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

Let $\mathrm{B}=3\mathrm{A}=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ -2& 2& -1\end{array}\right]$

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

$\begin{array}{l}\left[\begin{array}{ccc}\begin{array}{cc}1& 2\end{array}& 2& 1\\ \begin{array}{cc}-2& -1\end{array}& 2& -2\\ \begin{array}{cc}\begin{array}{c}2\\ 1\end{array}& \begin{array}{c}-2\\ 2\end{array}\end{array}& \begin{array}{c}1\\ 2\end{array}& \begin{array}{c}2\\ 1\end{array}\end{array}\right]\\ \mathrm{A}\mathrm{d}\mathrm{j}\left(\mathrm{B}\right)=\left[\begin{array}{ccc}3& 6& -6\\ 6& 3& 6\\ 6& -6& -3\end{array}\right]\\ {\mathrm{B}}^{-1}=\frac{1}{27}\times 3\times \left[\begin{array}{ccc}1& 2& -2\\ 2& 1& 2\\ 2& -2& -1\end{array}\right]\\ {\left(3\mathrm{A}\right)}^{-1}=\frac{1}{9}\left[3{\mathrm{A}}^{\mathrm{T}}\right]\\ {\mathrm{A}}^{-1}={\mathrm{A}}^{\mathrm{T}}\end{array}$

Explanation:

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

Now its transpose will be 

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

The product will be 

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

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

⇒ AA^T = I;

Concept:

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

For a square matrix A, A-1 = $\frac{1}{\left|\begin{array}{c}A\end{array}\right|}$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 = $\left[\begin{array}{ccc}\left|\begin{array}{cc}3& 0\\ 2& 1\end{array}\right|& \left|\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right|& \left|\begin{array}{cc}-1& 3\\ 0& 2\end{array}\right|\\ \left|\begin{array}{cc}2& -2\\ 2& 1\end{array}\right|& \left|\begin{array}{cc}1& -2\\ 0& 1\end{array}\right|& \left|\begin{array}{cc}1& 2\\ 0& 2\end{array}\right|\\ \left|\begin{array}{cc}2& -2\\ 3& 0\end{array}\right|& \left|\begin{array}{cc}1& -2\\ -1& 0\end{array}\right|& \left|\begin{array}{cc}1& 2\\ -1& 3\end{array}\right|\end{array}\right]$

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

C0-factors of A is 

C_A = $\left[\begin{array}{ccc}(-1{)}^{1+1}(3)& (-1{)}^{1+2}(-1)& (-1{)}^{1+3}(-2)\\ (-1{)}^{1+4}(6)& (-1{)}^{1+5}(1)& (-1{)}^{1+6}(2)\\ (-1{)}^{1+7}(6)& (-1{)}^{1+8}(-2)& (-1{)}^{1+9}(5)\end{array}\right]$

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

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 = $\frac{1}{\left|\begin{array}{c}A\end{array}\right|}$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:

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

$\left|A\right|=\left|\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right|$

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

∴ |A| = 1

Now,

$adj\left(A\right)=\left[\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right]$

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

​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 = $\frac{1}{\left|\begin{array}{c}A\end{array}\right|}$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 = $\left[\begin{array}{ccc}\left|\begin{array}{cc}3& 0\\ 2& 1\end{array}\right|& \left|\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right|& \left|\begin{array}{cc}-1& 3\\ 0& 2\end{array}\right|\\ \left|\begin{array}{cc}2& -2\\ 2& 1\end{array}\right|& \left|\begin{array}{cc}1& -2\\ 0& 1\end{array}\right|& \left|\begin{array}{cc}1& 2\\ 0& 2\end{array}\right|\\ \left|\begin{array}{cc}2& -2\\ 3& 0\end{array}\right|& \left|\begin{array}{cc}1& -2\\ -1& 0\end{array}\right|& \left|\begin{array}{cc}1& 2\\ -1& 3\end{array}\right|\end{array}\right]$

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

C0-factors of A is 

C_A = $\left[\begin{array}{ccc}(-1{)}^{1+1}(3)& (-1{)}^{1+2}(-1)& (-1{)}^{1+3}(-2)\\ (-1{)}^{1+4}(6)& (-1{)}^{1+5}(1)& (-1{)}^{1+6}(2)\\ (-1{)}^{1+7}(6)& (-1{)}^{1+8}(-2)& (-1{)}^{1+9}(5)\end{array}\right]$

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

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 = $\frac{1}{\left|\begin{array}{c}A\end{array}\right|}$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) = $\frac{1}{\mathrm{D}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{o}\mathrm{f}\mathrm{A}}$

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) = $\frac{1}{\mathrm{D}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{o}\mathrm{f}\mathrm{A}}$ = $\frac{1}{4}$

Additional Information

A^-1 = $\frac{\mathrm{a}\mathrm{d}\mathrm{j}\mathrm{A}}{\mathrm{D}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{o}\mathrm{f}\mathrm{A}}$

adj.A = Transverse of matrix ^{$\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]$}

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

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: $\left(\begin{array}{cc}3& 1\\ 7& 5\end{array}\right)$

Concept used:

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

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

Calculations:

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

⇒ $\left(\begin{array}{cc}5& -1\\ -7& 3\end{array}\right)$/ (15 - 7)

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

Hence, The correct option is 3.

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

$\therefore \left|A\right|=k^3$

$\begin{array}{l}adjA=\left(\begin{array}{ccc}{k}^2& 0& 0\\ 0& k^2& 0\\ 0& 0& k^2\end{array}\right)\\ A^{-1}=\frac{1}{\left|A\right|}\left(adjA\right)=\frac{1}{k^3}\left(\begin{array}{ccc}{k}^2& 0& 0\\ 0& k^2& 0\\ 0& 0& k^2\end{array}\right)\\ \frac{1}{k}\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)=\frac{1}{k}\mathrm{I}\end{array}$