The correct answer is Option (B). Explanation: $A A^{-1}=\mid$ $\operatorname{det}\left(A A^{-1}=I\right)$ $\operatorname{det}(A) \operatorname{det}\left(A^{-1}\right)=1$...

The correct answer is Option (B). Explanation: $\mid$ adj. $\left.A|=| A\right|^{n-1}=|A|^{2} \quad($ for $n=3)$

fig(1) A =     =  Now  =  =  L.H.S. =  =  =  =  =  =  = R.H.S. Now,   to find , multiplying  by              =   

fig(1)  A =      =  $A^{3}=A^{2} A=\left[\begin{array}{ccc}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2...

$A^{2}=A A=\left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}11 & 8 \\ 4 &...

$A^{2}=A A$ $A^{2}=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]=\left[\begin{array}{rr}8 & 5 \\ -5 &...

$A=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]$ $|\mathrm{A}|=\left|\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right|=1 \neq 0$ $\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}$...

Let A =    =    exists. A11 = , A12 = , A13 = , A21 = ,   A22 = , A23 = ,  A31 = , A32 = , A33 =   adj. A =   

Let A =    =    exists. A11 = ,  A12 = , A13 = , A21 = , A22 = , A23 = , A31 = , A32 = , A33 =   adj. A =   

Let $\mathrm{A}=\left[\begin{array}{ccc}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right]$ Therefore, $A^{-1}$ exists A11 = , A12 = , A13 = , A21 = , A22 = , A23 = ,...

Let $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$ Therefore, $A^{-1}$ exists Find adj A:  212.  \text { 212. }...

$\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$ Solution: Let $\mathrm{A}=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0...

Let $A=\left[ {{a}_{ij}} \right]$ be a square matrix of order n. The adjoint of a matrix A is the transporse of the cofactor matrix of A. It is denoted by adj A. Given...

Let $A=\left[\begin{array}{rr}2 & -2 \\ 4 & 3\end{array}\right]$ $|A|=\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]=14 \neq 0$ Since determinant of the matrix is not zero,...

. Let A =          =             adj. A =   A. (adj. A) =  =  =  ……….(i) Again  (adj. A). A =  =  =   ……….(ii) And  =  Also  =     ……….(iii)  From eq. (i), (ii) and (iii)  A. (adj. A) = (adj. A). A...

Let A =   adj. A =   A.(adj. A) =  =  =   …..(i) Again  (adj. A). A =  =  =     …..(ii) And  =  Again      …..(iii)  From eq. (i), (ii) and (iii)  A. (adj. A) = (adj. A). A...

Let $\mathrm{A}=\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right]$ Cofactors of the above matrix are $\mathrm{A}_{11}=+\left|\begin{array}{ll}3...

Let $A=$ $\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ Cofactors of the above matrix are 11 $A_{11}=4$ $A_{12}=-3$ $\mathrm{A}_{21}=-2$ $A_{22}=1$ adj....