Determinants & Inverse of Matrix Test - 1

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.

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

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

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

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 |.

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 \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_2 & y_2 & 1 \end{vmatrix}\)

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

On squaring both side, 

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

 ⇒ \(\dfrac{3a^4}{4}\)                  

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

Concept:

The adjoint of a square matrix is the transpose of its cofactor matrix.

The determinant of a square matrix is the same as the determinant of its transpose.

Determinant of the adjoint of matrix = (determinant of the matrix)\(\rm ^{n-1}\), where n = order of the matrix

⇒ |Adj A| =  |A|^n-1

Calculation:

Given that determinant of the matrix

 \(Δ=\rm \begin{vmatrix}a_1&b_1&c_1\\\ a_2 &b_2&c_2\\\ a_3&b_3&c_3\end{vmatrix}\) 

and order = n = 3

We know that determinant formed by replacing each of its elements by its cofactor = determinant of the transpose of adjoint

⇒ \(\rm \begin{vmatrix}A_1&B_1&C_1\\\ A_2 &B_2&C_2\\ A_3&B_3&C_3\end{vmatrix}\) = (Δ)^3-1

⇒ \(\rm \begin{vmatrix}A_1&B_1&C_1\\\ A_2 &B_2&C_2\\ A_3&B_3&C_3\end{vmatrix}\) = Δ²

∴ The value of the determinant is Δ2.

Concept:

Geometric Progression:

• The ratio between any two consecutive terms of a geometric progression is a fixed constant, called the common ratio (r) of the progression.

Logarithms:

• log a - log b = \(\rm \log {a\over b}\).

Determinants:

• A linear combination of the rows/columns does not affect the value of the determinant.
• If two rows/columns of a given matrix are interchanged, then the value of the determinant gets multiplied by -1.
• If a row/column of a given matrix is multiplied by a scalar k, then the value of the determinant is also multiplied by k.

Calculation:

Let the common ratio of the given G.P. be r.

Let D = \(\begin{vmatrix} \rm \ln a_1 & \rm \ln a_2 & \rm \ln a_3 \\ \rm \ln a_4 & \rm \ln a_5 & \rm \ln a_6 \\ \rm \ln a_7 & \rm \ln a_8 & \rm \ln a_9 \end{vmatrix}\).

Using C₁ → C₁ - C₂ and C2 → C2 - C3, we get:

⇒ D = \(\begin{vmatrix} \rm \ln a_1 -\ln a_2& \rm \ln a_2 -\ln a_3& \rm \ln a_3 \\ \rm \ln a_4 -\ln a_5& \rm \ln a_5 -\ln a_6& \rm \ln a_6 \\ \rm \ln a_7-\ln a_8 & \rm \ln a_8-\ln a_9 & \rm \ln a_9 \end{vmatrix}\)

⇒ D = \(\begin{vmatrix} \rm \ln {a_1 \over a_2}& \rm \ln {a_2 \over a_3}& \rm \ln a_3 \\ \rm \ln {a_4 \over a_5}& \rm \ln {a_5 \over a_6}& \rm \ln a_6 \\ \rm \ln {a_7 \over a_8} & \rm \ln {a_8 \over a_9} & \rm \ln a_9 \end{vmatrix}\)

⇒ D = \(\begin{vmatrix} \rm \ln r& \rm \ln r& \rm \ln a_3 \\ \rm \ln r& \rm \ln r& \rm \ln a_6 \\ \rm \ln r & \rm \ln r & \rm \ln a_9 \end{vmatrix}\)

Using C₁ → C₁ - C₂, we get:

⇒ D = \(\begin{vmatrix} 0& \rm \ln r& \rm \ln a_3 \\0& \rm \ln r& \rm \ln a_6 \\ 0 & \rm \ln r & \rm \ln a_9 \end{vmatrix}\)

Expanding along C₁, we get:

⇒ D = 0.

Given:

\(\left| {\;\begin{array}{*{20}{c}} {x + 2}&2&2\\ 2&{x + 2}&2\\ 2&2&{x + 2} \end{array}} \right|=0\)

C1 → C1 + C2 + C3

\(\left| {\begin{array}{*{20}{c}} {x + 6}&2&2\\ {x + 6}&{x + 2}&2\\ {x + 6}&2&{x + 2} \end{array}} \right|=0\)

R1 → R1 - R2

\(\left( {x + 6} \right)\left| {\begin{array}{*{20}{c}} 1&2&2\\ 1&{x + 2}&2\\ 1&2&{x + 2} \end{array}} \right|=0\)

\(\left( {x + 6} \right)\left| {\begin{array}{*{20}{c}} 0&{ - x}&0\\ 1&{x + 2}&2\\ 1&2&{x + 2} \end{array}} \right|=0\)

\(\left( {x + 6} \right)\left( x \right)\left| {\begin{array}{*{20}{c}} 1&2\\ 1&{x + 2} \end{array}} \right|=0\)

\(\left( {x + 6} \right) \cdot x \cdot x = 0\)

x = 0, 0, -6

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;