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: $$A^{adj} = \begin{bmatrix} 3 & -6 & 3 \\ -4 & 8 & -4 \\ 1 & -2 & 1 \end{bmatrix}$$ 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: $$\lambda_1 = 1, \quad \lambda_2 = 2.$$ 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: $$\begin{bmatrix} 0 & 2 & 3 \\ 4 & 4 & 6 \\ 7 & 8 & 8 \end{bmatrix}x = 0$$ This system has the following solution: $$x_1 = -2t, \quad x_2 = t, \quad x_3 = 0$$ where $$t$$ is a free parameter. Therefore, the eigenvector corresponding to $\lambda_1$ is: $$\mathbf{v}_1 = \begin{bmatrix} -2t \\ t \\ 0 \end{bmatrix}$$ Similarly, for $\lambda_2 = 2$, we have: $$\begin{bmatrix} -1 & 2 & 3 \\ 4 & 3 & 6 \\ 7 & 8 & 7 \end{bmatrix}x = 0$$ This system has the following solution: $$x_1 = t, \quad x_2 = -t, \quad x_3 = 0$$ where $$t$$ is a free parameter. Therefore, the eigenvector corresponding to $\lambda_2$ is: $$\mathbf{v}_2 = \begin{bmatrix} t \\ -t \\ 0 \end{bmatrix}$$

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 $$=\lbrace 1(5 - 6) - 2(4 - 8) + 3(32 - 40) \rbrace$$

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

$$A^{−1} = \frac{A^{adj}}{|A|}$$

$$= \frac{1}{(-17)} \begin{bmatrix} -3 & 6 & -3 \\ 10 & -16 & 10 \\ -14 & 24 & -14 \end{bmatrix}$$

$$X = \frac{1}{-17} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \times \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$$

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

$$X = \frac{1}{-17} \begin{bmatrix} (1×1) + (2×4) + (3×7) \\ (1×2) + (2×5) + (3×8) \\ (1×3) + (2×6) + (3×9) \end{bmatrix}$$

$$X =\frac{1}{-17} \begin{bmatrix} 19 \\ 32 \\ 45 \end{bmatrix} = \begin{bmatrix} -\frac{19}{17} \\ - \frac{32}{17} \\ -\frac{45}{17} \end{bmatrix}$$

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. $$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$$

$$=\begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -12 \end{bmatrix} \xrightarrow{R_2+3R_1 \rightarrow R_2} $$

$$=\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 0 & -6 & -12 \end{bmatrix} \xrightarrow{6R_2+R_3 \rightarrow R_3}$$

$$=\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$ 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: $$\begin{align*} |A| &= \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix} \\ &= 1 \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} - 2 \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} + 3 \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} \\ &= 1(45 - 48) - 2(36 - 42) + 3(32 - 35) \\ &= -3 - 12 - 9 \\ &= -24 \end{align*}$$

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 $$A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$$, the determinant is: $$|A| = (2)(5) - (3)(4) = 2$$

and the adjoint is: $$A^{adj} = \begin{bmatrix} 5 & -3 \\ -4 & 2 \end{bmatrix}$$

Therefore, the inverse of $$A$$ is: $$A^{-1} = \frac{1}{2} \begin{bmatrix} 5 & -3 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} \frac{5}{2} & -\frac{3}{2} \\ -2 & 1 \end{bmatrix}$$

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 $$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$$, the characteristic polynomial is: $$p(x) = det\begin{bmatrix} 1-x & 2 & 3 \\ 4 & 5-x & 6 \\ 7 & 8 & 9-x \end{bmatrix}$$

$$\Rightarrow p(x)=$$

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

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