SATHEE: Determinants Exercise 01

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.prev Determinants Exercise 02 next Continuity and Differentiability Miscellaneous ExercisesTable of Contents Question: Answer: Question: Answer: Question: Answer: Question: Answer: Question: Answer: Question: Answer: Question: Answer: Question: Answer: Question: Answer: Question: Answer: Question: Answer: Question: Answer: Question: Answer: Question: Answer: Question: Answer: Question: Answer: Question: Answer: Question: Answer:Engineering Preparation Chemistry 11th Chemistry 12th Mathematics 11th Mathematics 12th Physics 11th Physics 12th JEE Previous Year Questions JEE PYQ - Chapterwise JEE Tutorial Sessions JEE Syllabus JEE Mock Tests JEE Mentorship Medical Preparation Biology 11th Biology 12th Chemistry 11th Chemistry 12th Physics 11th Physics 12th NEET Previous Year Questions NEET PYQ - Chapterwise NEET Tutorial Sessions NEET Syllabus NEET Mock Tests NEET Mentorship Syllabus JEE Syllabus NEET Syllabus Test Platform JEE / Engineering Test NEET / Medical Test Sample Mock Test Mock Test for JEE Additional problems chapterwise Mentorship Mentorship for JEE Mentorship for NEET Important Resources Physics Formulas Chemistry Formulas Mathematics Formulas Useful Articles Sample Mock Tests Privacy Policy NCERT Books & Solutions NCERT Books for JEE NCERT Books for NEET NCERT Solutions for NEET NCERT Exemplar for JEE NCERT Exemplar for NEET NCERT Books for 9th-10th NCERT Exemplar for 9th-10th Other Resources Solve With Me NCERT Exercises Media Coverage Help Videos Games Arithmetical operation based Games© 2024 Copyright SATHEE Privacy Policy Powered by Prutor@IITK sathee@iitk.ac.in Media coverage of SATHEE Play Store App StoreAsk SATHEE Welcome to SATHEE ! Select from'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓 I'm relatively new and can sometimes make mistakes. If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning. To begin your journey now, click on " I understand " . Menu How to attempt Mock Tests? (JEE | NEET) How to register/create an account? (JEE | NEET) How to reset password? (JEE | NEET) What is the syllabus of JEE / NEET? How to register a school? What is SATHEE? How to download the SATHEE app? How to get your doubts cleared? How to become a doctor? How to learn more in less time? How to improve my studies? How to revise the syllabus? How to prepare for JEE exam? How to increase the learning speed? How to manage time? How to study concentrated? How to avoid distractions? How to overcome exam stress? Submitvar input = document.getElementById("chatmessage");
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