Mathematics Test - 15

Singular Matrix: 

• A square matrix that is not invertible is called singular matrix.
• A square matrix is singular if and only if its determinant is 0.
• In square matrix number of columns equal to the number of rows.
• A matrix is singular matrix if its det = 0
• If the determinant of a matrix is not equal to zero, then the matrix is called a non-singular matrix. i.e., det ≠ 0.
• det(AB) = det(A) × det(B)
• det(A) × det(B) = det(B) × det(A)
• \({\rm{det}}\left( {{{\rm{A}}^{ - 1}}} \right) = {\rm{\;}}\frac{1}{{{\rm{detA\;}}}} \Rightarrow \;{\rm{det}}\left( {{{\rm{A}}^{ - 1}}} \right) \times {\rm{\;det}}\left( {\rm{A}} \right) = 1\)

Concept:

Singular matrix: If the matrix is singular, then it's determinant = 0, and hence inverse does not exist. 

Given:

A is a singular matrix.

As we know, AA-1 = I

\( ⇒ {\rm{A}} × \left( {\frac{{{\rm{Adj\;A}}}}{{\det {\rm{A}}}}} \right) = {\rm{I}}\)

A (Adj A) = det A × I 

= 0

A (Adj A) is a null matrix

If A is an n x n non-singular matrix, then |Adj A| is |A|n-1

Concept:

Determinant of the product of matrices is equal to the product of determinants of the matrices.

i.e. |AB| = |A| |B|.

Calculation:

Given that \(\rm A=\begin{bmatrix} 1& 3\\ 5& 2\end{bmatrix}\) and \(\rm B=\begin{bmatrix} 3&2\\ 1& 3\end{bmatrix}\)

Now, |AB| = |A| |B| = (2 - 15)(9 - 2) = -91.

Additional Information:

The matrix product AB will be \(\rm \begin{bmatrix} 1& 3\\ 5& 2\end{bmatrix}\begin{bmatrix} 3& 2\\ 1& 3\end{bmatrix}\) = \(\begin{bmatrix} 3+3& 2+9\\ 15+2& 10+6\end{bmatrix}\) = \(\begin{bmatrix} 6& 11\\ 17& 16\end{bmatrix}\).

And |AB| = 96 - 187 = -91.

CONCEPT:

Since A is orthogonal,

⇒ AA^T = I

CALCULATION:

Given: AAT = I

⇒ |AA^T| = |I| 

⇒ |A||AT| =  |I| = 1

The determinant of the transpose of a square matrix is equal to the determinant of the matrix, that is, |A^T| = |A|

⇒ |A|² = 1

⇒ |A| = +1

Concept: 

If two linear equations

a1​x + b1y ​= c1​ and a2​x + b2y ​= c2​. Then,

(a) If a1​​/a2 = b1​​/b2 ​​= c1​​/c2 ​​, the system is consistent and has infinitely many solutions.

(b) If a1​​/a2 = b1​​/b2 ≠ c1​​/c2 the system has no solution and is inconsistent

Calculation:

The equations 2x - ky + 7 = 0 and 6x - 12y + 15 = 0

2/6 = - k/-12 ≠ 7/15

Using, 2/6 = - k/-12

⇒ k =  24/6 =  4

∴ At k =  4, The equations 2x - ky + 7 = 0 and 6x - 12y + 15 = 0 have no solution 

Concept:

The area of a triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃), is given by the expression

\(\rm Area = \frac{1}{2}\begin{vmatrix} x_{1} &x_{2} & 1\\ y_{1}& y_{2} &1 \\ z_{1}&z_{2} & 1 \end{vmatrix}\)  

Calculation:

Given, Area of triangle, A = 4 sq. unit and vertices (K, 0), (4, 0), (0, 2) 

since the area is always Positive But the determinant can be both positive and negative.  

∴ Δ = ± 4 . 

⇒ ± 4 = \(\rm = \frac{1}{2}\begin{vmatrix} k &0 & 1\\ 4& 0 &1 \\ 0&2 & 1 \end{vmatrix}\)  

⇒ ± 4  \(\rm = \frac{1}{2} \left [ k(0-2) - 0 (4-0)+1(8-0)\right ]\) 

⇒ ± 8 = -2k + 8 

So, 8 = -2k + 8 or  -8 = -2k +8 

⇒ k = 0 or  8  .

The correct option is 2.

Concept:

Two matrices A and B are said to be equal if:

• They are of same order
• The corresponding elements of the matrices are same.

Calculation:

Comparing the corresponding elements:

M₁₁ : a – b = -1 ⇒ b = a + 1

M₂₁ : 2a – b = 0 ⇒ b = 2a

⇒ 2a = a + 1 ⇒ a = 1, b = 2

M₁₂ : 2a + c = 5 ⇒ 2 × 1 + c = 5 ⇒ c = 3

M₂₂ : 3c + d = 13 ⇒ 3 × 3 + d = 13 ⇒ d = 4

Concept

Let the system of equations be,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

\(\Rightarrow \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_1}}&{{{\rm{b}}_1}}&{{{\rm{c}}_1}}\\ {{{\rm{a}}_2}}&{{{\rm{b}}_2}}&{{{\rm{c}}_2}}\\ {{{\rm{a}}_3}}&{{{\rm{b}}_3}}&{{{\rm{c}}_3}} \end{array}} \right]{\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {{{\rm{d}}_1}}\\ {{{\rm{d}}_2}}\\ {{{\rm{d}}_3}} \end{array}} \right]\)

⇒ AX = B

⇒ X = A^-1 B = \({\rm{\;}}\frac{{{\rm{adj\;}}\left( {\rm{A}} \right)}}{{\det {\rm{\;}}({\rm{A}})}}{\rm{\;B}}\)

⇒ If det (A) ≠ 0, system is consistent having unique solution.

Calculation:

Given system of linear equation are kx + y + z = 1, x + ky + z = 1 and x + y + kz = 1

Let A = \(\left[ {\begin{array}{*{20}{c}} {\rm{k}}&1&1\\ 1&{\rm{k}}&1\\ 1&1&{\rm{k}} \end{array}} \right]\)

det (A) = |A| = k (k² – 1) – 1(k -1) + 1 (1 – k)

⇒ |A| = k³ – k – k + 1 + 1 – k = k³ – 3k + 2

For unique solution,

det (A) ≠ 0

⇒ k³ – 3k + 2 ≠ 0

⇒ (k – 1) (k² + k - 2) ≠ 0

⇒ (k – 1) (k – 1) (k + 2) ≠ 0

∴ k ≠ 1 and k ≠ -2

Calculation:

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

R₁ = R₁ - R₂

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

R₂ = R₂ - R₃

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

Now, Expanding along R₁, we get

⇒ x [y(1 + z) - (-z)] - (-y) [0 - (-z)] + 0 = 0

⇒ x [y + yz + z] + y [z] = 0

⇒ xy + xyz + xz + yz = 0

Dividing by xyz

⇒ \(\rm {1\over z}+ 1 +{1\over y}+ {1\over x} = 0\)

⇒ \(\boldsymbol{\rm x^{-1}+ y^{-1}+ z^{-1} = -1}\)

Concept:

Let, A = [a_ij] be a square matrix.

To compute the minor, M_ij, and the cofactor, C_ij, we find the determinant of the given matrix with row i and column j removed.

So,  C_ij = (-1)^i+j M_ij , where, M_ij = Minor of matrix A

Calculation:

Given:

\( \begin{bmatrix} 0 & 4 & 6\\ -4 & 0&-2 \\ -6& 2 & 0 \end{bmatrix}\)

Removing the 2nd row and 3rd column we obtain the determinant as

\(\begin{vmatrix} 0 & 4\\ -6 & 2\end{vmatrix}\)

= 0 × 2 - (-6) × 4

= 24