CBSE Class 12 Mathematics Matrices Notes Set 03

Important Notes and Formulas

Definition

A rectangular arrangement of numbers (which may be real or complex numbers) in rows and columns, is called a matrix. This arrangement is enclosed by small ( ) or big [ ] brackets. The numbers are called the elements of the matrix or entries in the matrix. A matrix is represented by capital letters A, B, C etc. and its elements by small letters a, b, c, x, y etc. The following are some examples of matrices:
\( A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix} \),   \( B = \begin{bmatrix} 2 + i & -3 & 2 \\ 1 & -3 + i & -5 \end{bmatrix} \),   C = [1, 4, 9],   \( D = \begin{bmatrix} a \\ g \\ h \end{bmatrix} \),   E = [l]

Order of a Matrix

A matrix having m rows and n columns is called a matrix of order m × n or simply m × n matrix (read as 'an m by n matrix'). A matrix A of order m × n is usually written in the following manner
\( A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \ldots a_{1j} & \ldots a_{1n} \\ a_{21} & a_{22} & a_{23} & \ldots a_{2j} & \ldots a_{2n} \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ a_{i1} & a_{i2} & a_{i3} & \ldots a_{ij} & \ldots a_{in} \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ a_{m1} & a_{m2} & a_{m3} & \ldots a_{mj} & \ldots a_{mn} \end{bmatrix} \) or \( A = [a_{ij}]_{m \times n} \), where i = 1, 2, …, m; j = 1, 2, …, n
Here \( a_{ij} \) denotes the element of i^th row and j^th column. Example: order of matrix \( \begin{bmatrix} 3 & -1 & 5 \\ 6 & 2 & -7 \end{bmatrix} \) is 2 × 3.
Note: A matrix of order m × n contains mn elements. Every row of such a matrix contains n elements and every column contains m elements.

Equality of Matrices

Two matrices A and B are said to be equal matrices if they are of the same order and their corresponding elements are equal. Example: If \( A = \begin{bmatrix} 1 & 6 & 3 \\ 5 & 2 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{bmatrix} \) are equal matrices,
then \( a_1 = 1, a_2 = 6, a_3 = 3, b_1 = 5, b_2 = 2, b_3 = 1 \)

Types of Matrices

(1) Row matrix: A matrix is said to be a row matrix or row vector if it has only one row and any number of columns. Example: [5  0  3] is a row matrix of order 1 × 3 and [2] is a row matrix of order 1 × 1.

(2) Column matrix: A matrix is said to be a column matrix or column vector if it has only one column and any number of rows. Example: \( \begin{bmatrix} 2 \\ 3 \\ -6 \end{bmatrix} \) is a column matrix of order 3 × 1 and [2] is a column matrix of order 1 × 1. Observe that [2] is both a row matrix as well as a column matrix.

(3) Singleton matrix: If in a matrix there is only one element then it is called a singleton matrix.
Thus, \( A = [a_{ij}]_{m \times n} \) is a singleton matrix if m = n = 1. Example: [2], [3], [a], [−3] are singleton matrices.

(4) Null or zero matrix: If in a matrix all the elements are zero then it is called a zero matrix and it is generally denoted by O. Thus \( A = [a_{ij}]_{m \times n} \) is a zero matrix if \( a_{ij} = 0 \) for all i and j.
Example: [0], \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \), \( \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \), [0  0] are all zero matrices, but of different orders.

(5) Square matrix: If the number of rows and the number of columns in a matrix are equal, then it is called a square matrix. Thus \( A = [a_{ij}]_{m \times n} \) is a square matrix if m = n. Example: \( \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \) is a square matrix of order 3 × 3.
(i) If m ≠ n then the matrix is called a rectangular matrix.
(ii) The elements of a square matrix A for which i = j, i.e. \( a_{11}, a_{22}, a_{33}, \ldots, a_{nn} \) are called diagonal elements and the line joining these elements is called the principal diagonal or leading diagonal of matrix A.

(6) Diagonal matrix: If all elements except the principal diagonal in a square matrix are zero, it is called a diagonal matrix. Thus a square matrix \( A = [a_{ij}] \) is a diagonal matrix if \( a_{ij} = 0 \), when i ≠ j.
Example: \( \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix} \) is a diagonal matrix of order 3 × 3, which can be denoted by diag [2, 3, 4].
Note: Number of zeros in a diagonal matrix is given by \( n^2 - n \) where n is the order of the matrix.
A diagonal matrix of order n × n having \( d_1, d_2, \ldots, d_n \) as diagonal elements is denoted by diag \( [d_1, d_2, \ldots, d_n] \).

(7) Identity matrix: A square matrix in which elements in the main diagonal are all '1' and rest are all zero is called an identity matrix or unit matrix. Thus, the square matrix \( A = [a_{ij}] \) is an identity matrix, if \( a_{ij} = \begin{cases} 1, & \text{if } i = j \\ 0, & \text{if } i \neq j \end{cases} \)
We denote the identity matrix of order n by \( I_n \).
Example: [1], \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \) are identity matrices of order 1, 2 and 3 respectively.

(8) Scalar matrix: A square matrix whose all non-diagonal elements are zero and diagonal elements are equal is called a scalar matrix. Thus, if \( A = [a_{ij}] \) is a square matrix and \( a_{ij} = \begin{cases} \alpha, & \text{if } i = j \\ 0, & \text{if } i \neq j \end{cases} \), then A is a scalar matrix.
Example: [2], \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), \( \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \) are scalar matrices of order 1, 2 and 3 respectively.
Note: Unit matrix and null square matrices are also scalar matrices.

Addition and Subtraction of Matrices

If \( A = [a_{ij}]_{m \times n} \) and \( B = [b_{ij}]_{m \times n} \) are two matrices of the same order then their sum A + B is a matrix whose each element is the sum of corresponding elements, i.e. \( A + B = [a_{ij} + b_{ij}]_{m \times n} \)
Example: If \( A = \begin{bmatrix} 5 & 2 \\ 1 & 3 \\ 4 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 5 \\ 2 & 2 \\ 3 & 3 \end{bmatrix} \), then \( A + B = \begin{bmatrix} 5 + 1 & 2 + 5 \\ 1 + 2 & 3 + 2 \\ 4 + 3 & 1 + 3 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 3 & 5 \\ 7 & 4 \end{bmatrix} \)
Similarly, their subtraction A − B is defined as \( A - B = [a_{ij} - b_{ij}]_{m \times n} \),
i.e. in the above example \( A - B = \begin{bmatrix} 5 - 1 & 2 - 5 \\ 1 - 2 & 3 - 2 \\ 4 - 3 & 1 - 3 \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ -1 & 1 \\ 1 & -2 \end{bmatrix} \)
Note: Matrix addition and subtraction are possible only when the matrices are of the same order.

Properties of matrix addition: If A, B and C are matrices of the same order, then
(i) A + B = B + A (Commutative law)
(ii) (A + B) + C = A + (B + C) (Associative law)
(iii) A + O = O + A = A, where O is the zero matrix which is the additive identity of the matrix.
(iv) A + (−A) = O = (−A) + A, where (−A) is obtained by changing the sign of every element of A, which is the additive inverse of the matrix.
(v) \( \left. \begin{matrix} A + B = A + C \\ B + A = C + A \end{matrix} \right\rbrace \Rightarrow B = C \) (Cancellation law)

Scalar Multiplication of Matrices

Let \( A = [a_{ij}]_{m \times n} \) be a matrix and k be a number, then the matrix which is obtained by multiplying every element of A by k is called scalar multiplication of A by k and it is denoted by kA.
Thus, if \( A = [a_{ij}]_{m \times n} \), then \( kA = Ak = [ka_{ij}]_{m \times n} \). Example: If \( A = \begin{bmatrix} 2 & 4 \\ 3 & 1 \\ 4 & 6 \end{bmatrix} \), then \( 5A = \begin{bmatrix} 10 & 20 \\ 15 & 5 \\ 20 & 30 \end{bmatrix} \)
Properties of scalar multiplication: If A, B are matrices of the same order and λ, μ are any two scalars then
(i) λ(A + B) = λA + λB    (ii) (λ + μ)A = λA + μA
(iii) λ(μA) = (λμ)A = μ(λA)    (iv) (−λA) = −(λA) = λ(−A)
Note: All the laws of ordinary algebra hold for the addition or subtraction of matrices and their multiplication by scalars.

Multiplication of Matrices

Two matrices A and B are conformable for the product AB if the number of columns in A (pre-multiplier) is the same as the number of rows in B (post-multiplier). Thus, if \( A = [a_{ij}]_{m \times n} \) and \( B = [b_{ij}]_{n \times p} \) are two matrices of order m × n and n × p respectively, then their product AB is of order m × p and is defined as \( (AB)_{ij} = \sum_{r=1}^{n} a_{ir} b_{rj} \)
\( = [a_{i1}\ a_{i2} \ldots a_{in}] \begin{bmatrix} b_{1j} \\ b_{2j} \\ \vdots \\ b_{nj} \end{bmatrix} \) = (i^th row of A)(j^th column of B) .....(i), where i = 1, 2, …, m and j = 1, 2, …, p
Now we define the product of a row matrix and a column matrix.

Let \( A = [a_1\ a_2 \ldots a_n] \) be a row matrix and \( B = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix} \) be a column matrix.
Then \( AB = [a_1 b_1 + a_2 b_2 + \ldots + a_n b_n] \) ...(ii). Thus, from (i),
\( (AB)_{ij} \) = Sum of the products of elements of i^th row of A with the corresponding elements of j^th column of B.

Properties of matrix multiplication: If A, B and C are three matrices such that their product is defined, then
(i) AB ≠ BA (Generally not commutative)
(ii) (AB)C = A(BC) (Associative law)
(iii) IA = A = AI, where I is the identity matrix for matrix multiplication
(iv) A(B + C) = AB + AC (Distributive law)
(v) If AB = AC \( \not\Rightarrow \) B = C (Cancellation law is not applicable)
(vi) If AB = O, it does not mean that A = O or B = O; again, the product of two non-zero matrices may be a zero matrix.
Note: If A and B are two matrices such that AB exists, then BA may or may not exist.
The multiplication of two diagonal matrices is also a diagonal matrix and \( \text{diag}(a_1, a_2, \ldots, a_n) \times \text{diag}(b_1, b_2, \ldots, b_n) = \text{diag}(a_1 b_1, a_2 b_2, \ldots, a_n b_n) \)
The multiplication of two scalar matrices is also a scalar matrix.
If A and B are two matrices of the same order, then
(i) \( (A + B)^2 = A^2 + B^2 + AB + BA \)    (ii) \( (A - B)^2 = A^2 + B^2 - AB - BA \)
(iii) \( (A - B)(A + B) = A^2 - B^2 + AB - BA \)    (iv) \( (A + B)(A - B) = A^2 - B^2 - AB + BA \)
(v) A(−B) = (−A)B = −(AB)

Positive Integral Powers of a Matrix

The positive integral powers of a matrix A are defined only when A is a square matrix. Also then \( A^2 = A.A \), \( A^3 = A.A.A = A^2 A \). Also for any positive integers m, n,
(i) \( A^m A^n = A^{m+n} \)    (ii) \( (A^m)^n = A^{mn} = (A^n)^m \)
(iii) \( I^n = I, I^m = I \)    (iv) \( A^0 = I_n \), where A is a square matrix of order n.

Matrix Polynomial

Let \( f(x) = a_0 x^n + a_1 x^{n-1} + a_2 x^{n-2} + \ldots + a_{n-1} x + a_n \) be a polynomial and let A be a square matrix of order n. Then \( f(A) = a_0 A^n + a_1 A^{n-1} + a_2 A^{n-2} + \ldots + a_{n-1} A + a_n I_n \) is called a matrix polynomial.
Example: If \( f(x) = x^2 - 3x + 2 \) is a polynomial and A is a square matrix, then \( A^2 - 3A + 2I \) is a matrix polynomial.

Transpose of a Matrix

The matrix obtained from a given matrix A by changing its rows into columns or columns into rows is called the transpose of matrix A and is denoted by \( A^T \) or A′.
From the definition it is obvious that if the order of A is m × n, then the order of \( A^T \) is n × m.
Example: Transpose of matrix \( \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{bmatrix}_{2 \times 3} \) is \( \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \end{bmatrix}_{3 \times 2} \)
Properties of transpose: Let A and B be two matrices, then
(i) \( (A^T)^T = A \)
(ii) \( (A + B)^T = A^T + B^T \), A and B being of the same order
(iii) \( (kA)^T = kA^T \), k be any scalar (real or complex)
(iv) \( (AB)^T = B^T A^T \), A and B being conformable for the product AB
(v) \( (A_1 A_2 A_3 \ldots A_{n-1} A_n)^T = A_n^T A_{n-1}^T \ldots A_3^T A_2^T A_1^T \)
(vi) \( I^T = I \)

Determinant of a Matrix

If \( A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \) is a square matrix, then its determinant, denoted by |A| or Det (A), is defined as \( |A| = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \)
Properties of determinant of a matrix:
(i) |A| exists ⇔ A is a square matrix    (ii) |AB| = |A| |B|
(iii) \( |A^T| = |A| \)    (iv) \( |kA| = k^n |A| \), if A is a square matrix of order n
(v) If A and B are square matrices of the same order then |AB| = |BA|
(vi) If A is a skew-symmetric matrix of odd order then |A| = 0
(vii) If \( A = \text{diag}(a_1, a_2, \ldots, a_n) \) then \( |A| = a_1 a_2 \ldots a_n \)
(viii) \( |A|^n = |A^n| \), n ∈ N.

Special Types of Matrices

(1) Symmetric and skew-symmetric matrix
(i) Symmetric matrix: A square matrix \( A = [a_{ij}] \) is called a symmetric matrix if \( a_{ij} = a_{ji} \) for all i, j or \( A^T = A \). Example: \( \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix} \)
Note: Every unit matrix and square zero matrix are symmetric matrices.
Maximum number of different elements in a symmetric matrix is \( \frac{n(n + 1)}{2} \).
(ii) Skew-symmetric matrix: A square matrix \( A = [a_{ij}] \) is called a skew-symmetric matrix if \( a_{ij} = -a_{ji} \) for all i, j or \( A^T = -A \). Example: \( \begin{bmatrix} 0 & h & g \\ -h & 0 & f \\ -g & -f & 0 \end{bmatrix} \)
Note: All principal diagonal elements of a skew-symmetric matrix are always zero because for any diagonal element \( a_{ii} = -a_{ii} \Rightarrow a_{ii} = 0 \).

Properties of symmetric and skew-symmetric matrices:
(i) If A is a square matrix, then \( A + A^T, AA^T, A^T A \) are symmetric matrices, while \( A - A^T \) is a skew-symmetric matrix.
(ii) If A is a symmetric matrix, then \( -A, KA, A^T, A^n, A^{-1}, B^T AB \) are also symmetric matrices, where n ∈ N, K ∈ R and B is a square matrix of the same order as A.
(iii) If A is a skew-symmetric matrix, then
(a) \( A^{2n} \) is a symmetric matrix for n ∈ N,
(b) \( A^{2n+1} \) is a skew-symmetric matrix for n ∈ N,
(c) kA is also a skew-symmetric matrix, where k ∈ R,
(d) \( B^T AB \) is also a skew-symmetric matrix where B is a square matrix of the same order as A.
(iv) If A, B are two symmetric matrices, then
(a) A ± B, AB + BA are also symmetric matrices,
(b) AB − BA is a skew-symmetric matrix,
(c) AB is a symmetric matrix, when AB = BA.
(v) If A, B are two skew-symmetric matrices, then
(a) A ± B, AB − BA are skew-symmetric matrices,
(b) AB + BA is a symmetric matrix.
(vi) If A is a skew-symmetric matrix and C is a column matrix, then \( C^T AC \) is a zero matrix.
(vii) Every square matrix A can be uniquely expressed as the sum of a symmetric and a skew-symmetric matrix, i.e. \( A = \left[\frac{1}{2}(A + A^T)\right] + \left[\frac{1}{2}(A - A^T)\right] \).

(2) Singular and non-singular matrix: Any square matrix A is said to be non-singular if |A| ≠ 0, and a square matrix A is said to be singular if |A| = 0. Here |A| (or det(A) or simply det |A|) means the corresponding determinant of the square matrix A.
Example: \( A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} \), then \( |A| = \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = 10 - 12 = -2 \) ⇒ A is a non-singular matrix.

Adjoint of a Square Matrix

Let \( A = [a_{ij}] \) be a square matrix of order n and let \( C_{ij} \) be the cofactor of \( a_{ij} \) in A. Then the transpose of the matrix of cofactors of elements of A is called the adjoint of A and is denoted by adj A.
Thus, \( \text{adj } A = [C_{ij}]^T \Rightarrow (\text{adj } A)_{ij} = C_{ji} \) = cofactor of \( a_{ji} \) in A.
If \( A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \), then \( \text{adj } A = \begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix}^T = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix} \),
where \( C_{ij} \) denotes the cofactor of \( a_{ij} \) in A.
Example: \( A = \begin{bmatrix} p & q \\ r & s \end{bmatrix} \), \( C_{11} = s, C_{12} = -r, C_{21} = -q, C_{22} = p \)
∴ \( \text{adj } A = \begin{bmatrix} s & -r \\ -q & p \end{bmatrix}^T = \begin{bmatrix} s & -q \\ -r & p \end{bmatrix} \)
Note: The adjoint of a square matrix of order 2 can be easily obtained by interchanging the diagonal elements and changing the signs of the off-diagonal elements.

Properties of adjoint matrix: If A, B are square matrices of order n and \( I_n \) is the corresponding unit matrix, then
(i) \( A(\text{adj } A) = |A| I_n = (\text{adj } A)A \) (Thus A (adj A) is always a scalar matrix)
(ii) \( |\text{adj } A| = |A|^{n-1} \)    (iii) \( \text{adj}(\text{adj } A) = |A|^{n-2} A \)
(iv) \( |\text{adj}(\text{adj } A)| = |A|^{(n-1)^2} \)    (v) \( \text{adj}(A^T) = (\text{adj } A)^T \)
(vi) adj (AB) = (adj B)(adj A)    (vii) \( \text{adj}(A^m) = (\text{adj } A)^m \), m ∈ N
(viii) \( \text{adj}(kA) = k^{n-1}(\text{adj } A) \), k ∈ R    (ix) \( \text{adj}(I_n) = I_n \)
(x) adj (O) = O    (xi) A is symmetric ⇒ adj A is also symmetric.
(xii) A is diagonal ⇒ adj A is also diagonal.    (xiii) A is singular ⇒ |adj A| = 0

Inverse of a Matrix

A non-singular square matrix of order n is invertible if there exists a square matrix B of the same order such that \( AB = I_n = BA \).
In such a case, we say that the inverse of A is B and we write \( A^{-1} = B \).
The inverse of A is given by \( A^{-1} = \frac{1}{|A|}.\text{adj } A \)
The necessary and sufficient condition for the existence of the inverse of a square matrix A is that |A| ≠ 0.

Properties of inverse matrix: If A and B are invertible matrices of the same order, then
(i) \( (A^{-1})^{-1} = A \)    (ii) \( (A^T)^{-1} = (A^{-1})^T \)
(iii) \( (AB)^{-1} = B^{-1} A^{-1} \)    (iv) \( (A^k)^{-1} = (A^{-1})^k \), k ∈ N [In particular \( (A^2)^{-1} = (A^{-1})^2 \)]
(v) \( \text{adj}(A^{-1}) = (\text{adj } A)^{-1} \)    (vi) \( |A^{-1}| = \frac{1}{|A|} = |A|^{-1} \)
(vii) \( A = \text{diag}(a_1 a_2 \ldots a_n) \Rightarrow A^{-1} = \text{diag}(a_1^{-1} a_2^{-1} \ldots a_n^{-1}) \)
(viii) A is symmetric ⇒ \( A^{-1} \) is also symmetric.
(ix) A is diagonal, |A| ≠ 0 ⇒ \( A^{-1} \) is also diagonal.
(x) A is a scalar matrix ⇒ \( A^{-1} \) is also a scalar matrix.
(xi) Every invertible matrix possesses a unique inverse.
Note: (Cancellation law with respect to multiplication)
If A is a non-singular matrix, i.e. if |A| ≠ 0, then \( A^{-1} \) exists and \( AB = AC \Rightarrow A^{-1}(AB) = A^{-1}(AC) \)
\( \Rightarrow (A^{-1}A)B = (A^{-1}A)C \Rightarrow IB = IC \Rightarrow B = C \) ∴ AB = AC ⇒ B = C ⇔ |A| ≠ 0

System of Simultaneous Linear Equations

Consider the following system of m linear equations in n unknowns
\( a_{11}x_1 + a_{12}x_2 + \ldots + a_{1n}x_n = b_1 \)
\( a_{21}x_1 + a_{22}x_2 + \ldots + a_{2n}x_n = b_2 \)
… … … …
\( a_{m1}x_1 + a_{m2}x_2 + \ldots + a_{mn}x_n = b_m \)
The system of equations can be written in matrix form as \( \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{m1} & a_{m2} & \ldots & a_{mn} \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{bmatrix} \) or AX = B,
where \( A = \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{m1} & a_{m2} & \ldots & a_{mn} \end{bmatrix}_{m \times n} \), \( X = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}_{n \times 1} \) and \( B = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{bmatrix}_{m \times 1} \)
The m × n matrix A is called the coefficient matrix of the system of linear equations.

(1) Solution: A set of values of the variables \( x_1, x_2, \ldots, x_n \) which simultaneously satisfy all the equations is called a solution of the system of equations. Example: x = 2, y = −3 is a solution of the system of linear equations 3x + y = 3, 2x + y = 1, because 3(2) + (−3) = 3 and 2(2) + (−3) = 1.
(2) Consistent system: If the system of equations has one or more solutions, then it is said to be a consistent system of equations, otherwise it is an inconsistent system of equations. Example: the system of linear equations 2x + 3y = 5, 4x + 6y = 10 is consistent, because x = 1, y = 1 and x = 2, y = 1/3 are solutions of it.
However, the system of linear equations 2x + 3y = 5, 4x + 6y = 8 is inconsistent, because there is no set of values of x, y which satisfy the two equations simultaneously.
(3) Homogeneous and non-homogeneous system of linear equations: A system of equations AX = B is called a homogeneous system if B = O. Otherwise, it is called a non-homogeneous system of equations.
Example: The system of equations 2x + 3y = 0, 3x − y = 0 is a homogeneous system of linear equations whereas the system of equations given by 2x + 3y = 1, 3x − y = 5 is a non-homogeneous system of linear equations.

Solution of a Non-Homogeneous System of Linear Equations

Matrix method: Let AX = B be a system of n linear equations with n unknowns. If A is non-singular, then \( A^{-1} \) exists. ∴ \( AX = B \Rightarrow A^{-1}(AX) = A^{-1}B \), [pre-multiplying by \( A^{-1} \)]
\( \Rightarrow (A^{-1}A)X = A^{-1}B \), [by associativity]
\( \Rightarrow I_n X = A^{-1}B \Rightarrow X = A^{-1}B \). Thus, the system of equations AX = B has a solution given by \( X = A^{-1}B \).
Now, let \( X_1 \) and \( X_2 \) be two solutions of AX = B. Then, \( AX_1 = B \) and \( AX_2 = B \)
\( \Rightarrow AX_1 = AX_2 \Rightarrow A^{-1}(AX_1) = A^{-1}(AX_2) \Rightarrow (A^{-1}A)X_1 = (A^{-1}A)X_2 \Rightarrow I_n X_1 = I_n X_2 \Rightarrow X_1 = X_2 \).
Hence, the given system has a unique solution.
Thus, if A is a non-singular matrix, then the system of equations given by AX = B has a unique solution given by \( X = A^{-1}B \).
If A is a singular matrix, then the system of equations given by AX = B may be consistent with infinitely many solutions or it may be inconsistent also.

Criterion of consistency: Let AX = B be a system of n linear equations in n unknowns.
(i) If |A| ≠ 0, then the system is consistent and has a unique solution given by \( X = A^{-1}B \).
(ii) If |A| = 0 and (adj A) B = O, then the system is consistent and has infinitely many solutions.
(iii) If |A| = 0 and (adj A) B ≠ O, then the system is inconsistent.

Algorithm for solving a non-homogeneous system of linear equations: We shall give the algorithm for three equations in three unknowns. But it can be generalized to any number of equations.
Let AX = B be a non-homogeneous system of 3 linear equations in 3 unknowns. To solve this system of equations we proceed as follows:
Step I: Write the given system of equations in matrix form, AX = B and obtain A, B.
Step II: Find |A|.
Step III: If |A| ≠ 0, then write "the system is consistent with unique solution". Obtain the unique solution by the following procedure. Find \( A^{-1} \) by using \( A^{-1} = \frac{1}{|A|}\text{adj } A \) and obtain the unique solution given by \( X = A^{-1}B \).
Step IV: If |A| = 0, then write "the system is either consistent with infinitely many solutions or it is inconsistent". To distinguish these two, proceed as follows: Find (adj A) B.
If (adj A) B ≠ O, then write "the system is inconsistent".
If (adj A) B = O, then the system is consistent with infinitely many solutions. To find these solutions proceed as follows. Put z = k (any real number) and take any two equations out of the three equations. Solve these equations for x and y. Let the values of x and y be λ and μ respectively. Then x = λ, y = μ, z = k is the required solution, where λ and μ are functions of k.

Solved Examples

Example 1. A square matrix \( A = [a_{ij}] \) in which \( a_{ij} = 0 \) for i ≠ j and \( a_{ij} = k \) (constant) for i = j is called a
(a) Unit matrix
(b) Scalar matrix
(c) Null matrix
(d) Diagonal matrix
Answer:
Ans. (b) Scalar matrix
When \( a_{ij} = 0 \) for i ≠ j and \( a_{ij} \) is constant for i = j, then the matrix \( [a_{ij}]_{n \times n} \) is called a scalar matrix.

Example 2. If A, B are square matrices of order 3, A is non-singular and AB = O, then B is a
(a) Null matrix
(b) Singular matrix
(c) Unit matrix
(d) Non-singular matrix
Answer:
Ans. (a) Null matrix
AB = O when B is a null matrix.

Example 3. If \( A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} \), then \( A^2 = \) [Rajasthan PET 2001]
(a) \( \begin{bmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{bmatrix} \)
(b) \( \begin{bmatrix} \cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{bmatrix} \)
(c) \( \begin{bmatrix} \cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \end{bmatrix} \)
(d) \( \begin{bmatrix} -\cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & -\cos 2\alpha \end{bmatrix} \)
Answer:
Ans. (c) \( \begin{bmatrix} \cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \end{bmatrix} \)
Since \( A^2 = A.A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}\begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} \cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \end{bmatrix} \)

Example 4. If \( A = \begin{bmatrix} a & b \\ b & a \end{bmatrix} \) and \( A^2 = \begin{bmatrix} \alpha & \beta \\ \beta & \alpha \end{bmatrix} \), then [AIEEE 2003]
(a) \( \alpha = a^2 + b^2, \beta = ab \)
(b) \( \alpha = a^2 + b^2, \beta = 2ab \)
(c) \( \alpha = a^2 + b^2, \beta = a^2 - b^2 \)
(d) \( \alpha = 2ab, \beta = a^2 + b^2 \)
Answer:
Ans. (b) \( \alpha = a^2 + b^2, \beta = 2ab \)
\( A^2 = \begin{bmatrix} \alpha & \beta \\ \beta & \alpha \end{bmatrix} = \begin{bmatrix} a & b \\ b & a \end{bmatrix}\begin{bmatrix} a & b \\ b & a \end{bmatrix} = \begin{bmatrix} a^2 + b^2 & 2ab \\ 2ab & a^2 + b^2 \end{bmatrix} \). On comparing, we get \( \alpha = a^2 + b^2, \beta = 2ab \)

Example 5. If \( A = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix} \), n ∈ N, then \( A^{4n} \) equals [AMU 1992]
(a) \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
(b) \( \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix} \)
(c) \( \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \)
(d) \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
Answer:
Ans. (a) \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( A^2 = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix}\begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \), \( A^4 = A^2.A^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \); \( (A^4)^n = I^n = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

Example 6. If \( A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \), \( B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \) and \( (A + B)^2 = A^2 + B^2 \), then the values of a and b are [Kurukshetra CEE 2002]
(a) a = 4, b = 1
(b) a = 1, b = 4
(c) a = 0, b = 4
(d) a = 2, b = 4
Answer:
Ans. (b) a = 1, b = 4
We have \( (A + B)^2 = A^2 + B^2 + A.B + BA \)
∵ \( (A + B)^2 = A^2 + B^2 \) ⇒ AB + BA = O
∴ \( \begin{bmatrix} a - b & 2 \\ 2a - b & 3 \end{bmatrix} + \begin{bmatrix} a + 2 & -a - 1 \\ b - 2 & -b + 1 \end{bmatrix} = O \)
\( \begin{bmatrix} 2a + 2 - b & -a + 1 \\ 2a - 2 & 4 - b \end{bmatrix} = O \). On comparing, we get −a + 1 = 0 ⇒ a = 1 and 4 − b = 0 ⇒ b = 4

Example 7. The order of \( [x\ y\ z] \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} \) is [EAMCET 1994]
(a) 3 × 1
(b) 1 × 1
(c) 1 × 3
(d) 3 × 3
Answer:
Ans. (b) 1 × 1
Order will be (1 × 3)(3 × 3)(3 × 1) = (1 × 1)

Example 8. Let \( F(\alpha) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \). Then \( F(\alpha).F(\alpha') \) is equal to
(a) \( F(\alpha\alpha') \)
(b) \( F(\alpha / \alpha') \)
(c) \( F(\alpha + \alpha') \)
(d) \( F(\alpha - \alpha') \)
Answer:
Ans. (c) \( F(\alpha + \alpha') \)
We have \( F(\alpha) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \), \( F(\alpha') = \begin{bmatrix} \cos \alpha' & -\sin \alpha' & 0 \\ \sin \alpha' & \cos \alpha' & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( F(\alpha).F(\alpha') = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} \cos \alpha' & -\sin \alpha' & 0 \\ \sin \alpha' & \cos \alpha' & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos(\alpha + \alpha') & -\sin(\alpha + \alpha') & 0 \\ \sin(\alpha + \alpha') & \cos(\alpha + \alpha') & 0 \\ 0 & 0 & 1 \end{bmatrix} = F(\alpha + \alpha') \)

Example 9. For the matrix \( A = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 2 & 1 & 0 \end{bmatrix} \), which of the following is correct
(a) \( A^3 + 3A^2 - I = O \)
(b) \( A^3 - 3A^2 - I = O \)
(c) \( A^3 + 2A^2 - I = O \)
(d) \( A^3 - A^2 + I = O \)
Answer:
Ans. (b) \( A^3 - 3A^2 - I = O \)
\( A^2 = A.A = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 2 & 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 2 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 1 \\ 5 & 6 & 2 \\ 3 & 4 & 1 \end{bmatrix} \), \( A^3 = A^2.A = \begin{bmatrix} 2 & 3 & 1 \\ 5 & 6 & 2 \\ 3 & 4 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 2 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 7 & 9 & 3 \\ 15 & 19 & 6 \\ 9 & 12 & 4 \end{bmatrix} \)
\( A^3 - 3A^2 = \begin{bmatrix} 7 & 9 & 3 \\ 15 & 19 & 6 \\ 9 & 12 & 4 \end{bmatrix} - \begin{bmatrix} 6 & 9 & 3 \\ 15 & 18 & 6 \\ 9 & 12 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I \) ⇒ \( A^3 - 3A^2 - I = O \)

Example 10. If \( A = \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix} \), then the value of α for which \( A^2 = B \) is [IIT Screening 2003]
(a) 1
(b) −1
(c) 4
(d) No real values
Answer:
Ans. (d) No real values
\( A^2 = \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} \alpha^2 & 0 \\ \alpha + 1 & 1 \end{bmatrix} \) ∵ \( A^2 = B \) (given)
Then \( \begin{bmatrix} \alpha^2 & 0 \\ \alpha + 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix} \) ⇒ \( \alpha^2 = 1 \) and α + 1 = 5. Clearly no real value of α.

Example 11. If A and B are square matrices of the same order then [Pb. CET 1992; Roorkee 1995; MP PET 1990; Rajasthan PET 1992, 94]
(a) (AB)′ = A′B′
(b) (AB)′ = B′A′
(c) AB = O, if |A| = 0 or |B| = 0
(d) AB = O, if |A| = I or B = I
Answer:
Ans. (b) (AB)′ = B′A′
\( A = [a_{ij}]_{n \times n}, B = [b_{jk}]_{n \times n} \), \( AB = [a_{ij}]_{n \times n}[b_{jk}]_{n \times n} = [c_{ik}]_{n \times n} \), where \( c_{ik} = \sum_j a_{ij} b_{jk} \)
\( (AB)' = [c_{ik}]'_{n \times n} = [c_{ki}]_{n \times n} = [b_{kj}]_{n \times n}[a_{ji}]_{n \times n} = B'A' \)
Alternatively, let \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}_{2 \times 2} \), \( B = \begin{bmatrix} 1 & 3 \\ 0 & 4 \end{bmatrix}_{2 \times 2} \); \( AB = \begin{bmatrix} 1 & 11 \\ 3 & 25 \end{bmatrix} \)
\( (AB)' = \begin{bmatrix} 1 & 3 \\ 11 & 25 \end{bmatrix} \) .....(i) and \( B'A' = \begin{bmatrix} 1 & 0 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 11 & 25 \end{bmatrix} \) .....(ii)
From (i) and (ii), (AB)′ = B′A′

Example 12. If A, B are 3 × 2 order matrices and C is a 2 × 3 order matrix, then which of the following matrices is not defined [Rajasthan PET 1998]
(a) A′ + B
(b) B + C′
(c) A′ + C
(d) A′ + B′
Answer:
Ans. (a) A′ + B
Order of A is 3 × 2 and order of B is 3 × 2 and order of A′ is 2 × 3, then \( A' + B = [A']_{2 \times 3} + [B]_{3 \times 2} \) is not possible because the orders are not the same.

Example 13. The matrix \( \begin{bmatrix} 0 & 5 & -7 \\ -5 & 0 & 11 \\ 7 & -11 & 0 \end{bmatrix} \) is known as [Karnataka CET 2000]
(a) Upper triangular matrix
(b) Skew-symmetric matrix
(c) Symmetric matrix
(d) Diagonal matrix
Answer:
Ans. (b) Skew-symmetric matrix
In a skew-symmetric matrix, \( a_{ij} = -a_{ji} \) ∀ i, j = 1, 2, 3 and for j = i, \( a_{ii} = -a_{ii} \) ⇒ each \( a_{ii} = 0 \)
Hence the given matrix is a skew-symmetric matrix [∵ \( A^T = -A \)].

Example 14. The matrix \( \begin{bmatrix} 2 & \lambda & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \) is non-singular if [Kurukshetra CEE 2002]
(a) λ ≠ −2
(b) λ ≠ 2
(c) λ ≠ 3
(d) λ ≠ −3
Answer:
Ans. (a) λ ≠ −2
The given matrix \( A = \begin{bmatrix} 2 & \lambda & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \) is non-singular if |A| ≠ 0
⇒ \( |A| = 2(-9 + 8) - \lambda(3 - 4) + (-4)(2 - 3) \neq 0 \) (expanding along R₁)
⇒ −2 + λ + 4 ≠ 0 ⇒ λ + 2 ≠ 0 ⇒ λ ≠ −2

Example 15. If \( A = \begin{bmatrix} 4 & x + 2 \\ 2x - 3 & x + 1 \end{bmatrix} \) is symmetric, then x = [Karnataka CET 1994]
(a) 3
(b) 5
(c) 2
(d) 4
Answer:
Ans. (b) 5
For a symmetric matrix, \( A = A^T \Rightarrow \begin{bmatrix} 4 & 2x - 3 \\ x + 2 & x + 1 \end{bmatrix} = \begin{bmatrix} 4 & x + 2 \\ 2x - 3 & x + 1 \end{bmatrix} \) ⇒ 2x − 3 = x + 2 ⇒ x = 5

Example 16. If A and B are square matrices of order n × n, then \( (A - B)^2 \) is equal to [Karnataka CET 1999; Kerala (Engg.) 2002]
(a) \( A^2 - B^2 \)
(b) \( A^2 - 2AB + B^2 \)
(c) \( A^2 + 2AB + B^2 \)
(d) \( A^2 - AB - BA + B^2 \)
Answer:
Ans. (d) \( A^2 - AB - BA + B^2 \)
Given A and B are square matrices of order n × n, we know that \( (A - B)^2 = (A - B)(A - B) = A^2 - AB - BA + B^2 \)

Example 17. Matrix A is such that \( A^2 = 2A - I \), where I is the identity matrix. Then for n ≥ 2, \( A^n = \)
(a) nA − (n − 1)I
(b) nA − I
(c) \( 2^{n-1}A - (n - 1)I \)
(d) \( 2^{n-1}A - I \)
Answer:
Ans. (a) nA − (n − 1)I
We have \( A^2 = 2A - I \Rightarrow A^2.A = (2A - I)A \); \( A^3 = 2A^2 - IA = 2[2A - I] - IA \Rightarrow A^3 = 3A - 2I \)
Similarly \( A^4 = 4A - 3I \) and hence \( A^n = nA - (n - 1)I \)

Example 18. Let \( A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \), the only correct statement about the matrix A is [AIEEE 2004]
(a) \( A^2 = I \)
(b) A = (−1)I, where I is the unit matrix
(c) \( A^{-1} \) does not exist
(d) A is a zero matrix
Answer:
Ans. (a) \( A^2 = I \)
\( A^2 = A.A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I \). Also, \( A^{-1} \) exists as |A| = 1

Example 19. If \( A = \begin{bmatrix} 4 & 2 \\ 3 & 4 \end{bmatrix} \), then |adj A| is equal to [UPSEAT 2003]
(a) 16
(b) 10
(c) 6
(d) None of these
Answer:
Ans. (b) 10
\( \text{adj } A = \begin{bmatrix} 4 & -2 \\ -3 & 4 \end{bmatrix} \)
\( |\text{adj } A| = \begin{vmatrix} 4 & -2 \\ -3 & 4 \end{vmatrix} = 16 - 6 = 10 \)

Example 20. If matrix \( A = \begin{bmatrix} 3 & 2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{bmatrix} \) and \( A^{-1} = \frac{1}{K}\text{adj}(A) \), then K is [UPSEAT 2002]
(a) 7
(b) −7
(c) 1/7
(d) 11
Answer:
Ans. (d) 11
We know that \( A^{-1} = \frac{\text{adj}(A)}{|A|} \). We have \( A^{-1} = \frac{1}{K}\text{adj}(A) \), i.e. K = |A|
and \( K = \begin{vmatrix} 3 & 2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{vmatrix} = 3(3) - 2(1) + 4(1) = 9 - 2 + 4 = 11 \)

Example 21. The inverse of matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is [AMU 2001]
(a) \( \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)
(b) \( \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)
(c) \( \frac{1}{|A|}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
(d) \( \begin{bmatrix} b & -a \\ d & -c \end{bmatrix} \)
Answer:
Ans. (b) \( \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)
Here \( |A| = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \), \( \text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \). Hence \( A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)

Example 22. Let \( A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix} \) and \( 10.B = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix} \). If B is the inverse of matrix A, then α is [AIEEE 2004]
(a) 5
(b) −1
(c) 2
(d) −2
Answer:
Ans. (a) 5
We have, \( A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix} \), ∴ |A| = 1(4) + 1(5) + 1(1) = 10 and \( \text{adj}(A) = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \)
Then \( A^{-1} = \frac{1}{10}\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \)
According to the question, B is the inverse of matrix A. Hence α = 5

Example 23. Matrix \( A = \begin{bmatrix} 1 & 0 & -K \\ 2 & 1 & 3 \\ K & 0 & 1 \end{bmatrix} \) is invertible for [UPSEAT 2002]
(a) K = 1
(b) K = −1
(c) K = 0
(d) All real K
Answer:
Ans. (d) All real K
For invertible, |A| ≠ 0, i.e. \( \begin{vmatrix} 1 & 0 & -K \\ 2 & 1 & 3 \\ K & 0 & 1 \end{vmatrix} \neq 0 \)
⇒ 1(1) − K(−K) ≠ 0 ⇒ \( |A| = K^2 + 1 \neq 0 \), which is true for all real K.

Example 24. Let p be a non-singular matrix, \( I + p + p^2 + \ldots + p^n = O \) (O denotes the null matrix), then \( p^{-1} = \) [Orissa JEE 2002]
(a) \( p^n \)
(b) \( -p^n \)
(c) \( -(I + p + \ldots + p^n) \)
(d) None of these
Answer:
Ans. (a) \( p^n \)
We have, \( I + p + p^2 + \ldots + p^n = O \)
Multiplying both sides by \( p^{-1} \), \( p^{-1} + I + Ip + \ldots + p^{n-1}I = O.p^{-1} \)
\( p^{-1} + I(I + p + \ldots + p^{n-1}) = O \Rightarrow p^{-1} = -I(I + p + p^2 + \ldots + p^{n-1}) = -(-p^n) = p^n \)

Example 25. Let \( f(\alpha) = \begin{bmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \), where α ∈ R, then \( [f(\alpha)]^{-1} \) is equal to [AMU 2000]
(a) \( f(-\alpha) \)
(b) \( f(\alpha^{-1}) \)
(c) \( f(2\alpha) \)
(d) None
Answer:
Ans. (a) \( f(-\alpha) \)
\( |f(\alpha)| = \begin{vmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{vmatrix} = 1 \), adj of \( f(\alpha) = \begin{bmatrix} \cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( [f(\alpha)]^{-1} = \begin{bmatrix} \cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \) ......(i) and \( f(-\alpha) = \begin{bmatrix} \cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \) ......(ii)
From (i) and (ii), \( [f(\alpha)]^{-1} = f(-\alpha) \)

Example 26. If I is a unit matrix of order 10, then the determinant of I is equal to [Kerala (Engg.) 2002]
(a) 10
(b) 1
(c) 1/10
(d) 9
Answer:
Ans. (b) 1
Determinant of a unit matrix of any order = 1.

Example 27. If \( A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix} \) and \( |A^3| = 125 \), then α = [IIT Screening 2004]
(a) ±3
(b) ±2
(c) ±5
(d) 0
Answer:
Ans. (a) ±3
\( 125 = |A^3| = |A|^3 \Rightarrow |A| = 5 \) and \( |A| = \alpha^2 - 4 = 5 \Rightarrow \alpha^2 = 9 \Rightarrow \alpha = \pm 3 \)

Example 28. If |A| denotes the value of the determinant of the square matrix A of order 3, then |−2A| = [MP PET 1987, 89, 92, 2000]
(a) −8|A|
(b) 8|A|
(c) −2|A|
(d) None of these
Answer:
Ans. (a) −8|A|
We know that \( \det(-A) = (-1)^n \det A \), where n is the order of the square matrix.
If A is a square matrix of order 3, then n = 3. Hence \( |-2A| = (-2)^3|A| = -8|A| \).

Example 29. For how many value(s) of x in the closed interval [−4, −1] is the matrix \( \begin{bmatrix} 3 & -1 + x & 2 \\ 3 & -1 & x + 2 \\ x + 3 & -1 & 2 \end{bmatrix} \) singular [Karnataka CET 2002]
(a) 2
(b) 0
(c) 3
(d) 1
Answer:
Ans. (d) 1
\( \begin{vmatrix} 3 & x - 1 & 2 \\ 3 & -1 & x + 2 \\ x + 3 & -1 & 2 \end{vmatrix} = 0 \)
Expanding along R₁: \( 3[-2 + (x + 2)] - (x - 1)[6 - (x + 2)(x + 3)] + 2[-3 + (x + 3)] = 0 \)
⇒ \( 3x + (x - 1)(x^2 + 5x) + 2x = 0 \Rightarrow x^3 + 4x^2 = 0 \Rightarrow x^2(x + 4) = 0 \) ⇒ x = 0, −4
Hence only one value of x in the closed interval [−4, −1], i.e. x = −4

Example 30. Inverse of a diagonal matrix (if it exists) is a
(a) Skew-symmetric matrix
(b) Diagonal matrix
(c) Non-invertible matrix
(d) None of these
Answer:
Ans. (b) Diagonal matrix
Let \( A = \text{diag}(d_1, d_2, \ldots, d_n) \)
As A is invertible, therefore \( \det(A) \neq 0 \Rightarrow d_1, d_2, \ldots, d_n \neq 0 \Rightarrow d_i \neq 0 \) for i = 1, 2, 3, …, n
Here, the cofactor of each non-diagonal entry is 0 and the cofactor of \( a_{ii} \)
\( = (-1)^{i+i} \det[\text{diag}(d_1, \ldots, d_{i-1}, d_{i+1}, \ldots, d_n)] = d_1 d_2 \ldots d_{i-1} d_{i+1} \ldots d_n = \frac{d_1 d_2 \ldots d_n}{d_i} = \frac{|A|}{d_i} \)
\( A^{-1} = \frac{1}{|A|}(\text{adj } A) = \text{diag}\left(\frac{1}{d_1}, \frac{1}{d_2}, \ldots, \frac{1}{d_n}\right) \), which is a diagonal matrix

Example 31. If \( \begin{bmatrix} x + y & 2x + z \\ x - y & 2z + w \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ 0 & 10 \end{bmatrix} \), then the values of x, y, z, w are
(a) 2, 2, 3, 4
(b) 2, 3, 1, 2
(c) 3, 3, 0, 1
(d) None of these
Answer:
Ans. (a) 2, 2, 3, 4
We have \( \begin{bmatrix} x + y & 2x + z \\ x - y & 2z + w \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ 0 & 10 \end{bmatrix} \)
x + y = 4, 2x + z = 7, x − y = 0 and 2z + w = 10 ⇒ x = 2 and y = 2, z = 3, w = 4

Example 32. The system of linear equations x + y + z = 2, 2x + y − z = 3, 3x + 2y + kz = 4 has a unique solution if [EAMCET 1994]
(a) k ≠ 0
(b) −1 < k < 1
(c) −2 < k < 2
(d) k = 0
Answer:
Ans. (a) k ≠ 0
The given system of equations has a unique solution if \( \begin{vmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{vmatrix} \neq 0 \)
⇒ 1(k + 2) − 1(2k + 3) + 1(4 − 3) ≠ 0 ⇒ −k ≠ 0 ⇒ k ≠ 0