If $A=\left[\begin{array}{lll}1 & a & 0 \\ 0 & 1 & b \\ 0 & 0 & 1\end{array}\right]$, then the deter...

To find the determinant of $ S = I - A + A^2 - A^3 + A^4 - \cdots + A^{2020} $, we start by considering the properties of the matrix $ A $.Given the matrix $ A $:$ A = \begin{bmatrix} 1 & a & 0 \\ 0 & 1 & b \\ 0 & 0 & 1 \end{bmatrix} $The determinant of $ A $ is computed as:$ |A| = 1 \times 1 \times 1 = 1 $Since the determinant of $ A $ is 1, for any power $ n $:$ |A^n| = |A|^n = 1^n = 1 $Thus, regardless of the power $ n $, the determinant of $ A^n $ remains 1.Now, for the expression $ S = I - A + A^2 - A^3 + \cdots + A^{2020} $, we note the alternating signs. The determinant of $ S $ can be computed by the linearity property of determinants, combining each term's determinant:$ \operatorname{det}(S) = \det(I) - \det(A) + \det(A^2) - \det(A^3) + \cdots + \det(A^{2020}) $Substituting the known determinants (all equal to 1):$ \operatorname{det}(S) = 1 - 1 + 1 - 1 + \cdots + 1 $Since there are 2020 terms and they alternate between 1 and -1, if you sum these terms, every pair of terms (consisting of a +1 and a -1) cancels out, resulting in:$ \operatorname{det}(S) = 1 $Thus, the determinant of $ S $ is 1.

The number of skew-symmetric matrices $A=\left[a_i j\right]_{3 \times 3}$, where $a_i j \in\{-3,-2,-...

A skew-symmetric matrix $A$ is defined as a square matrix that satisfies the condition $A^T=$ $-A$. For a $3 \times 3$ skew-symmetric matrix, this implies:$$ A=\left[\begin{array}{ccc} 0 & a_{12} & a_{13} \\ -a_{12} & 0 & a_{23} \\ -a_{13} & -a_{23} & 0 \end{array}\right] $$All $a_{11}, a_{22}$, and $a_{33}$ are all zero.The elements $a_{12}, a_{13}$, and $a_{23}$ can be any value from the set $\{-3,-2,-1,0,1,2,3\}$. The matrix is completely determined by these off-diagonal elements and the remaining elements get adjusted accordingly.There are 3 off-diagonal elements ( $a_{12}, a_{13}, a_{23}$ ), and each of these elements can take 7 possible values. Therefore, the total number of such matrices is:$$ 7 \times 7 \times 7=7^3 $$

Let $A$ be the matrix $\left[\begin{array}{ccc}\cos \theta & 0 & -\sin \theta \\ 1 & 1 & 1 \\ \sin \...

Given the matrix $ A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 1 & 1 & 1 \\ \sin \theta & 0 & \cos \theta \end{bmatrix} $, we will determine the determinant of $ A^k $ for any natural number $ k $.The determinant of matrix $ A $ can be calculated as follows:$ \begin{aligned} \text{det}(A) &= \cos \theta \cdot (1 \cdot \cos \theta - 1 \cdot 0) \\ &\quad - 0 \cdot (1 \cdot \cos \theta - 1 \cdot \sin \theta) \\ &\quad + (-\sin \theta) \cdot (1 \cdot 0 - 1 \cdot \sin \theta). \end{aligned} $Simplifying each term, we get:$ \begin{aligned} \text{det}(A) &= \cos \theta \cdot \cos \theta - \sin \theta \cdot (-\sin \theta) \\ &= \cos^2 \theta + \sin^2 \theta. \end{aligned} $Using the Pythagorean identity, we know that $\cos^2 \theta + \sin^2 \theta = 1$.Therefore, $\text{det}(A) = 1$.Since the determinant of a product of matrices is the product of their determinants, we have:$ \text{det}(A^k) = (\text{det}(A))^k = 1^k = 1. $Thus, for any natural number $ k $, the determinant of $ A^k $ is 1.

Let $M$ be a $3 \times 3$ matrix with real entries such that$$ \left\{\left[\begin{array}{l} x_1 \\ ...

Firstly,$$ \begin{array}{cc} x_1+x_2=x_2+x_3 & x_1+x_2=0 \\ \Rightarrow x_1=x_3 & \Rightarrow x_1=-x_2 \end{array} $$So let,$$ x_1=x_3=-x_2=x $$Let,$$ \begin{gathered} M=\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right] \\ {\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right]\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right]=\left[\begin{array}{l} a x_1+b x_2+c x_3 \\ d x_1+e x_2+f x_3 \\ g x_1+h x_2+i x_3 \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]} \end{gathered} $$Which can be simplified to,$$ \left[\begin{array}{c} x(a-b+c) \\ x(d-e+f) \\ x(g-h+i) \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] $$$$ \begin{aligned} &a+c=b\ &\text { }\\ &d+f=e\\ &g+i=f \end{aligned} $$$$ \begin{aligned} &\text { Now you can see, }\\ &\begin{gathered} C_1+C_3=C_2 \end{gathered} \end{aligned} $$Now using the properties of determinants, $C_2 \rightarrow C_2-C_1-C_3$ will not change the value of the determinant:$$ C_2 \rightarrow C_2-\left(C_1+C_3\right)=0 $$Now upon expanding along $C_2$ :$$ \begin{gathered} |M|=\left|\begin{array}{ccc} a & \downarrow 0 & c \\ d & 0 & f \\ g & 0 & i \end{array}\right|=0 \\ \operatorname{det}(M)=0 \end{gathered} $$

Let$$ A=\left\{x \in \mathbf{R} \left\lvert\,-31<\operatorname{det}\left[\begin{array}{cc} 3 x-1 & 2...

$ A = \left\{x \in \mathbb{R} \;\middle|\; -31 < \det\begin{bmatrix} 3x - 1 & 2\\ -2 & 5 \end{bmatrix} \le 29 \right\} $Step 1. Compute the determinant$ \det\begin{bmatrix} 3x - 1 & 2\\ -2 & 5 \end{bmatrix} = (3x - 1)(5) - (2)(-2) $Simplify:$ = 15x - 5 + 4 = 15x - 1 $So the condition becomes:$ -31 < 15x - 1 \le 29 $Step 2. Solve the inequalitiesLeft side:$ -31 < 15x - 1 $Add 1 to both sides:$ -30 < 15x $Divide by 15:$ -2 < x $Right side:$ 15x - 1 \le 29 $Add 1 to both sides:$ 15x \le 30 $Divide by 15:$ x \le 2 $Step 3. Combine$ -2 < x \le 2 $Thus,$ A = (-2, 2] $✅Correct Option: A.$A = (-2, 2]$

Matrices And Determinants - IAT-IISER Maths PYQs

Q: Let$$ A=\left\{x \in \mathbf{R} \left\lvert\,-31<\operatorname{det}\left[\begin{array}{cc} 3 x-1 & 2...

$ A = \left\{x \in \mathbb{R} \;\middle|\; -31 < \det\begin{bmatrix} 3x - 1 & 2\\ -2 & 5 \end{bmatrix} \le 29 \right\} $Step 1. Compute the determinant$ \det\begin{bmatrix} 3x - 1 & 2\\ -2 & 5 \end{bmatrix} = (3x - 1)(5) - (2)(-2) $Simplify:$ = 15x - 5 + 4 = 15x - 1 $So the condition becomes:$ -31 < 15x - 1 \le 29 $Step 2. Solve the inequalitiesLeft side:$ -31 < 15x - 1 $Add 1 to both sides:$ -30 < 15x $Divide by 15:$ -2 < x $Right side:$ 15x - 1 \le 29 $Add 1 to both sides:$ 15x \le 30 $Divide by 15:$ x \le 2 $Step 3. Combine$ -2 < x \le 2 $Thus,$ A = (-2, 2] $✅Correct Option: A.$A = (-2, 2]$

Given the matrix

A%20%3D%20%5Cbegin%7Bbmatrix%7D%20%5Ccos%20%5Ctheta%20%26%200%20%26%20-%5Csin%20%5Ctheta%20%5C%5C%201%20%26%201%20%26%201%20%5C%5C%20%5Csin%20%5Ctheta%20%26%200%20%26%20%5Ccos%20%5Ctheta%20%5Cend%7Bbmatrix%7D

, we will determine the determinant of

A%5Ek

for any natural number

k

.The determinant of matrix

A

can be calculated as follows:

%5Cbegin%7Baligned%7D%20%5Ctext%7Bdet%7D(A)%20%26%3D%20%5Ccos%20%5Ctheta%20%5Ccdot%20(1%20%5Ccdot%20%5Ccos%20%5Ctheta%20-%201%20%5Ccdot%200)%20%5C%5C%20%26%5Cquad%20-%200%20%5Ccdot%20(1%20%5Ccdot%20%5Ccos%20%5Ctheta%20-%201%20%5Ccdot%20%5Csin%20%5Ctheta)%20%5C%5C%20%26%5Cquad%20%2B%20(-%5Csin%20%5Ctheta)%20%5Ccdot%20(1%20%5Ccdot%200%20-%201%20%5Ccdot%20%5Csin%20%5Ctheta).%20%5Cend%7Baligned%7D

Simplifying each term, we get:

%5Cbegin%7Baligned%7D%20%5Ctext%7Bdet%7D(A)%20%26%3D%20%5Ccos%20%5Ctheta%20%5Ccdot%20%5Ccos%20%5Ctheta%20-%20%5Csin%20%5Ctheta%20%5Ccdot%20(-%5Csin%20%5Ctheta)%20%5C%5C%20%26%3D%20%5Ccos%5E2%20%5Ctheta%20%2B%20%5Csin%5E2%20%5Ctheta.%20%5Cend%7Baligned%7D

Using the Pythagorean identity, we know that

%5Ccos%5E2%20%5Ctheta%20%2B%20%5Csin%5E2%20%5Ctheta%20%3D%201

.Therefore,

%5Ctext%7Bdet%7D(A)%20%3D%201

.Since the determinant of a product of matrices is the product of their determinants, we have:

%5Ctext%7Bdet%7D(A%5Ek)%20%3D%20(%5Ctext%7Bdet%7D(A))%5Ek%20%3D%201%5Ek%20%3D%201.

Thus, for any natural number

k

, the determinant of

A%5Ek

is 1.

Firstly,$

%5Cbegin%7Barray%7D%7Bcc%7D%20x_1%2Bx_2%3Dx_2%2Bx_3%20%26%20x_1%2Bx_2%3D0%20%5C%5C%20%5CRightarrow%20x_1%3Dx_3%20%26%20%5CRightarrow%20x_1%3D-x_2%20%5Cend%7Barray%7D

So%20let%2C

x_1%3Dx_3%3D-x_2%3Dx

Let%2C

%5Cbegin%7Bgathered%7D%20M%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Blll%7D%20a%20%26%20b%20%26%20c%20%5C%5C%20d%20%26%20e%20%26%20f%20%5C%5C%20g%20%26%20h%20%26%20i%20%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%20%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Blll%7D%20a%20%26%20b%20%26%20c%20%5C%5C%20d%20%26%20e%20%26%20f%20%5C%5C%20g%20%26%20h%20%26%20i%20%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bl%7D%20x_1%20%5C%5C%20x_2%20%5C%5C%20x_3%20%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bl%7D%20a%20x_1%2Bb%20x_2%2Bc%20x_3%20%5C%5C%20d%20x_1%2Be%20x_2%2Bf%20x_3%20%5C%5C%20g%20x_1%2Bh%20x_2%2Bi%20x_3%20%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bl%7D%200%20%5C%5C%200%20%5C%5C%200%20%5Cend%7Barray%7D%5Cright%5D%7D%20%5Cend%7Bgathered%7D

Which%20can%20be%20simplified%20to%2C

%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D%20x(a-b%2Bc)%20%5C%5C%20x(d-e%2Bf)%20%5C%5C%20x(g-h%2Bi)%20%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bl%7D%200%20%5C%5C%200%20%5C%5C%200%20%5Cend%7Barray%7D%5Cright%5D

%5Cbegin%7Baligned%7D%20%26a%2Bc%3Db%5C%20%26%5Ctext%20%7B%20%7D%5C%5C%20%26d%2Bf%3De%5C%5C%20%26g%2Bi%3Df%20%5Cend%7Baligned%7D

%5Cbegin%7Baligned%7D%20%26%5Ctext%20%7B%20Now%20you%20can%20see%2C%20%7D%5C%5C%20%26%5Cbegin%7Bgathered%7D%20C_1%2BC_3%3DC_2%20%5Cend%7Bgathered%7D%20%5Cend%7Baligned%7D

Now%20using%20the%20properties%20of%20determinants%2C

C_2 \rightarrow C_2-C_1-C_3

will%20not%20change%20the%20value%20of%20the%20determinant%3A

C_2%20%5Crightarrow%20C_2-%5Cleft(C_1%2BC_3%5Cright)%3D0

Now%20upon%20expanding%20along

C_2

%3A

%5Cbegin%7Bgathered%7D%20%7CM%7C%3D%5Cleft%7C%5Cbegin%7Barray%7D%7Bccc%7D%20a%20%26%20%5Cdownarrow%200%20%26%20c%20%5C%5C%20d%20%26%200%20%26%20f%20%5C%5C%20g%20%26%200%20%26%20i%20%5Cend%7Barray%7D%5Cright%7C%3D0%20%5C%5C%20%5Coperatorname%7Bdet%7D(M)%3D0%20%5Cend%7Bgathered%7D

Matrices And Determinants

High-Yield Trend

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6 MCQs

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Matrices And Determinants

High-Yield Trend

Questions 6 MCQs

Official Solution

Official Solution

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Official Solution

Official Solution

Questions
6 MCQs