Samacheer Kalvi Class 11 Business Maths Solutions Chapter 1 Matrices and Determinants Exercise 1.1

Question 1. Find the minors and cofactors of all the elements of the following determinants.
(i) \( \left|\begin{array}{cc} 5 & 20 \\ 0 & -1 \end{array}\right| \)
(ii) \( \left|\begin{array}{rrr} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{array}\right| \)
Answer:
(i) For the determinant \( \left|\begin{array}{cc} 5 & 20 \\ 0 & -1 \end{array}\right| \):
Minor of 5 = \( M_{11} = -1 \)
Minor of 20 = \( M_{12} = 0 \)
Minor of 0 = \( M_{21} = 20 \)
Minor of -1 = \( M_{22} = 5 \)
Cofactor of 5 = \( A_{11} = (-1)^{1+1} M_{11} = 1 \times -1 = -1 \)
Cofactor of 20 = \( A_{12} = (-1)^{1+2} M_{12} = -1 \times 0 = 0 \)
Cofactor of 0 = \( A_{21} = (-1)^{2+1} M_{21} = -1 \times 20 = -20 \)
Cofactor of -1 = \( A_{22} = (-1)^{2+2} M_{22} = 1 \times 5 = 5 \)

(ii) For the determinant \( \left|\begin{array}{rrr} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{array}\right| \):
Minor of 1 is \( M_{11} = \left|\begin{array}{rr} -1 & 2 \\ 5 & 2 \end{array}\right| = (-1)(2) - (2)(5) = -2 - 10 = -12 \)
Minor of -3 is \( M_{12} = \left|\begin{array}{rr} 4 & 2 \\ 3 & 2 \end{array}\right| = (4)(2) - (2)(3) = 8 - 6 = 2 \)
Minor of 2 is \( M_{13} = \left|\begin{array}{rr} 4 & -1 \\ 3 & 5 \end{array}\right| = (4)(5) - (-1)(3) = 20 + 3 = 23 \)
Minor of 4 is \( M_{21} = \left|\begin{array}{rr} -3 & 2 \\ 5 & 2 \end{array}\right| = (-3)(2) - (2)(5) = -6 - 10 = -16 \)
Minor of -1 is \( M_{22} = \left|\begin{array}{rr} 1 & 2 \\ 3 & 2 \end{array}\right| = (1)(2) - (2)(3) = 2 - 6 = -4 \)
Minor of 2 is \( M_{23} = \left|\begin{array}{rr} 1 & -3 \\ 3 & 5 \end{array}\right| = (1)(5) - (-3)(3) = 5 + 9 = 14 \)
Minor of 3 is \( M_{31} = \left|\begin{array}{rr} -3 & 2 \\ -1 & 2 \end{array}\right| = (-3)(2) - (2)(-1) = -6 + 2 = -4 \)
Minor of 5 is \( M_{32} = \left|\begin{array}{rr} 1 & 2 \\ 4 & 2 \end{array}\right| = (1)(2) - (2)(4) = 2 - 8 = -6 \)
Minor of 2 is \( M_{33} = \left|\begin{array}{rr} 1 & -3 \\ 4 & -1 \end{array}\right| = (1)(-1) - (-3)(4) = -1 + 12 = 11 \)
Cofactor of 1 is \( A_{11} = (-1)^{1+1} M_{11} = 1 \times -12 = -12 \)
Cofactor of -3 is \( A_{12} = (-1)^{1+2} M_{12} = -1 \times 2 = -2 \)
Cofactor of 2 is \( A_{13} = (-1)^{1+3} M_{13} = 1 \times 23 = 23 \)
Cofactor of 4 is \( A_{21} = (-1)^{2+1} M_{21} = -1 \times -16 = 16 \)
Cofactor of -1 is \( A_{22} = (-1)^{2+2} M_{22} = 1 \times -4 = -4 \)
Cofactor of 2 is \( A_{23} = (-1)^{2+3} M_{23} = -1 \times 14 = -14 \)
Cofactor of 3 is \( A_{31} = (-1)^{3+1} M_{31} = 1 \times -4 = -4 \)
Cofactor of 5 is \( A_{32} = (-1)^{3+2} M_{32} = -1 \times -6 = 6 \)
Cofactor of 2 is \( A_{33} = (-1)^{3+3} M_{33} = 1 \times 11 = 11 \)
In simple words: To find a minor, cover the row and column of an element and calculate the determinant of the remaining part. For a cofactor, multiply the minor by either 1 or -1 based on its position in the matrix.

🎯 Exam Tip: Remember that minors are just the determinant of the submatrix, while cofactors include a sign based on the position \((-1)^{i+j}\).

Question 2. Evaluate \( \left|\begin{array}{rrr} 3 & -2 & 4 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right| \)
Answer: To evaluate the determinant, we expand along the first row:
\( \left|\begin{array}{rrr} 3 & -2 & 4 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right| = 3 \left|\begin{array}{cc} 0 & 1 \\ 2 & 3 \end{array}\right| - (-2) \left|\begin{array}{cc} 2 & 1 \\ 1 & 3 \end{array}\right| + 4 \left|\begin{array}{cc} 2 & 0 \\ 1 & 2 \end{array}\right| \)
Now, we calculate the 2x2 determinants:
\( = 3((0)(3) - (1)(2)) + 2((2)(3) - (1)(1)) + 4((2)(2) - (0)(1)) \)
\( = 3(0 - 2) + 2(6 - 1) + 4(4 - 0) \)
\( = 3(-2) + 2(5) + 4(4) \)
\( = -6 + 10 + 16 \)
\( = 4 + 16 \)
\( = 20 \)
The value of the determinant is 20. This method helps reduce a larger determinant into smaller, easier-to-solve ones.
In simple words: To find the value of a 3x3 determinant, pick a row or column. Then, for each number in that row/column, multiply it by its cofactor and add them up.

🎯 Exam Tip: Choose the row or column with the most zeros to simplify calculations, as terms multiplied by zero become zero.

Question 3. Solve: \( \left|\begin{array}{lll} 2 & x & 3 \\ 4 & 1 & 6 \\ 1 & 2 & 7 \end{array}\right|=0 \)
Answer: We need to find the value of x for which the determinant is zero. Let's expand the determinant along the first row:
\( 2 \left|\begin{array}{cc} 1 & 6 \\ 2 & 7 \end{array}\right| - x \left|\begin{array}{cc} 4 & 6 \\ 1 & 7 \end{array}\right| + 3 \left|\begin{array}{cc} 4 & 1 \\ 1 & 2 \end{array}\right| = 0 \)
Now, calculate the 2x2 determinants:
\( 2((1)(7) - (6)(2)) - x((4)(7) - (6)(1)) + 3((4)(2) - (1)(1)) = 0 \)
\( 2(7 - 12) - x(28 - 6) + 3(8 - 1) = 0 \)
\( 2(-5) - x(22) + 3(7) = 0 \)
\( -10 - 22x + 21 = 0 \)
\( -22x + 11 = 0 \)
Now, we solve for x:
\( -22x = -11 \)
\( x = \frac{-11}{-22} \)
\( x = \frac{1}{2} \)
So, the value of x that makes the determinant zero is \( \frac{1}{2} \). Solving such equations helps us understand how matrix values affect the determinant.
In simple words: Set the determinant to zero, then work it out like a regular equation to find the unknown letter, x.

🎯 Exam Tip: Double-check your arithmetic, especially when dealing with negative signs and large numbers, as a small error can lead to a wrong answer.

Question 4. Find |AB| if \( A = \left[\begin{array}{rr} 3 & -1 \\ 2 & 1 \end{array}\right] \) and \( B = \left[\begin{array}{rr} 3 & 0 \\ 1 & -2 \end{array}\right] \)
Answer: First, we need to find the product of the two matrices, AB:
\( AB = \left[\begin{array}{rr} 3 & -1 \\ 2 & 1 \end{array}\right] \left[\begin{array}{rr} 3 & 0 \\ 1 & -2 \end{array}\right] \)
To multiply matrices, we multiply rows by columns:
\( AB = \left[\begin{array}{ll} (3)(3) + (-1)(1) & (3)(0) + (-1)(-2) \\ (2)(3) + (1)(1) & (2)(0) + (1)(-2) \end{array}\right] \)
\( AB = \left[\begin{array}{ll} 9 - 1 & 0 + 2 \\ 6 + 1 & 0 - 2 \end{array}\right] \)
\( AB = \left[\begin{array}{rr} 8 & 2 \\ 7 & -2 \end{array}\right] \)
Now, we find the determinant of the product matrix AB:
\( |AB| = \left|\begin{array}{rr} 8 & 2 \\ 7 & -2 \end{array}\right| \)
\( |AB| = (8)(-2) - (2)(7) \)
\( |AB| = -16 - 14 \)
\( |AB| = -30 \)
The determinant of the product of two matrices is equal to the product of their individual determinants, which can be a quicker way to verify this type of answer.
In simple words: First, multiply matrix A by matrix B to get a new matrix AB. Then, find the determinant of this new AB matrix.

🎯 Exam Tip: Remember the property \( |AB| = |A| \cdot |B| \). You can calculate \( |A| \) and \( |B| \) separately and then multiply them to get the same result as calculating \( |AB| \) directly, which can act as a useful check.

Question 5. Solve: \( \left|\begin{array}{rrr} 7 & 4 & 11 \\ -3 & 5 & x \\ -x & 3 & 1 \end{array}\right|=0 \)
Answer: We need to solve for x when the determinant is equal to zero. Let's expand the determinant along the first row:
\( 7 \left|\begin{array}{cc} 5 & x \\ 3 & 1 \end{array}\right| - 4 \left|\begin{array}{cc} -3 & x \\ -x & 1 \end{array}\right| + 11 \left|\begin{array}{cc} -3 & 5 \\ -x & 3 \end{array}\right| = 0 \)
Now, we calculate the 2x2 determinants:
\( 7((5)(1) - (x)(3)) - 4((-3)(1) - (x)(-x)) + 11((-3)(3) - (5)(-x)) = 0 \)
\( 7(5 - 3x) - 4(-3 + x^2) + 11(-9 + 5x) = 0 \)
Now, distribute the numbers and simplify the equation:
\( 35 - 21x + 12 - 4x^2 - 99 + 55x = 0 \)
Combine like terms:
\( -4x^2 + (-21x + 55x) + (35 + 12 - 99) = 0 \)
\( -4x^2 + 34x - 52 = 0 \)
To make it easier, we can divide the entire equation by -2:
\( \frac{-4x^2}{-2} + \frac{34x}{-2} - \frac{52}{-2} = 0 \)
\( 2x^2 - 17x + 26 = 0 \)
Now, we factor the quadratic equation. We look for two numbers that multiply to \( (2)(26) = 52 \) and add up to -17. These numbers are -13 and -4.
\( 2x^2 - 13x - 4x + 26 = 0 \)
Factor by grouping:
\( x(2x - 13) - 2(2x - 13) = 0 \)
\( (2x - 13)(x - 2) = 0 \)
Set each factor to zero to find the possible values for x:
\( 2x - 13 = 0 \quad \text{or} \quad x - 2 = 0 \)
\( 2x = 13 \quad \text{or} \quad x = 2 \)
\( x = \frac{13}{2} \quad \text{or} \quad x = 2 \)
The values of x for which the determinant is zero are \( \frac{13}{2} \) and 2. Quadratic equations often yield two solutions.
In simple words: When the determinant equals zero, set up the equation and simplify it to find x. You might get two answers for x from a quadratic equation.

🎯 Exam Tip: When solving a quadratic equation, remember to check both possible values of x in the original equation if time permits, to ensure accuracy.

Question 6. Evaluate: \( \left|\begin{array}{lll} 1 & a & a^{2}-bc \\ 1 & b & b^{2}-ca \\ 1 & c & c^{2}-ab \end{array}\right| \)
Answer: We need to evaluate the given determinant. We can use determinant properties to simplify the calculation.
Let \( \Delta = \left|\begin{array}{lll} 1 & a & a^{2}-bc \\ 1 & b & b^{2}-ca \\ 1 & c & c^{2}-ab \end{array}\right| \)
We can split the third column into two parts:
\( \Delta = \left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right| + \left|\begin{array}{lll} 1 & a & -bc \\ 1 & b & -ca \\ 1 & c & -ab \end{array}\right| \)
Now, take out -1 from the third column of the second determinant:
\( \Delta = \left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right| - \left|\begin{array}{lll} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{array}\right| \)
Multiply \( R_1 \) by a, \( R_2 \) by b, \( R_3 \) by c in the second determinant, and divide the determinant by abc to keep its value unchanged:
\( \Delta = \left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right| - \frac{1}{abc} \left|\begin{array}{lll} a & a^{2} & abc \\ b & b^{2} & abc \\ c & c^{2} & abc \end{array}\right| \)
Take out abc from \( C_3 \) of the second determinant:
\( \Delta = \left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right| - \frac{abc}{abc} \left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right| \)
Now, swap \( C_1 \) and \( C_3 \) in the second determinant. This changes the sign of the determinant:
\( \Delta = \left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right| - (-1) \left|\begin{array}{lll} 1 & a^{2} & a \\ 1 & b^{2} & b \\ 1 & c^{2} & c \end{array}\right| \)
Swap \( C_2 \) and \( C_3 \) in the second determinant again. This changes the sign back:
\( \Delta = \left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right| + (-1)(-1) \left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right| \)
\( \Delta = \left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right| + \left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right| \)
This means the two determinants are identical, but with opposite signs initially due to the operations. Therefore, their sum is zero. Using determinant properties can often simplify complex determinant calculations to zero.
\( \Delta = 0 \)
In simple words: Break the determinant into two smaller ones. Use rules like swapping columns or multiplying rows to make the two smaller determinants the same but with opposite signs. When you add them, the total will be zero.

🎯 Exam Tip: Look for opportunities to use determinant properties like linearity or row/column operations to simplify the determinant before expanding, as this can greatly reduce calculation time and error.

Question 7. Prove that \( \left|\begin{array}{lll} \frac{1}{a} & bc & b+c \\ \frac{1}{b} & ca & c+a \\ \frac{1}{c} & ab & a+b \end{array}\right|=0 \)
Answer: We need to prove that the given determinant equals zero. Let's use properties of determinants.
Let \( \Delta = \left|\begin{array}{lll} \frac{1}{a} & bc & b+c \\ \frac{1}{b} & ca & c+a \\ \frac{1}{c} & ab & a+b \end{array}\right| \)
To remove the fractions, we can multiply \( R_1 \) by a, \( R_2 \) by b, and \( R_3 \) by c. To keep the determinant value the same, we must also divide the entire determinant by abc:
\( \Delta = \frac{1}{abc} \left|\begin{array}{lll} a \cdot \frac{1}{a} & a \cdot bc & a(b+c) \\ b \cdot \frac{1}{b} & b \cdot ca & b(c+a) \\ c \cdot \frac{1}{c} & c \cdot ab & c(a+b) \end{array}\right| \)
\( \Delta = \frac{1}{abc} \left|\begin{array}{lll} 1 & abc & ab+ac \\ 1 & abc & bc+ba \\ 1 & abc & ca+cb \end{array}\right| \)
Now, we can take out abc from the second column (\( C_2 \)):
\( \Delta = \frac{abc}{abc} \left|\begin{array}{lll} 1 & 1 & ab+ac \\ 1 & 1 & bc+ba \\ 1 & 1 & ca+cb \end{array}\right| \)
So, \( \Delta = \left|\begin{array}{lll} 1 & 1 & ab+ac \\ 1 & 1 & bc+ba \\ 1 & 1 & ca+cb \end{array}\right| \)
Since the first column (\( C_1 \)) and the second column (\( C_2 \)) are identical, the value of the determinant is 0. This is a fundamental property of determinants. Whenever two columns (or rows) are identical, the determinant is zero.
In simple words: We can make two columns of the determinant exactly the same by multiplying rows and taking out common factors. When two columns are identical, the determinant always equals zero.

🎯 Exam Tip: Remember that if any two rows or two columns of a determinant are identical (or proportional), the value of the determinant is zero. This property is a powerful shortcut.

Question 8. Prove that \( \left|\begin{array}{ccc} -a^{2} & ab & ac \\ ab & -b^{2} & bc \\ ac & bc & -c^{2} \end{array}\right|=4 a^{2} b^{2} c^{2} \)
Answer: We need to prove that the given determinant equals \( 4a^2b^2c^2 \).
Let \( \Delta = \left|\begin{array}{ccc} -a^{2} & ab & ac \\ ab & -b^{2} & bc \\ ac & bc & -c^{2} \end{array}\right| \)
First, we can take out common factors from each row. Take 'a' from \( R_1 \), 'b' from \( R_2 \), and 'c' from \( R_3 \):
\( \Delta = abc \left|\begin{array}{ccc} -a & b & c \\ a & -b & c \\ a & b & -c \end{array}\right| \)
Next, we can take out common factors from each column. Take 'a' from \( C_1 \), 'b' from \( C_2 \), and 'c' from \( C_3 \):
\( \Delta = abc \cdot abc \left|\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right| \)
\( \Delta = a^2b^2c^2 \left|\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right| \)
Now, we perform row operations to simplify the determinant. Apply \( R_2 \rightarrow R_2 + R_1 \) and \( R_3 \rightarrow R_3 + R_1 \):
\( \Delta = a^2b^2c^2 \left|\begin{array}{ccc} -1 & 1 & 1 \\ 1+(-1) & -1+1 & 1+1 \\ 1+(-1) & 1+1 & -1+1 \end{array}\right| \)
\( \Delta = a^2b^2c^2 \left|\begin{array}{ccc} -1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 2 & 0 \end{array}\right| \)
Now, expand the determinant along the first column (\( C_1 \)):
\( \Delta = a^2b^2c^2 \left( -1 \left|\begin{array}{cc} 0 & 2 \\ 2 & 0 \end{array}\right| - 0 \left|\begin{array}{cc} 1 & 1 \\ 2 & 0 \end{array}\right| + 0 \left|\begin{array}{cc} 1 & 1 \\ 0 & 2 \end{array}\right| \right) \)
\( \Delta = a^2b^2c^2 \left( -1 ((0)(0) - (2)(2)) \right) \)
\( \Delta = a^2b^2c^2 \left( -1 (0 - 4) \right) \)
\( \Delta = a^2b^2c^2 (-1)(-4) \)
\( \Delta = 4a^2b^2c^2 \)
Thus, the determinant is proven to be equal to \( 4a^2b^2c^2 \). Factoring out common terms and using row operations simplifies the evaluation of complex determinants.
In simple words: First, take out common letters from each row and column. Then, use row operations to create zeros, making the determinant easier to solve. Finally, calculate the remaining part to prove the identity.

🎯 Exam Tip: Always look for common factors in rows or columns first, as taking them out simplifies the numbers inside the determinant significantly before proceeding with row/column operations or expansion.