ID: 218 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 1: If \[ A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \] then \(A^{-1}\) is

• A. \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}
• B. \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}
• C. \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
• D. \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

ID: 223 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 2: A is a square matrix of order 2 such that det(A)=4 , then det(4adjA) is

• A. 16
• B. 64
• C. 256
• D. 512

ID: 227 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 3: Let \[ A = \begin{pmatrix} 1 & -2 & -1 \\ 0 & 4 & -1 \\ -3 & 2 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} -2 \\ -5 \\ -7 \end{pmatrix}, \quad C = \begin{pmatrix} 9 & 8 & 7 \end{pmatrix}, \] which of the following is defined?

• A. Only AB
• B. Only AC
• C. Only BA
• D. All AB, AC and BA

View Solution

✅ Manual Solution:

The problem is to determine which matrix multiplication is defined. A product of two matrices $P \times Q$ is defined if and only if the number of columns of $P$ equals the number of rows of $Q$.

The order of Matrix A is 3×3

The order of Matrix B is 3×1 and

The order of Matrix C is 1×3

Checking Products

1. AB: $A_{3\times \mathbf{3}}\times B_{\mathbf{3}\times 1}$ $\to 3=3$ . Defined.

2. AC: $A_{3\times \mathbf{3}}\times C_{\mathbf{1}\times 3}$ $\to 3\ne 1$ . Not Defined.

3. BA: $B_{3\times \mathbf{1}}\times A_{\mathbf{3}\times 3}$ $\to 1\ne 3$ . Not Defined.

Hence The correct option is Only AB.

ID: 232 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 4: If \[ A = \begin{pmatrix} 7 & 0 & x \\ 0 & 7 & 0 \\ 0 & 0 & y \end{pmatrix} \] is a scalar matrix, then \(y^x\) is equal to

• A. \(0\)
• B. \(1\)
• C. \(7\)
• D. \(\pm 7\)

View Solution

✅ Manual Solution:

A scalar matrix is a diagonal matrix where all the elements on the main diagonal are equal, and all non-diagonal elements are zero.

Here \(x=0\) and \(y=7\), So \(y^x\) is equal to 1

ID: 234 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 5: If \(A\) and \(B\) are invertible matrices, then which of the following is \underline{not correct}?

• A. \((A + B)^{-1} = B^{-1} + A^{-1}\)
• B. \((AB)^{-1} = B^{-1} A^{-1}\)
• C. \(\text{adj}(A) = |A| A^{-1}\)
• D. \(|A^{-1}| = |A|^{-1}\)

ID: 243 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 6: If \(A=\begin{bmatrix}1&12&4y\\ 6x&5&2x\\ 8x&4&6\end{bmatrix}\) is a symmetric matrix, then \((2x+y)\) is

• A. \(-8\)
• B. 0
• C. 6
• D. 8

ID: 247 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 7: Which of the following can be both a symmetric and skew-symmetric matrix ?

• A. Unit Matrix
• B. Diagonal Matrix
• C. Null Matrix
• D. Row Matrix

ID: 249 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 8: Four friends Abhay, Bina, Chhaya and Devesh were asked to simplify \(4~AB+3(AB+BA)-4~BA,\) where A and B are both matrices of order \(2\times2\).

It is known that \(A\ne B\ne I\) and \(A^{-1}\ne B\).

Their answers are given as:

Abhay: \(6 AB\),

Bina : \(7 AB-BA\),

Chhaya: \(8 AB\),

Devesh: \(7 BA - AB\).

Who answered it correctly?

• A. Abhay
• B. Bina
• C. Chhaya
• D. Devesh

View Solution

✅ Manual Solution:

\(4~AB+3(AB+BA)-4~BA\)

=\(4~AB+3AB+3BA-4~BA\)

\(=7AB-BA\)

Hence Bina answered it correctly

ID: 256 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 9: If A and B are square matrices of order m such that \(A^{2}-B^{2}=(A-B)(A+B),\) then which of the following is always correct?

• A. \(A=B\)
• B. \(AB=BA\)
• C. \(A=0\) or \(B=0\)
• D. \(A=I\) or \(B=I\)

View Solution

✅ Manual Solution:

The identity $A^2 - B^2 = (A - B)(A + B)$ holds for square matrices $A$ and $B$ only if they commute, i.e., $AB = BA$. This is because matrix multiplication is not generally commutative.

ID: 296 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 10: If M and N are square matrices of order 3 such that det \((M)=m\) and \(MN=mI,\) then det (N) is equal to:

• A. -1
• B. 1
• C. \(-m^{2}\)
• D. \(m^{2}\)

View Solution

✅ Manual Solution:

We use two key properties of determinants:

1. Product Rule: \(\text{det}(AB) = \text{det}(A) \cdot \text{det}(B)\)
2. Scalar Multiplication Rule: \(\text{det}(c A) = c^n \cdot \text{det}(A)\) (where \(n\) is the order of the matrix)

We take the determinant of both sides of the equation \(MN = mI\):

\[\text{det}(MN) = \text{det}(mI)\]

Since the order \(n=3\) and \(\text{det}(I) = 1\), the right side simplifies to \(\text{det}(mI) = m^3 \cdot 1 = m^3\).

Substitute \(\text{det}(M) = m\) and \(\text{det}(mI) = m^3\) we get :

\[m \cdot \text{det}(N) = m^3\] \[\mathbf{\text{det}(N) = m^2}\]

ID: 310 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 11: If \(A=\begin{bmatrix}1&2&3\\ -4&3&7\end{bmatrix}\) and \(B=\begin{bmatrix}4&3\\ -1&2\\ 0&5\end{bmatrix},\) then the correct statement is:

• A. Only AB is defined.
• B. Only BA is defined.
• C. AB and BA, both are defined.
• D. AB and BA, both are not defined.

ID: 311 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 12: If \(\begin{vmatrix}2x&5\\ 12&x\end{vmatrix}=\begin{vmatrix}6&-5\\ 4&3\end{vmatrix}\) then the value of x is:

• A. 3
• B. 7
• C. \(\pm7\)
• D. \(\pm3\)

ID: 313 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 13: If \(A=[a_{ij}]\) is a \(3\times3\) diagonal matrix such that \(a_{11}=1\), \(a_{22}=5\) and \(a_{33}=-2\), then \(|A|\) is:

• A. 0
• B. -10
• C. 10
• D. 1

ID: 315 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 14: If \(\begin{bmatrix}4+x&x-1\\ -2&3\end{bmatrix}\) is a singular matrix, then the value of x is:

• A. 0
• B. 1
• C. -2
• D. 4

ID: 326 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 15: Let both \(AB^{\prime}\) and \(B^{\prime}A\) be defined for matrices A and B. If order of A is \(n\times m\), then the order of B is:

• A. \(n\times n\)
• B. \(n\times m\)
• C. \(m\times m\)
• D. \(m\times n\)

ID: 327 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 16: If \(A=\begin{bmatrix}-1&0&0\\ 0&3&0\\ 0&0&5\end{bmatrix},\) then A is a/an:

• A. scalar matrix
• B. identity matrix
• C. symmetric matrix
• D. skew-symmetric matrix

ID: 329 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 17: Sum of two skew-symmetric matrices of same order is always a/an:

• A. skew-symmetric matrix
• B. symmetric matrix
• C. null matrix
• D. identity matrix

ID: 346 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 18: What is the total number of possible matrices of order \(3\times3\) with each entry as \(\sqrt{2}\) or \(\sqrt{3}\)?

• A. 9
• B. 512
• C. 615
• D. 64

ID: 347 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 19: The matrix \(A=\begin{bmatrix}\sqrt{3}&0&0\\ 0&\sqrt{2}&0\\ 0&0&\sqrt{5}\end{bmatrix}\) is a/an:

• A. scalar matrix
• B. identity matrix
• C. null matrix
• D. symmetric matrix

ID: 348 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 20: If A and B are two square matrices each of order 3 with \(|A|=3\) and \(|B|=5\), then \(|2AB|\) is:

• A. 30
• B. 120
• C. 15
• D. 225

ID: 349 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 21: Let A be a square matrix of order 3. If \(|A|=5\), then \(|\operatorname{adj} A|\) is:

• A. 5
• B. 125
• C. 25
• D. -5

ID: 350 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 22: If \(\begin{bmatrix}2x-1&3x\\ 0&y^{2}-1\end{bmatrix}=\begin{bmatrix}x+3&12\\ 0&35\end{bmatrix},\) then the value of \((x-y)\) is :

• A. 2 or 10
• B. 2 or 10
• C. 2 or - 10
• D. -2 or - 10

ID: 471 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 23: If a matrix has 36 elements, the number of possible orders it can have, is:

• A. 13
• B. 3
• C. 5
• D. 9

ID: 474 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 24: If \(\begin{bmatrix}x+y&2\\ 5&xy\end{bmatrix}=\begin{bmatrix}6&2\\ 5&8\end{bmatrix},\) then the value of \((\frac{24}{x}+\frac{24}{y})\) is:

• A. 7
• B. 6
• C. 8
• D. 18

ID: 481 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 25: \(\begin{vmatrix}x+1&x-1\\ x^{2}+x+1&x^{2}-x+1\end{vmatrix}\) is equal to:

• A. \(2x^{3}\)
• B. 2
• C. 0
• D. \(2x^{3}-2\)

ID: 487 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 26: If A and B are two non-zero square matrices of same order such that \((A+B)^{2}=A^{2}+B^{2}\) then :

• A. \(AB=O\)
• B. \(AB=-BA\)
• C. \(BA=O\)
• D. \(AB=BA\)

ID: 488 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 27: If the sum of all the elements of a \(3\times3\) scalar matrix is 9, then the product of all its elements is:

• A. 0
• B. 9
• C. 27
• D. 729

ID: 490 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 28: If \(\begin{vmatrix}-a&b&c\\ a&-b&c\\ a&b&-c\end{vmatrix}= kabc,\) then the value of k is:

• A. 0
• B. 1
• C. 2
• D. 4

ID: 499 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 29: If \(A=[a_{ij}]\) be a \(3\times3\) matrix, where \(a_{ij}=i-3j\), then which of the following is false ?

• A. \(a_{11}\lt0\)
• B. \(a_{12}+a_{21}=-6\)
• C. \(a_{13}\gt a_{31}\)
• D. \(a_{31}=0\)

ID: 504 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 30: If \(F(x)=\begin{bmatrix}\cos~x&-\sin~x&0\\ \sin~x&\cos~x&0\\ 0&0&1\end{bmatrix}\) and \([F(x)]^{2}=F(kx)\), then the value of k is :

• A. 1
• B. 2
• C. 0
• D. -2

ID: 506 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 31: If \(A=[a_{ij}]\) is an identity matrix, then which of the following is true ?

• A. \(a_{ij}=\begin{cases}0,&if~i=j\\ 1,&if~i\ne j\end{cases}\)
• B. \(a_{ij}=1,\forall i,j\)
• C. \(a_{ij}=0,\forall i,j\)
• D. \(a_{ij}=\begin{cases}0,&if~i\ne j\\ 1,&if~i=j\end{cases}\)

ID: 508 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 32: Let \(A=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\) be a square matrix such that adj \(A=A\) Then, \((a+b+c+d)\) is equal to :

• A. 2a
• B. 2b
• C. 2c
• D. 0

ID: 517 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 33: If A and B are two skew symmetric matrices, then \((AB+BA)\) is :

• A. a skew symmetric matrix
• B. a symmetric matrix
• C. a null matrix
• D. an identity matrix

ID: 518 Type: Mcq Year: 2024

Question 34: If \(\begin{vmatrix}1&3&1\\ k&0&1\\ 0&0&1\end{vmatrix}=\pm6,\) then the value of k is:

• A. 2
• B. -2
• C. \(\pm2\)
• D. \(\mp2\)

View AI Solution

🤖 AI Generated Solution:

**Correct Option if MCQ:** C **Reasoning:** * \(1(0-0) - 3(k-0) + 1(0-0) = -3k\) * \(-3k = \pm6\) * \(k = \mp2\) (which is \(\pm2\))

ID: 523 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 35: If \(A=\begin{bmatrix}2&0&0\\ 0&3&0\\ 0&0&5\end{bmatrix},\) then \(A^{-1}\) is:

• A. \([\begin{matrix}\frac{1}{2}&0&0\\ 0&3&0\\ 0&0&\frac{1}{5}\end{matrix}]\)
• B. \(30[\begin{matrix}\frac{1}{2}&0&0\\ 0&\frac{1}{3}&0\\ 0&0&\frac{1}{5}\end{matrix}]\)
• C. \(\frac{1}{30}[\begin{matrix}2&0&0\\ 0&3&0\\ 0&0&5\end{matrix}]\)
• D. \(\frac{1}{30}[\begin{matrix}\frac{1}{2}&0&0\\ 0&\frac{1}{3}&0\\ 0&0&\frac{1}{5}\end{matrix}]\)

ID: 524 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 36: \(If\begin{bmatrix}a&c&0\\ b&d&0\\ 0&0&5\end{bmatrix}\) is a scalar matrix, then the value of \(a+2b+3c+4d\) is:

• A. 0
• B. 5
• C. 10
• D. 25

ID: 525 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 37: Given that \(A^{-1}=\frac{1}{7}\begin{bmatrix}2&1\\ -3&2\end{bmatrix}\) matrix A is

• A. \(7[\begin{matrix}2&-1\\ 3&2\end{matrix}]\)
• B. \([\begin{matrix}2&-1\\ 3&2\end{matrix}]\)
• C. \(\frac{1}{7}[\begin{matrix}2&-1\\ 3&2\end{matrix}]\)
• D. \(\frac{1}{49}[\begin{matrix}2&-1\\ 3&2\end{matrix}]\)

ID: 526 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 38: \(If~A=[\begin{matrix}2&1\\ -4&-2\end{matrix}].\) then the value of \(I-A+A^{2}-A^{3}+...is\): (1)

• A. \([\begin{matrix}-1&-1\\ 4&3\end{matrix}]\)
• B. \([\begin{matrix}3&1\\ -4&-1\end{matrix}]\)
• C. \([\begin{matrix}0&0\\ 0&0\end{matrix}]\)
• D. \([\begin{matrix}1&0\\ 0&1\end{matrix}]\)

ID: 527 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 39: \(If~A=\begin{bmatrix}-2&0&0\\ 1&2&3\\ 5&1&-1\end{bmatrix},\) then the value of | A (adj. A) | is:

• A. 100 I
• B. 10 I
• C. 10
• D. 1000

ID: 528 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 40: Given that \([\begin{matrix}1&x\end{matrix}]\begin{bmatrix}4&0\\ -2&0\end{bmatrix}=0,\) the value of x is:

• A. -4
• B. -2
• C. 2
• D. 4

ID: 543 Type: Mcq Year: 2024

Question 41: If \(A=\begin{bmatrix}a&c&-1\\ b&0&5\\ 1&-5&0\end{bmatrix}\) is a skew-symmetric matrix, then the value of \(2a-(b+c)\) is:

• A. 0
• B. 1
• C. -10
• D. 10

View AI Solution

🤖 AI Generated Solution:

A matrix $A$ is skew-symmetric if its transpose is equal to its negative, i.e., $A^T = -A$. This implies two main properties for the elements of the matrix: 1. All diagonal elements are zero ($a_{ii} = 0$). 2. Elements symmetric with respect to the main diagonal are opposite in sign ($a_{ij} = -a_{ji}$). Given $A=\begin{bmatrix}a&c&-1\\ b&0&5\\ 1&-5&0\end{bmatrix}$ is a skew-symmetric matrix. From property 1, the diagonal elements must be zero: $a_{11} = a = 0$ From property 2, for $a_{12}$ and $a_{21}$: $c = -b \implies b+c=0$ Now substitute the values of $a$ and $b+c$ into the expression $2a-(b+c)$: $2a-(b+c) = 2(0) - (0) = 0 - 0 = 0$ The final answer is $\boxed{0}$.

ID: 544 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 42: If A is a square matrix of order 3 such that the value of \(|adj\cdot A|=8,\) then the value of \(|A^{T}|\) is: (1)

• A. \(\sqrt{2}\)
• B. \(-\sqrt{2}\)
• C. 8
• D. \(2\sqrt{2}\)

ID: 545 Type: Mcq Year: 2024

Question 43: If inverse of matrix \(\begin{bmatrix}1&3&3\\ 1&\lambda&3\\ 1&3&4\end{bmatrix}\) is the matrix \(\begin{bmatrix}7&-3&-3\\ -1&1&0\\ -1&0&1\end{bmatrix},\) then value of \(\lambda\) is:

• A. -4
• B. 1
• C. 3
• D. 4

View AI Solution

🤖 AI Generated Solution:

Let the given matrix be $A = \begin{bmatrix}1&3&3\\ 1&\lambda&3\\ 1&3&4\end{bmatrix}$ and its inverse be $A^{-1} = \begin{bmatrix}7&-3&-3\\ -1&1&0\\ -1&0&1\end{bmatrix}.$ We know that the product of a matrix and its inverse is the identity matrix, i.e., $A A^{-1} = I.$ $\begin{bmatrix}1&3&3\\ 1&\lambda&3\\ 1&3&4\end{bmatrix} \begin{bmatrix}7&-3&-3\\ -1&1&0\\ -1&0&1\end{bmatrix} = \begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}.$ Consider the element in the second row and first column of the product $(A A^{-1})_{21}$: $(1)(7) + (\lambda)(-1) + (3)(-1) = 0$ $7 - \lambda - 3 = 0$ $4 - \lambda = 0$ $\lambda = 4.$ Alternatively, consider the element $(A A^{-1})_{22}$: $(1)(-3) + (\lambda)(1) + (3)(0) = 1$ $-3 + \lambda = 1$ $\lambda = 4.$ The final answer is $\boxed{4}$.

ID: 546 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 44: If \(\begin{bmatrix}x&2&0\end{bmatrix}\begin{bmatrix}5\\ -1\\ x\end{bmatrix}=\begin{bmatrix}3&1\end{bmatrix}\begin{bmatrix}-2\\ x\end{bmatrix},\) then value of x is:

• A. -1
• B. 0
• C. 1
• D. 2

ID: 547 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 45: Find the matrix \(A^{2}\), where \(A=[a_{ij}]\) is a \(2\times2\) matrix whose elements are given by \(a_{ij}=\) maximum (i, j) - minimum (i, j):

• A. \([\begin{matrix}0&0\\ 0&0\end{matrix}]\)
• B. \([\begin{matrix}1&0\\ 0&1\end{matrix}]\)
• C. \([\begin{matrix}0&1\\ 1&0\end{matrix}]\)
• D. \([\begin{matrix}1&1\\ 1&1\end{matrix}]\)

ID: 258 Type: Assertion-reason Source: AISSCE(Board Exam) Year: 2025

Question 46:

Assertion (A): \(A=\) diag [3 5 2] is a scalar matrix of order \(3\times3.\)

Reason (R) : If a diagonal matrix has all non-zero elements equal, it is known as a scalar matrix.

ID: 362 Type: Very short (VSA) Source: AISSCE(Board Exam) Year: 2025

Question 47: Let A and B be two square matrices of order 3 such that det (A) \(=3\) and det (B) \(=-4\). Find the value of det (-6AB).

View AI Solution

🤖 AI Generated Solution:

To find the value of $\det(-6AB)$, we will use the properties of determinants. **Given:** * A and B are two square matrices of order 3. * $\det (A) = 3$ * $\det (B) = -4$ **Properties of Determinants:** For square matrices A and B of order $n$, and a scalar $k$: 1. $\det(kA) = k^n \det(A)$ 2. $\det(AB) = \det(A) \det(B)$ **Solution:** Here, the order of the matrices, $n=3$. We need to find $\det(-6AB)$. Applying the first property, with $k=-6$ and considering the matrix as $(AB)$: $$ \det(-6AB) = (-6)^3 \det(AB) $$ Now, we need to find $\det(AB)$. Using the second property: $$ \det(AB) = \det(A) \det(B) $$ Substitute the given values of $\det(A)$ and $\det(B)$: $$ \det(AB) = (3) \times (-4) $$ $$ \det(AB) = -12 $$ Now, substitute this value back into the expression for $\det(-6AB)$: $$ \det(-6AB) = (-6)^3 \times (-12) $$ $$ \det(-6AB) = (-216) \times (-12) $$ Performing the multiplication: $$ \det(-6AB) = 2592 $$ **Final Answer:** The value of $\det (-6AB)$ is $\boxed{2592}$.

ID: 363 Type: Very short (VSA) Source: AISSCE(Board Exam) Year: 2025

Question 48: Let A and B be two square matrices of order 3 such that det (A) \(=3\) and det (B) \(=-4\). Find the value of det (-6AB).

ID: 448 Type: Very short (VSA) Source: AISSCE(Board Exam) Year: 2025

Question 49: If \(A=\begin{bmatrix}2&3\\ -1&2\end{bmatrix}\), then show that \(A^{2}-4A+7I=0\).

View AI Solution

🤖 AI Generated Solution:

**Question:** If \(A=\begin{bmatrix}2&3\\ -1&2\end{bmatrix}\), then show that \(A^{2}-4A+7I=0\). **Solution:** Given the matrix \(A=\begin{bmatrix}2&3\\ -1&2\end{bmatrix}\). We need to show that \(A^{2}-4A+7I=0\), where \(I\) is the identity matrix of order 2 and \(0\) is the zero matrix of order 2. **Step 1: Calculate \(A^2\)** $$A^2 = A \times A = \begin{bmatrix}2&3\\ -1&2\end{bmatrix} \begin{bmatrix}2&3\\ -1&2\end{bmatrix}$$ $$A^2 = \begin{bmatrix}(2)(2)+(3)(-1) & (2)(3)+(3)(2) \\ (-1)(2)+(2)(-1) & (-1)(3)+(2)(2)\end{bmatrix}$$ $$A^2 = \begin{bmatrix}4-3 & 6+6 \\ -2-2 & -3+4\end{bmatrix}$$ $$A^2 = \begin{bmatrix}1 & 12 \\ -4 & 1\end{bmatrix}$$ **Step 2: Calculate \(4A\)** $$4A = 4 \begin{bmatrix}2&3\\ -1&2\end{bmatrix}$$ $$4A = \begin{bmatrix}4 \times 2 & 4 \times 3 \\ 4 \times -1 & 4 \times 2\end{bmatrix}$$ $$4A = \begin{bmatrix}8 & 12 \\ -4 & 8\end{bmatrix}$$ **Step 3: Calculate \(7I\)** The identity matrix \(I\) for a $2 \times 2$ matrix is \(\begin{bmatrix}1&0\\ 0&1\end{bmatrix}\). $$7I = 7 \begin{bmatrix}1&0\\ 0&1\end{bmatrix}$$ $$7I = \begin{bmatrix}7 \times 1 & 7 \times 0 \\ 7 \times 0 & 7 \times 1\end{bmatrix}$$ $$7I = \begin{bmatrix}7 & 0 \\ 0 & 7\end{bmatrix}$$ **Step 4: Substitute the calculated values into the expression \(A^{2}-4A+7I\)** $$A^{2}-4A+7I = \begin{bmatrix}1 & 12 \\ -4 & 1\end{bmatrix} - \begin{bmatrix}8 & 12 \\ -4 & 8\end{bmatrix} + \begin{bmatrix}7 & 0 \\ 0 & 7\end{bmatrix}$$ Perform the subtraction first: $$\begin{bmatrix}1 & 12 \\ -4 & 1\end{bmatrix} - \begin{bmatrix}8 & 12 \\ -4 & 8\end{bmatrix} = \begin{bmatrix}1-8 & 12-12 \\ -4-(-4) & 1-8\end{bmatrix}$$ $$ = \begin{bmatrix}-7 & 0 \\ -4+4 & -7\end{bmatrix}$$ $$ = \begin{bmatrix}-7 & 0 \\ 0 & -7\end{bmatrix}$$ Now, add \(7I\): $$\begin{bmatrix}-7 & 0 \\ 0 & -7\end{bmatrix} + \begin{bmatrix}7 & 0 \\ 0 & 7\end{bmatrix} = \begin{bmatrix}-7+7 & 0+0 \\ 0+0 & -7+7\end{bmatrix}$$ $$ = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$$ This is the zero matrix of order 2, denoted by \(0\). Therefore, \(A^{2}-4A+7I=0\). Hence, it is shown that \(A^{2}-4A+7I=0\).

ID: 432 Type: Short (SA) Source: AISSCE(Board Exam) Year: 2025

Question 50: Let \(A=\begin{bmatrix}1\\ 4\\ -2\end{bmatrix}\) and \(C=\begin{bmatrix}3&4&2\\ 12&16&8\\ -6&-8&-4\end{bmatrix}\). Find the matrix B if \(AB=C\).

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**Solution:** **1. Given Matrices:** We are given the matrices: $$A=\begin{bmatrix}1\\ 4\\ -2\end{bmatrix} \quad \text{and} \quad C=\begin{bmatrix}3&4&2\\ 12&16&8\\ -6&-8&-4\end{bmatrix}$$ We are also given the matrix equation \(AB=C\). **2. Determine the Dimensions of Matrix B:** Let the dimension of matrix B be \(m \times n\). Matrix A has dimensions \(3 \times 1\). Matrix C has dimensions \(3 \times 3\). For the product \(AB\) to be defined, the number of columns in A must be equal to the number of rows in B. Number of columns in A = 1. Therefore, the number of rows in B, \(m\), must be 1. The dimension of the product matrix \(AB\) will be (rows of A) \(\times\) (columns of B). Dimension of \(AB = 3 \times n\). Since \(AB=C\), the dimension of \(AB\) must be equal to the dimension of C. So, \(3 \times n = 3 \times 3\). This implies \(n=3\). Thus, matrix B must be a \(1 \times 3\) matrix. **3. Represent Matrix B:** Let matrix B be represented as: $$B=\begin{bmatrix}b_1&b_2&b_3\end{bmatrix}$$ **4. Perform the Matrix Multiplication \(AB\):** Now, we multiply A by B: $$AB = \begin{bmatrix}1\\ 4\\ -2\end{bmatrix} \begin{bmatrix}b_1&b_2&b_3\end{bmatrix}$$ $$AB = \begin{bmatrix}1 \cdot b_1 & 1 \cdot b_2 & 1 \cdot b_3 \\ 4 \cdot b_1 & 4 \cdot b_2 & 4 \cdot b_3 \\ -2 \cdot b_1 & -2 \cdot b_2 & -2 \cdot b_3\end{bmatrix}$$ $$AB = \begin{bmatrix}b_1 & b_2 & b_3 \\ 4b_1 & 4b_2 & 4b_3 \\ -2b_1 & -2b_2 & -2b_3\end{bmatrix}$$ **5. Equate \(AB\) with \(C\):** We are given \(AB=C\), so we equate the product with the given matrix C: $$\begin{bmatrix}b_1 & b_2 & b_3 \\ 4b_1 & 4b_2 & 4b_3 \\ -2b_1 & -2b_2 & -2b_3\end{bmatrix} = \begin{bmatrix}3&4&2\\ 12&16&8\\ -6&-8&-4\end{bmatrix}$$ **6. Solve for the Elements of B:** By comparing the corresponding elements of the two matrices, we get a system of equations: From the first row: $$b_1 = 3$$ $$b_2 = 4$$ $$b_3 = 2$$ Let's verify these values with the other rows to ensure consistency: From the second row: $$4b_1 = 12 \implies 4(3) = 12 \implies 12 = 12 \quad \text{(Consistent)}$$ $$4b_2 = 16 \implies 4(4) = 16 \implies 16 = 16 \quad \text{(Consistent)}$$ $$4b_3 = 8 \implies 4(2) = 8 \implies 8 = 8 \quad \text{(Consistent)}$$ From the third row: $$-2b_1 = -6 \implies -2(3) = -6 \implies -6 = -6 \quad \text{(Consistent)}$$ $$-2b_2 = -8 \implies -2(4) = -8 \implies -8 = -8 \quad \text{(Consistent)}$$ $$-2b_3 = -4 \implies -2(2) = -4 \implies -4 = -4 \quad \text{(Consistent)}$$ All equations are consistent with the values \(b_1=3\), \(b_2=4\), and \(b_3=2\). **7. Final Answer:** Therefore, the matrix B is: $$B=\begin{bmatrix}3&4&2\end{bmatrix}$$ The final answer is $\boxed{\begin{bmatrix}3&4&2\end{bmatrix}}$.

ID: 456 Type: Short (SA) Source: AISSCE(Board Exam) Year: 2025

Question 51: Let \(2x+5y-1=0\) and \(3x+2y-7=0\) represent the equations of two lines on which the ants are moving on the ground. Using matrix method, find a point common to the paths of the ants.

ID: 457 Type: Short (SA) Source: AISSCE(Board Exam) Year: 2025

Question 52: A shopkeeper sells 50 Chemistry, 60 Physics and 35 Maths books on day I and sells 40 Chemistry, 45 Physics and 50 Maths books on day II. If the selling price for each such subject book is ?150 (Chemistry), ? 175 (Physics) and ? 180 (Maths), then find his total sale in two days, using matrix method. If cost price of all the books together is 35,000, what profit did he earn after the sale of two days?

ID: 274 Type: Long (LA) Source: AISSCE(Board Exam) Year: 2025

Question 53: Given \(A=\begin{bmatrix}-4&4&4\\ -7&1&3\\ 5&-3&-1\end{bmatrix}\) and \(B=\begin{bmatrix}1&-1&1\\ 1&-2&-2\\ 2&1&3\end{bmatrix},\) find \(AB\). Hence, solve the system of linear equations: \(x-y+z=4\), \(x-2y-2z=9\), \(2x+y+3z=1\).

ID: 283 Type: Long (LA) Source: AISSCE(Board Exam) Year: 2025

Question 54: If \(A=\begin{bmatrix}1&2&0\\ -2&-1&-2\\ 0&-1&1\end{bmatrix},\) then find \(A^{-1}\). Hence, solve the system of linear equations: \(x-2y=10\), \(2x-y-z=8\), \(-2y+z=7\).

ID: 379 Type: Long (LA) Source: AISSCE(Board Exam) Year: 2025

Question 55: If A is a \(3\times3\) invertible matrix, show that for any scalar \(k\ne0.\), \((kA)^{-1}=\frac{1}{k}A^{-1}.\) Hence calculate \((3A)^{-1}\) where \(A=\begin{bmatrix}2&-1&1\\ -1&2&-1\\ 1&-1&2\end{bmatrix}.\)

ID: 420 Type: Long (LA) Source: AISSCE(Board Exam) Year: 2025

Question 56: A furniture workshop produces three types of furniture chairs, tables and beds each day. On a particular day the total number of furniture pieces produced is 45. It was also found that production of beds exceeds that of chairs by 8, while the total production of beds and chairs together is twice the production of tables. Determine the units produced of each type of furniture, using **matrix method**.

ID: 644 Type: Long (LA) Source: AISSCE(Board Exam) Year: 2024

Question 57: If \(A=\begin{bmatrix}1&2&-3\\ 2&0&-3\\ 1&2&0\end{bmatrix},\) then find \(A^{-1}\) and hence solve the following system of equations: \(x+2y-3z=1\), \(2x-3z=2\), \(x+2y=3\).

ID: 645 Type: Long (LA) Source: AISSCE(Board Exam) Year: 2024

Question 58: If \(A=\begin{bmatrix}1&-2&0\\ 2&-1&-1\\ 0&-2&1\end{bmatrix},\) find \(A^{-1}\) and use it to solve the following system of equations: \(x-2y=10\), \(2x-y-z=8\), \(-2y+z=7\).

ID: 650 Type: Long (LA) Source: AISSCE(Board Exam) Year: 2024

Question 59: Solve the following system of equations, using matrices: \(\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4\), \(\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1\), \(\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2\) where x, y, \(z\ne0\)

ID: 654 Type: Long (LA) Source: AISSCE(Board Exam) Year: 2024

Question 60: Find the product of the matrices \(\begin{bmatrix}1&2&-3\\ 2&3&2\\ 3&-3&-4\end{bmatrix}\begin{bmatrix}-6&17&13\\ 14&5&-8\\ -15&9&-1\end{bmatrix}\) and hence solve the system of linear equations: \(x+2y-3z=-4\), \(2x+3y+2z=2\), \(3x-3y-4z=11\).

ID: 656 Type: Long (LA) Source: AISSCE(Board Exam) Year: 2024

Question 61: If \(A=\begin{bmatrix}1&\cot~x\\ -\cot~x&1\end{bmatrix}\) show that \(A^{\prime}A^{-1}=\begin{bmatrix}-\cos~2x&-\sin~2x\\ \sin~2x&-\cos~2x\end{bmatrix}\)

ID: 657 Type: Long (LA) Source: AISSCE(Board Exam) Year: 2024

Question 62: If \(A=\begin{bmatrix}-1&a&2\\ 1&2&x\\ 3&1&1\end{bmatrix}\) and \(A^{-1}=\begin{bmatrix}1&-1&1\\ -8&7&-5\\ b&y&3\end{bmatrix},\) find the value of \((a+x)-(b+y)\).

ID: 666 Type: Long (LA) Source: AISSCE(Board Exam) Year: 2024

Question 63: If \(A=[\begin{matrix}2&1&-3\\ 3&2&1\\ 1&2&-1\end{matrix}],\) find \(A^{-1}\) and hence solve the following system of equations: \(2x+y-3z=13\), \(3x+2y+z=4\), \(x+2y-z=8\).