Mathematics Test - 2

Calculation:

A and B is non - singular matrices of order n x n 

|A| ≠  0 and |B| ≠  0     ....(i)

A and B are of the same order, so AB is defined and is on the same order. 

Thus,

|AB| = |A||B|

|AB| ≠  0             [Using equ (i)]

Thus AB is non-singular.

CONCEPT:

In general the determinant of a square matrix can be evaluated along any row (column) and it is equal to sum of the product of elements in any row and there corresponding co-factor.

i.e \(\left| A \right| = \;\mathop \sum \limits_{j = 1}^n {a_{ij}}\;{C_{ij}},\;for\;some\;i \in N\),

where Cij is the co-factor.

CALCULATION:

Here, it is given that matrix A is of order 3

As we know that,  the determinant of a square matrix can be evaluated along any row (column) and it is equal to sum of the product of elements in any row and there corresponding co-factor i.e \(\left| A \right| = \;\mathop \sum \limits_{j = 1}^n {a_{ij}}\;{C_{ij}},\;for\;some\;i \in N\),

where Cij is the co-factor.

As we can see that, in each option except the option A all the elements of a row(column) are multiplied with its corresponding co-factor.

Hence, option A is the correct answer.

Since A is a scalar matrix, we have \(A = \left( {\begin{array}{*{20}{c}} k&0&0\\ 0&k&0\\ 0&0&k \end{array}} \right)\)

\(\therefore \left| A \right| = {k^3}\)

\(\begin{array}{l} adj\;A = \left( {\begin{array}{*{20}{c}} {{k^2}}&0&0\\ 0&{{k^2}}&0\\ 0&0&{{k^2}} \end{array}} \right)\\ {A^{ - 1}} = \frac{1}{{\left| A \right|}}\left( {adj\;A} \right) = \frac{1}{{{k^3}}}\left( {\begin{array}{*{20}{c}} {{k^2}}&0&0\\ 0&{{k^2}}&0\\ 0&0&{{k^2}} \end{array}} \right)\\ \frac{1}{k}\left( {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right) = \frac{1}{k}{\rm{I}} \end{array}\)

Concept:

Solving a \(3\times 3\) determinant:

A \(3\times 3\) determinant can be solved as follows:

\(\begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i \end{vmatrix} = a(ei-fh)-b(di-fg)+c(dh-ge)\)

Calculation:

Simplify the given determinant as follows:

\(\begin{align*} \begin{vmatrix} a & b & 0\\ 0 & a & b\\ b & 0 & a \end{vmatrix} &= 0\\ a(a^2-0) - b(0-b^2) +0(0-ab) &= 0\\ a^3+b^3 &= 0\\ a^3 &= -b^3\\ \dfrac{a^3}{b^3} &= -1 \end{align*}\)

Therefore, the given condition holds only if \(\dfrac{a}{b}\) is one of the cube roots of -1.

Formula used:

\(\rm A = \begin{vmatrix} a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3}\ \end{vmatrix}\)

det(A) = \(\rm a_{1}\begin{vmatrix} b_{2} & b_{3}\\ c_{2} & c_{3} \end{vmatrix} - a_{2} \begin{vmatrix} b_{1} & b_{3}\\ c_{1} & c_{3} \end{vmatrix} + a_{3} \begin{vmatrix} b_{1} & b_{2}\\ c_{1} & c_{2} \end{vmatrix}\)

= a₁(b₂ c₃ - b₃ c₂) - a₂(b₁c₃ - b₃c₁) + a₃ (b₁c₂ - b2c1)

Calculation:

\(\begin{vmatrix} cos C & tan A & 0\\ sin B & 0 & -tanA\\ 0 & sinB & cos C \end{vmatrix}\)

= cos C (0 + tan A sin B) - tan A (sin B cos C) 

= tan A sin B cos C - tan A sin B cos C

= 0

Concepts:

If A and B are square matrices of order n, then

|K × AB|= Kⁿ x {| A| × | B |}

Where

 K is any real number and a scalar.

n is the order of matrix A or B as both have the same order

Calculation:

Given:  A and B be (3 x 3) matrices with |A| = 4 and | B | = 3

As A and B be (3 x 3) matrices then n = 3

 By using above formula, we get

   | 2AB| = 2³ x 3 × 4 = 8 × 3 × 4 = 96.

∴ | 2AB| = 96 Additional Information

The important properties of the determinant of a matrix are given below:

1.      | A | = | A’ | for any square matrix A, where A^’ is the transpose matrix of A.

2.    If in any square matrix (say A)  of order ‘N’, either two rows or two columns are identical, then | A | = 0.

3.    If |λA| = λ^N  × | A |, when  A is the square matrix of Order ‘N’ and λ is any scalar.

4.    If A and B are two square matrices of the same order, then |AB |= | A |× | B |.

Explanation:

Given matrix \(A = \frac {1}{3} \left( {\begin{array}{*{20}{c}} 1&{2}&2\\ 2&{1}&{-2}\\ -2&2&{-1} \end{array}} \right)\)

Now its transpose will be 

\(A^T = \frac {1}{3} \left( {\begin{array}{*{20}{c}} 1&{2}&{-2}\\ 2&{1}&{2}\\ 2&{-2}&{-1} \end{array}} \right)\)

The product will be 

AA^T = \(\frac {1}{3} \left( {\begin{array}{*{20}{c}} 1&{2}&2\\ 2&{1}&{-2}\\ -2&2&{-1} \end{array}} \right) \cdot \frac {1}{3} \left( {\begin{array}{*{20}{c}} 1&{2}&{-2}\\ 2&{1}&{2}\\ 2&{-2}&{-1} \end{array}} \right)\)

Here, AA^T = \(\frac {1}{9} \left( {\begin{array}{*{20}{c}} 9&{0}&{0}\\ 0&{9}&{0}\\ 0&{0}&{9} \end{array}} \right)\)

⇒ AA^T = I;

Concept:

Area of equilateral triangle = \(\frac{\sqrt{3}}{4}\)×a2

Calculation:

Given: The co-ordinates of its vertices are (x1, y1); (x2, y2): (x3, y3) then the square of the determinant \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}\)

⇒ (△ ABC ) = \(\dfrac{1}{2}\) \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}\) = ( \(\frac{\sqrt{3}}{4}\)) ×a²

On squaring both side, 

⇒ \(\dfrac{1}{4}\)  \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}^2\) = \(\dfrac{3a^4}{16}\)

⇒ \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}^2\) = \(\dfrac{3a^4}{4}\)

Concept:

Let \(A = \left[ {{a_{ij}}} \right]\) be a square matrix.

Steps for obtaining Inverse of the matrix 

1. We need to calculate cofactors of all elements

\(\Rightarrow {{\rm{C}}_{{\rm{ij}}}}{\rm{\;or\;}}{{\rm{A}}_{{\rm{ij}}}} = {\left( { - 1} \right)^{{\rm{i}} + {\rm{j}}}}{\rm{\;}}{{\rm{M}}_{{\rm{ij}}}}\)

Here, M_ij = Minor of matrix A

2. Transpose of the cofactor is called Adjoint of the matrix

⇒ Adj (A) = C^T

3. If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\) then determinant of A is given by: |A| = (a­₁₁ × a₂₂) – (a₁₂ – a₂₁).

4. Inverse of matrix A = A^-1 = \(\frac{{{\rm{Adj\;}}\left( {\rm{A}} \right)}}{{\left| {\rm{A}} \right|}}\)

Calculation:

Given:

 A = \(\left[ {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta }\\ {\sin \theta }&{\cos \theta } \end{array}} \right]\)

⇒ Det (A) = |A| = (cos θ × cos θ) – (sin θ × -sin θ) = cos² θ + sin² θ = 1

Now find the Cofactors of matrix A,

A₁₁ = (-1)¹⁺¹ cos θ = cos θ

A₁₂ = (-1)¹⁺² (sin θ) = -sin θ

A₂₁ = (-1)²⁺¹ (-sin θ) = sin θ

A₂₂ = (-1)²⁺² cos θ = cos θ

Cofactors of matrix A = \(= {\rm{\;}}{{\rm{A}}_{{\rm{ij}}}} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\cos {\rm{\theta }}}&{- \sin {\rm{\theta }}}\\ { \sin {\rm{\theta }}}&{\cos {\rm{\theta }}} \end{array}} \right]\)

Now, Adj (A) = \({\left[ {{{\rm{A}}_{{\rm{ij}}}}} \right]^{\rm{T}}} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\cos {\rm{\theta }}}&{ \sin {\rm{\theta }}}\\ {-\sin {\rm{\theta }}}&{\cos {\rm{\theta }}} \end{array}} \right]\)

Hence A^-1 = \(\frac{{{\rm{Adj\;}}\left( {\rm{A}} \right)}}{{\left| {\rm{A}} \right|}} = \;\frac{{\left[ {\begin{array}{*{20}{c}} {\cos \theta }&{ \sin \theta }\\ {-\sin \theta }&{\cos \theta } \end{array}} \right]}}{1} = \;\left[ {\begin{array}{*{20}{c}} {\cos \theta }&{ \sin \theta }\\ {-\sin \theta }&{\cos \theta } \end{array}} \right]\)

∴ Option 1 is correct.

Shortcut Method to finding Adj (A)

Let A = \(\left[ {\begin{array}{*{20}{c}} {\rm{a}}&b\\ {\rm{c}}&{\rm{d}} \end{array}} \right]\)

Adj A = \(\left[ {\begin{array}{*{20}{c}} {\rm{d}}&{ - b}\\ { - {\rm{c}}}&{\rm{a}} \end{array}} \right]\)

Note: Interchange the diagonal elements and change the sign of remaining elements.

Concept:

Properties of determinants:​

For a n×n matrix A, det(kA) = kn det(A).

Calculation:

Given:

|A| = 5

k = 5

From the properties of the determinants, we know that |KA| = Kn |A|, where n is the order of the determinant.

Here, n = 2, therefore, the answer is K2 |A|.

|5A| = 52|A|

|5A| = 52 × 5 = 125