If A = begin{bmatrix} 1 & 2 & 3 1 & 1 & a 2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The determinant of matrix $A$ is given by $|A| = 1(7 - 4a) - 2(7 - 2a) + 3(4 - 2) = 7 - 4a - 14 + 4a + 6 = -1$. Since $B = A^{-1}$,we know that $B

Vedclass · Accepted Answer

$a = 5, b = -7$

Vedclass · Answer

(D) The determinant of matrix $A$ is given by $|A| = 1(7 - 4a) - 2(7 - 2a) + 3(4 - 2) = 7 - 4a - 14 + 4a + 6 = -1$. Since $B = A^{-1}$,we know that $B = \frac{1}{|A|} 	ext{adj}(A)$. Given $|A| = -1$,the element $B_{ij} = \frac{C_{ji}}{|A|} = -C_{ji}$,where $C_{ji}$ is the cofactor of the element $A_{ji}$. For element $B_{13} = b$,we have $b = -C_{31}$. The cofactor $C_{31} = (-1)^{3+1} \begin{vmatrix} 2 & 3 \ 1 & a \end{vmatrix} = 2a - 3$. Thus,$b = -(2a - 3) = 3 - 2a$,which implies $2a + b = 3$. For element $B_{21} = -3$,we have $-3 = -C_{12}$. The cofactor $C_{12} = (-1)^{1+2} \begin{vmatrix} 1 & a \ 2 & 7 \end{vmatrix} = -(7 - 2a) = 2a - 7$. Thus,$-3 = -(2a - 7) = 7 - 2a$,which implies $2a = 10$,so $a = 5$. Substituting $a = 5$ into $2a + b = 3$,we get $2(5) + b = 3$,so $10 + b = 3$,which gives $b = -7$.

If $A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & a \\ 2 & 4 & 7 \end{bmatrix}$ and $B = \begin{bmatrix} 13 & 2 & b \\ -3 & -1 & 2 \\ -2 & 0 & 1 \end{bmatrix}$ where matrix $B$ is the inverse of matrix $A$,then the values of $a$ and $b$ are:

Similar Questions

Given $A = \begin{bmatrix} x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z \end{bmatrix}$,$xyz = 60$ and $8x + 4y + 3z = 20$,then $A \cdot (\text{adj } A)$ is equal to

If $A$ is an invertible matrix of order $2$,then $\det(A^{-1})$ is equal to

Let $A = \begin{bmatrix} x + \lambda & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda \end{bmatrix}$,then $A^{-1}$ exists if

Find the adjoint of the matrix: $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

Let $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta \end{bmatrix}$,then the inverse of $A$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module