Matrices - Class 12 Mathematics - Chapter 3 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Matrices | NCERT | Mathematics | Class 12
If $A$ is a matrix of order 3 and $|A|=8$, then $|Adj|A\rvert=$
A) 1
B) 2
C) $2^{5}$
D) $2^{6}$
The correct answer is D) (2^{6}).
To find the determinant of the adjugate matrix ( \text{adj}(A) ) when given ( |A| = 8 ) and knowing that ( A ) is a matrix of order 3, you use the property: $$ A \cdot \text{adj}(A) = \begin{bmatrix} |A| & 0 & 0 \ 0 & |A| & 0 \ 0 & 0 & |A| \end{bmatrix} $$ This implies: $$ |A| \cdot |\text{adj}(A)| = \left|\begin{bmatrix} |A| & 0 & 0 \ 0 & |A| & 0 \ 0 & 0 & |A| \end{bmatrix}\right| $$ The determinant of a diagonal matrix is the product of the diagonal elements. Thus: $$ |A| \cdot |\text{adj}(A)| = |A|^3 $$
Given that ( |A| = 8 ), we then have: $$ 8 \cdot |\text{adj}(A)| = 8^3 $$ or $$ |\text{adj}(A)| = \frac{8^3}{8} = 8^2 = 64 $$ Since ( 8 ) can be written as ( 2^3 ), we rewrite ( 8^2 ) as: $$ |\text{adj}(A)| = (2^3)^2 = 2^6 $$
Thus ( |\text{adj}(A)| = 2^6 ), which corresponds to answer choice D) (2^{6}).
If $A \times A$, i.e., $A^{2} = 1,$ then $A$ is said to be an involutary matrix.
A) True
B) False
Solution
The correct answer is Option A: True.
Recall the definition of an involutary matrix: a matrix $A$ is considered involutary if it satisfies two key conditions:
- $A$ must be a square matrix.
- The product $A \times A$ (or $A^2$) should equal the identity matrix $I$.
In this case, because it is stated that $A^2 = I$, we can deduce:
- $A$ is a square matrix since we can compute $A^2$.
- It satisfies the condition $A^2 = I$.
Thus, given these points, $A$ fulfills the criteria of being an involutary matrix, making the statement True.
$A$, $B$, $C$ are three square matrices of order $n$ such that $A = B + C$ and $C$ is a nilpotent matrix with index 2. If $B$ and $C$ commute, then $A^{10}$ is equal to:
A) $B^{10} + 12B^{9}C$
B) $B^{10} + 10B^{9}C$
C) $B^{10} + 12B^{10}C$
D) $B^{10} + 10B^{10}C$
Solution
The correct answer is B) $B^{10} + 10B^{9}C$.
Given that $C$ is a nilpotent matrix with index 2, this implies: $$ C^2 = 0 $$
Also, it's given that $B$ and $C$ commute, indicating: $$ BC = CB $$
Expanding $A$ which is $A = B + C$, we calculate $A^2$: $$ A^2 = (B+C)^2 = B^2 + BC + CB + C^2 = B^2 + 2BC + C^2 $$ Since $C^2 = 0$, we simplify to: $$ A^2 = B^2 + 2BC $$
Further, we calculate $A^3$: $$ A^3 = (B^2 + 2BC)(B + C) = B^3 + B^2C + 2BCB + 2BC^2 $$ Using the commutative property of $B$ and $C$, and $C^2=0$, we simplify to: $$ A^3 = B^3 + 3B^2C $$
Similarly, for higher powers of $A$, specifically $A^{10}$, it is generally found by the pattern: $$ A^n = B^n + nB^{n-1}C $$
Thus, evaluating for $n=10$: $$ A^{10} = B^{10} + 10B^9C $$
Therefore, the solution to $A^{10}$ based on the given conditions and calculations is $B^{10} + 10B^9C$.
In a third-order matrix $\mathrm{A}$, $a_{ij}$ denotes the element in the $i$-th row and $j$-th column. $$ \begin{array}{l} \text{If } a_{ij}=0 \text{ for } i=j \ =1 \text{ for } i>j \ =-1 \text{ for } i<j \end{array} $$
Then the matrix is
A) skew symmetric
B) symmetric
C) not invertible
D) non-singular
The matrix $A$ is defined as: $$ \begin{aligned} A = \begin{bmatrix} 0 & -1 & -1 \ 1 & 0 & -1 \ 1 & 1 & 0 \end{bmatrix} \end{aligned} $$
Criteria for Skew Symmetry:
- A matrix $A$ is skew-symmetric if $A^T = -A$, where $A^T$ is the transpose of $A$.
- For matrix $A$, we have: $$ A^T = \begin{bmatrix} 0 & 1 & 1 \ -1 & 0 & 1 \ -1 & -1 & 0 \end{bmatrix} $$ and clearly, $A^T = -A$. Thus, $A$ is skew-symmetric.
Singular or Non-invertible:
- A matrix is non-invertible (or singular) if its determinant is zero.
- The determinant of $A$ is: $$ \begin{aligned} |A| & = \left|\begin{array}{ccc} 0 & -1 & -1 \ 1 & 0 & -1 \ 1 & 1 & 0 \end{array}\right| = 0 \end{aligned} $$ Since the determinant $|A|$ is zero, $A$ is non-invertible.
Thus, the correct options are:
- A) skew-symmetric
- C) not invertible
Let $A$ and $B$ be $3 \times 3$ matrices of real numbers, where $A$ is symmetric, $B$ is skew-symmetric, and $(A+B)(A-B)=(A-B)(A+B)$. If $(AB)^{t}=(-1)^{k} AB$, where $(AB)^{t}$ is the transpose of matrix $AB$, then the possible values of $k$ are:
A. 1
B. 3
C. 5
D. All of these
The correct answer is D. All of these.
Starting from the commutation relation $(A+B)(A-B) = (A-B)(A+B)$, we simplify as follows:
$$ A^2 - AB + BA - B^2 = A^2 + AB - BA - B^2 $$
From the above, after simplifying and cancelling terms, we arrive at:
$$ AB - BA = 0 \quad \Rightarrow \quad AB = BA $$
This establishes that $A$ and $B$ commute.
Given the property of the transpose of the product $(AB)^t = (-1)^k AB$, we expand the left-hand side using $(XY)^t = Y^tX^t$ for any matrices $X$ and $Y$:
$$ (AB)^t = B^t A^t $$
Since $A$ is symmetric and $B$ is skew-symmetric, $A^t = A$ and $B^t = -B$. Substituting these into the equation gives us:
$$ -B A = (-1)^k AB $$
Multiplying both sides by $-1$ gives:
$$ BA = (-1)^{k+1} AB $$
Using the fact that $AB = BA$ (from commutation relation established earlier), we can rewrite it as:
$$ AB = (-1)^{k+1} AB $$
This equation leads to:
$$ (-1)^{k+1} = 1 $$
This means that $k+1$ must be an even number, thus $k$ must be odd.
Possible values of the odd integer $k$ that fit this condition are 1, 3, 5, and extrapolating, any odd positive integer. Thus, all the values given in the options are possible choices for $k$, leading to the answer D. All of these.
Matrix $X$ is such that $X^{2} = 2X - I$, where $I$ is the identity matrix. Then for $n \geq 2$, $X^{n}$ is equal to
A) $nX - (n-1)$
B) $nX - 1$
C) $2^{n}X - (n-1)$
D) $2^{n}X - 1$
The correct option is A) $nX - (n-1)I$.
We start with the given equation: $$ X^2 = 2X - I $$
We can proceed by induction. For $n = 2$, we indeed have $X^2 = 2X - I$ which matches with the induction base case $2X - 1I$.
Now, assume for induction that $X^n = nX - (n-1)I$. We need to show that $X^{n+1} = (n+1)X - nI$.
Using the induction hypothesis and the initial equation, we compute: $$ X^{n+1} = X \cdot X^n = X(nX - (n-1)I) = nX^2 - (n-1)X $$
Substitute $X^2$ using the given equation: $$ n(2X - I) - (n-1)X = 2nX - nI - (n-1)X = (n+1)X - nI $$
This confirms the pattern $$ X^{n} = nX - (n-1)I $$ is valid for all $n \geq 2$. Hence, the correct answer is option A) $nX - (n-1)I$.
If $A$ and $B$ are symmetric matrices of the same order and $X = AB + BA$ and $Y = AB - BA$, then $(XY)^{\top}$ is equal to
A $XY$ B $\mathrm{YX}$
C $-YX$
D None of these
The correct answer is C, $-YX$.
Given that $A$ and $B$ are symmetric matrices: $$ A^\top = A \quad \text{and} \quad B^\top = B. $$ For the matrix $X = AB + BA$, its transpose is: $$ X^\top = (AB + BA)^\top = (AB)^\top + (BA)^\top = B^\top A^\top + A^\top B^\top = BA + AB = X. $$ For the matrix $Y = AB - BA$, its transpose is: $$ Y^\top = (AB - BA)^\top = (AB)^\top - (BA)^\top = B^\top A^\top - A^\top B^\top = BA - AB = -Y. $$ Thus, $X$ is symmetric and $Y$ is skew-symmetric.
Now, to find $(XY)^\top$, we use the property $(XY)^\top = Y^\top X^\top$: $$ (XY)^\top = Y^\top X^\top = (-Y)X = -YX. $$ Therefore, the expression $(XY)^\top$ equals $-YX$. This aligns with option C.
If $A$ and $B$ are symmetric matrices, prove that $AB - BA$ is a skew-symmetric matrix.
To prove that $AB - BA$ is a skew-symmetric matrix when $A$ and $B$ are symmetric matrices, recall that a matrix $M$ is symmetric if $M^T = M$ and skew-symmetric if $M^T = -M$. Knowing that $A$ and $B$ are symmetric, we have:
$$ A^T = A \quad \text{and} \quad B^T = B $$
Now, consider the transpose of the expression $AB - BA$:
$$ (AB - BA)^T = (AB)^T - (BA)^T $$
Using the property of transpose over product $(XY)^T = Y^T X^T$, the equation simplifies to:
$$ (AB)^T - (BA)^T = B^T A^T - A^T B^T $$
Substituting back the symmetries of $A$ and $B$ (since $A^T = A$ and $B^T = B$):
$$ B^T A^T - A^T B^T = BA - AB $$
Observe that $BA - AB = -(AB - BA)$. Thus, we conclude:
$$ (AB - BA)^T = -(AB - BA) $$
Therefore, the matrix $AB - BA$ is skew-symmetric, since its transpose is the negative of itself:
$$ (AB - BA)^T = -(AB - BA) $$
This proves that $AB - BA$ is indeed a skew-symmetric matrix.
If $A=\left(a_{ij}\right){2\times 2}$, where $a{ij}=i+j$, then $A$ is equal to
(A) $\left[\begin{array}{ll}1 & 1 \ 2 & 2\end{array}\right]$ (B) $\left[\begin{array}{ll}1 & 2 \ 1 & 2\end{array}\right]$ (C) $\left[\begin{array}{ll}1 & 2 \ 3 & 4\end{array}\right]$ (D) $\left[\begin{array}{ll}2 & 3 \ 3 & 4\end{array}\right]$
The matrix $A = (a_{ij}){2 \times 2}$ is defined such that each element $a{ij} = i + j$. We can determine the values of $a_{ij}$ by substituting the row number for $i$ and the column number for $j$. Let's fill in the matrix:
- For $a_{11}$ (first row, first column): $$ a_{11} = 1 + 1 = 2 $$
- For $a_{12}$ (first row, second column): $$ a_{12} = 1 + 2 = 3 $$
- For $a_{21}$ (second row, first column): $$ a_{21} = 2 + 1 = 3 $$
- For $a_{22}$ (second row, second column): $$ a_{22} = 2 + 2 = 4 $$
Putting these values into the matrix form: $$ A = \left[\begin{array}{cc} 2 & 3 \ 3 & 4 \end{array}\right] $$
Comparing this matrix with the given options, the correct answer is (D) $\left[\begin{array}{ll}2 & 3 \ 3 & 4\end{array}\right]$.
In a factory, three types of toothbrushes are manufactured every day. On a certain day, the production is 570. The production of toothbrushes of the second kind exceeds the production of toothbrushes of the first kind by 100. Also, the total production of toothbrushes of the first and second kind is four times the production of the third kind.
If $x$, $y$, and $z$ denote the production of toothbrushes of three kinds, respectively, then using the matrix method, the algebraic representation of the given condition is:
(a) $\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 1 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 570 \\ 100 \\ 0 \end{bmatrix} $
(b) $\begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 1 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 570 \\ 100 \\ 0 \end{bmatrix} $
(c) $\begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 1 & -4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 570 \\ 100 \\ 0 \end{bmatrix} $
(d) $\begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 1 & -4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 570 \\ -100 \\ 0 \end{bmatrix} $
Here's a clear and structured explanation of how to represent the given conditions using a matrix method:
Problem Setup:
There are three types of toothbrushes manufactured: $x$, $y$, and $z$ represent the daily production of these types, respectively.
Total production is 570.
The production of the second type exceeds the production of the first type by 100 units.
The sum of the production of the first and second types is four times the production of the third type.
Translating the Situation into Equations:
Total Production: $$ x + y + z = 570 $$ This states that the sum of all three types of toothbrushes produced equals 570.
Production Difference: $$ y = x + 100 $$ Rearranging this, we get: $$ x - y = -100 $$ This represents that the production of the second type ($y$) exceeds that of the first type ($x$) by 100 units.
Production Ratio: $$ x + y = 4z $$ Rearranging, we have: $$ x + y - 4z = 0 $$ This equation shows that the sum of the production of the first and second types is four times the production of the third type ($z$).
Matrix Representation:
To use the matrix method, we can represent these equations in the form $\mathbf{A}\mathbf{x} = \mathbf{b}$.
Matrix $\mathbf{A}$ (Coefficient Matrix): $$\begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 1 & -4 \end{bmatrix} $$
Vector $\mathbf{x}$ (Variable Vector): $$ \begin{bmatrix} x \\ y \\ x \end{bmatrix} $$
Vector $\mathbf{b}$ (Constants Vector): $$ \begin{bmatrix} 570 \\ -100 \\ 0 \end{bmatrix} $$
The equations can be solved using matrix operations to find the values of $x$, $y$, and $z$, which provide the daily production rates of the three types of toothbrushes.
Final Answer:
Option (D) is correct as it aligns with the matrix representation we derived: $$ \mathbf{A}\mathbf{x} = \mathbf{b} $$
$$ \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 1 & -4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 570 \\ -100 \\ 0 \end{bmatrix} $$
This detailed setup and solution will assist in understanding the matrix method application in representing and solving a system of linear equations in real-world scenarios.
$\begin{bmatrix} -2x+y \\ x+y+z \\ x+y \end{bmatrix} = \begin{bmatrix} -3 \\ 3 \\ 3 \end{bmatrix}$, then the respective value of $x$, $y$ and $z$ are:
A) 0, 1, 2
B) 1, 2, 0
C) 2, 1, 0
D) 2, 0, 1
Let's solve the given system of equations step-by-step:
Equation 1:
$$ -2x + y = -3 $$
Equation 2:
$$ x + y + z = 3 $$
Equation 3:
$$ x + y = 3 $$
First, subtract Equation 3 from Equation 2: $$ (x + y + z) - (x + y) = 3 - 3 \ z = 0 $$
We find that $z = 0$.
Now using Equation 3: $$ x + y = 3 $$ Substitute the value of $z$ into Equation 2: $$ x + y + 0 = 3 $$ This simplifies back to $x + y = 3$, which we already know.
Now our main task is to find $x$ and $y$. We can use Equation 1 to express $y$ in terms of $x$: $$ y = 2x - 3 $$
Substitute $y$ from above into Equation 3: $$ x + (2x - 3) = 3 \\ 3x - 3 = 3 \\ 3x = 6 \\ x = 2 $$
Now substitute $x = 2$ back into the expression for $y$: $$ y = 2(2) - 3 \\ y = 4 - 3 \\ y = 1 $$
Therefore, the values are $x = 2$, $y = 1$, and $z = 0$. The correct option is C) 2, 1, 0.
If $A^{3} - A^{2} - 3A - I_{3} = O$ and $A^{2} = A$, then the inverse of matrix A is:
A. 2A B. $-3I$ C. $-\frac{1}{5A}$
Given the equation $$ A^{3} - A^{2} - 3A - I_{3} = O $$ and knowing that $$ A^{2} = A, $$ we can simplify the initial equation. Since $A^2 = A$, we replace $A^3$ in the equation by $A^2 \cdot A = A \cdot A = A^2$. By substituting back, the equation becomes: $$ A^2 - A^2 - 3A - I_3 = 0. $$ Simplifying this, we have: $$ -3A - I_3 = 0. $$ Rearranging gives: $$ 3A = -I_3. $$
To find the inverse of $A$, $A^{-1}$, we can multiply both sides of the equation by $\frac{1}{3}$: $$ A = -\frac{1}{3}I_3, $$ implying that $$ A^{-1} = -3I_3, $$ because the inverse of $\frac{1}{3}$ times the identity matrix $I_3$ is simply $-3$ times the identity matrix, as multiplying $A$ by $A^{-1}$ must yield the identity matrix $I_3$.
Therefore, the inverse of matrix $A$ is Option B: $-3I$.
If $A=\left[\begin{array}{cc}1 & 2 \ 3 & -1\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 3 \ -1 & 1\end{array}\right]$, then matrix $A B$ is a:
A. singular matrix
B. identity matrix
C. null matrix
D. non-singular matrix
To determine the nature of matrix $AB$ given matrices $A$ and $B$, we use the matrix multiplication rule. Given:
$$ A=\begin{bmatrix} 1 & 2 \ 3 & -1 \end{bmatrix}, \quad B=\begin{bmatrix} 1 & 3 \ -1 & 1 \end{bmatrix} $$
Performing the multiplication step-by-step:
For the element at the first row, first column of $AB$: $$ (1 \times 1) + (2 \times -1) = 1 - 2 = -1 $$
For the element at the first row, second column of $AB$: $$ (1 \times 3) + (2 \times 1) = 3 + 2 = 5 $$
For the element at the second row, first column of $AB$: $$ (3 \times 1) + (-1 \times -1) = 3 + 1 = 4 $$
For the element at the second row, second column of $AB$: $$ (3 \times 3) + (-1 \times 1) = 9 - 1 = 8 $$
So the product matrix $AB$ is:
$$ AB=\begin{bmatrix} -1 & 5 \ 4 & 8 \end{bmatrix} $$
Next, to ascertain the nature of $AB$ (whether it is singular or non-singular), we compute its determinant:
$$ \text{det}(AB) = (-1 \times 8) - (5 \times 4) = -8 - 20 = -28 $$
A matrix is singular if its determinant is $0$. Here, $\text{det}(AB) = -28 \neq 0$, which implies that $AB$ is non-singular.
In conclusion, the correct answer is D. non-singular matrix.
Simplify:
Matrix Component | Expression |
---|---|
First Component: $\cos \theta$ | $\begin{bmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{bmatrix}$ |
Second Component: $\sin \theta$ | $\begin{bmatrix} \sin \theta & -\cos \theta \ \cos \theta & \sin \theta \end{bmatrix}$ |
This table separates the two matrix components, with each associated with the respective trigonometric coefficient.
A. $\begin{bmatrix} 1 & 0 \ 1 & 0 \end{bmatrix}$
B. $\begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$
C. $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$
D. $\begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix}$
Let's break it down:
Step 1: Multiplication of Scalar and Matrix
First, for each matrix, multiply each element by the scalar outside the matrix:
$\cos\theta$ multiplying the first matrix: $$ \begin{bmatrix} \cos\theta\cos\theta & \cos\theta\sin\theta \ -\cos\theta\sin\theta & \cos\theta\cos\theta \end{bmatrix} = \begin{bmatrix} \cos^2\theta & \cos\theta\sin\theta \ -\cos\theta\sin\theta & \cos^2\theta \end{bmatrix} $$
$\sin\theta$ multiplying the second matrix: $$ \begin{bmatrix} \sin\theta\sin\theta & -\sin\theta\cos\theta \ \sin\theta\cos\theta & \sin\theta\sin\theta \end{bmatrix} = \begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta \ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix} $$
Step 2: Addition of the Two Resulting Matrices
Now, add the corresponding elements of these two matrices:
$$ \begin{bmatrix} \cos^2\theta & \cos\theta\sin\theta \ -\cos\theta\sin\theta & \cos^2\theta \end{bmatrix} + \begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta \ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix} = \begin{bmatrix} \cos^2\theta + \sin^2\theta & \cos\theta\sin\theta - \sin\theta\cos\theta \ -\cos\theta\sin\theta + \sin\theta\cos\theta & \cos^2\theta + \sin^2\theta \end{bmatrix} $$
Step 3: Simplify Using Trigonometric Identities
Using the Pythagorean identity, $\cos^2\theta + \sin^2\theta = 1$, and knowing that $\cos\theta\sin\theta - \sin\theta\cos\theta = 0$, the expression simplifies to:
$$ \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} $$
This is the identity matrix of size 2x2. Therefore, the simplified value of the given expression is:
$$ \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} $$
Based on the given options, the answer is:
C. $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$
If the system of linear equations $x-y=0$, $2x+3y+4z=17$ and $y+2z=7$ are written in matrix form as $PX=Q$, then the value of $|P|$ is:
A) 6
B) 8
C) 14
D) 20
To find the determinant $|P|$ of the matrix $P$ that corresponds to the given system of equations, let's first express these equations in matrix form:
The first equation is $x - y = 0$,
The second equation is $2x + 3y + 4z = 17$,
The third equation is $y + 2z = 7$.
By aligning the coefficients of $x$, $y$, and $z$ from each equation into rows, we can create the matrix $P$:
$$ P = \begin{bmatrix} 1 & -1 & 0 \ 2 & 3 & 4 \ 0 & 1 & 2 \ \end{bmatrix} $$
The corresponding matrix form of the equations is $P \mathbf{X} = \mathbf{Q}$, where $P$ is as above, $\mathbf{X}$ is the column matrix of variables $\begin{bmatrix} x \ y \ z \end{bmatrix}$, and $\mathbf{Q}$ is the constant matrix $\begin{bmatrix} 0 \ 17 \ 7 \end{bmatrix}$.
To find $|P|$, the determinant of matrix $P$, we use the determinant formula for a 3x3 matrix:
$$ |P| = a(ei − fh) − b(di − fg) + c(dh − eg) $$
For matrix $P$:
$a = 1$, $b = -1$, $c = 0$,
$d = 2$, $e = 3$, $f = 4$,
$g = 0$, $h = 1$, $i = 2$.
Substituting these values in: $$ |P| = 1(3 \cdot 2 - 4 \cdot 1) - (-1)(2 \cdot 2 - 4 \cdot 0) + 0(2 \cdot 1 - 3 \cdot 0) $$
Simplify: $$ |P| = 1(6 - 4) + 1(4 - 0) + 0 $$ $$ |P| = 1 \cdot 2 + 1 \cdot 4 $$ $$ |P| = 2 + 4 $$ $$ |P| = 6 $$
Therefore, the value of $|P|$ is 6. The correct answer is:
A) 6
The adjoint of matrix $\begin{bmatrix} 3 & 2 \\ -1 & 4 \end{bmatrix}$ is:
A $\begin{bmatrix} 4 & 1 \\ -2 & 3 \end{bmatrix}$
B $\begin{bmatrix} 4 & -2 \\ 1 & 3 \end{bmatrix}$
C $\begin{bmatrix} 4 & -1 \\ 2 & 3 \end{bmatrix}$
D $\begin{bmatrix} 4 & 2 \\ -1 & 3 \end{bmatrix}$
To find the adjoint of the matrix $$ \begin{bmatrix} 3 & 2 \\ -1 & 4 \end{bmatrix} $$ we first need to understand the process of calculating an adjoint for a $2 \times 2$ matrix.
Given a $2 \times 2$ matrix $$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$ the adjoint is obtained as follows:
Swap the elements $a$ and $d$.
Change the signs of elements $b$ and $c$.
Applying this procedure to our matrix:
The original matrix elements are: $$ a=3, \quad b=2, \quad c=-1, \quad d=4. $$
After swapping $a$ and $d$, and changing the signs of $b$ and $c$, we get:
$$\begin{bmatrix} 4 & -2 \\ 1 & 3 \end{bmatrix}$$
Thus, the adjoint of the matrix is: $$\begin{bmatrix} 4 & -2 \\ 1 & 3 \end{bmatrix}$$
The answer to the question is: Option B $\left[\begin{array}{cc}4 & -2 \\ 1 & 3\end{array}\right]$. This matrix correctly reflects the changes in position, and sign switching as per the rules for finding the adjoint of a $2 \times 2$ matrix.
If $\mathrm{A}$ is a matrix of order $3 \times 3$ and $|A|=4$, then the value of $|\operatorname{adj} \mathrm{A}|$ is:
A) 4
B) 16
C) 64
D) 12
To solve the problem, we need to find the value of $|\operatorname{adj} \mathrm{A}|$ given that $\mathrm{A}$ is a $3 \times 3$ matrix and $|A|=4$.
For any square matrix $\mathrm{A}$, the relationship between the determinant of $\mathrm{A}$ and its adjugate (adjoint) $\operatorname{adj}(\mathrm{A})$ is given by: $$ |\operatorname{adj}(\mathrm{A})| = |A|^{n-1} $$ where $n$ is the order of the square matrix.
For the case given:
$\mathrm{A}$ is a $3 \times 3$ matrix, hence $n = 3$.
$|A| = 4$.
Substituting these values into the formula: $$ |\operatorname{adj}(\mathrm{A})| = 4^{3-1} = 4^2 = 16 $$
Thus, the value of $|\operatorname{adj} \mathrm{A}|$ is 16. Therefore, the correct answer is B) 16.
From the matrix equation $AB = AC$, we conclude that $B = C$ provided:
A is a symmetric matrix,
B is a singular matrix,
C is a skew-symmetric matrix,
D is a non-singular matrix.
In the matrix equation $AB = AC$, the conclusion that $B = C$ holds true under specific conditions related to the matrix $A$. Let's go through a detailed explanation of how this works:
When given $AB = AC$, we can attempt to simplify or manipulate this equation to isolate $B$ and $C$:
Subtract $AC$ from both sides: $$ AB - AC = 0 $$
Factor out $A$ from the left side: $$ A(B - C) = 0 $$
This result implies that either $A = 0$ or $(B - C) = 0$. If $A \neq 0$, for $B = C$ to be surely concluded without any additional information on $B$ or $C$, the matrix $A$ must have another specific property.
Matrix $A$ must be invertible: To further isolate $B - C$, we must multiply both sides by the inverse of $A$. This is valid only if the inverse of $A$ exists. The existence of $A^{-1}$ implies that $A$ is a non-singular matrix (determinant of $A$ is not zero). The operations would be:
$$ A^{-1}(A(B - C)) = A^{-1}0 $$
$$ (A^{-1}A)(B-C) = 0 $$
$$ I(B - C) = 0 $$
$$ B - C = 0 $$
$$ B = C $$
Here, $I$ is the identity matrix. From the above steps, it is clear that we required $A$ to be invertible to ensure $B = C$. This need for $A$ to be non-singular determines the conditions under which $B = C$ can be conclusively said from $AB = AC$.
Further information about matrix properties in provided options:
A is a symmetric matrix: This property relates to the form of $A$ but does not guarantee invertibility.
B is a singular matrix: This discusses properties about $B$ not $A$.
C is a skew-symmetric matrix: This again deals with the nature of $C$, not $A$.
D is a non-singular matrix: As discussed, the non-singularity of $A$ (invertibility) is crucial for concluding $B = C$ from $AB = AC$.
Given these considerations, the most suitable answer to the question is option D. This is because a non-singular (invertible) matrix $A$ ensures that moving from $AB = AC$ to $B = C$ is valid, aligning directly with the required manipulation for proving $B = C$.
Adjoint of the matrix $N\left[\begin{array}{ccc}-4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{array}\right]$ is:
A $\mathrm{N}$
B $2 \mathrm{~N}$
C $-\mathrm{N}$
D None of these
The correct option is $\mathbf{A}$: $\mathrm{N}$.
Given the matrix
$$ N = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix} $$
We need to determine the adjoint of the matrix $N$.
To find the adjoint, we first need the cofactors of the matrix $N$. The cofactors are determined as follows:
$c_{11} = -4$
$c_{12} = 1$
$c_{13} = 4$
$c_{21} = -3$
$c_{22} = 0$
Other cofactors can be calculated similarly.
The adjoint matrix $\operatorname{adj} N$ is the transpose of the cofactor matrix. However, in this particular case, upon calculation, the adjoint ends up being the same as the original matrix $N$. Therefore,
$$ \operatorname{adj} N = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix} = N $$
Thus, the adjoint of the given matrix $N$ is indeed $\mathbf{N}$, confirming that the correct option is $\mathbf{A}$.
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