Maharashtra Board Class 11 Maths Part 1 Chapter 6 Determinants 6.1 Solutions

Question 1. Evaluate the following determinants:
(i) \(\begin{vmatrix} 4 & 7 \\ -7 & 0 \end{vmatrix}\)
(ii) \(\begin{vmatrix} 3 & -5 & 2 \\ 1 & 8 & 9 \\ 3 & 7 & 0 \end{vmatrix}\)
(iii) \(\begin{vmatrix} 1 & i & 3 \\ -i & 2 & 5 \\ 3 & 2 & i^4 \end{vmatrix}\)
(iv) \(\begin{vmatrix} 5 & 5 & 5 \\ 5 & 4 & 4 \\ 5 & 4 & 8 \end{vmatrix}\)
(v) \(\begin{vmatrix} 2i & 3 \\ 4 & -i \end{vmatrix}\)
(vi) \(\begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}\)
(vii) \(\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}\)
(viii) \(\begin{vmatrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{vmatrix}\)
Answer:
(i) \(\begin{vmatrix} 4 & 7 \\ -7 & 0 \end{vmatrix}\)
= 4(0) - (-7)(7)
= 0 + 49
= 49
(ii) \(\begin{vmatrix} 3 & -5 & 2 \\ 1 & 8 & 9 \\ 3 & 7 & 0 \end{vmatrix}\)
= 3(0 - 63) - (-5)(0 - 27) + 2(7 - 24)
= 3(-63) + 5(-27) + 2(-17)
= -189 - 135 - 34
= -358
(iii) \(\begin{vmatrix} 1 & i & 3 \\ -i & 2 & 5 \\ 3 & 2 & i^4 \end{vmatrix}\)
...[\(i^2 = -1, i^4 = 1\)]
= 1 \(\begin{vmatrix} 2 & 5 \\ 2 & i^4 \end{vmatrix}\) - \(i \begin{vmatrix} -i & 5 \\ 3 & i^4 \end{vmatrix}\) + 3 \(\begin{vmatrix} -i & 2 \\ 3 & 2 \end{vmatrix}\)
= 1(2(1) - 10) - i(-i(1) - 15) + 3(-2i - 6)
= 1(2 - 10) - i(-i - 15) + 3(-2i - 6)
= -8 + \(i^2\) + 15i - 6i - 18
= \(i^2\) - 26 + 9i
= -1 - 26 + 9i ...[:. \(i^2 = -1\)]
= -27 + 9i
(iv) \(\begin{vmatrix} 5 & 5 & 5 \\ 5 & 4 & 4 \\ 5 & 4 & 8 \end{vmatrix}\)
= 5(4 \(\times\) 8 - 4 \(\times\) 4) - 5(5 \(\times\) 8 - 4 \(\times\) 5) + 5(5 \(\times\) 4 - 4 \(\times\) 5)
= 5(32 - 16) - 5(40 - 20) + 5(20 - 20)
= 5(16) - 5(20) + 5(0)
= 80 - 100
= -20
(v) \(\begin{vmatrix} 2i & 3 \\ 4 & -i \end{vmatrix}\)
= 2i(-i) - 3(4)
= -2\(i^2\) - 12
= -2(-1) - 12 ...[:. \(i^2 = -1\)]
= 2 - 12
= -10
(vi) \(\begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}\)
= 3(1 \(\times\) 1 - (-2) \(\times\) 3) - (-4)(1 \(\times\) 1 - (-2) \(\times\) 2) + 5(1 \(\times\) 3 - 1 \(\times\) 2)
= 3(1 + 6) + 4(1 + 4) + 5(3 - 2)
= 3(7) + 4(5) + 5(1)
= 21 + 20 + 5
= 46
(vii) \(\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}\)
= a(bc - \(f^2\)) - h(hc - gf) + g(hf - gb)
= abc - a\(f^2\) - \(h^2\)c + fgh + fgh - g\(b^2\)
= abc + 2fgh - a\(f^2\) - b\(g^2\) - c\(h^2\)
(viii) \(\begin{vmatrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{vmatrix}\)
= 0 \(\begin{vmatrix} 0 & -c \\ c & 0 \end{vmatrix}\) - a \(\begin{vmatrix} -a & -c \\ b & 0 \end{vmatrix}\) + (-b) \(\begin{vmatrix} -a & 0 \\ b & c \end{vmatrix}\)
= 0 - a(0 + bc) - b(-ac - 0)
= -a(bc) - b(-ac)
= -abc + abc
= 0
In simple words: This question requires evaluating different types of determinants, including 2x2 and 3x3 matrices, as well as those involving complex numbers (i) and algebraic terms. The process involves specific rules for expanding determinants, such as cofactor expansion.

🎯 Exam Tip: Pay close attention to signs when expanding determinants, especially with negative terms or complex numbers like 'i'. Simplify expressions carefully to avoid calculation errors.

Question 2. Find the value(s) of x, if
(i) \(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}\)
(ii) \(\begin{vmatrix} 2 & 1 & x+1 \\ -1 & 3 & -4 \\ 0 & -5 & 3 \end{vmatrix} = 0\)
(iii) \(\begin{vmatrix} x-1 & x & x-2 \\ 0 & x-2 & x-3 \\ 0 & 0 & x-3 \end{vmatrix} = 0\)
Answer:
(i) Given: \(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}\)
10 - 12 = 5x - 6x
-2 = -x
x = 2
Check:
L.H.S. = \(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}\) = 10 - 12 = -2
R.H.S. = \(\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}\) = \(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}\) = 10 - 12 = -2
Thus, our answer is correct.
(ii) Given: \(\begin{vmatrix} 2 & 1 & x+1 \\ -1 & 3 & -4 \\ 0 & -5 & 3 \end{vmatrix} = 0\)
2 \(\begin{vmatrix} 3 & -4 \\ -5 & 3 \end{vmatrix}\) - 1 \(\begin{vmatrix} -1 & -4 \\ 0 & 3 \end{vmatrix}\) + (x+1) \(\begin{vmatrix} -1 & 3 \\ 0 & -5 \end{vmatrix}\) = 0
2(9 - 20) - 1(-3 - 0) + (x+1)(5 - 0) = 0
2(-11) - 1(-3) + (x+1)(5) = 0
-22 + 3 + 5x + 5 = 0
5x = 14
x = \(\frac{14}{5}\)
(iii) Given: \(\begin{vmatrix} x-1 & x & x-2 \\ 0 & x-2 & x-3 \\ 0 & 0 & x-3 \end{vmatrix} = 0\)
This is an upper triangular matrix, so the determinant is the product of its diagonal elements.
(x - 1)((x - 2)(x - 3) - 0) - x(0 - 0) + (x - 2)(0 - 0) = 0
(x - 1)(x - 2)(x - 3) = 0
So, x - 1 = 0 or x - 2 = 0 or x - 3 = 0
Therefore, x = 1 or x = 2 or x = 3
In simple words: This question involves solving equations where one or more determinants are set equal to a value. For 2x2 determinants, we cross-multiply and subtract. For 3x3 determinants, we use cofactor expansion. For triangular matrices, the determinant is simply the product of the diagonal elements.

🎯 Exam Tip: Remember that for an upper or lower triangular matrix, the determinant is the product of its diagonal entries. This can save significant time in calculations. Always double-check your arithmetic, especially with negative numbers.

Question 3. Solve the following equations.
(i) \(\begin{vmatrix} x & 2 & 2 \\ 2 & x & 2 \\ 2 & 2 & x \end{vmatrix} = 0\)
(ii) \(\begin{vmatrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & 5x^2 \end{vmatrix} = 0\)
Answer:
(i) Given: \(\begin{vmatrix} x & 2 & 2 \\ 2 & x & 2 \\ 2 & 2 & x \end{vmatrix} = 0\)
x \(\begin{vmatrix} x & 2 \\ 2 & x \end{vmatrix}\) - 2 \(\begin{vmatrix} 2 & 2 \\ 2 & x \end{vmatrix}\) + 2 \(\begin{vmatrix} 2 & x \\ 2 & 2 \end{vmatrix}\) = 0
x(\(x^2\) - 4) - 2(2x - 4) + 2(4 - 2x) = 0
x(\(x^2\) - 4) - 2(2x - 4) - 2(2x - 4) = 0
x(x + 2)(x - 2) - 4(2x - 4) = 0
x(x + 2)(x - 2) - 8(x - 2) = 0
(x - 2)[x(x + 2) - 8] = 0
(x - 2)(\(x^2\) + 2x - 8) = 0
(x - 2)(\(x^2\) + 4x - 2x - 8) = 0
(x - 2)(x + 4)(x - 2) = 0
(x - 2)\(^2\)(x + 4) = 0
So, (x - 2)\(^2\) = 0 or x + 4 = 0
Therefore, x - 2 = 0 or x = -4
x = 2 or x = -4
(ii) Given: \(\begin{vmatrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & 5x^2 \end{vmatrix} = 0\)
1((-2)(\(5x^2\)) - 5(2x)) - 4(1(\(5x^2\)) - 5(1)) + 20(1(2x) - (-2)(1)) = 0
1(-10\(x^2\) - 10x) - 4(5\(x^2\) - 5) + 20(2x + 2) = 0
-10\(x^2\) - 10x - 20\(x^2\) + 20 + 40x + 40 = 0
-30\(x^2\) + 30x + 60 = 0
Dividing throughout by (-30):
\(x^2\) - x - 2 = 0
\(x^2\) - 2x + x - 2 = 0
x(x - 2) + 1(x - 2) = 0
(x - 2)(x + 1) = 0
So, x - 2 = 0 or x + 1 = 0
Therefore, x = 2 or x = -1
In simple words: This problem involves finding the values of 'x' that satisfy the given determinant equations. We expand the determinants using the appropriate rules and then solve the resulting polynomial equations. For part (i), simplifying the determinant leads to a cubic equation that can be factored. For part (ii), it simplifies to a quadratic equation.

🎯 Exam Tip: When solving determinant equations, try to factor out common terms or use row/column operations to simplify the determinant before expansion. Remember to solve the resulting algebraic equation completely to find all possible values of x.

Question 4. Find the value of x, if
\(\begin{vmatrix} x & -1 & 2 \\ 2x & 1 & -3 \\ 3 & -4 & 5 \end{vmatrix} = 29\)
Answer:
Given: \(\begin{vmatrix} x & -1 & 2 \\ 2x & 1 & -3 \\ 3 & -4 & 5 \end{vmatrix} = 29\)
x(1 \(\times\) 5 - (-3) \(\times\) (-4)) - (-1)(2x \(\times\) 5 - (-3) \(\times\) 3) + 2(2x \(\times\) (-4) - 1 \(\times\) 3) = 29
x(5 - 12) + 1(10x + 9) + 2(-8x - 3) = 29
-7x + 10x + 9 - 16x - 6 = 29
-13x + 3 = 29
-13x = 26
x = -2
In simple words: To find the value of x, we expand the given 3x3 determinant using cofactor expansion, set it equal to 29, and then solve the resulting linear equation for x.

🎯 Exam Tip: Be very careful with arithmetic signs when expanding determinants, especially when dealing with negative coefficients and subtraction. A small error can lead to an incorrect final value for x.

Question 5. Find x and y if \(\begin{vmatrix} 4i & i & 2i \\ 1 & 3i^2 & 4 \\ 5 & -3 & i \end{vmatrix} = x + iy\), where \(i = \sqrt{-1}\).
Answer:
Given: \(\begin{vmatrix} 4i & i & 2i \\ 1 & 3i^2 & 4 \\ 5 & -3 & i \end{vmatrix} = x + iy\)
We know that \(i^2 = -1\).
So the determinant becomes: \(\begin{vmatrix} 4i & i & 2i \\ 1 & 3(-1) & 4 \\ 5 & -3 & i \end{vmatrix} = \begin{vmatrix} 4i & i & 2i \\ 1 & -3 & 4 \\ 5 & -3 & i \end{vmatrix}\)
Expand the determinant:
\(4i \begin{vmatrix} -3 & 4 \\ -3 & i \end{vmatrix}\) - \(i \begin{vmatrix} 1 & 4 \\ 5 & i \end{vmatrix}\) + \(2i \begin{vmatrix} 1 & -3 \\ 5 & -3 \end{vmatrix}\)
= 4i((-3)(i) - 4(-3)) - i(1(i) - 4(5)) + 2i(1(-3) - (-3)(5))
= 4i(-3i + 12) - i(i - 20) + 2i(-3 + 15)
= -12\(i^2\) + 48i - \(i^2\) + 20i + 24i
= -13\(i^2\) + 92i
Substitute \(i^2 = -1\):
= -13(-1) + 92i
= 13 + 92i
We are given that the determinant equals x + iy.
Comparing with x + iy, we get
x = 13, y = 92
In simple words: This problem asks us to evaluate a determinant with complex numbers and then equate the result to the complex number form 'x + iy' to find the real and imaginary parts. The key is to remember that \(i^2 = -1\) and correctly perform the determinant expansion and complex number arithmetic.

🎯 Exam Tip: When working with complex numbers in determinants, carefully substitute \(i^2 = -1\) and combine real and imaginary terms separately. Ensure that the final answer is presented in the standard x + iy form for correct comparison.