CBSE Class 12 Mathematics Matrices VBQs Set A

Algebra of Matrices

BASIC CONCEPTS

Matrix: A matrix is a rectangular arrangement of numbers or functions arranged into a fixed number of rows and columns. A matrix is written inside brackets [ ]. Each entry in a matrix is called an element of the matrix.

Order of Matrix: The dimension or order of matrix is defined by the number of rows and columns of that matrix. By conversion the dimension or order of a matrix is given by: No. of rows \(\times\) No. of columns. If a matrix have \(m\) rows and \(n\) columns then its order (dimension) is written as \(m \times n\) and read as \(m\) by \(n\).

Row Matrix: A matrix having one row and any number of column is called a row matrix. In other words, matrix of order \(1 \times n\) is always a row matrix.
e.g., \([a, b, c, d]_{1 \times 4}\) is a row matrix.

Column Matrix: A matrix having any number of rows but only one column is called column matrix. In other words, a matrix of order \(m \times 1\) is always a column matrix.
e.g., \(\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}_{4 \times 1}\) is a column matrix.

Square Matrix: A matrix in which the number of rows is equal to the number of columns, say \(n\), is called a square matrix of order \(n\).

Diagonal Elements: The elements \(a_{ij}\) of a square matrix \(A = [a_{ij}]_{n \times n}\) for which \(i = j\), i.e., the elements \(a_{11}, a_{22}, \dots, a_{nn}\) are called the diagonal elements and the line along which the diagonal elements lie, is called the principal diagonal or leading diagonal.

Diagonal Matrix: A square matrix \([a_{ij}]\) is said to be a diagonal matrix if \(a_{ij} = 0\) for \(i \neq j\). In other words, a square matrix is said to be a diagonal matrix, if its element not on principal diagonal are zero.

Scalar Matrix: A square matrix \(A = [a_{ij}]_{n \times n}\) is called a scalar matrix, if:
(i) \(a_{ij} = 0 \forall i \neq j\) and (ii) \(a_{ii} = c \forall i\), where \(c \neq 0\).
In other words, a square matrix is said to be scalar, if it is a diagonal matrix and entries on its principal diagonal are equal.

Identity Matrix: A square matrix in which all non diagonal elements are zero and all diagonal elements are equal to 1 is called identity matrix.
i.e., \(I = [a_{ij}]_{m \times n}\) is an identity matrix if \(a_{ij} = 0 \forall i \neq j\) and \(a_{ij} = 1 \forall i = j\).
For example, \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}_{3 \times 3}\) is an identity matrix.

Null or Zero Matrix: A matrix whose all elements are zero is called a null matrix or a zero matrix i.e., \(A = [a_{ij}]_{m \times n}\) is null matrix if \(a_{ij} = 0, \forall i, j\).

Upper and Lower Triangular Matrices: A square matrix \(A = [a_{ij}]\) is called:
(i) an upper triangular matrix, if \(a_{ij} = 0 \forall i > j\), i.e., all entries below principal diagonal are zero.
(ii) a lower triangular matrix, if \(a_{ij} = 0 \forall i < j\), i.e., all entries above principal diagonal are zero.

Equality of Matrices: Two matrices \(A = [a_{ij}]_{m \times n}\) and \(B = [b_{ij}]_{m \times n}\) of the same order are equal, if \(a_{ij} = b_{ij} \forall i = 1, 2, \dots, m\) and \(j = 1, 2, \dots, n\).

Sum of Matrices: If \(A = [a_{ij}]_{m \times n}\) and \(B = [b_{ij}]_{m \times n}\) are two matrices of the same order \(m \times n\), then their sum \(A + B\) is an \(m \times n\) matrix such that \((A + B)_{ij} = a_{ij} + b_{ij} \forall i = 1, 2, \dots, m\) and \(j = 1, 2, \dots, n\).
Following are the properties of matrix addition:

• Commutativity: If \(A\) and \(B\) are two matrices of the same order, then \(A + B = B + A\)
• Associativity: If \(A, B\) and \(C\) are three matrices of the same order, then \((A + B) + C = A + (B + C)\)
• Existence of Identity: The null matrix is the identity element for matrix addition i.e., \(A + O = A + O = A\)
• Existence of Inverse: For every matrix \(A = [a_{ij}]_{m \times n}\) there exists a matrix \(-A = [-a_{ij}]_{m \times n}\) such that \(A + (-A) = O = (-A) + A\)
• Cancellation Laws: If \(A, B\) and \(C\) are three matrices of the same order, then \(A + B = A + C \Rightarrow B = C\) and \(B + A = C + A \Rightarrow B = C\)

Scalar Multiplication: Let \(A = [a_{ij}]\) be an \(m \times n\) matrix and \(k\) be any number called a scalar. Then, the matrix obtained by multiplying every element of \(A\) by \(k\) is called the scalar multiple of \(A\) by \(k\) and is denoted by \(kA\). Thus, \(kA = [ka_{ij}]_{m \times n}\).
Following are the properties of scalar multiplication:

•  \(k(A + B) = kA + kB\)
•  \((k + l)A = kA + lA\)
•  \((kl)A = k(lA) = l(kA)\)
•  \((-k)A = -(kA) = k(-A)\)
•  \(1A = A\)
•  \((-1)A = -A\)

Note that a scalar matrix can be obtained by multiplying an identity matrix by a scalar.

Subtraction of Matrices: If \(A\) and \(B\) are two matrices of the same order, then \(A - B = A + (-B)\).

Multiplication of Matrices: Two matrices \(A\) and \(B\) are said to be defined for multiplication, if the number of columns of \(A\) (pre multiplier) is equal to the number of rows of \(B\) (post-multiplier). For example, if the order of \(A\) (pre-multiplier) is \(m \times n\) and the order of \(B\) (post-multiplier) is \(n \times p\) then \(A\) and \(B\) is defined for multiplication and order of product of \(A\) and \(B\) denoted by \(AB\) is \(m \times p\).
i.e., \(A_{m \times n} \times B_{n \times p} = AB_{m \times p}\)

Definition of Product: Let \(A = [a_{ij}]_{m \times n}\) and \(B = [b_{jk}]_{n \times p}\) be two matrices then product of \(A\) and \(B\) denoted by \(AB\) is given as \(AB = [c_{ik}]_{m \times p}\) where, \(c_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + \dots + a_{in}b_{nj} = \sum_{r=1}^n a_{ir}b_{rj}\) [\(\forall 1 \le i \le m\) and \(1 \le j \le p\)]. Here, \(A\) is pre-multiplier or pre-factor \(B\) is post-multiplier or post-factor.

Properties of Matrix Multiplication:

• Matrix multiplication is not commutative.
• Matrix multiplication is associative i.e., \((AB)C = A(BC)\) wherever both sides of the equality are defined.
• Matrix multiplication is distributive over matrix addition i.e., \(A(B + C) = AB + AC\) and \((B + C)A = BA + CA\) wherever both sides of the equality are defined.
• If \(A\) is an \(m \times n\) matrix, then \(I_m A = A = A I_n\)
• If \(A\) is an \(m \times n\) matrix and \(O\) is a null matrix, then \(A_{m \times n} O_{n \times p} = O_{m \times p}\) and \(O_{p \times m} \times A_{m \times n} = O_{p \times n}\) i.e., the product of a matrix with a null matrix is a null matrix.
• In matrix multiplication the product of two non-zero matrices may be a 'zero-matrix' i.e., \(AB = 0\), does not imply that at least one of the \(A\) or \(B\) should be zero.

Powers of Square Matrix: If \(A\) is a square matrix, then we define \(A^1 = A\) and \(A^{n+1} = A^n \cdot A\).

Matrix Polynomial: If \(A\) is a square matrix and \(a_0, a_1, \dots, a_n\) are constants, then \(a_0 A^n + a_1 A^{n-1} + a_2 A^{n-2} + \dots + a_{n-1} A + a_n I\) is called a matrix polynomial.

Transpose of a Matrix: Let \(A = [a_{ij}]\) be an \(m \times n\) matrix. Then, the transpose of \(A\), denoted by \(A^T\), is an \(n \times m\) matrix such that \((A^T)_{ij} = a_{ji} \forall i = 1, 2, \dots, n; j = 1, 2, \dots, m\). i.e., the matrix obtained by interchanging rows into columns, of a given matrix \(A\) is called the transpose of \(A\) and is denoted by \(A^T\) or \(A'\).
Following are the properties of transpose of a matrix:

• \((A^T)^T = A\)
• \((A + B)^T = A^T + B^T\)
• \((kA)^T = k A^T\)
• \((AB)^T = B^T A^T\)
• \((ABC)^T = C^T B^T A^T\)

Symmetric Matrix: A square matrix \(A = [a_{ij}]\) is called a symmetric matrix, if \(a_{ij} = a_{ji} \forall i, j\) i.e., \(A = A^T\).

Skew Symmetric Matrix: A square matrix \(A = [a_{ij}]\) is called a skew symmetric matrix, if \(a_{ij} = -a_{ji} \forall i, j\) i.e., \(A^T = -A\).

Properties of Skew-Symmetric Matrix: All main diagonal elements of a skew-symmetric matrix are zero.

Theorem: Every square matrix can be uniquely expressed as the sum of a symmetric and a skew-symmetric matrix.

Theorem: All positive integral powers of a symmetric matrix are symmetric.

Theorem: All odd positive integral powers of a skew-symmetric matrix are skew-symmetric.

Selected NCERT Questions

Question. Find the values of \(x, y\) and \(z\) if \(\begin{bmatrix} x + y + z \\ x + z \\ y + z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}\). 
Answer: We have, \(x + y + z = 9, x + z = 5\) and \(y + z = 7\). On solving these equations, we get \(x = 2, y = 4, z = 3\).

Question. Find the values of \(a, b, c\) and \(d\) if \(\begin{bmatrix} a - b & 2a + c \\ 2a - b & 3c + d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}\). 
Answer: Given: \(\begin{bmatrix} a - b & 2a + c \\ 2a - b & 3c + d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}\)
\(\Rightarrow a - b = -1\) …(i)
\(2a + c = 5\) …(ii)
\(2a - b = 0\) …(iii)
\(3c + d = 13\) …(iv)
From (iii), \(2a = b \Rightarrow a = b/2\)
Putting in (i), we get \(b/2 - b = -1 \Rightarrow -b/2 = -1 \Rightarrow b = 2\)
(i) \(\Rightarrow a - 2 = -1 \Rightarrow a = 1\)
(ii) \(\Rightarrow 2(1) + c = 5 \Rightarrow c = 5 - 2 = 3\)
(iv) \(\Rightarrow 3(3) + d = 13 \Rightarrow d = 13 - 9 = 4\)
i.e., \(a = 1, b = 2, c = 3, d = 4\)

Question. If \( x\begin{bmatrix} 2 \\ 3 \end{bmatrix} + y\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \), find the value of x and y.
Answer: Given, \( \begin{bmatrix} 2x \\ 3x \end{bmatrix} + \begin{bmatrix} -y \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \Rightarrow \begin{bmatrix} 2x - y \\ 3x + y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \). On comparing corresponding elements, we get \( 2x - y = 10 \) ...(i) and \( 3x + y = 5 \) ...(ii). Adding (i) and (ii), we get \( 5x = 15 \Rightarrow x = 3 \). On putting \( x = 3 \) in (ii), we get \( 9 + y = 5 \Rightarrow y = 5 - 9 \Rightarrow y = -4 \).

Question. Two farmers Ramkishan and Gurcharan Singh cultivates only three varieties of rice namely Basmati, Permal and Naura. The sale (in ₹) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices A and B.
September Sales (in ₹)
\( A = \begin{bmatrix} 10,000 & 20,000 & 30,000 \\ 50,000 & 30,000 & 10,000 \end{bmatrix} \begin{matrix} \text{Ramkishan} \\ \text{Gurchanran Singh} \end{matrix} \)
October Sales (in ₹)
\( B = \begin{bmatrix} 5,000 & 10,000 & 6,000 \\ 20,000 & 10,000 & 10,000 \end{bmatrix} \begin{matrix} \text{Ramkishan} \\ \text{Gurchanran Singh} \end{matrix} \)
(i) Find the combined sales in September and October for each farmer in each variety.
(ii) Find the decrease in sales from September to October.
(iii) If both farmers receive 2% profit on gross sales, compute the profit for each farmer and for each variety sold in October.
Answer: (i) Combined sales in September and October for each farmer in each variety is given by \( A + B = \begin{bmatrix} 15,000 & 30,000 & 36,000 \\ 70,000 & 40,000 & 20,000 \end{bmatrix} \).
(ii) Change in sales from September to October is given by \( A - B = \begin{bmatrix} 5,000 & 10,000 & 24,000 \\ 30,000 & 20,000 & 0 \end{bmatrix} \).
(iii) 2% of B = \( \frac{2}{100} \times B = 0.02 \times B \). Profit = \( 0.02 \begin{bmatrix} 5,000 & 10,000 & 6,000 \\ 20,000 & 10,000 & 10,000 \end{bmatrix} = \begin{bmatrix} 100 & 200 & 120 \\ 400 & 200 & 200 \end{bmatrix} \). Thus, in October Ramkishan receives ₹ 100, ₹ 200 and ₹ 120 as profit, and Gurcharan Singh receives profit of ₹ 400, ₹ 200 and ₹ 200.

Question. If \( A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \), prove that \( A^3 - 6A^2 + 7A + 2I = 0 \).
Answer: \( A^2 = A.A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \).
\( A^3 = A^2.A = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 21 & 12 & 34 \\ 0 & 8 & 0 \\ 34 & 23 & 55 \end{bmatrix} \). Wait, Correct calculation: \( A^3 = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} \).
LHS = \( A^3 - 6A^2 + 7A + 2I = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} - \begin{bmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix} + \begin{bmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{bmatrix} + \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \). Hence proved.

Question. If \( A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), then find k so that \( A^2 = kA - 2I \).
Answer: \( A^2 = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} \).
Now, \( A^2 = kA - 2I \Rightarrow \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = k \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix} \).
Equating corresponding elements: \( 1 = 3k - 2 \Rightarrow 3k = 3 \Rightarrow k = 1 \).

Question. Express the matrix \( A = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \) as the sum of a symmetric and a skew symmetric matrix.
Answer: Let \( P = \frac{1}{2}(A + A') = \begin{bmatrix} 3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{bmatrix} \) which is symmetric, and \( Q = \frac{1}{2}(A - A') = \begin{bmatrix} 0 & 5/2 & 3/2 \\ -5/2 & 0 & 3 \\ -3/2 & -3 & 0 \end{bmatrix} \) which is skew-symmetric. Then \( A = P + Q \).

Question. For matrix \( A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix} \), verify that (i) (A + A') is a symmetric matrix. (ii) (A - A') is a skew symmetric matrix.
Answer: \( A' = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix} \).
(i) \( A + A' = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix} \). Since \( (A + A')' = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix} = A + A' \), it is symmetric.
(ii) \( A - A' = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \). Since \( (A - A')' = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = -(A - A') \), it is a skew symmetric matrix.

Question. Find the matrix 'X' so that \( X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} \).
Answer: Let \( X \) have order \( 2 \times 2 \). Let \( X = \begin{bmatrix} x & y \\ z & w \end{bmatrix} \).
Then \( \begin{bmatrix} x & y \\ z & w \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} x+4y & 2x+5y & 3x+6y \\ z+4w & 2z+5w & 3z+6w \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} \).
Solving \( x+4y = -7 \) and \( 2x+5y = -8 \), we get \( x = 1, y = -2 \).
Solving \( z+4w = 2 \) and \( 2z+5w = 4 \), we get \( z = 2, w = 0 \).
Hence, \( X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix} \).

Question. If \( A = \begin{bmatrix} 0 & -\tan \alpha/2 \\ \tan \alpha/2 & 0 \end{bmatrix} \) and I is the identity matrix of order 2, then show that: \( I + A = (I - A) \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \).
Answer: LHS = \( I + A = \begin{bmatrix} 1 & -\tan \alpha/2 \\ \tan \alpha/2 & 1 \end{bmatrix} \).
RHS = \( (I - A) \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} = \begin{bmatrix} 1 & \tan \alpha/2 \\ -\tan \alpha/2 & 1 \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \).
Substituting \( \sin \alpha = \frac{2 \tan \alpha/2}{1 + \tan^2 \alpha/2} \) and \( \cos \alpha = \frac{1 - \tan^2 \alpha/2}{1 + \tan^2 \alpha/2} \), we simplify RHS to get \( \begin{bmatrix} 1 & -\tan \alpha/2 \\ \tan \alpha/2 & 1 \end{bmatrix} \). Hence, LHS = RHS.

Question. If \( F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \), then show that F(x).F(y) = F(x + y).
Answer: LHS = \( F(x) . F(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} \cos x \cos y - \sin x \sin y & -\cos x \sin y - \sin x \cos y & 0 \\ \sin x \cos y + \cos x \sin y & -\sin x \sin y + \cos x \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix} = F(x + y) = \text{RHS} \).

Question. Show that the matrix \( B^T AB \) is symmetric or skew-symmetric according as A is symmetric or skew-symmetric.
Answer: Case I: Let A be a symmetric matrix. Then \( A^T = A \). \( (B^T AB)^T = B^T A^T (B^T)^T = B^T AB \). \( \therefore B^T AB \) is a symmetric matrix. Case II: Let A be a skew-symmetric matrix. Then, \( A^T = -A \). \( (B^T AB)^T = B^T A^T (B^T)^T = B^T (-A)B = -B^T AB \). \( \therefore B^T AB \) is a skew-symmetric matrix.

Question. If \( A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} \), then prove that \( A^n = \begin{bmatrix} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix} \), then \( n \in N \).
Answer: We prove by principle of mathematical induction. Let \( P(n): A^n = \begin{bmatrix} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix} \). For \( n=1 \), \( A^1 = [3^0] = [1] \), true. Assume true for \( n=k \). Now, \( A^{k+1} = A.A^k = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \end{bmatrix} = \begin{bmatrix} 3.3^{k-1} & 3.3^{k-1} & 3.3^{k-1} \\ 3.3^{k-1} & 3.3^{k-1} & 3.3^{k-1} \\ 3.3^{k-1} & 3.3^{k-1} & 3.3^{k-1} \end{bmatrix} = \begin{bmatrix} 3^k & 3^k & 3^k \\ 3^k & 3^k & 3^k \\ 3^k & 3^k & 3^k \end{bmatrix} \). Hence, by PMI true for all \( n \in N \).

Multiple Choice Questions

Question. If \( A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \), then \( A^2 \) is equal to
(a) \( \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \)
(b) \( \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \)
(c) \( \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} \)
(d) \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
Answer: (d)

Question. If \( A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \), then \( A + A' = I \), if the value of \(\alpha\) is
(a) \( \pi/6 \)
(b) \( \pi/3 \)
(c) \( \pi \)
(d) \( 3\pi/2 \)
Answer: (b)

Question. If A and B are square matrices of the same order, then (A + B)(A – B) is equal to
(a) \( A^2 – B^2 \)
(b) \( A^2 – BA – AB – B^2 \)
(c) \( A^2 – B^2 + BA – AB \)
(d) \( A^2 – BA + B^2 + AB \)
Answer: (c)

Question. Total number of possible matrices of order 3 × 3 with each entry 2 or 0 is
(a) 9
(b) 27
(c) 81
(d) 512
Answer: (d)

Question. The matrix \( \begin{bmatrix} 0 & -5 & 8 \\ 5 & 0 & 12 \\ -8 & -12 & 0 \end{bmatrix} \) is a
(a) diagonal matrix
(b) symmetric matrix
(c) skew symmetric matrix
(d) scalar matrix
Answer: (c)

Question. Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k respectively. The restriction on n, k and p so that PY + WY will be defined are
(a) k = 3, p = n
(b) k is arbitrary, p = 2
(c) p is arbitrary, k = 3
(d) k = 2, p = 3
Answer: (a)

Question. If n = p, then the order of the matrix 7X – 5Z is
(a) p × 2
(b) 2 × n
(c) n × 3
(d) p × n
Answer: (b)

Question. If \( \begin{bmatrix} 2x + y & 4x \\ 5x - 7 & 4x \end{bmatrix} = \begin{bmatrix} 7 & 7y-13 \\ y & x+6 \end{bmatrix} \), then the value of x and y is
(a) x = 3, y = 1
(b) x = 2, y = 3
(c) x = 2, y = 4
(d) x = 3, y = 3
Answer: (b)

Question. If A is matrix of order m × n and B is a matrix such that AB' and B'A are both defined, the order of matrix B is
(a) m × m
(b) n × n
(c) n × m
(d) m × n
Answer: (c)

Question. If A and B are matrices of same order, then (AB' – BA') is a
(a) skew-symmetric matrix
(b) null matrix
(c) symmetric matrix
(d) unit matrix
Answer: (a)

Question. If \(A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\) is such that \(A^2 = I\), then
(a) \(1 + \alpha^2 + \beta\gamma = 0\)
(b) \(1 - \alpha^2 + \beta\gamma = 0\)
(c) \(1 - \alpha^2 - \beta\gamma = 0\)
(d) \(1 + \alpha^2 - \beta\gamma = 0\)
Answer: (c)

Question. On using elementary column operations \(C_2 \to C_2 - 2C_1\) in the following matrix equation \(\begin{bmatrix} 1 & -3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix}\), we have:
(a) \(\begin{bmatrix} 1 & -5 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -2 & 2 \end{bmatrix} \begin{bmatrix} 3 & -5 \\ 2 & 0 \end{bmatrix}\)
(b) \(\begin{bmatrix} 1 & -5 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 3 & 0 \\ 5 & 2 \end{bmatrix}\)
(c) \(\begin{bmatrix} 1 & 2 \\ 5 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}\)
(d) \(\begin{bmatrix} 1 & -5 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 3 & -5 \\ 2 & 0 \end{bmatrix}\)
Answer: (d)

Question. If \(A = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}\), then \(A^n\) (where \(n \in N\)) equals
(a) \(\begin{bmatrix} 1 & na \\ 0 & 1 \end{bmatrix}\)
(b) \(\begin{bmatrix} 1 & n^2a \\ 0 & 1 \end{bmatrix}\)
(c) \(\begin{bmatrix} 1 & na \\ 0 & 0 \end{bmatrix}\)
(d) \(\begin{bmatrix} n & na \\ 0 & n \end{bmatrix}\)
Answer: (a)

Question. If \(A\) is square matrix such that \(A^2 = I\), then \((A - I)^3 + (A + I)^3 - 7A\) is equal to
(a) \(A\)
(b) \(I - A\)
(c) \(I + A\)
(d) \(3A\)
Answer: (a)

Question. If the matrix \(AB\) is zero, then
(a) It is not necessary that either \(A = O\) or \(B = O\)
(b) \(A = O\) or \(B = O\)
(c) \(A = O\) and \(B = O\)
(d) All the statements are wrong
Answer: (a)

Question. If \(A = \begin{bmatrix} 2 & -1 & 3 \\ -4 & 5 & 1 \end{bmatrix}\) and \(B = \begin{bmatrix} 2 & 3 \\ 4 & -2 \\ 1 & 5 \end{bmatrix}\), then
(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.
Answer: (c)

Question. If \(A\) and \(B\) are symmetric matrices of the same order, then \((AB' - BA')\) is a
(a) Skew symmetric matrix
(b) Null matrix
(c) Symmetric matrix
(d) None of these
Answer: (a)

Question. The matrix \(\begin{bmatrix} 0 & 5 & -7 \\ -5 & 0 & 11 \\ 7 & -11 & 0 \end{bmatrix}\) is
(a) a skew-symmetric matrix
(b) a symmetric matrix
(c) a diagonal matrix
(d) an upper triangular matrix.
Answer: (a)

Question. If \(A\) and \(B\) are two matrices of the order \(3 \times m\) and \(3 \times n\) respectively and \(m = n\), then the order of matrix \((5A - 2B)\) is
(a) \(m \times 3\)
(b) \(3 \times 3\)
(c) \(m \times n\)
(d) \(3 \times n\)
Answer: (d)

Question. If \(A = \begin{bmatrix} 5 & x \\ y & 0 \end{bmatrix}\) and \(A = A'\), then
(a) \(x = 0, y = 5\)
(b) \(x + y = 5\)
(c) \(x = y\)
(d) none of these
Answer: (c)