JEE Mains (Engineering)

1. To find the inverse of a matrix, you can use the adjoint formula. The adjoint of a matrix is the transpose of its cofactor matrix. For the given matrix $$A$$, the adjoint is: $$A^{adj}=\left[\begin{array}{ccc}3& -6& 3\\ -4& 8& -4\\ 1& -2& 1\end{array}\right]$$ Therefore, the inverse of $$A$$ is: $$A^{-1}=\frac{A^{adj}}{|A|}$$ where $$|A|$$ is the determinant of $$A$$. In this case, the determinant is $$|A|=0$$, so the matrix is not invertible.

2. To find the eigenvalues and eigenvectors of a matrix, you can use the characteristic polynomial. The characteristic polynomial of matrix $$A$$ is: $$p(x)=det(A-xI)=(1-x)(4-x)-2(3-x)=x^2-5x+2$$ The roots of the characteristic polynomial are the eigenvalues of the matrix. In this case, the eigenvalues are: $${\lambda }_1=1,{\lambda }_2=2.$$ To find the eigenvectors, you can solve the system of linear equations $(A-\lambda I)x=0$ for each eigenvalue. For ${\lambda }_1=1$, we have: $$\left[\begin{array}{ccc}0& 2& 3\\ 4& 4& 6\\ 7& 8& 8\end{array}\right]x=0$$ This system has the following solution: $$x_1=-2t,x_2=t,x_3=0$$ where $$t$$ is a free parameter. Therefore, the eigenvector corresponding to ${\lambda }_1$ is: $${\mathbf{v}}_1=\left[\begin{array}{c}-2t\\ t\\ 0\end{array}\right]$$ Similarly, for ${\lambda }_2=2$, we have: $$\left[\begin{array}{ccc}-1& 2& 3\\ 4& 3& 6\\ 7& 8& 7\end{array}\right]x=0$$ This system has the following solution: $$x_1=t,x_2=-t,x_3=0$$ where $$t$$ is a free parameter. Therefore, the eigenvector corresponding to ${\lambda }_2$ is: $${\mathbf{v}}_2=\left[\begin{array}{c}t\\ -t\\ 0\end{array}\right]$$

3. To solve the system of linear equations, you can use the following formula $$X=A^{-1}B$$

Here the inverse of (A) is not possible as the determinant of matrix is zero.

$$A^{-1}=\frac{A^{adj}}{|A|}$$

Determinant of Matrix A $$=\{1(5-6)-2(4-8)+3(32-40)\}$$

$$=-1+8-24=-17\ne 0$$

$$A^{-1}=\frac{A^{adj}}{|A|}$$

$$=\frac{1}{(-17)}\left[\begin{array}{ccc}-3& 6& -3\\ 10& -16& 10\\ -14& 24& -14\end{array}\right]$$

$$X=\frac{1}{-17}\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]\times \left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$$

We observe that the answer obtained are not integral. It is because 1/17 is an irrational number.

$$X=\frac{1}{-17}\left[\begin{array}{c}(1\times 1)+(2\times 4)+(3\times 7)\\ (1\times 2)+(2\times 5)+(3\times 8)\\ (1\times 3)+(2\times 6)+(3\times 9)\end{array}\right]$$

$$X=\frac{1}{-17}\left[\begin{array}{c}19\\ 32\\ 45\end{array}\right]=\left[\begin{array}{c}-\frac{19}{17}\\ -\frac{32}{17}\\ -\frac{45}{17}\end{array}\right]$$

4. To find the rank of a matrix, you can use row reduction. Row reduction is a process of transforming a matrix into an equivalent matrix in which all the elements below the main diagonal are zero and the elements on the main diagonal are all 1. $$A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$$

$$=\left[\begin{array}{ccc}1& 2& 3\\ 0& -3& -6\\ 0& -6& -12\end{array}\right]\stackrel{R_2+3R_1\to R_2}{\to }$$

$$=\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 0& -6& -12\end{array}\right]\stackrel{6R_2+R_3\to R_3}{\to }$$

$$=\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$$ The rank of the matrix is the number of nonzero rows in the reduced row echelon form. In this case, the rank is 2.

5. To find the determinant of a matrix, you can use the Laplace expansion. The Laplace expansion is a method for computing the determinant of a matrix by expanding it along a row or column. For the given matrix $$A$$, the determinant is: $$\begin{array}{rl}|A|& =\left|\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right|\\ & =1\left|\begin{array}{cc}5& 6\\ 8& 9\end{array}\right|-2\left|\begin{array}{cc}4& 6\\ 7& 9\end{array}\right|+3\left|\begin{array}{cc}4& 5\\ 7& 8\end{array}\right|\\ & =1(45-48)-2(36-42)+3(32-35)\\ & =-3-12-9\\ & =-24\end{array}$$

CBSE Board Exams (Science)

1. To find the inverse of a matrix, you can use the formula: $$A^{-1}=\frac{1}{|A|}{A}^{adj}$$ where $$|A|$$ is the determinant of $$A$$ and $$A^{adj}$$ is the adjoint of $$A$$.

For the given matrix $$A=\left[\begin{array}{cc}2& 3\\ 4& 5\end{array}\right]$$, the determinant is: $$|A|=(2)(5)-(3)(4)=2$$

and the adjoint is: $$A^{adj}=\left[\begin{array}{cc}5& -3\\ -4& 2\end{array}\right]$$

Therefore, the inverse of $$A$$ is: $$A^{-1}=\frac{1}{2}\left[\begin{array}{cc}5& -3\\ -4& 2\end{array}\right]=\left[\begin{array}{cc}\frac{5}{2}& -\frac{3}{2}\\ -2& 1\end{array}\right]$$

2. To find the eigenvalues and eigenvectors of a matrix, you can use the characteristic polynomial: $$p(x)=det(A-xI)=0$$ where $$I$$ is the identity matrix. For the given matrix $$A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$$, the characteristic polynomial is: $$p(x)=det\left[\begin{array}{ccc}1-x& 2& 3\\ 4& 5-x& 6\\ 7& 8& 9-x\end{array}\right]$$

$$\Rightarrow p(x)=$$

$$(1-x)\{(5-x)(9-x)-(48-x)\}-2\{4(9-x)-6(7)\}+$$ $$(3)\{4(5-x)-(48-x)\}$$

$$(x-1)\lbrace (5-x)(9-x