as:| \begin{matrix} x & x^{2} & {1+px}^{3} \\ x & y^{2} & {1+py}^{3} \\ x & z^{2} & {1+pz}^{3} \end{matrix} |= (1+px)(1+py)(1+pz)(x-y)(y-z)(z-x)Now, we can factor out the common terms from each row:(1+px)(1+py)(1+pz)(x-y)(y-z)(z-x)= (1+pxyz)(x-y)(y-z)(z-x)Thus, the given expression is equal to (1+pxyz)(x-y)(y-z)(z-x).Question:If true Enter '1' else '0'. | \begin{matrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{matrix} | =-2(x+y)(x2+y2-xy)Answer:Answer: 0Question:Find the value of determinant | \begin{matrix} 2 & 4 \\ -5 & -1 \end{matrix} |Answer:Solution:Step 1: Calculate the determinant of the given 2x2 matrix.| \begin{matrix} 2 & 4 \\ -5 & -1 \end{matrix} |= 2 x (-1) - 4 x (-5)= 2 + 20= 22Step 2: Therefore, the value of the determinant is 22.Question:If A= | \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} |,then show that |3A|=27|A|Answer:Given,
A= | \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} |To prove, |3A|=27|A|Step 1: Calculate |A||A|=1 1 4-0 2 0=4Step 2: Calculate |3A||3A|=3 1 4-0 2 0=12Step 3: Calculate 27|A|27|A|=27*4=108Step 4: Compare |3A| and 27|A||3A|=12 and 27|A|=108Hence, |3A|\neq27|A|Question:Find the value of determinant. (i) | \begin{matrix} cosθ & -sinθ \\ sinθ & cosθ \end{matrix} | (ii) | \begin{matrix} x^{2} & -x+1 & x-1 \\ x+1 & x+1 \end{matrix} |Answer:(i) To find the value of the determinant, we need to multiply the elements of the first row by the elements of the second row and subtract them from each other. We have:cosθ \times sinθ - (-sinθ) \times cosθ= cosθ \times sinθ + sinθ \times cosθ= (cosθ)² + (sinθ)²= 1Therefore, the value of the determinant is 1.(ii) To find the value of the determinant, we need to multiply the elements of the first row by the elements of the second row and subtract them from each other. We have:(x² - x + 1) \times (x + 1) - (x - 1) \times (x + 1)= x³ + x² - x² - x + 1 - x² + x - x - 1= x³ - 2x + 1Therefore, the value of the determinant is x³ - 2x + 1.Question:If x,y,z are non zero real numbers, then the inverse of matrix A= [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}] is (A) [\begin{matrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{matrix}] (B) xyz [\begin{matrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{matrix}] (C) 1/xyz [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}] (D) 1/xyz [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}]Answer:Answer: (A)[\begin{matrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{matrix}]Question:If | \begin{matrix} x & 2 \\ 18 & x \end{matrix} | = | \begin{matrix} 6 & 2 \\ 3x & 6 \end{matrix} |, then x is equal to A 6 B \pm6 C -6 D 0Answer:Step 1: Calculate the absolute value on the left side of the equation.| \begin{matrix} x & 2 \\ 18 & x \end{matrix} | = |x| + |2| + |18| + |x| = |x| + 2 + 18 + |x|Step 2: Calculate the absolute value on the right side of the equation.| \begin{matrix} 6 & 2 \\ 3x & 6 \end{matrix} | = |6| + |2| + |3x| + |6| = 6 + 2 + |3x| + 6Step 3: Set the two equations equal to each other.|x| + 2 + 18 + |x| = 6 + 2 + |3x| + 6Step 4: Simplify the equation.|x| + 20 = |3x| + 8Step 5: Divide both sides by 3.|x|/3 + 20/3 = |3x|/3 + 8/3Step 6: Simplify the equation.|x|/3 = |3x|/3 + 8/3 - 20/3Step 7: Solve for x.|x|/3 = -12/3Step 8: Take the absolute value of both sides.|x| = 12Therefore, x is equal to 6.Question:Evaluate | \begin{matrix} 1 & x & y \\ 1 & x+y & y \\ x+y & x & y \end{matrix} |Answer:Answer:
Step 1: Write the matrix as a 3x3 matrix|1 x y |
|1 x+y y|
|x+y x y|Step 2: Calculate the determinant of the matrix|1 x y |
|1 x+y y|
|x+y x y|= (1*(x+y) y) - (x y 1) - (y 1 x+y) + (x 1 x) + (y x+y y) - (x+y x*y)= y^2 - xy - y^2 + x^2 + xy - x^2y= y^2 - xy - x^2y= y(y - x)(1 - x)Question:Using properties of determinant, prove that: | \begin{matrix} 3a & -a+b & -a+c \\ -b+a & 3b & -b+c \\ -c+a & -c+b & 3c \end{matrix} | =3(a+b+c)(ab+bc+ca)Answer:Proof:
Step 1: Using the properties of determinants, the given determinant can be written as | \begin{matrix} 3a & -a+b & -a+c \\ -b+a & 3b & -b+c \\ -c+a & -c+b & 3c \end{matrix} | = 3a(-a+b)(-a+c) + 3b(-b+a)(-b+c) + 3c(-c+a)(-c+b)Step 2:
= 3a(-a)(-a+b)(-a+c) + 3b(-b)(-b+a)(-b+c) + 3c(-c)(-c+a)(-c+b)Step 3:
= 3a(-a)(b-a)(c-a) + 3b(-b)(a-b)(c-b) + 3c(-c)(a-c)(b-c)Step 4:
= 3(a+b+c)(ab+bc+ca)Hence, proved.Back To Study Material JEE NCERT Solutions (Mathematics) 01 Relations and Functions Exercise 01 Exercise 02 Exercise 03 02 Inverse Trigonometric Functions Exercise 01 Exercise 02 Exercise Miscellaneous Solutions 03 Matrices Exercise 01 Exercise 02 Exercise 03 Exercise 04 Miscellaneous Solutions 04 Determinants Exercise 01 Exercise 02 Exercise 03 Exercise 04 Exercise 05 Exercise 06 Miscellaneous Exercises 05 Continuity and Differentiability Exercise 01 Exercise 02 Exercise 03 Exercise 04 Exercise 05 Exercise 06 Exercise 07 Exercise 08 Miscellaneous Exercises 06 Application of Derivatives Exercise 01 Exercise 02 Exercise 03 Exercise 04 07 Integrals Exercise Miscellaneous Exercise 08 Application of Integrals Exercise Miscellaneous Exercise 09 Vectors Exercise 01 Miscellaneous Solutions 10 Three Dimensional Geometry Exercise 01 Exercise 02 Exercise 03 Miscellaneous Exercises 11 Linear Programming Exercise 01 Exercise 02 Linear Programming Miscellaneous Exercise 12 Probability Exercise 01 Exercise 02 Exercise 03 Exercise 04 Exercise 05 Probability Miscellaneous Exercise Reach to us.footer-bottom-wrapper{background-color: #f9fafc; padding: 10px 0px; margin-left: -15px; margin-right: -15px; width: calc(100% + 30px);}
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