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

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

Question

(C) First,we calculate the determinant of $A$: $|A| = 1(1-3) - 2(2-1) + 3(6-1) = 1(-2) - 2(1) + 3(5) = -2 - 2 + 15 = 11$. Next,we calculate the determ

Vedclass · Accepted Answer

$198$

Vedclass · Answer

(C) First,we calculate the determinant of $A$: $|A| = 1(1-3) - 2(2-1) + 3(6-1) = 1(-2) - 2(1) + 3(5) = -2 - 2 + 15 = 11$. Next,we calculate the determinant of $B$: $|B| = 2(4-8) - 3(6-4) + 4(12-4) = 2(-4) - 3(2) + 4(8) = -8 - 6 + 32 = 18$. Since $|AB| = |A| 	imes |B|$,we have $|AB| = 11 	imes 18 = 198$. Using the property of the adjoint matrix,$|\operatorname{Adj}(M)| = |M|^{n-1}$ where $n$ is the order of the matrix. Here $n=3$,so $|\operatorname{Adj}(AB)| = |AB|^{3-1} = |AB|^2$. Thus,$\sqrt{|\operatorname{Adj}(AB)|} = \sqrt{|AB|^2} = |AB| = 198$.

If $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{bmatrix}$,then $\sqrt{|\operatorname{Adj}(AB)|} = $

Similar Questions

If $A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}$,then $A^T \cdot A^{-1} = $

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:

If $A$ is a singular matrix,then $\text{adj } A$ is

${\left[ {\begin{array}{*{20}{c}}{ - 6}&5\\{ - 7}&6\end{array}} \right]^{ - 1}}$ =

The inverse matrix of $\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module