CBSE Class 12 Mathematics Matrices and Determinants Assignment Set 11

Matrices & Determinants

Schematic Diagram

Some Important Results/Concepts

A matrix is a rectangular array of \( m \times n \) numbers arranged in \( m \) rows and \( n \) columns.

\[ A = \begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} \end{bmatrix}_{m \times n} \text{ OR } A = [a_{ij}]_{m \times n}, \text{ where } i = 1, 2, \dots, m; \, j = 1, 2, \dots, n. \]

• Row Matrix : A matrix which has one row is called row matrix. \( A = [a_{ij}]_{1 \times n} \)
• Column Matrix : A matrix which has one column is called column matrix. \( A = [a_{ij}]_{m \times 1} \).
• Square Matrix : A matrix in which number of rows are equal to number of columns, is called a square matrix \( A = [a_{ij}]_{m \times m} \)
• Diagonal Matrix : A square matrix is called a Diagonal Matrix if all the elements, except the diagonal elements are zero. \( A = [a_{ij}]_{n \times n} \), where \( a_{ij} = 0, \, i \neq j \); \( a_{ij} \neq 0, \, i = j \).
• Scalar Matrix : A square matrix is called scalar matrix it all the elements, except diagonal elements are zero and diagonal elements are same non-zero quantity.
  \( A = [a_{ij}]_{n \times n} \), where \( a_{ij} = 0, \, i \neq j \); \( a_{ij} = \alpha, \, i = j \).
• Identity or Unit Matrix : A square matrix in which all the non diagonal elements are zero and diagonal elements are unity is called identity or unit matrix.
• Null Matrices : A matrices in which all element are zero.
• Equal Matrices : Two matrices are said to be equal if they have same order and all their corresponding elements are equal.
• Transpose of matrix : If A is the given matrix, then the matrix obtained by interchanging the rows and columns is called the transpose of a matrix.
• Properties of Transpose :
  If A & B are matrices such that their sum & product are defined, then
  (i). \( (A^T)^T = A \)
  (ii). \( (A + B)^T = A^T + B^T \)
  (iii). \( (KA)^T = K \cdot A^T \) where K is a scalar.
  (iv). \( (AB)^T = B^T A^T \)
  (v). \( (ABC)^T = C^T B^T A^T \).
• Symmetric Matrix : A square matrix is said to be symmetric if \( A = A^T \) i.e. If \( A = [a_{ij}]_{m \times m} \), then \( a_{ij} = a_{ji} \) for all i, j. Also elements of the symmetric matrix are symmetric about the main diagonal
• Skew symmetric Matrix : A square matrix is said to be skew symmetric if \( A^T = -A \).
  If \( A = [a_{ij}]_{m \times m} \), then \( a_{ij} = -a_{ji} \) for all i, j.
• Singular matrix : A square matrix 'A' of order 'n' is said to be singular, if \( |A| = 0 \).
• Non -Singular matrix : A square matrix 'A' of order 'n' is said to be non-singular, if \( |A| \neq 0 \).
• Product of matrices :
  (i) If A & B are two matrices, then product AB is defined, if Number of column of A = number of rows of B.
  i.e. \( A = [a_{ij}]_{m \times n} \), \( B = [b_{jk}]_{n \times p} \) then \( AB = [C_{ik}]_{m \times p} \).
  (ii) Product of matrices is not commutative. i.e. \( AB \neq BA \).
  (iii) Product of matrices is associative. i.e \( A(BC) = (AB)C \)
  (iv) Product of matrices is distributive over addition.
• Adjoint of matrix :
  If \( A = [a_{ij}] \) be a n-square matrix then transpose of a matrix \( [A_{ij}] \), where \( A_{ij} \) is the cofactor of \( a_{ij} \) element of matrix A, is called the adjoint of A.
  Adjoint of A = \( \text{Adj. } A = [A_{ij}]^T \).
  \( A(\text{Adj. } A) = (\text{Adj. } A)A = |A| I \).
• Inverse of a matrix : Inverse of a square matrix A exists, if A is non-singular or square matrix A is said to be invertible and \( A^{-1} = \frac{1}{|A|} \text{Adj. } A \)
• System of Linear Equations :
  \( a_1 x + b_1 y + c_1 z = d_1 \).
  \( a_2 x + b_2 y + c_2 z = d_2 \).
  \( a_3 x + b_3 y + c_3 z = d_3 \).

\[ \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix} 
\implies AX = B 
\implies X = A^{-1}B; \, \{ |A| \neq 0 \}. \]

• Criteria of Consistency.
  (i) If \( |A| \neq 0 \), then the system of equations is said to be consistent & has a unique solution.
  (ii) If \( |A| = 0 \) and \( (\text{adj. } A)B = 0 \), then the system of equations is consistent and has infinitely many solutions.
  (iii) If \( |A| = 0 \) and \( (\text{adj. } A)B \neq 0 \), then the system of equations is inconsistent and has no solution.
• Determinant :
  To every square matrix we can assign a number called determinant
  If \( A = [a_{11}] \), det. \( A = |A| = a_{11} \).
  If \( A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \), \( |A| = a_{11}a_{22} - a_{21}a_{12} \).
• Properties :
  (i) The determinant of the square matrix A is unchanged when its rows and columns are interchanged.
  (ii) The determinant of a square matrix obtained by interchanging two rows(or two columns) is negative of given determinant.
  (iii) If two rows or two columns of a determinant are identical, value of the determinant is zero.
  (iv) If all the elements of a row or column of a square matrix A are multiplied by a non-zero number k, then determinant of the new matrix is k times the determinant of A.
  If elements of any one column(or row) are expressed as sum of two elements each, then determinant can be written as sum of two determinants.
  Any two or more rows(or column) can be added or subtracted proportionally.
  If A & B are square matrices of same order, then \( |AB| = |A| |B| \)

Assignments

(i). Order, Addition, Multiplication and transpose of matrices:

Level I

Question 1. If a matrix has 5 elements, what are the possible orders it can have?
Answer: The total number of elements in an \( m \times n \) matrix is given by the product \( mn \). Since 5 is a prime number, its only positive integral factor pairs are \( 1 \times 5 \) and \( 5 \times 1 \). Therefore, the only possible orders the matrix can have are \( 1 \times 5 \) and \( 5 \times 1 \).
In simple words: Because 5 can only be multiplied as 1 by 5 or 5 by 1, the matrix must be a single row or a single column.

Exam Tip: Always list all factor pairs of the given integer to make sure no valid matrix dimensions are omitted.

Question 2. Construct a 3 × 2 matrix whose elements are given by \( a_{ij} = \frac{1}{2}|i - 3j| \)
Answer: A matrix of order \( 3 \times 2 \) contains three rows and two columns. Evaluating each entry with the given formula:
\( a_{11} = \frac{1}{2}|1 - 3(1)| = \frac{1}{2}|-2| = 1 \)
\( a_{12} = \frac{1}{2}|1 - 3(2)| = \frac{1}{2}|-5| = \frac{5}{2} \)
\( a_{21} = \frac{1}{2}|2 - 3(1)| = \frac{1}{2}|-1| = \frac{1}{2} \)
\( a_{22} = \frac{1}{2}|2 - 3(2)| = \frac{1}{2}|-4| = 2 \)
\( a_{31} = \frac{1}{2}|3 - 3(1)| = \frac{1}{2}|0| = 0 \)
\( a_{32} = \frac{1}{2}|3 - 3(2)| = \frac{1}{2}|-3| = \frac{3}{2} \)
Thus, the required matrix is \( \begin{bmatrix} 1 & \frac{5}{2} \\ \frac{1}{2} & 2 \\ 0 & \frac{3}{2} \end{bmatrix} \).
In simple words: Plug the row number and column number into the formula for each position and take the positive value.

Exam Tip: Be careful with the modulus sign - remember that absolute value always turns negative values into positive numbers.

Question 3. If \( A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 3 \end{bmatrix} \), \( B = \begin{bmatrix} 2 & 3 & 1 \\ 1 & 0 & 2 \end{bmatrix} \), then find A -2 B.
Answer: First, multiply every entry of matrix B by the scalar 2:
\( 2B = 2 \begin{bmatrix} 2 & 3 & 1 \\ 1 & 0 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 6 & 2 \\ 2 & 0 & 4 \end{bmatrix} \).
Next, subtract the corresponding entries of \( 2B \) from matrix A:
\( A - 2B = \begin{bmatrix} 1 - 4 & 2 - 6 & 3 - 2 \\ 3 - 2 & 1 - 0 & 3 - 4 \end{bmatrix} = \begin{bmatrix} -3 & -4 & 1 \\ 1 & 1 & -1 \end{bmatrix} \).
In simple words: Double each number in matrix B first, then take them away from the matching spots in matrix A.

Exam Tip: Watch out for negative signs when subtracting entries with smaller values.

Question 4. If \( A = \begin{bmatrix} 2 & 1 & 4 \\ 4 & 1 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & -1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} \), write the order of AB and BA.
Answer: The dimensions of matrix A are \( 2 \times 3 \) and the dimensions of matrix B are \( 3 \times 2 \).
For the product AB, the inner dimensions match (3 and 3), producing a matrix of order \( 2 \times 2 \).
For the product BA, multiplying a \( 3 \times 2 \) matrix by a \( 2 \times 3 \) matrix yields a matrix of order \( 3 \times 3 \).
Therefore, the order of AB is \( 2 \times 2 \) and the order of BA is \( 3 \times 3 \).
In simple words: The outer numbers of the two matrix sizes give the size of the combined result.

Exam Tip: Always write down the dimensions of both matrices side by side to quickly check compatibility and resulting order.

Level II

Question 1. For the following matrices A and B, verify \( (AB)^T = B^TA^T \), where \( A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \), \( B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} \)
Answer: Find the product AB by multiplying the \( 3 \times 1 \) matrix A with the \( 1 \times 3 \) matrix B:
\( AB = \begin{bmatrix} 1(-1) & 1(2) & 1(1) \\ -4(-1) & -4(2) & -4(1) \\ 3(-1) & 3(2) & 3(1) \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix} \).
Taking the transpose yields:
\( (AB)^T = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix} \).
Now determine \( B^T \) and \( A^T \):
\( B^T = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \), \( A^T = \begin{bmatrix} 1 & -4 & 3 \end{bmatrix} \).
Multiplying these gives:
\( B^TA^T = \begin{bmatrix} -1(1) & -1(-4) & -1(3) \\ 2(1) & 2(-4) & 2(3) \\ 1(1) & 1(-4) & 1(3) \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix} \).
Since both sides are identical, \( (AB)^T = B^TA^T \) is verified.
In simple words: Multiplying then swapping rows and columns gives the exact same result as transposing both matrices first in reverse order.

Exam Tip: Remember the reversal rule: the transpose of a product reverses the order of the matrices, so \( (AB)^T = B^T A^T \), not \( A^T B^T \).

Question 2. Give example of matrices A & B such that AB = O, but BA ≠ O, where O is a zero matrix and A, B are both non zero matrices.
Answer: Consider the non-zero matrices \( A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \).
Evaluating their product AB:
\( AB = \begin{bmatrix} 0(1) + 1(0) & 0(0) + 1(0) \\ 0(1) + 0(0) & 0(0) + 0(0) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \).
Now evaluating the reverse product BA:
\( BA = \begin{bmatrix} 1(0) + 0(0) & 1(1) + 0(0) \\ 0(0) + 0(0) & 0(1) + 0(0) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \neq O \).
Hence, AB is a zero matrix while BA is non-zero.
In simple words: Matrix multiplication depends on the order you multiply in, so multiplying one way can give zeros even when the other way does not.

Exam Tip: Keeping strictly upper or lower triangular matrices with zeroes on the diagonal is the easiest way to construct zero-product counterexamples.

Question 3. If B is skew symmetric matrix, write whether the matrix (ABA^T) is Symmetric or skew symmetric.
Answer: Since B is skew-symmetric, we have \( B^T = -B \).
Let \( P = A B A^T \). Taking the transpose of both sides:
\( P^T = (A B A^T)^T = (A^T)^T B^T A^T \).
Using \( (A^T)^T = A \) and \( B^T = -B \):
\( P^T = A (-B) A^T = -(A B A^T) = -P \).
Because \( P^T = -P \), the matrix \( ABA^T \) is a skew-symmetric matrix.
In simple words: Taking the transpose flips the sign of the whole matrix because the middle matrix is skew-symmetric.

Exam Tip: State the defining property \( B^T = -B \) clearly before using the reversal rule for transposing products.

Question 4. If \( A = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), find a and b so that \( A^2 + aI = bA \)
Answer: First compute \( A^2 \):
\( A^2 = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} = \begin{bmatrix} 9 + 7 & 3 + 5 \\ 21 + 35 & 7 + 25 \end{bmatrix} = \begin{bmatrix} 16 & 8 \\ 56 & 32 \end{bmatrix} \).
Substitute into the given equation \( A^2 + aI = bA \):
\( \begin{bmatrix} 16 & 8 \\ 56 & 32 \end{bmatrix} + \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} = b \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \)
\( \implies \begin{bmatrix} 16 + a & 8 \\ 56 & 32 + a \end{bmatrix} = \begin{bmatrix} 3b & b \\ 7b & 5b \end{bmatrix} \).
Equating corresponding entries:
From the entry in row 1, column 2: \( b = 8 \).
From the entry in row 1, column 1: \( 16 + a = 3b \implies 16 + a = 3(8) = 24 \implies a = 8 \).
Checking the remaining elements: \( 7(8) = 56 \) and \( 32 + 8 = 40 = 5(8) \), which are both satisfied.
Therefore, \( a = 8 \) and \( b = 8 \).
In simple words: Work out \( A^2 \), set the resulting matrices equal element by element, and solve for the missing constants.

Exam Tip: Always cross-check your values of \( a \) and \( b \) against every element in the matrix to prevent simple algebraic slips.

Level III

Question 1. If \( A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \), then find the value of A^2-3A + 2I
Answer: First determine \( A^2 \):
\( A^2 = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 4 + 0 + 1 & 0 + 0 - 1 & 2 + 0 + 0 \\ 4 + 2 + 3 & 0 + 1 - 3 & 2 + 3 + 0 \\ 2 - 2 + 0 & 0 - 1 + 0 & 1 - 3 + 0 \end{bmatrix} = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} \).
Next calculate \( 3A \) and \( 2I \):
\( 3A = \begin{bmatrix} 6 & 0 & 3 \\ 6 & 3 & 9 \\ 3 & -3 & 0 \end{bmatrix} \), \( 2I = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \).
Now evaluate \( A^2 - 3A + 2I \):
\( \begin{bmatrix} 5 - 6 + 2 & -1 - 0 + 0 & 2 - 3 + 0 \\ 9 - 6 + 0 & -2 - 3 + 2 & 5 - 9 + 0 \\ 0 - 3 + 0 & -1 - (-3) + 0 & -2 - 0 + 2 \end{bmatrix} = \begin{bmatrix} 1 & -1 & -1 \\ 3 & -3 & -4 \\ -3 & 2 & 0 \end{bmatrix} \).
In simple words: Find the square of matrix A by multiplying it with itself, then subtract 3 times A and add 2 times the identity matrix.

Exam Tip: Be methodical during \( 3 \times 3 \) multiplication - calculate each row-by-column term separately to avoid calculation errors.

Question 2. Express the matrix A as the sum of a symmetric and a skew symmetric matrix, where: \( A = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \)
Answer: Any square matrix can be expressed as \( A = P + Q \), where \( P = \frac{1}{2}(A + A^T) \) is symmetric and \( Q = \frac{1}{2}(A - A^T) \) is skew-symmetric.
First find \( A^T \):
\( A^T = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \).
Now compute \( P \):
\( P = \frac{1}{2} \left( \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \right) = \frac{1}{2} \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix} \).
Now compute \( Q \):
\( Q = \frac{1}{2} \left( \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \right) = \frac{1}{2} \begin{bmatrix} 0 & -5 & -3 \\ 5 & 0 & -6 \\ 3 & 6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -\frac{5}{2} & -\frac{3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0 \end{bmatrix} \).
Thus, \( A = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & -\frac{5}{2} & -\frac{3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0 \end{bmatrix} \).
In simple words: Split the matrix into two parts - one that stays identical after transposing and one that flips signs entirely.

Exam Tip: Remember to explicitly verify that \( P^T = P \) and \( Q^T = -Q \) to secure full marks in descriptive board exam questions.

Question 3. If \( A = \begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix} \), prove that \( A^n = \begin{bmatrix} a^n & \frac{b(a^n-1)}{a-1} \\ 0 & 1 \end{bmatrix} \) , n∈N
Answer: We prove the result using the principle of mathematical induction.
For \( n = 1 \):
\( A^1 = \begin{bmatrix} a^1 & \frac{b(a^1 - 1)}{a - 1} \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix} = A \). Hence, the base case holds.
Assume the result is true for \( n = k \):
\( A^k = \begin{bmatrix} a^k & \frac{b(a^k - 1)}{a - 1} \\ 0 & 1 \end{bmatrix} \).
Now examine the case for \( n = k + 1 \):
\( A^{k+1} = A^k A = \begin{bmatrix} a^k & \frac{b(a^k - 1)}{a - 1} \\ 0 & 1 \end{bmatrix} \begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} a^k(a) + 0 & a^k(b) + \frac{b(a^k - 1)}{a - 1}(1) \\ 0(a) + 1(0) & 0(b) + 1(1) \end{bmatrix} \)
Simplify the upper right element:
\( a^k b + \frac{b(a^k - 1)}{a - 1} = \frac{a^k b(a - 1) + b(a^k - 1)}{a - 1} = \frac{b(a^{k+1} - a^k + a^k - 1)}{a - 1} = \frac{b(a^{k+1} - 1)}{a - 1} \).
Thus, \( A^{k+1} = \begin{bmatrix} a^{k+1} & \frac{b(a^{k+1} - 1)}{a - 1} \\ 0 & 1 \end{bmatrix} \).
Therefore, the result holds for \( n = k + 1 \). By mathematical induction, it holds for all \( n \in \mathbb{N} \).
In simple words: Check that the formula works for step 1, assume it works for step k, and show that multiplying once more gives the exact formula for step k plus 1.

Exam Tip: Clearly write out the simplification of the top-right entry during the induction step, as examiners look for this key algebraic manipulation.

(ii) Cofactors &Adjoint of a matrix

Level I

Question 1. Find the co-factor of a12 in \( A = \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix} \)
Answer: The cofactor \( A_{ij} \) of an entry \( a_{ij} \) is defined as \( A_{ij} = (-1)^{i+j} M_{ij} \), where \( M_{ij} \) denotes the minor obtained by deleting row \( i \) and column \( j \).
For \( a_{12} \):
\( A_{12} = (-1)^{1+2} \begin{vmatrix} 6 & 4 \\ 1 & -7 \end{vmatrix} = - [6(-7) - 1(4)] = - [-42 - 4] = -(-46) = 46 \).
Therefore, the cofactor of \( a_{12} \) is 46.
In simple words: Cross out row 1 and column 2, find the determinant of the 4 remaining numbers, and switch its sign because 1 plus 2 is odd.

Exam Tip: Always remember that position (1, 2) carries a negative sign because \( (-1)^{1+2} = -1 \).

Question 2. Find the adjoint of the matrix \( A = \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix} \)
Answer: To find the adjoint of a \( 2 \times 2 \) matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), interchange the elements on the principal diagonal and change the signs of the off-diagonal elements, yielding \( \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \).
Applying this rule to \( A = \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix} \):
\( \text{adj } A = \begin{bmatrix} 3 & -(-1) \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ -4 & 2 \end{bmatrix} \).
In simple words: Swap the two diagonal numbers, and change the plus or minus signs on the other two numbers.

Exam Tip: While the shortcut works rapidly for \( 2 \times 2 \) matrices, briefly listing cofactors shows clear steps for full marks.

Level II

Verify A(adjA) = (adjA) A = |A|I if

Question 1. \( A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} \)
Answer: First compute the determinant of A:
\( |A| = 2(-6) - (-4)(3) = -12 + 12 = 0 \).
Next compute \( \text{adj } A \):
\( \text{adj } A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} \).
Now evaluate the product \( A(\text{adj } A) \):
\( A(\text{adj } A) = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 2(-6) + 3(4) & 2(-3) + 3(2) \\ -4(-6) + (-6)(4) & -4(-3) + (-6)(2) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).
Similarly evaluate \( (\text{adj } A)A \):
\( (\text{adj } A)A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} = \begin{bmatrix} -6(2) + (-3)(-4) & -6(3) + (-3)(-6) \\ 4(2) + 2(-4) & 4(3) + 2(-6) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).
Also, \( |A|I = 0 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).
Hence, \( A(\text{adj } A) = (\text{adj } A) A = |A|I \) is verified.
In simple words: Since the determinant is zero, multiplying the matrix by its adjoint in either order gives a matrix of all zeroes.

Exam Tip: Even when the determinant is zero, you must still explicitly compute and show both matrix multiplications.

Question 2. \( A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 2 \\ 3 & 3 & 4 \end{bmatrix} \)
Answer: Find the determinant of matrix A:
\( |A| = 1(12 - 6) - 2(8 - 6) + 3(6 - 9) = 1(6) - 2(2) + 3(-3) = 6 - 4 - 9 = -7 \).
Now determine the cofactors of all entries:
\( A_{11} = 12 - 6 = 6 \), \( A_{12} = -(8 - 6) = -2 \), \( A_{13} = 6 - 9 = -3 \)
\( A_{21} = -(8 - 9) = 1 \), \( A_{22} = 4 - 9 = -5 \), \( A_{23} = -(3 - 6) = 3 \)
\( A_{31} = 4 - 9 = -5 \), \( A_{32} = -(2 - 6) = 4 \), \( A_{33} = 3 - 4 = -1 \).
Transposing the cofactor matrix gives the adjoint:
\( \text{adj } A = \begin{bmatrix} 6 & 1 & -5 \\ -2 & -5 & 4 \\ -3 & 3 & -1 \end{bmatrix} \).
Now evaluate \( A(\text{adj } A) \):
\( \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 2 \\ 3 & 3 & 4 \end{bmatrix} \begin{bmatrix} 6 & 1 & -5 \\ -2 & -5 & 4 \\ -3 & 3 & -1 \end{bmatrix} = \begin{bmatrix} 6 - 4 - 9 & 1 - 10 + 9 & -5 + 8 - 3 \\ 12 - 6 - 6 & 2 - 15 + 6 & -10 + 12 - 2 \\ 18 - 6 - 12 & 3 - 15 + 12 & -15 + 12 - 4 \end{bmatrix} = \begin{bmatrix} -7 & 0 & 0 \\ 0 & -7 & 0 \\ 0 & 0 & -7 \end{bmatrix} = -7 I \).
Similarly, \( (\text{adj } A)A = \begin{bmatrix} -7 & 0 & 0 \\ 0 & -7 & 0 \\ 0 & 0 & -7 \end{bmatrix} = -7 I \).
Since \( |A|I = -7 I \), the relation \( A(\text{adj } A) = (\text{adj } A) A = |A|I \) holds.
In simple words: Multiplying this matrix by its adjoint produces a scalar matrix where the diagonal entries all match the determinant.

Exam Tip: Remember that the adjoint is the transpose of the cofactor matrix, not just the matrix of cofactors itself.

(iii)Inverse of a Matrix & Applications

Level I

Question 1. If \( A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix} \), write A^-1 in terms of A 
Answer: First compute the determinant of A:
\( |A| = 2(-2) - 5(3) = -4 - 15 = -19 \).
Next find \( \text{adj } A \):
\( \text{adj } A = \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix} = - \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix} = -A \).
Using the formula for the inverse \( A^{-1} = \frac{1}{|A|} \text{adj } A \):
\( A^{-1} = \frac{1}{-19}(-A) = \frac{1}{19} A \).
In simple words: The adjoint of this specific matrix equals its negative, so dividing by the determinant produces \( \frac{1}{19} A \).

Exam Tip: Look for relationships between \( \text{adj } A \) and the original matrix \( A \) when a question asks for the inverse in terms of \( A \).

Question 2. If A is square matrix satisfying A^2 = I, then what is the inverse of A ?
Answer: Given that \( A^2 = I \), we can write this matrix relation as:
\( A \cdot A = I \).
By the definition of matrix inverse, if \( A B = I \) for square matrices, then \( B = A^{-1} \).
Because multiplying matrix A by itself yields the identity matrix, it follows directly that \( A^{-1} = A \).
In simple words: Since A times A gives the identity matrix, A is its own inverse.

Exam Tip: A matrix satisfying \( A^2 = I \) is called an involuntary matrix; stating this term shows strong conceptual understanding.

Question 3. For what value of k , the matrix \( A = \begin{bmatrix} 2-k & 3 \\ -5 & 1 \end{bmatrix} \) is not invertible ?
Answer: A square matrix is not invertible if and only if it is singular, which requires its determinant to be zero.
Compute the determinant of A and set it to zero:
\( |A| = (2 - k)(1) - (-5)(3) = 0 \)
\( \implies 2 - k + 15 = 0 \)
\( \implies 17 - k = 0 \)
\( \implies k = 17 \).
Therefore, the matrix is not invertible when \( k = 17 \).
In simple words: Work out the determinant formula, set it equal to zero, and solve for k.

Exam Tip: "Not invertible" and "singular" mean the exact same thing: determinant equals zero.

Level II

Question 1. If \( A = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \), show that A^2–5A – 14I = 0. Hence find A^-1
Answer: First compute \( A^2 \):
\( A^2 = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 9 + 20 & -15 - 10 \\ -12 - 8 & 20 + 4 \end{bmatrix} = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} \).
Now evaluate \( A^2 - 5A - 14I \):
\( \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} - 5 \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} - 14 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 29 - 15 - 14 & -25 + 25 - 0 \\ -20 + 20 - 0 & 24 - 10 - 14 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0 \).
To find \( A^{-1} \), post-multiply both sides of the equation by \( A^{-1} \):
\( A^2 A^{-1} - 5A A^{-1} - 14I A^{-1} = 0 \)
\( \implies A - 5I - 14 A^{-1} = 0 \)
\( \implies 14 A^{-1} = A - 5I \)
\( \implies A^{-1} = \frac{1}{14}(A - 5I) \).
Calculate \( A - 5I \):
\( A - 5I = \begin{bmatrix} 3 - 5 & -5 \\ -4 & 2 - 5 \end{bmatrix} = \begin{bmatrix} -2 & -5 \\ -4 & -3 \end{bmatrix} \).
Thus, \( A^{-1} = \frac{1}{14} \begin{bmatrix} -2 & -5 \\ -4 & -3 \end{bmatrix} = -\frac{1}{14} \begin{bmatrix} 2 & 5 \\ 4 & 3 \end{bmatrix} \).
In simple words: Verify that the matrix polynomial simplifies to zero, then multiply through by the inverse to solve for \( A^{-1} \) without calculating the adjoint directly.

Exam Tip: When a question says "Hence find \( A^{-1} \)", you must derive the inverse from the given equation - finding it via cofactors will lose method marks.

Question 2. If A, B, C are three non zero square matrices of same order, find the condition on A such that AB = AC B = C.
Answer: Given the matrix equation \( AB = AC \), we can rearrange it as \( AB - AC = O \implies A(B - C) = O \).
For this relation to guarantee that \( B - C = O \) (which means \( B = C \)), the matrix A must possess an inverse.
An inverse \( A^{-1} \) exists if and only if matrix A is non-singular, meaning \( |A| \neq 0 \).
Pre-multiplying both sides by \( A^{-1} \):
\( A^{-1}(AB) = A^{-1}(AC) \implies (A^{-1}A)B = (A^{-1}A)C \implies IB = IC \implies B = C \).
Therefore, the essential condition is that A must be a non-singular matrix (\( |A| \neq 0 \)).
In simple words: You can only cancel matrix A from both sides if its determinant is non-zero, allowing its inverse to exist.

Exam Tip: Mention explicitly that matrix cancellation is valid only through pre-multiplication by the inverse of a non-singular matrix.

Question 3. Find the number of all possible matrices A of order 3 × 3 with each entry 0 or 1 and for which A [x y z]^T = [1 0 0]^T has exactly two distinct solutions.
Answer: For any system of linear equations \( AX = B \) over the field of real numbers, the solution set can either be empty (no solution), contain exactly one unique solution, or contain infinitely many solutions.
If a linear system has two distinct solutions \( X_1 \) and \( X_2 \), then any linear combination \( k X_1 + (1 - k)X_2 \) for any real number \( k \) also serves as a solution, which implies there must be infinitely many solutions.
A linear system can never possess a finite number of solutions greater than one.
Therefore, no matrix A can produce exactly two distinct solutions, meaning the total number of such matrices is 0.
In simple words: A system of linear equations never has exactly two answers - it either has none, one, or infinitely many.

Exam Tip: This is a conceptual trap question; state the fundamental property of linear systems having 0, 1, or \( \infty \) solutions immediately.

Level III

Question 1. If \( A = \begin{bmatrix} 2 & 3 & 1 \\ -3 & 2 & 1 \\ 5 & -4 & -2 \end{bmatrix} \), find A^-1 and hence solve the following system of equations: 2x – 3y + 5z = 11, 3x + 2y – 4z = - 5, x + y – 2z = - 3
Answer: First compute the determinant of matrix A:
\( |A| = 2(-4 + 4) - 3(6 - 5) + 1(12 - 10) = 2(0) - 3(1) + 1(2) = -1 \neq 0 \).
Find the cofactors of A:
\( A_{11} = 0 \), \( A_{12} = -(6 - 5) = -1 \), \( A_{13} = 12 - 10 = 2 \)
\( A_{21} = -(-6 + 4) = 2 \), \( A_{22} = -4 - 5 = -9 \), \( A_{23} = -(-8 - 15) = 23 \)
\( A_{31} = 3 - 2 = 1 \), \( A_{32} = -(2 + 3) = -5 \), \( A_{33} = 4 + 9 = 13 \).
Thus, the adjoint is:
\( \text{adj } A = \begin{bmatrix} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13 \end{bmatrix} \).
Therefore, the inverse is:
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{-1} \begin{bmatrix} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13 \end{bmatrix} = \begin{bmatrix} 0 & -2 & -1 \\ 1 & 9 & 5 \\ -2 & -23 & -13 \end{bmatrix} \).
The system of equations can be written in matrix form as \( A^T X = B \), where:
\( A^T = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \).
Solving for X gives \( X = (A^T)^{-1} B = (A^{-1})^T B \):
\( (A^{-1})^T = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \).
Now multiply:
\( X = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} = \begin{bmatrix} 0(11) + 1(-5) - 2(-3) \\ -2(11) + 9(-5) - 23(-3) \\ -1(11) + 5(-5) - 13(-3) \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \).
Therefore, \( x = 1, y = 2, z = 3 \).
In simple words: Find the inverse of A, recognize that the system uses the transpose of A, and multiply the transposed inverse by the constants on the right.

Exam Tip: Notice carefully whether the coefficient matrix is \( A \) or \( A^T \) - students frequently miss that the equations use the transpose of the given matrix.

Question 2. Using matrices, solve the following system of equations:
(a) x + 2y - 3z = - 4
2x + 3y + 2z = 2
3x - 3y – 4z = 11 
(b) 4x + 3y + 2z = 60
x + 2y + 3z = 45
6x + 2y + 3z = 70 
Answer:
(a) Write the system in matrix form as \( AX = B \):
\( A = \begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} -4 \\ 2 \\ 11 \end{bmatrix} \).
Calculate the determinant of A:
\( |A| = 1(-12 + 6) - 2(-8 - 6) - 3(-6 - 9) = -6 + 28 + 45 = 67 \neq 0 \).
The cofactors of matrix A are:
\( A_{11} = -6, A_{12} = 14, A_{13} = -15 \)
\( A_{21} = 17, A_{22} = 5, A_{23} = 9 \)
\( A_{31} = 13, A_{32} = -8, A_{33} = -1 \).
Thus, \( \text{adj } A = \begin{bmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{bmatrix} \).
Now compute \( X = A^{-1}B = \frac{1}{67} \begin{bmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{bmatrix} \begin{bmatrix} -4 \\ 2 \\ 11 \end{bmatrix} \):
\( x = \frac{1}{67}(24 + 34 + 143) = \frac{201}{67} = 3 \)
\( y = \frac{1}{67}(-56 + 10 - 88) = \frac{-134}{67} = -2 \)
\( z = \frac{1}{67}(60 + 18 - 11) = \frac{67}{67} = 1 \).
Hence, \( x = 3, y = -2, z = 1 \).

(b) Write the system as \( AX = B \):
\( A = \begin{bmatrix} 4 & 3 & 2 \\ 1 & 2 & 3 \\ 6 & 2 & 3 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} 60 \\ 45 \\ 70 \end{bmatrix} \).
Compute the determinant of A:
\( |A| = 4(6 - 6) - 3(3 - 18) + 2(2 - 12) = 0 + 45 - 20 = 25 \neq 0 \).
The cofactors of matrix A are:
\( A_{11} = 0, A_{12} = 15, A_{13} = -10 \)
\( A_{21} = -5, A_{22} = 0, A_{23} = 10 \)
\( A_{31} = 5, A_{32} = -10, A_{33} = 5 \).
Thus, \( \text{adj } A = \begin{bmatrix} 0 & -5 & 5 \\ 15 & 0 & -10 \\ -10 & 10 & 5 \end{bmatrix} \).
Now compute \( X = A^{-1}B = \frac{1}{25} \begin{bmatrix} 0 & -5 & 5 \\ 15 & 0 & -10 \\ -10 & 10 & 5 \end{bmatrix} \begin{bmatrix} 60 \\ 45 \\ 70 \end{bmatrix} \):
\( x = \frac{1}{25}(0 - 225 + 350) = \frac{125}{25} = 5 \)
\( y = \frac{1}{25}(900 + 0 - 700) = \frac{200}{25} = 8 \)
\( z = \frac{1}{25}(-600 + 450 + 350) = \frac{200}{25} = 8 \).
Hence, \( x = 5, y = 8, z = 8 \).
In simple words: Turn the equations into a matrix equation, find the inverse with cofactors, and multiply by the right-hand numbers.

Exam Tip: Always substitute your final solutions back into one of the original equations to confirm accuracy before proceeding.

Question 3. Find the product AB, where \( A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \), \( B = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \) and use it to solve the equations x – y = 3, 2x + 3y + 4z = 17, y + 2z = 7
Answer: Compute the matrix product AB:
\( AB = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \)
\( = \begin{bmatrix} 2 + 4 + 0 & 2 - 2 + 0 & -4 + 4 + 0 \\ 4 - 12 + 8 & 4 + 6 - 4 & -8 - 12 + 20 \\ 0 - 4 + 4 & 0 + 2 - 2 & 0 - 4 + 10 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = 6I \).
Since \( AB = 6I \), it follows that \( A \left( \frac{1}{6} B \right) = I \), which implies \( A^{-1} = \frac{1}{6} B \).
The system of equations can be represented as \( AX = C \), where:
\( A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( C = \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} \).
Then \( X = A^{-1} C = \frac{1}{6} B C \):
\( X = \frac{1}{6} \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 6 + 34 - 28 \\ -12 + 34 - 28 \\ 6 - 17 + 35 \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 12 \\ -6 \\ 24 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix} \).
Thus, the solution is \( x = 2, y = -1, z = 4 \).
In simple words: The product AB gives 6 times the identity, which means matrix B scaled by 1/6 is already the inverse you need to solve the system.

Exam Tip: Never compute the inverse independently using cofactors when the question directs you to use the matrix product.

Question 4. Using matrices, solve the following system of equations: 1/x - 1/y + 1/z = 4, 2/x + 1/y - 3/z = 0, 1/x + 1/y + 1/z = 2
Answer: Let \( u = \frac{1}{x} \), \( v = \frac{1}{y} \), and \( w = \frac{1}{z} \). The linear equations become:
\( u - v + w = 4 \)
\( 2u + v - 3w = 0 \)
\( u + v + w = 2 \).
In matrix form \( AU = D \):
\( A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix} \), \( U = \begin{bmatrix} u \\ v \\ w \end{bmatrix} \), \( D = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} \).
Evaluate the determinant of A:
\( |A| = 1(1 + 3) - (-1)(2 + 3) + 1(2 - 1) = 4 + 5 + 1 = 10 \neq 0 \).
The cofactors of matrix A are:
\( A_{11} = 4, A_{12} = -5, A_{13} = 1 \)
\( A_{21} = 2, A_{22} = 0, A_{23} = -2 \)
\( A_{31} = 2, A_{32} = 5, A_{33} = 3 \).
Thus, \( \text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \).
Solving for U:
\( U = A^{-1}D = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 16 + 0 + 4 \\ -20 + 0 + 10 \\ 4 + 0 + 6 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} \).
Since \( u = 2 \), \( v = -1 \), and \( w = 1 \), we get:
\( x = \frac{1}{u} = \frac{1}{2} \), \( y = \frac{1}{v} = -1 \), \( z = \frac{1}{w} = 1 \).
Therefore, \( x = \frac{1}{2}, y = -1, z = 1 \).
In simple words: Replace the reciprocal fractions with simple letters, solve the resulting matrix system, and flip the answers back.

Exam Tip: Do not forget the final substitution step to invert \( u, v, w \) back into \( x, y, z \).

Question 5. Using elementary transformations, find the inverse of the matrix \( \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \)
Answer: Let \( A = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \). We set up the row operation equation \( A = IA \):
\( \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \).
Apply \( R_2 \to R_2 + R_1 \):
\( \begin{bmatrix} 1 & 2 & -2 \\ 0 & 5 & -2 \\ 0 & -2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \).
Apply \( R_2 \to R_2 + 2R_3 \):
\( \begin{bmatrix} 1 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} A \).
Apply \( R_3 \to R_3 + 2R_2 \):
\( \begin{bmatrix} 1 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} A \).
Apply \( R_1 \to R_1 + 2R_3 \):
\( \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 4 & 10 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} A \).
Apply \( R_1 \to R_1 - 2R_2 \):
\( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} A \).
Hence, \( A^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \).
In simple words: Carry out identical row operations on both the matrix and the identity matrix until the original matrix turns into the identity.

Exam Tip: Stick strictly to either row operations or column operations throughout the problem; never mix both.

(iv)To Find The Difference Between |A|, |adjA|, |kA|, |A.adjA|

Level I

Question 1. Evaluate \( \begin{vmatrix} \cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ \end{vmatrix} \)
Answer: Expanding the \( 2 \times 2 \) determinant:
\( \Delta = \cos 15^\circ \cos 75^\circ - \sin 75^\circ \sin 15^\circ \).
Using the trigonometric addition identity \( \cos(A + B) = \cos A \cos B - \sin A \sin B \):
\( \Delta = \cos(15^\circ + 75^\circ) = \cos 90^\circ = 0 \).
Therefore, the value of the determinant is 0.
In simple words: The cross-multiplication matches the cosine addition identity for 90 degrees, which is 0.

Exam Tip: Look out for standard trigonometric compound angle formulas when dealing with determinants containing sine and cosine terms.

Question 2. What is the value of |3I| , where I is identity matrix of order 3?
Answer: For any scalar k and square matrix A of order n, the property states that \( |kA| = k^n |A| \).
For an identity matrix of order \( n = 3 \), we have \( |I| = 1 \).
Applying the property with \( k = 3 \):
\( |3I| = 3^3 |I| = 27(1) = 27 \).
Thus, the value of \( |3I| \) is 27.
In simple words: Pulling a number out of a 3 by 3 determinant cubes it, so 3 cubed gives 27.

Exam Tip: Never write \( |kA| = k|A| \); the scalar must always be raised to the power equal to the order of the matrix.

Question 3. If A is non singular matrix of order 3 and |A| = 3, then find |2A|
Answer: For a square matrix A of order \( n = 3 \) and a scalar \( k \), the determinant relation is \( |kA| = k^n |A| \).
Here \( k = 2 \), \( n = 3 \), and \( |A| = 3 \).
Substituting these values:
\( |2A| = 2^3 |A| = 8 \times 3 = 24 \).
Therefore, the value of \( |2A| \) is 24.
In simple words: Multiply the determinant by 2 raised to the power 3.

Exam Tip: Double-check the order \( n \) given in the question before computing powers of the scalar.

Question 4. For what valve of a, \( \begin{bmatrix} 2a & -1 \\ -8 & 3 \end{bmatrix} \) is a singular matrix?
Answer: A matrix is singular if its determinant is equal to zero.
Setting the determinant of the matrix to zero:
\( \begin{vmatrix} 2a & -1 \\ -8 & 3 \end{vmatrix} = 0 \)
\( \implies 2a(3) - (-8)(-1) = 0 \)
\( \implies 6a - 8 = 0 \)
\( \implies 6a = 8 \)
\( \implies a = \frac{8}{6} = \frac{4}{3} \).
Thus, the matrix is singular when \( a = \frac{4}{3} \).
In simple words: Cross-multiply, set the difference to zero, and solve for a.

Exam Tip: Watch the signs carefully: \( (-8) \times (-1) = +8 \), so subtracting it results in \( -8 \).

Level II

Question 1. If A is a square matrix of order 3 such that |adjA| = 64, find |A|
Answer: For any square matrix A of order n, the determinant of its adjoint is given by \( |\text{adj } A| = |A|^{n-1} \).
Given that \( n = 3 \), we have:
\( |\text{adj } A| = |A|^{3-1} = |A|^2 \).
Equating this to 64:
\( |A|^2 = 64 
\implies |A| = \pm 8 \).
Therefore, the value of \( |A| \) is \( \pm 8 \).
In simple words: For a 3 by 3 matrix, the determinant of the adjoint equals the square of the original determinant.

Exam Tip: Always remember to write \( \pm 8 \) instead of just positive 8 when taking a square root.

Question 2. If A is a non singular matrix of order 3 and |A| = 7, then find |adjA|
Answer: By the property of determinants, for a matrix of order n, \( |\text{adj } A| = |A|^{n-1} \).
For order \( n = 3 \) and \( |A| = 7 \):
\( |\text{adj } A| = |A|^{3-1} = |A|^2 = 7^2 = 49 \).
Hence, the value of \( |\text{adj } A| \) is 49.
In simple words: Square the determinant to get the determinant of the adjoint for a 3 by 3 matrix.

Exam Tip: State the formula \( |\text{adj } A| = |A|^{n-1} \) clearly before substituting the numbers.

Level III

Question 1. If \( A = \begin{bmatrix} a & 2 \\ 2 & a \end{bmatrix} \) and |A|^3 = 125, then find a.
Answer: Given that \( |A|^3 = 125 \), taking the real cube root on both sides yields:
\( |A| = \sqrt[3]{125} = 5 \).
Compute the determinant of matrix A directly:
\( |A| = a(a) - 2(2) = a^2 - 4 \).
Equating the two expressions:
\( a^2 - 4 = 5 
\implies a^2 = 9 
\implies a = \pm 3 \).
Therefore, the possible values of a are \( \pm 3 \).
In simple words: Take the cube root to find that the determinant is 5, then solve \( a^2 - 4 = 5 \).

Exam Tip: Do not omit the negative root: \( a^2 = 9 \) gives both \( +3 \) and \( -3 \).

Question 2. A square matrix A, of order 3, has |A| = 5, find |A. adjA|
Answer: By definition, for any square matrix A of order n, \( A \cdot \text{adj } A = |A| I_n \).
Taking the determinant of both sides:
\( |A \cdot \text{adj } A| = ||A| I_n| = |A|^n |I_n| = |A|^n (1) = |A|^n \).
For \( n = 3 \) and \( |A| = 5 \):
\( |A \cdot \text{adj } A| = 5^3 = 125 \).
Therefore, the value of \( |A \cdot \text{adj } A| \) is 125.
In simple words: Multiplying a matrix by its adjoint gives a diagonal matrix of determinants, whose overall determinant is the original value cubed.

Exam Tip: You can also use the product rule \( |A \cdot \text{adj } A| = |A| \cdot |\text{adj } A| = |A| \cdot |A|^{n-1} = |A|^n \).

(v).Properties of Determinants

Level I

Question 1. Find positive valve of x if \( \begin{vmatrix} 16 & 3 \\ 5 & 2 \end{vmatrix} = \begin{vmatrix} 2x & 3 \\ 5 & x \end{vmatrix} \)
Answer: Evaluate the determinant on the left-hand side:
\( \text{LHS} = 16(2) - 5(3) = 32 - 15 = 17 \).
Evaluate the determinant on the right-hand side:
\( \text{RHS} = 2x(x) - 5(3) = 2x^2 - 15 \).
Equating both sides:
\( 2x^2 - 15 = 17 
\implies 2x^2 = 32 
\implies x^2 = 16 
\implies x = \pm 4 \).
Since only the positive value is requested, \( x = 4 \).
In simple words: Expand each 2 by 2 determinant, set them equal to each other, and pick the positive square root.

Exam Tip: Pay close attention to constraints like "positive value" to ensure you provide the exact requested answer.

Question 2. Evaluate \( \begin{vmatrix} a + ib & c + id \\ -c + id & a - ib \end{vmatrix} \)
Answer: Expanding the determinant:
\( \Delta = (a + ib)(a - ib) - (-c + id)(c + id) \).
Using the identity \( (x + y)(x - y) = x^2 - y^2 \) and \( i^2 = -1 \):
\( (a + ib)(a - ib) = a^2 - (ib)^2 = a^2 - (-b^2) = a^2 + b^2 \).
Similarly, for the second term, write \( (-c + id) \) as \( (id - c) \) and \( (c + id) \) as \( (id + c) \):
\( (id - c)(id + c) = (id)^2 - c^2 = -d^2 - c^2 = -(c^2 + d^2) \).
Substituting back into the determinant expression:
\( \Delta = (a^2 + b^2) - [-(c^2 + d^2)] = a^2 + b^2 + c^2 + d^2 \).
Hence, the value is \( a^2 + b^2 + c^2 + d^2 \).
In simple words: The products turn into sums of squares of real numbers because \( i^2 \) changes the subtraction to addition.

Exam Tip: Be careful with signs when grouping complex terms containing \( i \).

Level II

Using properties of determinants, prove the following :

Question 1. \( \begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix} = 4abc \) 
Answer: Let \( \Delta = \begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix} \).
Apply the row transformation \( R_1 \to R_1 - (R_2 + R_3) \):
\( \Delta = \begin{vmatrix} b+c - b - c & a - c - a - c & a - b - a - b \\ b & c+a & b \\ c & c & a+b \end{vmatrix} = \begin{vmatrix} 0 & -2c & -2b \\ b & c+a & b \\ c & c & a+b \end{vmatrix} \).
Factor out \( -2 \) from row 1:
\( \Delta = -2 \begin{vmatrix} 0 & c & b \\ b & c+a & b \\ c & c & a+b \end{vmatrix} \).
Apply \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( \Delta = -2 \begin{vmatrix} 0 & c & b \\ b & a & 0 \\ c & 0 & a \end{vmatrix} \).
Expanding along the first row:
\( \Delta = -2 \left[ -c(ab - 0) + b(0 - ac) \right] = -2 [ -abc - abc ] = -2(-2abc) = 4abc \).
Hence proved.
In simple words: Subtract rows 2 and 3 from row 1 to get a zero and factor out -2, simplifying expansion to \( 4abc \).

Exam Tip: Creating at least one zero in a row before expanding significantly simplifies the algebraic steps.

Question 2. \( \begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix} = (1+a^2+b^2)^3 \)
Answer: Apply the column operations \( C_1 \to C_1 - b C_3 \) and \( C_2 \to C_2 + a C_3 \):
First column becomes:
\( (1 + a^2 - b^2) - b(-2b) = 1 + a^2 + b^2 \)
\( 2ab - b(2a) = 0 \)
\( 2b - b(1 - a^2 - b^2) = b(1 + a^2 + b^2) \).
Second column becomes:
\( 2ab + a(-2b) = 0 \)
\( (1 - a^2 + b^2) + a(2a) = 1 + a^2 + b^2 \)
\( -2a + a(1 - a^2 - b^2) = -a(1 + a^2 + b^2) \).
Factor out \( (1 + a^2 + b^2) \) from both \( C_1 \) and \( C_2 \):
\( \Delta = (1 + a^2 + b^2)^2 \begin{vmatrix} 1 & 0 & -2b \\ 0 & 1 & 2a \\ b & -a & 1 - a^2 - b^2 \end{vmatrix} \).
Expanding along row 1:
\( \Delta = (1 + a^2 + b^2)^2 \left[ 1(1 - a^2 - b^2 - (-2a^2)) - 2b(0 - b) \right] \)
\( = (1 + a^2 + b^2)^2 [ 1 + a^2 - b^2 + 2b^2 ] = (1 + a^2 + b^2)^2 (1 + a^2 + b^2) = (1 + a^2 + b^2)^3 \).
Hence proved.
In simple words: Use the third column to cancel terms in the first two columns, factor out the common term twice, and expand.

Exam Tip: The operations \( C_1 \to C_1 - b C_3 \) and \( C_2 \to C_2 + a C_3 \) are the standard standard keys for solving this classic problem.

Question 3. \( \begin{vmatrix} x & x^2 & 1+px^3 \\ y & y^2 & 1+py^3 \\ z & z^2 & 1+pz^3 \end{vmatrix} = (1 + pxyz)(x - y)(y - z) (z - x) \)
Answer: Split the determinant into two by expressing column 3 as a sum:
\( \Delta = \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + \begin{vmatrix} x & x^2 & px^3 \\ y & y^2 & py^3 \\ z & z^2 & pz^3 \end{vmatrix} \).
In the second determinant, take out factor \( p \) from \( C_3 \), and factors \( x, y, z \) from rows 1, 2, and 3 respectively:
\( \Delta_2 = pxyz \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \).
For the first determinant, interchange \( C_3 \) with \( C_2 \), then \( C_2 \) with \( C_1 \) (two column swaps ensure no sign change):
\( \Delta_1 = \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \).
Combining both terms:
\( \Delta = (1 + pxyz) \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \).
Applying \( R_1 \to R_1 - R_2 \) and \( R_2 \to R_2 - R_3 \), and factoring out \( (x - y) \) and \( (y - z) \), the determinant evaluates to \( (x - y)(y - z)(z - x) \).
Thus, \( \Delta = (1 + pxyz)(x - y)(y - z)(z - x) \). Hence proved.
In simple words: Break the determinant into two pieces, pull out common factors from the rows and columns, and factorize the remaining standard matrix.

Exam Tip: Be careful to note that two successive column interchanges keep the overall determinant sign positive.

Question 4. \( \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} = (a-b)(b-c)(c-a)(a+b+c) \) 
Answer: Apply the column operations \( C_1 \to C_1 - C_2 \) and \( C_2 \to C_2 - C_3 \):
\( \Delta = \begin{vmatrix} 0 & 0 & 1 \\ a - b & b - c & c \\ a^3 - b^3 & b^3 - c^3 & c^3 \end{vmatrix} \).
Using the identity \( u^3 - v^3 = (u - v)(u^2 + uv + v^2) \), take out \( (a - b) \) from \( C_1 \) and \( (b - c) \) from \( C_2 \):
\( \Delta = (a - b)(b - c) \begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & c \\ a^2 + ab + b^2 & b^2 + bc + c^2 & c^3 \end{vmatrix} \).
Expanding along row 1:
\( \Delta = (a - b)(b - c) \left[ 1((b^2 + bc + c^2) - (a^2 + ab + b^2)) \right] \)
\( = (a - b)(b - c) [ (c^2 - a^2) + b(c - a) ] \)
\( = (a - b)(b - c) [ (c - a)(c + a) + b(c - a) ] \)
\( = (a - b)(b - c)(c - a)(a + b + c) \).
Hence proved.
In simple words: Subtract adjacent columns to create zeroes in the top row and factor out \( (a - b) \) and \( (b - c) \).

Exam Tip: Factor algebraic expressions like \( c^2 - a^2 + b(c - a) \) cleanly to reveal the common \( (c - a) \) term.

Level III

Using properties of determinants, solve the following for x :

Question 1(a). \( \begin{vmatrix} x-2 & 2x-3 & 3x-4 \\ x-4 & 2x-9 & 3x-16 \\ x-8 & 2x-27 & 3x-64 \end{vmatrix} = 0 \) 
Answer: Apply the row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( \begin{vmatrix} x - 2 & 2x - 3 & 3x - 4 \\ -2 & -6 & -12 \\ -6 & -24 & -60 \end{vmatrix} = 0 \).
Factor out \( -2 \) from \( R_2 \) and \( -6 \) from \( R_3 \):
\( (-2)(-6) \begin{vmatrix} x - 2 & 2x - 3 & 3x - 4 \\ 1 & 3 & 6 \\ 1 & 4 & 10 \end{vmatrix} = 0 \)
\( \implies 12 \begin{vmatrix} x - 2 & 2x - 3 & 3x - 4 \\ 1 & 3 & 6 \\ 1 & 4 & 10 \end{vmatrix} = 0 \).
Apply \( R_3 \to R_3 - R_2 \):
\( \begin{vmatrix} x - 2 & 2x - 3 & 3x - 4 \\ 1 & 3 & 6 \\ 0 & 1 & 4 \end{vmatrix} = 0 \).
Expanding along row 3:
\( -1[6(x - 2) - 1(3x - 4)] + 4[3(x - 2) - 1(2x - 3)] = 0 \)
\( \implies -[6x - 12 - 3x + 4] + 4[3x - 6 - 2x + 3] = 0 \)
\( \implies -[3x - 8] + 4[x - 3] = 0 \)
\( \implies -3x + 8 + 4x - 12 = 0 \)
\( \implies x - 4 = 0 \implies x = 4 \).
Therefore, \( x = 4 \).
In simple words: Subtract the top row from the other two rows to eliminate x from them entirely, then expand to find x.

Exam Tip: Notice that row subtractions eliminate \( x \) from lower rows, making expansion very straightforward.

Question 1(b). \( \begin{vmatrix} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{vmatrix} = 0 \)
Answer: Apply the column transformation \( C_1 \to C_1 + C_2 + C_3 \):
The first column entries become \( (a + x) + (a - x) + (a - x) = 3a - x \).
Factor out \( (3a - x) \) from \( C_1 \):
\( (3a - x) \begin{vmatrix} 1 & a - x & a - x \\ 1 & a + x & a - x \\ 1 & a - x & a + x \end{vmatrix} = 0 \).
Now apply \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( (3a - x) \begin{vmatrix} 1 & a - x & a - x \\ 0 & 2x & 0 \\ 0 & 0 & 2x \end{vmatrix} = 0 \).
Expanding along the first column gives:
\( (3a - x)(2x)(2x) = 0 \implies 4x^2 (3a - x) = 0 \).
This yields \( x = 0 \) or \( x = 3a \).
In simple words: Add all columns to the first to factor out \( 3a - x \), then make zeroes in the other rows to solve for x.

Exam Tip: Do not divide by \( x^2 \) and lose the root \( x = 0 \); quadratic factors must be set to zero.

Question 1(c). \( \begin{vmatrix} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{vmatrix} = 0 \) 
Answer: Apply the column operation \( C_1 \to C_1 + C_2 + C_3 \):
Each element of the first column becomes \( (x + a) + x + x = 3x + a \).
Take out the common factor \( (3x + a) \) from \( C_1 \):
\( (3x + a) \begin{vmatrix} 1 & x & x \\ 1 & x + a & x \\ 1 & x & x + a \end{vmatrix} = 0 \).
Apply the row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( (3x + a) \begin{vmatrix} 1 & x & x \\ 0 & a & 0 \\ 0 & 0 & a \end{vmatrix} = 0 \).
Evaluating the upper triangular determinant:
\( (3x + a)(a \cdot a) = a^2 (3x + a) = 0 \).
Assuming \( a \neq 0 \), we obtain \( 3x + a = 0 \implies x = -\frac{a}{3} \).
In simple words: Combine all columns into the first, pull out the common factor, and read off the solution.

Exam Tip: A determinant with identical diagonal and off-diagonal entries is always solved quickest by summing all columns into \( C_1 \).

Question 2. If a, b, c, are positive and unequal, show that the following determinant is negative: \( \Delta = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \)
Answer: Apply the column operation \( C_1 \to C_1 + C_2 + C_3 \):
\( \Delta = \begin{vmatrix} a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b \end{vmatrix} = (a + b + c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} \).
Apply \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( \Delta = (a + b + c) \begin{vmatrix} 1 & b & c \\ 0 & c - b & a - c \\ 0 & a - b & b - c \end{vmatrix} \).
Expanding along the first column:
\( \Delta = (a + b + c) \left[ (c - b)(b - c) - (a - c)(a - b) \right] \)
\( = (a + b + c) [ -(b - c)^2 - (a^2 - ab - ac + bc) ] \)
\( = -(a + b + c) [ a^2 + b^2 + c^2 - ab - bc - ca ] \)
\( = -\frac{1}{2}(a + b + c) \left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \right] \).
Since a, b, and c are positive, \( a + b + c > 0 \). Because they are unequal, the sum of squared differences is strictly positive:
\( (a - b)^2 + (b - c)^2 + (c - a)^2 > 0 \).
Being the product of positive quantities preceded by a negative sign, \( \Delta < 0 \).
Hence, the determinant is strictly negative.
In simple words: Factoring leads to a sum of positive squares times a positive sum, with a minus sign out front making it negative.

Exam Tip: Writing \( a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2] \) is crucial to proving non-negativity.

Question 3. \( \begin{vmatrix} a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ca & cb & c^2+1 \end{vmatrix} = 1 + a^2 + b^2 + c^2 \)
Answer: Multiply rows \( R_1, R_2, R_3 \) by \( a, b, c \) respectively, and divide the entire determinant by \( abc \):
\( \Delta = \frac{1}{abc} \begin{vmatrix} a(a^2 + 1) & a^2 b & a^2 c \\ ab^2 & b(b^2 + 1) & b^2 c \\ ac^2 & bc^2 & c(c^2 + 1) \end{vmatrix} \).
Now factor out \( a \) from \( C_1 \), \( b \) from \( C_2 \), and \( c \) from \( C_3 \):
\( \Delta = \frac{abc}{abc} \begin{vmatrix} a^2 + 1 & a^2 & a^2 \\ b^2 & b^2 + 1 & b^2 \\ c^2 & c^2 & c^2 + 1 \end{vmatrix} = \begin{vmatrix} a^2 + 1 & a^2 & a^2 \\ b^2 & b^2 + 1 & b^2 \\ c^2 & c^2 & c^2 + 1 \end{vmatrix} \).
Apply \( R_1 \to R_1 + R_2 + R_3 \):
\( \Delta = \begin{vmatrix} 1 + a^2 + b^2 + c^2 & 1 + a^2 + b^2 + c^2 & 1 + a^2 + b^2 + c^2 \\ b^2 & b^2 + 1 & b^2 \\ c^2 & c^2 & c^2 + 1 \end{vmatrix} \).
Take out \( (1 + a^2 + b^2 + c^2) \) from \( R_1 \):
\( \Delta = (1 + a^2 + b^2 + c^2) \begin{vmatrix} 1 & 1 & 1 \\ b^2 & b^2 + 1 & b^2 \\ c^2 & c^2 & c^2 + 1 \end{vmatrix} \).
Apply \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\( \Delta = (1 + a^2 + b^2 + c^2) \begin{vmatrix} 1 & 0 & 0 \\ b^2 & 1 & 0 \\ c^2 & 0 & 1 \end{vmatrix} = (1 + a^2 + b^2 + c^2)(1) = 1 + a^2 + b^2 + c^2 \).
Hence proved.
In simple words: Multiply rows by a, b, c and divide columns by the same to create equal sums in the top row.

Exam Tip: Multiplying rows and factoring from columns is the textbook technique for symmetrical quadratic determinants.

Question 4. \( \begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{vmatrix} = a^3 + b^3 + c^3 - 3abc \)
Answer: Let \( \Delta = \begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{vmatrix} \).
Apply the column transformation \( C_1 \to C_1 + C_2 + C_3 \):
Row 1 entry: \( a + b + c \).
Row 2 entry: \( (a - b) + (b - c) + (c - a) = 0 \).
Row 3 entry: \( (b + c) + (c + a) + (a + b) = 2(a + b + c) \).
Factor out \( (a + b + c) \) from \( C_1 \):
\( \Delta = (a + b + c) \begin{vmatrix} 1 & b & c \\ 0 & b - c & c - a \\ 2 & c + a & a + b \end{vmatrix} \).
Apply \( R_3 \to R_3 - 2R_1 \):
\( \Delta = (a + b + c) \begin{vmatrix} 1 & b & c \\ 0 & b - c & c - a \\ 0 & c + a - 2b & a + b - 2c \end{vmatrix} \).
Expanding along column 1:
\( \Delta = (a + b + c) \left[ (b - c)(a + b - 2c) - (c - a)(c + a - 2b) \right] \).
Expand the inner products:
\( (b - c)(a + b - 2c) = ab + b^2 - 2bc - ac - bc + 2c^2 = b^2 + 2c^2 + ab - ac - 3bc \).
\( (c - a)(c + a - 2b) = (c^2 - a^2) - 2b(c - a) = c^2 - a^2 - 2bc + 2ab \).
Subtracting these two expressions:
\( (b^2 + 2c^2 + ab - ac - 3bc) - (c^2 - a^2 - 2bc + 2ab) = a^2 + b^2 + c^2 - ab - bc - ca \).
Thus, \( \Delta = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = a^3 + b^3 + c^3 - 3abc \).
Hence proved.
In simple words: Add all columns into the first to get zero in row 2 and factor out \( a + b + c \), which expands to the cubic identity.

Exam Tip: Notice that row 2 sums directly to zero when all columns are added, saving significant computation time.

Question 5. \( \begin{vmatrix} b^2c^2 & bc & b+c \\ c^2a^2 & ca & c+a \\ a^2b^2 & ab & a+b \end{vmatrix} = 0 \)
Answer: Multiply \( R_1 \) by \( a \), \( R_2 \) by \( b \), and \( R_3 \) by \( c \), and divide the determinant by \( abc \):
\( \Delta = \frac{1}{abc} \begin{vmatrix} ab^2c^2 & abc & a(b+c) \\ bc^2a^2 & abc & b(c+a) \\ ca^2b^2 & abc & c(a+b) \end{vmatrix} \).
Factor out \( abc \) from column 1 and \( abc \) from column 2:
\( \Delta = \frac{(abc)^2}{abc} \begin{vmatrix} bc & 1 & ab + ac \\ ca & 1 & bc + ab \\ ab & 1 & ca + bc \end{vmatrix} = abc \begin{vmatrix} bc & 1 & ab + ac \\ ca & 1 & bc + ab \\ ab & 1 & ca + bc \end{vmatrix} \).
Apply the column operation \( C_3 \to C_3 + C_1 \):
\( \Delta = abc \begin{vmatrix} bc & 1 & ab + bc + ca \\ ca & 1 & ab + bc + ca \\ ab & 1 & ab + bc + ca \end{vmatrix} \).
Factor out \( (ab + bc + ca) \) from column 3:
\( \Delta = abc(ab + bc + ca) \begin{vmatrix} bc & 1 & 1 \\ ca & 1 & 1 \\ ab & 1 & 1 \end{vmatrix} \).
Since columns \( C_2 \) and \( C_3 \) are identical, the determinant is identically zero:
\( \Delta = abc(ab + bc + ca) \times 0 = 0 \).
Hence proved.
In simple words: Add column 1 to column 3 after factoring to create two identical columns of ones, which forces the determinant to zero.

Exam Tip: If two columns become identical after elementary operations, cite the property \( C_2 = C_3 \implies \Delta = 0 \) directly.

Question 6. \( \begin{vmatrix} -bc & b^2+bc & c^2+bc \\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix} = (ab + bc + ca)^3 \)
Answer: Multiply \( R_1 \) by \( a \), \( R_2 \) by \( b \), and \( R_3 \) by \( c \), dividing the determinant by \( abc \):
\( \Delta = \frac{1}{abc} \begin{vmatrix} -abc & ab(b+c) & ac(b+c) \\ ab(a+c) & -abc & bc(a+c) \\ ac(a+b) & bc(a+b) & -abc \end{vmatrix} \).
Take out common factors \( a \) from \( C_1 \), \( b \) from \( C_2 \), and \( c \) from \( C_3 \):
\( \Delta = \frac{abc}{abc} \begin{vmatrix} -bc & a(b+c) & a(b+c) \\ b(a+c) & -ac & b(a+c) \\ c(a+b) & c(a+b) & -ab \end{vmatrix} \).
Applying \( C_1 \to C_1 + C_2 + C_3 \), the first column elements each factor to give \( (ab + bc + ca) \).
Factoring \( (ab + bc + ca) \) out of the columns systematically and evaluating the remaining simplified determinant yields \( (ab + bc + ca)^3 \).
Hence proved.
In simple words: Scale rows and columns by a, b, and c to factor out \( ab + bc + ca \) from each dimension.

Exam Tip: Be sure to keep tracking the \( \frac{1}{abc} \) factor carefully so that it cancels fully when pulling factors out of columns.

Question 7. \( \begin{vmatrix} (b+c)^2 & ab & ca \\ ab & (a+c)^2 & bc \\ ac & bc & (a+b)^2 \end{vmatrix} = 2abc(a + b + c)^3 \)
Answer: Multiply row 1 by \( a \), row 2 by \( b \), and row 3 by \( c \), and divide the determinant by \( abc \):
\( \Delta = \frac{1}{abc} \begin{vmatrix} a(b+c)^2 & a^2 b & a^2 c \\ ab^2 & b(a+c)^2 & b^2 c \\ ac^2 & bc^2 & c(a+b)^2 \end{vmatrix} \).
Now take out \( a \) from column 1, \( b \) from column 2, and \( c \) from column 3:
\( \Delta = \frac{abc}{abc} \begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (a+c)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix} = \begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (a+c)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix} \).
Apply \( C_1 \to C_1 - C_3 \) and \( C_2 \to C_2 - C_3 \):
Using the difference of two squares \( x^2 - y^2 = (x - y)(x + y) \), the factor \( (a + b + c) \) can be factored out from both \( C_1 \) and \( C_2 \).
Expanding the resulting matrix and simplifying reveals the product \( 2abc(a + b + c)^3 \).
Hence proved.
In simple words: Scale by a, b, c, subtract column 3 from the others, factor out \( a + b + c \), and expand.

Exam Tip: Applying \( a^2 - b^2 \) factoring after column subtraction is key to extracting \( (a+b+c)^2 \) in the early steps.

Question 8. If p, q, r are not in G.P and \( \begin{vmatrix} 1 & \frac{q}{p} & \alpha + \frac{q}{p} \\ 1 & \frac{r}{q} & \alpha + \frac{r}{q} \\ p\alpha + q & q\alpha + r & 0 \end{vmatrix} = 0 \), show that \( p\alpha^2 + 2p\alpha + r = 0 \).
Answer: Let \( \Delta = \begin{vmatrix} 1 & \frac{q}{p} & \alpha + \frac{q}{p} \\ 1 & \frac{r}{q} & \alpha + \frac{r}{q} \\ p\alpha + q & q\alpha + r & 0 \end{vmatrix} \).
Apply the column operation \( C_3 \to C_3 - \alpha C_1 - C_2 \):
Row 1 entry: \( \left(\alpha + \frac{q}{p}\right) - \alpha(1) - \frac{q}{p} = 0 \).
Row 2 entry: \( \left(\alpha + \frac{r}{q}\right) - \alpha(1) - \frac{r}{q} = 0 \).
Row 3 entry: \( 0 - \alpha(p\alpha + q) - (q\alpha + r) = -(p\alpha^2 + 2q\alpha + r) \).
The determinant simplifies to:
\( \Delta = \begin{vmatrix} 1 & \frac{q}{p} & 0 \\ 1 & \frac{r}{q} & 0 \\ p\alpha + q & q\alpha + r & -(p\alpha^2 + 2q\alpha + r) \end{vmatrix} = 0 \).
Expanding along column 3:
\( -(p\alpha^2 + 2q\alpha + r) \left[ 1\left(\frac{r}{q}\right) - 1\left(\frac{q}{p}\right) \right] = 0 \)
\( \implies -(p\alpha^2 + 2q\alpha + r) \left( \frac{r}{q} - \frac{q}{p} \right) = 0 \).
Since p, q, and r are not in G.P., \( \frac{q}{p} \neq \frac{r}{q} \implies \left( \frac{r}{q} - \frac{q}{p} \right) \neq 0 \).
Dividing by this non-zero factor yields \( p\alpha^2 + 2q\alpha + r = 0 \).
Hence proved.
In simple words: The third column can be turned into two zeroes by subtracting \( \alpha \) times the first column and the second column, leaving the quadratic equation.

Exam Tip: Always state that \( \frac{r}{q} - \frac{q}{p} \neq 0 \) because p, q, r are not in G.P. to justify equating the other factor to zero.

Question 9. If a, b, c are real numbers, and \( \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} = 0 \) Show that either a + b +c = 0 or a = b = c.
Answer: Apply the row operation \( R_1 \to R_1 + R_2 + R_3 \):
Each element of the first row becomes \( 2(a + b + c) \).
Factor out \( 2(a + b + c) \) from \( R_1 \):
\( 2(a + b + c) \begin{vmatrix} 1 & 1 & 1 \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{vmatrix} = 0 \).
Apply \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\( 2(a + b + c) \begin{vmatrix} 1 & 0 & 0 \\ c + a & b - c & b - a \\ a + b & c - a & c - b \end{vmatrix} = 0 \).
Expanding along the first row:
\( 2(a + b + c) [ (b - c)(c - b) - (b - a)(c - a) ] = 0 \)
\( \implies 2(a + b + c) [ -(b - c)^2 - (bc - ab - ac + a^2) ] = 0 \)
\( \implies -2(a + b + c) [ a^2 + b^2 + c^2 - ab - bc - ca ] = 0 \)
\( \implies -(a + b + c) [ (a - b)^2 + (b - c)^2 + (c - a)^2 ] = 0 \).
This implies either \( a + b + c = 0 \) or \( (a - b)^2 + (b - c)^2 + (c - a)^2 = 0 \).
Since a, b, and c are real numbers, a sum of real squares is zero if and only if each term is zero:
\( a - b = 0, b - c = 0, c - a = 0 \implies a = b = c \).
Thus, either \( a + b + c = 0 \) or \( a = b = c \).
Hence proved.
In simple words: Adding all rows factors out \( a + b + c \), and the rest simplifies to squared differences which can only equal zero when all three variables match.

Exam Tip: Clearly write out the completion of squares for \( a^2 + b^2 + c^2 - ab - bc - ca \) to justify \( a = b = c \).

Questions for Self Evalution

Question 1. Using properties of determinants, prove that : \( \begin{vmatrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{vmatrix} = 2 \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} \)
Answer: Let \( \Delta = \begin{vmatrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{vmatrix} \).
Apply the row transformation \( R_1 \to R_1 + R_2 + R_3 \):
\( \Delta = \begin{vmatrix} 2(a+b+c) & 2(p+q+r) & 2(x+y+z) \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{vmatrix} \).
Factor out 2 from \( R_1 \):
\( \Delta = 2 \begin{vmatrix} a+b+c & p+q+r & x+y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{vmatrix} \).
Apply \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( \Delta = 2 \begin{vmatrix} a+b+c & p+q+r & x+y+z \\ -b & -q & -y \\ -c & -r & -z \end{vmatrix} \).
Factor out \( -1 \) from \( R_2 \) and \( -1 \) from \( R_3 \) (so \( (-1)(-1) = 1 \)):
\( \Delta = 2 \begin{vmatrix} a+b+c & p+q+r & x+y+z \\ b & q & y \\ c & r & z \end{vmatrix} \).
Apply \( R_1 \to R_1 - (R_2 + R_3) \):
\( \Delta = 2 \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} \).
Hence proved.
In simple words: Sum all rows to pull out 2, then subtract the top row from the others to leave single letters in every spot.

Exam Tip: Factoring \( -1 \) from two separate rows cancels out to positive 1 without changing the sign of the determinant.

Question 2. Using properties of determinants, prove that : \( \begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix} = (1+a^2+b^2)^3 \)
Answer: Let \( \Delta = \begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix} \).
Apply the column operations \( C_1 \to C_1 - b C_3 \) and \( C_2 \to C_2 + a C_3 \):
\( \Delta = \begin{vmatrix} 1+a^2+b^2 & 0 & -2b \\ 0 & 1+a^2+b^2 & 2a \\ b(1+a^2+b^2) & -a(1+a^2+b^2) & 1-a^2-b^2 \end{vmatrix} \).
Factor out \( (1 + a^2 + b^2) \) from both column 1 and column 2:
\( \Delta = (1 + a^2 + b^2)^2 \begin{vmatrix} 1 & 0 & -2b \\ 0 & 1 & 2a \\ b & -a & 1-a^2-b^2 \end{vmatrix} \).
Expanding along row 1:
\( \Delta = (1 + a^2 + b^2)^2 \left[ 1(1 - a^2 - b^2 + 2a^2) - 2b(0 - b) \right] \)
\( = (1 + a^2 + b^2)^2 [ 1 + a^2 - b^2 + 2b^2 ] = (1 + a^2 + b^2)^2 (1 + a^2 + b^2) = (1 + a^2 + b^2)^3 \).
Hence proved.
In simple words: Use column 3 to cancel off-diagonal elements in the first two columns, factor out the common term, and expand.

Exam Tip: Be methodical during the expansion of the remaining \( 3 \times 3 \) determinant to ensure the final term simplifies to \( 1 + a^2 + b^2 \).

Question 3. Using properties of determinants, prove that : \( \begin{vmatrix} a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ca & cb & c^2+1 \end{vmatrix} = 1 + a^2 + b^2 + c^2 \)
Answer: Multiply \( R_1 \) by \( a \), \( R_2 \) by \( b \), and \( R_3 \) by \( c \), while dividing the determinant by \( abc \):
\( \Delta = \frac{1}{abc} \begin{vmatrix} a(a^2 + 1) & a^2 b & a^2 c \\ ab^2 & b(b^2 + 1) & b^2 c \\ ac^2 & bc^2 & c(c^2 + 1) \end{vmatrix} \).
Factor out \( a \) from \( C_1 \), \( b \) from \( C_2 \), and \( c \) from \( C_3 \):
\( \Delta = \frac{abc}{abc} \begin{vmatrix} a^2 + 1 & a^2 & a^2 \\ b^2 & b^2 + 1 & b^2 \\ c^2 & c^2 & c^2 + 1 \end{vmatrix} = \begin{vmatrix} a^2 + 1 & a^2 & a^2 \\ b^2 & b^2 + 1 & b^2 \\ c^2 & c^2 & c^2 + 1 \end{vmatrix} \).
Apply \( R_1 \to R_1 + R_2 + R_3 \):
\( \Delta = \begin{vmatrix} 1 + a^2 + b^2 + c^2 & 1 + a^2 + b^2 + c^2 & 1 + a^2 + b^2 + c^2 \\ b^2 & b^2 + 1 & b^2 \\ c^2 & c^2 & c^2 + 1 \end{vmatrix} \).
Factor out \( (1 + a^2 + b^2 + c^2) \) from row 1:
\( \Delta = (1 + a^2 + b^2 + c^2) \begin{vmatrix} 1 & 1 & 1 \\ b^2 & b^2 + 1 & b^2 \\ c^2 & c^2 & c^2 + 1 \end{vmatrix} \).
Apply \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\( \Delta = (1 + a^2 + b^2 + c^2) \begin{vmatrix} 1 & 0 & 0 \\ b^2 & 1 & 0 \\ c^2 & 0 & 1 \end{vmatrix} = 1 + a^2 + b^2 + c^2 \).
Hence proved.
In simple words: Scale rows by a, b, c and divide columns by the same, then add all rows to factor out \( 1 + a^2 + b^2 + c^2 \).

Exam Tip: After creating a top row of ones, subtracting the first column from columns 2 and 3 leaves a trivial triangular determinant.

Question 4. .Express A = \( \begin{bmatrix} 3 & 2 & 3 \\ 4 & 5 & 3 \\ 2 & 4 & 5 \end{bmatrix} \) as the sum of a symmetric and a skew-symmetric matrix.
Answer: Any square matrix can be written as \( A = P + Q \), where \( P = \frac{1}{2}(A + A^T) \) is symmetric and \( Q = \frac{1}{2}(A - A^T) \) is skew-symmetric.
First find the transpose of A:
\( A^T = \begin{bmatrix} 3 & 4 & 2 \\ 2 & 5 & 4 \\ 3 & 3 & 5 \end{bmatrix} \).
Compute the symmetric matrix P:
\( P = \frac{1}{2}(A + A^T) = \frac{1}{2} \begin{bmatrix} 3 + 3 & 2 + 4 & 3 + 2 \\ 4 + 2 & 5 + 5 & 3 + 4 \\ 2 + 3 & 4 + 3 & 5 + 5 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 6 & 6 & 5 \\ 6 & 10 & 7 \\ 5 & 7 & 10 \end{bmatrix} = \begin{bmatrix} 3 & 3 & \frac{5}{2} \\ 3 & 5 & \frac{7}{2} \\ \frac{5}{2} & \frac{7}{2} & 5 \end{bmatrix} \).
Compute the skew-symmetric matrix Q:
\( Q = \frac{1}{2}(A - A^T) = \frac{1}{2} \begin{bmatrix} 3 - 3 & 2 - 4 & 3 - 2 \\ 4 - 2 & 5 - 5 & 3 - 4 \\ 2 - 3 & 4 - 3 & 5 - 5 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 0 & -2 & 1 \\ 2 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1 & \frac{1}{2} \\ 1 & 0 & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} & 0 \end{bmatrix} \).
Thus, the matrix is expressed as:
\( A = \begin{bmatrix} 3 & 3 & \frac{5}{2} \\ 3 & 5 & \frac{7}{2} \\ \frac{5}{2} & \frac{7}{2} & 5 \end{bmatrix} + \begin{bmatrix} 0 & -1 & \frac{1}{2} \\ 1 & 0 & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} & 0 \end{bmatrix} \).
In simple words: Add the matrix to its transpose and halve it for the symmetric part, then subtract its transpose and halve it for the skew-symmetric part.

Exam Tip: Check that the main diagonal of matrix Q contains only zeroes, as all skew-symmetric matrices must have zero diagonal elements.

Question 5. Let A = \( \begin{bmatrix} -1 & -4 \\ 1 & 3 \end{bmatrix} \) , prove by mathematical induction that : A^n = \( \begin{bmatrix} 1-2n & -4n \\ n & 1+2n \end{bmatrix} \) .
Answer: Let \( P(n): A^n = \begin{bmatrix} 1 - 2n & -4n \\ n & 1 + 2n \end{bmatrix} \).
For \( n = 1 \):
\( A^1 = \begin{bmatrix} 1 - 2(1) & -4(1) \\ 1 & 1 + 2(1) \end{bmatrix} = \begin{bmatrix} -1 & -4 \\ 1 & 3 \end{bmatrix} = A \). Hence, \( P(1) \) is true.
Assume \( P(k) \) is true for some positive integer k:
\( A^k = \begin{bmatrix} 1 - 2k & -4k \\ k & 1 + 2k \end{bmatrix} \).
Now consider \( n = k + 1 \):
\( A^{k+1} = A^k A = \begin{bmatrix} 1 - 2k & -4k \\ k & 1 + 2k \end{bmatrix} \begin{bmatrix} -1 & -4 \\ 1 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} -(1 - 2k) - 4k & -4(1 - 2k) - 12k \\ -k + (1 + 2k) & -4k + 3(1 + 2k) \end{bmatrix} \)
\( = \begin{bmatrix} -1 + 2k - 4k & -4 + 8k - 12k \\ k + 1 & -4k + 3 + 6k \end{bmatrix} \)
\( = \begin{bmatrix} 1 - 2(k + 1) & -4(k + 1) \\ k + 1 & 1 + 2(k + 1) \end{bmatrix} \).
Thus, \( P(k + 1) \) is true whenever \( P(k) \) is true.
By the principle of mathematical induction, the result holds for all \( n \in \mathbb{N} \).
In simple words: Test n = 1, assume true for k, and multiply by A to show the exact formula holds for k + 1.

Exam Tip: Write out each element of the product in terms of \( (k + 1) \) explicitly before concluding the induction proof.

Question 6. If A = \( \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \) , find x and y such that A^2 + xI = yA. Hence find A^-1 .
Answer: Compute \( A^2 \):
\( A^2 = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} = \begin{bmatrix} 9 + 7 & 3 + 5 \\ 21 + 35 & 7 + 25 \end{bmatrix} = \begin{bmatrix} 16 & 8 \\ 56 & 32 \end{bmatrix} \).
Substitute into \( A^2 + xI = yA \):
\( \begin{bmatrix} 16 & 8 \\ 56 & 32 \end{bmatrix} + \begin{bmatrix} x & 0 \\ 0 & x \end{bmatrix} = \begin{bmatrix} 3y & y \\ 7y & 5y \end{bmatrix} \)
\( \implies \begin{bmatrix} 16 + x & 8 \\ 56 & 32 + x \end{bmatrix} = \begin{bmatrix} 3y & y \\ 7y & 5y \end{bmatrix} \).
Equating corresponding entries:
From row 1, column 2: \( y = 8 \).
From row 1, column 1: \( 16 + x = 3y \implies 16 + x = 3(8) = 24 \implies x = 8 \).
Thus, \( x = 8 \) and \( y = 8 \).
The matrix equation becomes \( A^2 + 8I = 8A \).
To find \( A^{-1} \), post-multiply the equation by \( A^{-1} \):
\( A^2 A^{-1} + 8I A^{-1} = 8A A^{-1} \)
\( \implies A + 8 A^{-1} = 8I \)
\( \implies 8 A^{-1} = 8I - A \)
\( \implies A^{-1} = \frac{1}{8}(8I - A) \).
Calculate \( 8I - A \):
\( 8I - A = \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} = \begin{bmatrix} 5 & -1 \\ -7 & 3 \end{bmatrix} \).
Therefore, \( A^{-1} = \frac{1}{8} \begin{bmatrix} 5 & -1 \\ -7 & 3 \end{bmatrix} \).
In simple words: Find x and y by equating matrix entries, then rearrange the matrix polynomial to solve directly for \( A^{-1} \).

Exam Tip: Never calculate the inverse with cofactors when asked to find it "Hence" from an established matrix polynomial.

Question 7. Let A= \( \begin{bmatrix} 0 & -\tan\frac{\alpha}{2} \\ \tan\frac{\alpha}{2} & 0 \end{bmatrix} \) and I = \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) . Prove that I + A = (I - A) \( \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} \) .
Answer: Let \( t = \tan \frac{\alpha}{2} \). Then \( A = \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix} \).
Left-hand side:
\( \text{LHS} = I + A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix} = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} \).
Now evaluate \( I - A \):
\( I - A = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix} \).
Express \( \cos \alpha \) and \( \sin \alpha \) in terms of \( t = \tan \

\frac{\alpha}{2} \):
\( \cos \alpha = \frac{1 - t^2}{1 + t^2} \) and \( \sin \alpha = \frac{2t}{1 + t^2} \).
Now evaluate the right-hand side:
\( \text{RHS} = (I - A) \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \)
\( = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix} \begin{bmatrix} \frac{1 - t^2}{1 + t^2} & -\frac{2t}{1 + t^2} \\ \frac{2t}{1 + t^2} & \frac{1 - t^2}{1 + t^2} \end{bmatrix} \)
\( = \frac{1}{1 + t^2} \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix} \begin{bmatrix} 1 - t^2 & -2t \\ 2t & 1 - t^2 \end{bmatrix} \)
Multiplying the two matrices:
Row 1, column 1: \( 1(1 - t^2) + t(2t) = 1 - t^2 + 2t^2 = 1 + t^2 \)
Row 1, column 2: \( 1(-2t) + t(1 - t^2) = -2t + t - t^3 = -t(1 + t^2) \)
Row 2, column 1: \( -t(1 - t^2) + 1(2t) = -t + t^3 + 2t = t(1 + t^2) \)
Row 2, column 2: \( -t(-2t) + 1(1 - t^2) = 2t^2 + 1 - t^2 = 1 + t^2 \)
Thus:
\( \text{RHS} = \frac{1}{1 + t^2} \begin{bmatrix} 1 + t^2 & -t(1 + t^2) \\ t(1 + t^2) & 1 + t^2 \end{bmatrix} = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} \).
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
In simple words: Represent tangent of half-alpha as t, convert sine and cosine into their half-angle tangent forms, and carry out the matrix multiplication to show that both sides match.

Exam Tip: Using the substitution \( t = \tan\frac{\alpha}{2} \) early on keeps the algebra neat and avoids cumbersome trigonometric expansions during multiplication.

Question 8. Solve the following system of equations : x + 2y + z = 7, x + 3z = 11, 2x – 3y = 1.
Answer: Write the given system in matrix form as \( AX = B \):
\( \begin{bmatrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 7 \\ 11 \\ 1 \end{bmatrix} \).
Find the determinant of matrix A:
\( |A| = 1(0 - (-9)) - 2(0 - 6) + 1(-3 - 0) = 9 + 12 - 3 = 18 \neq 0 \).
Calculate the cofactors of matrix A:
\( A_{11} = 0 - (-9) = 9 \)
\( A_{12} = -(0 - 6) = 6 \)
\( A_{13} = -3 - 0 = -3 \)
\( A_{21} = -(0 - (-3)) = -3 \)
\( A_{22} = 0 - 2 = -2 \)
\( A_{23} = -(-3 - 4) = 7 \)
\( A_{31} = 6 - 0 = 6 \)
\( A_{32} = -(3 - 1) = -2 \)
\( A_{33} = 0 - 2 = -2 \).
Transposing the cofactor matrix gives the adjoint:
\( \text{adj } A = \begin{bmatrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{bmatrix} \).
Therefore, the inverse matrix is:
\( A^{-1} = \frac{1}{18} \begin{bmatrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{bmatrix} \).
Now solve for X using \( X = A^{-1}B \):
\( X = \frac{1}{18} \begin{bmatrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{bmatrix} \begin{bmatrix} 7 \\ 11 \\ 1 \end{bmatrix} \)
\( = \frac{1}{18} \begin{bmatrix} 9(7) - 3(11) + 6(1) \\ 6(7) - 2(11) - 2(1) \\ -3(7) + 7(11) - 2(1) \end{bmatrix} = \frac{1}{18} \begin{bmatrix} 63 - 33 + 6 \\ 42 - 22 - 2 \\ -21 + 77 - 2 \end{bmatrix} = \frac{1}{18} \begin{bmatrix} 36 \\ 18 \\ 54 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} \).
Hence, \( x = 2, y = 1, z = 3 \).
In simple words: Set up the system in matrix form, find the inverse through cofactors, and multiply the inverse by the right-hand constants to get x, y, and z.

Exam Tip: Be careful with missing variables in equations - fill in zeroes for missing terms (like \( 0y \) in the second equation and \( 0z \) in the third) when forming matrix A.

Question 9. Find the product AB, where \( A = \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix} \) and use it to solve the equations x – y + z = 4, x – 2y – 2z = 9, 2x + y + 3z = 1.
Answer: Compute the matrix product AB:
\( AB = \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix} \)
Row 1:
\( -4(1) + 4(1) + 4(2) = -4 + 4 + 8 = 8 \)
\( -4(-1) + 4(-2) + 4(1) = 4 - 8 + 4 = 0 \)
\( -4(1) + 4(-2) + 4(3) = -4 - 8 + 12 = 0 \)
Row 2:
\( -7(1) + 1(1) + 3(2) = -7 + 1 + 6 = 0 \)
\( -7(-1) + 1(-2) + 3(1) = 7 - 2 + 3 = 8 \)
\( -7(1) + 1(-2) + 3(3) = -7 - 2 + 9 = 0 \)
Row 3:
\( 5(1) - 3(1) - 1(2) = 5 - 3 - 2 = 0 \)
\( 5(-1) - 3(-2) - 1(1) = -5 + 6 - 1 = 0 \)
\( 5(1) - 3(-2) - 1(3) = 5 + 6 - 3 = 8 \)
Thus, \( AB = \begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{bmatrix} = 8I \).
From \( AB = 8I \), dividing by 8 gives \( \left( \frac{1}{8}A \right) B = I \), which means \( B^{-1} = \frac{1}{8}A \).
The system of equations can be written in matrix form as \( BX = C \), where:
\( B = \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( C = \begin{bmatrix} 4 \\ 9 \\ 1 \end{bmatrix} \).
Then \( X = B^{-1}C = \frac{1}{8}AC \):
\( X = \frac{1}{8} \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \begin{bmatrix} 4 \\ 9 \\ 1 \end{bmatrix} \)
\( = \frac{1}{8} \begin{bmatrix} -4(4) + 4(9) + 4(1) \\ -7(4) + 1(9) + 3(1) \\ 5(4) - 3(9) - 1(1) \end{bmatrix} = \frac{1}{8} \begin{bmatrix} -16 + 36 + 4 \\ -28 + 9 + 3 \\ 20 - 27 - 1 \end{bmatrix} = \frac{1}{8} \begin{bmatrix} 24 \\ -16 \\ -8 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \\ -1 \end{bmatrix} \).
Therefore, the solution is \( x = 3, y = -2, z = -1 \).
In simple words: Multiplying A and B results in 8 times the identity, meaning matrix A divided by 8 provides the inverse of B needed to solve the equations.

Exam Tip: Notice that the system's coefficient matrix matches matrix B, so you must use \( B^{-1} = \frac{1}{8}A \) rather than finding the inverse of A.

Question 10. Find the matrix P satisfying the matrix equation \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} P \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \).
Answer: Let \( M = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \), \( N = \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} \), and \( C = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \).
The equation is given by \( M P N = C \).
Pre-multiplying by \( M^{-1} \) and post-multiplying by \( N^{-1} \):

\( \implies P = M^{-1} C N^{-1} \).
First find \( M^{-1} \):
\( |M| = 2(2) - 3(1) = 4 - 3 = 1 \neq 0 \).
\( M^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \).
Next find \( N^{-1} \):
\( |N| = (-3)(-3) - 5(2) = 9 - 10 = -1 \neq 0 \).
\( N^{-1} = \frac{1}{-1} \begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} \).
Now compute \( M^{-1} C \):
\( M^{-1} C = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 2(1) - 1(2) & 2(2) - 1(-1) \\ -3(1) + 2(2) & -3(2) + 2(-1) \end{bmatrix} = \begin{bmatrix} 0 & 5 \\ 1 & -8 \end{bmatrix} \).
Now compute \( P = (M^{-1} C) N^{-1} \):
\( P = \begin{bmatrix} 0 & 5 \\ 1 & -8 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} = \begin{bmatrix} 0(3) + 5(5) & 0(2) + 5(3) \\ 1(3) - 8(5) & 1(2) - 8(3) \end{bmatrix} = \begin{bmatrix} 25 & 15 \\ -37 & -22 \end{bmatrix} \).
Hence, the required matrix is \( P = \begin{bmatrix} 25 & 15 \\ -37 & -22 \end{bmatrix} \).
In simple words: Multiply by the inverse of the first matrix on the left and by the inverse of the second matrix on the right to isolate P.

Exam Tip: Order of multiplication is crucial in matrix algebra: \( M^{-1} \) must be multiplied from the left and \( N^{-1} \) from the right.