Class 12 Mathematics Chapter 4 Determinants

Key Concept:Equation of Line

b) $$y = 2x – 1$$
[Solution Description]
The slope m of the line joining two points (x_1,y_1) and (x_2,y_2) is given by:

$$
m = \frac{y_2 – y_1}{x_2 – x_1}
$$

Substituting in the values:

$$
m = \frac{11 – 3}{6 – 2} = 2
$$

Using the point-slope form of the equation of a line: $$y – y_1 = m(x – x_1)$$.

Substituting m = 2, x_1 = 2, and y_1 = 3:

$$
y – 3 = 2(x – 2)
$$

Expanding this equation gives:

$$
y = 2x – 4 + 3 = 2x – 1
$$

Therefore, the equation of the line is $$y = 2x – 1$$.

Key Concept:Expansion Along Row

c) 3
[Solution Description] To find the determinant, we expand along the second row:

$$
|A| = -(3) \cdot \left|\begin{array}{cc}
0 & -2 \\
7 & 8
\end{array}\right| + 4 \cdot \left|\begin{array}{cc}
1 & -2 \\
6 & 8
\end{array}\right| – 5 \cdot \left|\begin{array}{cc}
1 & 0 \\
6 & 7
\end{array}\right|
$$

Calculating each minor:

$$
\left|\begin{array}{cc} 0 & -2 \\ 7 & 8 \end{array}\right| = (0)(8) – (-2)(7) = 14
$$

$$
\left|\begin{array}{cc} 1 & -2 \\ 6 & 8 \end{array}\right| = (1)(8) – (-2)(6) = 20
$$

$$
\left|\begin{array}{cc} 1 & 0 \\ 6 & 7 \end{array}\right| = (1)(7) – (0)(6) = 7
$$

Substituting back into the equation:

$$
|A| = -(3)(14) + 4(20) – 5(7) = -42 + 80 – 35 = 3
$$
Thus, the determinant is 3.

Key Concept:Cofactor Sign

b) Negative
[Solution Description] The sign of the cofactor for an element at position $$(i,j)$$ is given by $$(-1)^{i+j}$$. For $$a_{32}, i=3$$ and $$j=2$$, so $$(-1)^{3+2} = (-1)^5 = -1$$. Thus, the sign is negative.

Key Concept:Real World Application

c) By allowing calculation of displacements when inverted.
[Solution Description] When dealing with systems represented by matrices like D, particularly in equilibrium conditions, it’s often necessary to solve for unknowns such as displacements. The inverse of the matrix helps in solving equations of the form DX = F, where X is the displacement vector and F is the force vector.

If D is invertible, we can use the relation $$D^{-1} = \frac{1}{|D|} adj(D)$$.

First, calculate the determinant:
$$
|D| = (2)(3) – (-1)(-1) = 6 – 1 = 5
$$

Then, the adjoint of D is:
$$
adj(D) = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}
$$

Thus, the inverse is:
$$
D^{-1} = \frac{1}{5} \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} \frac{3}{5} & \frac{1}{5} \\ \frac{1}{5} & \frac{2}{5} \end{bmatrix}
$$

This allows us to find $$X = D^{-1}F$$, which calculates the displacements from known forces.

Key Concept:Determinant Property

b) 9
[Solution Description]
The property states that $$|\text{adj}(E)| = |E|^{n-1}$$ for a matrix of order  n . Here  n = 3 , so:
$$
|\text{adj}(E)| = |3|^2 = 9
$$

Key Concept:System Representation

a) $$\begin{bmatrix} 4 & 1 \\ 3 & -2 \end{bmatrix}$$
[Solution Description] To represent the system 4x + y = 9 and 3x – 2y = 5, we need a matrix that captures the coefficients of x and y.

Thus, the correct matrix is:
$$
\begin{bmatrix} 4 & 1 \\ 3 & -2 \end{bmatrix}
$$

Key Concept:Vertex Substitution

a) 3
[Solution Description]
To find the area, substitute the coordinates into the formula:
$$
\Delta = \frac{1}{2} \left| 1(6-8) + 4(8-2) + 7(2-6) \right|
$$
$$
= \frac{1}{2} \left| 1(-2) + 4(6) + 7(-4) \right|
$$
$$
= \frac{1}{2} \left| -2 + 24 – 28 \right|
$$
$$
= \frac{1}{2} \left| -6 \right| = \frac{1}{2} \times 6 = 3
$$
So, the area is 3 square units.

Key Concept:Zero Sum Property

a) True
[Solution Description]
First, identify the elements $$a_{11} = 1$$, $$a_{12} = 2$$, $$a_{13} = 3$$.

Now calculate the cofactors of the third-row elements:
$$
A_{31} = (-1)^{3+1} \begin{vmatrix} 2 & 3 \\ 5 & 6 \end{vmatrix} = \begin{vmatrix} 2 & 3 \\ 5 & 6 \end{vmatrix} = 2 \times 6 – 3 \times 5 = 12 – 15 = -3
$$
$$
A_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 3 \\ 4 & 6 \end{vmatrix} = -\begin{vmatrix} 1 & 3 \\ 4 & 6 \end{vmatrix} = -(1 \times 6 – 3 \times 4) = -(6 – 12) = 6
$$
$$
A_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 2 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} 1 & 2 \\ 4 & 5 \end{vmatrix} = 1 \times 5 – 2 \times 4 = 5 – 8 = -3
$$

Now verify the zero sum property:
$$
1 \times (-3) + 2 \times 6 + 3 \times (-3) = -3 + 12 – 9 = 0
$$

The expression indeed evaluates to zero, verifying the property.

Key Concept:Zero Determinant

b) 0
[Solution Description] The determinant of a $$2 \times 2$$ matrix $$C = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$$ is computed as $$|C| = a_{11}a_{22} – a_{12}a_{21}$$. For matrix C, this becomes:
$$
|C| = 1 \cdot 6 – 2 \cdot 3 = 6 – 6 = 0.
$$

Key Concept:Identify Element

c) 3
[Solution Description] The minor $$M_{21}$$ is associated with the element found in the second row and first column. Thus, it corresponds to the element 3.

Key Concept:Basic Determinant

b) 2
[Solution Description] The formula for the determinant of a 2×2 matrix $$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$$ is given by $$|A| = a_{11}a_{22} – a_{21}a_{12}$$. Substituting the values gives:
$$
|A| = 3 \times 4 – 2 \times 5 = 12 – 10 = 2
$$
Therefore, the determinant is 2.

Key Concept:Single Element Determinant

b) 7
[Solution Description]
The determinant of a $$1 \times 1$$ matrix $$ A = [a_{11}] $$ is simply the value a_{11}.

Therefore, the determinant of the matrix $$ A = [7] $$ is 7.

Key Concept:Notation Clarification

b) Determinant of B
[Solution Description] The symbol $$|B|$$ represents the determinant of matrix B. It is a common misconception that it might represent modulus, but in the context of matrices, it refers to the determinant.

Key Concept:Basic Definition

d) All of the above
[Solution Description] The determinant of a matrix can be denoted by $$|A|,\text{det } A$$, or $$\Delta$$. These are standard notations used to indicate the determinant.

Key Concept:Zero Determinant

a) Yes

[Solution Description] To determine if the determinant is zero, calculate $$|C| = a_{11}a_{22} – a_{21}a_{12}$$. Here $$a_{11} = 1$$, $$a_{12} = 2$$, $$a_{21} = 2$$, $$a_{22} = 4$$. Thus,

$$|C| = 1 \times 4 – 2 \times 2 = 4 – 4 = 0$$

Since the determinant is zero, matrix $$C$$ has a determinant of zero.

Key Concept:Conceptual Application

a) 0

[Solution Description]

The determinant for a matrix of order one $$E = [x]$$ is $$\text{det}(E) = x$$. Given that this determinant equals zero, we have:

x = 0

Thus, the value of x must be zero.

Key Concept:Area of Triangle

c) 3

[Solution Description]

The area of the triangle can be calculated using the formula:

$$\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|$$

Substituting the given points:

$$x_1 = 1, \, y_1 = 2$$

$$x_2 = 3, y_2 = -4$$

$$x_3 = -1, y_3 = 5$$

The area calculation becomes:

$$\text{Area} = \frac{1}{2} \left| 1(-4-5) + 3(5-2) + (-1)(2+4) \right|$$

$$\frac{1}{2} \left| 1(-9) + 3(3) + (-1)(6) \right|$$

$$\frac{1}{2} \left| -9 + 9 – 6 \right|$$

$$\frac{1}{2} \times 6$$

$$= 3$$

Thus, the area of the triangle is 3 square units.

Key Concept:Area Calculation, Complex Determinant Evaluation

d) Any combination maximizing deviation from P or Q

[Solution Description] The area of a triangle with vertices $$(x_1, y_1), (x_2, y_2), (x_3, y_3)$$ can be calculated using determinants as:

$$\text{Area} = \frac{1}{2} \left| \begin{vmatrix} x_1 – y_1 – 1 \end{vmatrix} \right|$$

$$x_2 – y_2 – 1$$

$$\left| \begin{vmatrix} x_3 – y_3 – 1 \end{vmatrix} \right|$$

For points $$P(1, 4), Q(5, 6), R(a, b)$$,

$$\text{Area} = \frac{1}{2} \left| \begin{vmatrix} 1 & -4 & -1 \end{vmatrix} \right|$$

$$5 – 6 – 1$$

$$\left| a – b – 1 \right|$$

Calculate determinant:

$$\begin{vmatrix} 1 & -4 & -1 \end{vmatrix}$$

$$5 – 6 – 1$$

$$a – b – 1$$

$$1(6-b) – 4(5-a) + 1(5b-6a)$$

$$6 – b – 20 + 4a + 5b – 6a$$

$$= 5b – 2a – 14$$

To maximize the area, set the derivative $$\frac{d}{da}(5b – 2a – 14) = 0, \quad \frac{d}{db}(5b – 2a – 14) = 0$$
. Solving gives no dependency of one variable over another so choose any numbers for maximal distance centered at origin.

In conclusion, there is no fixed point for maxima except maximizing absolute differences from P or Q.

Key Concept:Advanced Determinant Expansion, Real-World Application

b) 136

[Solution Description]

To solve this problem, we expand the determinant along the first column:

$$D = 3 \begin{vmatrix} -5 & -7 \\ 2 & 0 \end{vmatrix} – 2 \begin{vmatrix} 1 & -4 \\ 2 & 0 \end{vmatrix} + 6 \begin{vmatrix} 1 & -4 \\ -5 & -7 \end{vmatrix}$$

Calculating each minor:

$$[\begin{vmatrix} -5 & -7 \\ 2 & 0 \end{vmatrix} = (-5)(0) – (-7)(2) = -14, \quad \begin{vmatrix} 1 & -4 \\ 2 & 0 \end{vmatrix}]$$
$$[= (1)(0) – (-4)(2) = -8, \quad\begin{vmatrix}1 & -4 \\-5 & -7\end{vmatrix} = (1)(-7) – (-4)(-5) = -7 – 20 = -27]$$

Substituting back,

$$D = 3(-14) – 2(-8) + 6(27) = -42 + 16 + 162 = 136$$

Thus, the value of the determinant is 136. This signifies that the volume of the parallelepiped formed by these vectors is 136 cubic units.

Key Concept:Area of Triangle

b) 6

[Solution Description]

The formula for the area of a triangle with vertices $$(x_1, y_1)$$, $$(x_2, y_2)$$, and $$(x_3, y_3)$$ is given by:

$$\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|$$

Substituting the given points $$(1, 0)$$, $$(4, 3)$$, and $$(-1, 2)$$ into the formula, we get:

$$\text{Area} = \frac{1}{2} \left| 1(3-2) + 4(2-0) + (-1)(0-3) \right|$$

Simplifying gives:

$$\text{Area} = \frac{1}{2} \left| 1 + 8 + 3 \right| = \frac{1}{2} \times 12 = 6$$

Thus, the area of the triangle is 6 square units.

Key Concept:Real-World Application, System Consistency

a) Apples: 10.5, Bananas: 12.75, Oranges: 6.25

[Solution Description]

Formulate the problem in terms of matrices. Let the prices per kg of apples, bananas, and oranges be a, b, and c respectively. Then, the system becomes:

$$[2a + 3b + c = 32, \quad a + b + 2c = 17, \quad a + 2b + 3c = 26]$$

Representing this in matrix form: $$AX = B$$, where:

$$A = \begin{bmatrix} 2 & -3 & -1 \\ 1 & -1 & -2 \\ 1 & -2 & -3 \end{bmatrix}, \quad B = \begin{bmatrix} 32 \\ 17 \\ 26 \end{bmatrix}$$

Calculate determinant of A:

$$\text{det}(A) = 2(1 \cdot 3 – 2 \cdot 2) – 3(1 \cdot 3 – 2 \cdot 1) + 1(1 \cdot 2 – 1 \cdot 1)

= 2(-1) – 3(1) + 1(1) = -2 – 3 + 1 = -4$$

Since $$\text{det}(A) \neq 0$$, A is invertible.

Find the inverse of A, denoted $$A^{-1}$$:

$$A^{-1} = \frac{1}{-4} \begin{bmatrix} 1 & -4 & -3 \\ 1 & -5 & -4 \\ -1 & 5 & -3 \end{bmatrix}$$

Solve for X using the inverse:

$$X = A^{-1}B = \frac{1}{-4} \begin{bmatrix} 1 & -4 & 3 \\ 1 & -5 & -4 \\ -1 & -5 & -3 \end{bmatrix} \begin{bmatrix} 32 \\ 17 \\ 26 \end{bmatrix}$$

Calculating step-by-step:

$$X = \frac{1}{-4} \begin{bmatrix} 42 \\ 51 \\ -25 \end{bmatrix}$$

$$\begin{bmatrix} -10.5 & -12.75 & 6.25 \end{bmatrix}$$

Thus, the cost per kg of apples is 10.5, bananas is 12.75, and oranges is 6.25.