Inverse of a Matrix MCQ Quiz in मल्याळम - Objective Question with Answer for Inverse of a Matrix - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Let \({\rm{B}} = 3{\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&2&2\\ 2&1&{ - 2}\\ { - 2}&2&{ - 1} \end{array}} \right]\)

\(\left| {\rm{B}} \right| = \left( {8 - 1 + 8} \right) - \left( { - 4 - 4 - 4} \right) = 15 + 12 = 27\)

\(\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} 1&2 \end{array}}&2&1\\ {\begin{array}{*{20}{c}} { - 2}&{ - 1} \end{array}}&2&{ - 2}\\ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} 2\\ 1 \end{array}}&{\begin{array}{*{20}{c}} { - 2}\\ 2 \end{array}} \end{array}}&{\begin{array}{*{20}{c}} 1\\ 2 \end{array}}&{\begin{array}{*{20}{c}} 2\\ 1 \end{array}} \end{array}} \right]\\ {\rm{Adj}}\left( {\rm{B}} \right) = \left[ {\begin{array}{*{20}{c}} 3&6&{ - 6}\\ 6&3&6\\ 6&{ - 6}&{ - 3} \end{array}} \right]\\ {{\rm{B}}^{ - 1}} = \frac{1}{{27}} \times 3 \times \left[ {\begin{array}{*{20}{c}} 1&2&{ - 2}\\ 2&1&2\\ 2&{ - 2}&{ - 1} \end{array}} \right]\\ {\left( {3{\rm{A}}} \right)^{ - 1}} = \frac{1}{9}\left[ {3{\rm{\;}}{{\rm{A}}^{\rm{T}}}} \right]\\ {{\rm{A}}^{ - 1}}{\rm{\;}} = {\rm{\;}}{{\rm{A}}^{\rm{T}}} \end{array}\)

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:

The inverse of square matrix A exists only when |A| ≠ 0

For a square matrix A, A-1 = \(\frac{1}{\begin{vmatrix} A \end{vmatrix}}\)adj(A), where adj(A) is the adjoint of A.

Solution:

Given, \(A=\left[\begin{array}{rrr}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & 2 & 1\end{array}\right]\)

Determinant lAl = \(\left[\begin{array}{rrr}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & 2 & 1\end{array}\right]\)

= 1(3 - 0) - 2(- 1 - 0) - 2(- 2 - 0)

= 9 

∴ Inverse of matrix A exists.

In order to find adjoint of A ,lets find minor of matrix of A

M_A = \(\begin{bmatrix} \begin{vmatrix} 3 & 0\\ 2 & 1 \end{vmatrix} & \begin{vmatrix} -1 &0 \\ 0& 1 \end{vmatrix} & \begin{vmatrix} -1 & 3\\ 0 & 2 \end{vmatrix}\\ \begin{vmatrix} 2 & -2\\ 2 & 1 \end{vmatrix} & \begin{vmatrix} 1 & -2\\ 0 & 1 \end{vmatrix} & \begin{vmatrix} 1 & 2\\ 0 & 2 \end{vmatrix}\\ \begin{vmatrix} 2 &-2 \\ 3 & 0 \end{vmatrix} & \begin{vmatrix} 1 & -2\\ -1 & 0 \end{vmatrix} & \begin{vmatrix} 1 & 2\\ -1 & 3 \end{vmatrix} \end{bmatrix}\)

= \(\left[\begin{array}{rrr}3 & -1 & -2 \\ 6 & 1 & 2 \\ 6 & -2 & 5\end{array}\right]\)

C0-factors of A is 

C_A = \(\begin{bmatrix} (-1)^{1+1} (3) & (-1)^{1+2} (-1) &(-1)^{1+3} (-2)\\ (-1)^{1+4}(6) & (-1)^{1+5}(1) & (-1)^{1+6}(2) \\ (-1)^{1+7}(6) & (-1)^{1+8}(-2) & (-1)^{1+9}(5) \end{bmatrix}\)

= \(\left[\begin{array}{rrr}3 & 1 & -2 \\ -6 & 1 & -2 \\ 6 & 2 & 5\end{array}\right]\)

Taking Transpose,

C_A^T = \(\left[\begin{array}{rrr}3 & -6 & 6 \\ 1 & 1 & 2 \\ -2 & -2 & 5\end{array}\right]\)=adj(A)

As we know that,

A^-1 = \(\frac{1}{\begin{vmatrix} A \end{vmatrix}}\)adj(A)

= \(\frac{1}{9}\left[\begin{array}{rrr}3 & -6 & 6 \\ 1 & 1 & 2 \\ -2 & -2 & 5\end{array}\right]\) = \(\left[\begin{array}{rrr}\frac{1}{3} & \frac{-2}{3} & \frac{2}{3} \\ \frac{1}{9} & \frac{1}{9} & \frac{2}{9} \\ \frac{-2}{9} & \frac{-2}{9} & \frac{5}{9}\end{array}\right]\)

∴ The inverse of A is, A-1 = \(\left[\begin{array}{rrr}\frac{1}{3} & \frac{-2}{3} & \frac{2}{3} \\ \frac{1}{9} & \frac{1}{9} & \frac{2}{9} \\ \frac{-2}{9} & \frac{-2}{9} & \frac{5}{9}\end{array}\right]\)

The correct answer is Option 4.

Explanation:

\({A^{ - 1}} = \frac{{adj\left( A \right)}}{{\left| A \right|}}\;\)

\(\left| A \right| = \left| {\begin{array}{*{20}{c}} 0&1&0\\ 0&0&1\\ 1&0&0 \end{array}} \right|\)

|A|= 0 (0 - 0) – 1 (0 - 1) + 0 (0 - 0)

∴ |A| = 1

Now,

\(adj\left( A \right) = \left[ {\begin{array}{*{20}{c}} 0&0&1\\ 1&0&0\\ 0&1&0 \end{array}} \right]\)

\(\therefore {A^{ - 1}} = \frac{1}{{\left| A \right|}}adj\left( A \right)\) 

​a number

A determinant is a real number associated with every square matrix.

Concept:

The inverse of square matrix A exists only when |A| ≠ 0

For a square matrix A, A-1 = \(\frac{1}{\begin{vmatrix} A \end{vmatrix}}\)adj(A), where adj(A) is the adjoint of A.

Solution:

Given, \(A=\left[\begin{array}{rrr}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & 2 & 1\end{array}\right]\)

Determinant lAl = \(\left[\begin{array}{rrr}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & 2 & 1\end{array}\right]\)

= 1(3 - 0) - 2(- 1 - 0) - 2(- 2 - 0)

= 9 

∴ Inverse of matrix A exists.

In order to find adjoint of A ,lets find minor of matrix of A

M_A = \(\begin{bmatrix} \begin{vmatrix} 3 & 0\\ 2 & 1 \end{vmatrix} & \begin{vmatrix} -1 &0 \\ 0& 1 \end{vmatrix} & \begin{vmatrix} -1 & 3\\ 0 & 2 \end{vmatrix}\\ \begin{vmatrix} 2 & -2\\ 2 & 1 \end{vmatrix} & \begin{vmatrix} 1 & -2\\ 0 & 1 \end{vmatrix} & \begin{vmatrix} 1 & 2\\ 0 & 2 \end{vmatrix}\\ \begin{vmatrix} 2 &-2 \\ 3 & 0 \end{vmatrix} & \begin{vmatrix} 1 & -2\\ -1 & 0 \end{vmatrix} & \begin{vmatrix} 1 & 2\\ -1 & 3 \end{vmatrix} \end{bmatrix}\)

= \(\left[\begin{array}{rrr}3 & -1 & -2 \\ 6 & 1 & 2 \\ 6 & -2 & 5\end{array}\right]\)

C0-factors of A is 

C_A = \(\begin{bmatrix} (-1)^{1+1} (3) & (-1)^{1+2} (-1) &(-1)^{1+3} (-2)\\ (-1)^{1+4}(6) & (-1)^{1+5}(1) & (-1)^{1+6}(2) \\ (-1)^{1+7}(6) & (-1)^{1+8}(-2) & (-1)^{1+9}(5) \end{bmatrix}\)

= \(\left[\begin{array}{rrr}3 & 1 & -2 \\ -6 & 1 & -2 \\ 6 & 2 & 5\end{array}\right]\)

Taking Transpose,

C_A^T = \(\left[\begin{array}{rrr}3 & -6 & 6 \\ 1 & 1 & 2 \\ -2 & -2 & 5\end{array}\right]\)=adj(A)

As we know that,

A^-1 = \(\frac{1}{\begin{vmatrix} A \end{vmatrix}}\)adj(A)

= \(\frac{1}{9}\left[\begin{array}{rrr}3 & -6 & 6 \\ 1 & 1 & 2 \\ -2 & -2 & 5\end{array}\right]\) = \(\left[\begin{array}{rrr}\frac{1}{3} & \frac{-2}{3} & \frac{2}{3} \\ \frac{1}{9} & \frac{1}{9} & \frac{2}{9} \\ \frac{-2}{9} & \frac{-2}{9} & \frac{5}{9}\end{array}\right]\)

∴ The inverse of A is, A-1 = \(\left[\begin{array}{rrr}\frac{1}{3} & \frac{-2}{3} & \frac{2}{3} \\ \frac{1}{9} & \frac{1}{9} & \frac{2}{9} \\ \frac{-2}{9} & \frac{-2}{9} & \frac{5}{9}\end{array}\right]\)

Concept:

Determinant of (A^-1) = \(\frac{1}{\rm Determinant~of~A}\)

Where, Determinant of A = 1(4 × 1 - 5 × 0) - 2(0 - 0) + 3(0 - 0) = 4

Calculation:

Given:

Determinant of A = 4 

So, Determinant of (A-1) = \(\frac{1}{\rm Determinant~of~A}\) = \(\frac{1}{4}\)

Additional Information

A^-1 = \(\frac{\rm adj A}{\rm Determinant~of~A}\)

adj.A = Transverse of matrix ^{\(\left[ {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}&{{A_{13}}}\\ {{A_{21}}}&{{A_{22}}}&{{A_{23}}}\\ {{A_{31}}}&{{A_{32}}}&{{A_{33}}} \end{array}} \right]\)}

A₁₁ = A_ij = (-1)ⁿ × (cross multiplication of remaining column and row) = (-1)ⁿ × \(\left( {\begin{array}{*{20}{c}} {{A_{22}}}&{{A_{23}}}\\ {{A_{32}}}&{{A_{33}}} \end{array}} \right)\)

Where, n = (i + j)

We follow the same procedure for the A₁₁, A₁₂, A₁₃, .......as so on. Also find the determinant of matrix A. Then applies the given formula after that we finally obtain the A^-1.

Concept:

Apply element-wise inversion i.e multiply by the inverse of the same element to make it an identity matrix until the desired objective is reached.

Calculation:

Let (BA)^-1 = K

Multiply by the BA on both sides, in the same order

(BA)(BA)^-1 = (BA)K

I = BAK   {∵ (A)(A^-1) = I}

Multiply B^-1 on both the sides

B^-1I = B^-1BAK

B^-1 = AK

Multiply A^-1 on both sides

A^-1B^-1 = A^-1AK

A^-1B^-1 = K

so, (BA)^-1 = A^-1B^-1  {∵ (BA)^-1 = K}

Given: \(\left( {\begin{array}{*{20}{c}} 3&1\\ 7&5 \end{array}} \right)\)

Concept used:

The inverse of matrix (A) = adjoint of A / det(A) 

or \(\left( {\begin{array}{*{20}{c}} a&b\\ c&d \end{array}} \right)\)^(-1) = \(\left( {\begin{array}{*{20}{c}} d&-b\\ -c&a \end{array}} \right)\)/ (Determinant A)

Calculations:

The inverse of \(\left( {\begin{array}{*{20}{c}} 3&1\\ 7&5 \end{array}} \right)\) = \(\left( {\begin{array}{*{20}{c}} 5&-1\\ -7&3 \end{array}} \right)\)/ (Determinant of given matrix)

⇒ \(\left( {\begin{array}{*{20}{c}} 5&-1\\ -7&3 \end{array}} \right)\)/ (15 - 7)

⇒ 1/8 ×  \(\left( {\begin{array}{*{20}{c}} 5&-1\\ -7&3 \end{array}} \right)\)

Hence, The correct option is 3.

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

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

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