78828
If \(A=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]\) and \(B=\left[\begin{array}{ll}1 & 0 \\ 5 & 1\end{array}\right]\), then value of \(\alpha\) for which \(\mathrm{A}^{2}=\mathrm{B}\), is
1 1
2 -1
3 4
4 No real values
Explanation:
(D) : Given that, \(\mathrm{A}=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]\) and \(\mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 5 & 1\end{array}\right]\) \(\Rightarrow\left[\begin{array}{cc} \alpha & 0 \\ 1 & 1 \end{array}\right]\left[\begin{array}{ll} \alpha & 0 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 5 & 1 \end{array}\right]\) \(\Rightarrow\left[\begin{array}{cc} \alpha^{2} & 0 \\ \alpha+1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 5 & 1 \end{array}\right]\) On comparing corresponding elements on both the side we get - \(\Rightarrow \alpha^{2}=1, \alpha+1=5\) \(\Rightarrow \alpha= \pm 1, \alpha=4\) \(\because\) There is no common value \(\therefore\) There is no real value of \(\alpha\) for which \(\mathrm{A}^{2}=\mathrm{B}\)
BITSAT-2005
Matrix and Determinant
78829
Let \(A=\left(\begin{array}{ccc}1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6\end{array}\right)\), then the values of \(t\) for which inverse of \(A\) does not exist
1 \(-2,1\)
2 3,2
3 \(2,-3\)
4 \(3,-1\)
Explanation:
(C) : Given that, \(A=\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6 \end{array}\right]\) We know that inverse of A does not exist only when \(|\mathrm{A}|=0\) \(\left|\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & \mathrm{t} \\ 4 & 7-\mathrm{t} & -6 \end{array}\right|=0\) \(1(-30-t(7-t))-3(-12-4 t)+2(2(7-t)-20)=0\) \(-30-7 \mathrm{t}+\mathrm{t}^{2}+36+12 \mathrm{t}+2(14-2 \mathrm{t}-20)=0\) \(-30-7 \mathrm{t}+\mathrm{t}^{2}+36+12 \mathrm{t}+2(-2 \mathrm{t}-6)=0\) \(-30-7 \mathrm{t}+\mathrm{t}^{2}+36+12 \mathrm{t}-12-4 \mathrm{t}=0\) \(\mathrm{t}^{2}+\mathrm{t}-6=0\) \(\mathrm{t}^{2}+3 \mathrm{t}-2 \mathrm{t}-6=0\) \(\mathrm{t}(\mathrm{t}+3)-2(\mathrm{t}+3)=0\) \((\mathrm{t}+3)(\mathrm{t}-2)=0\) \(\mathrm{t}=2,-3\)
VITEEE-2006
Matrix and Determinant
78830
If matrix \(A=\left[\begin{array}{ccc}3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1\end{array}\right]\) and \(A^{-1}=\frac{1}{k} \operatorname{adj}(A)\), then \(\mathrm{k}\) is
1 7
2 -7
3 15
4 -11
Explanation:
(C) : We have, \(A=\left[\begin{array}{ccc} 3 & -2 & 4 \tag{i}\\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{array}\right]\) And \(\quad \mathrm{A}^{-1}=\frac{1}{\mathrm{k}} \operatorname{adj}(\mathrm{A})\) Also we know that, \(A^{-1}=\frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}\) On comparing (i) and (ii) we get- \(\frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}=\frac{1}{\mathrm{k}} \operatorname{adj}(\mathrm{A})\) \(|\mathrm{A}|=\mathrm{k}\) \(|\mathrm{A}|=\left|\begin{array}{ccc}3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1\end{array}\right|\) Hence, \(|\mathrm{A}|=3(2+1)+2(1+0)+4(1-0)\) \(|\mathrm{A}|=3(3)+2(1)+4(1)\) \(|\mathrm{A}|=9+2+4\) \(|\mathrm{~A}|=15\) \(|\mathrm{~A}|=\mathrm{k}=15\)
Matrix and Determinant
78831
If the matrix \(A=\left|\begin{array}{ccc}1 & 3 & 1 \\ -1 & 2 & -3 \\ 0 & 1 & 2\end{array}\right|\) then adj (adj \(\left.A\right)\) is equal to
78828
If \(A=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]\) and \(B=\left[\begin{array}{ll}1 & 0 \\ 5 & 1\end{array}\right]\), then value of \(\alpha\) for which \(\mathrm{A}^{2}=\mathrm{B}\), is
1 1
2 -1
3 4
4 No real values
Explanation:
(D) : Given that, \(\mathrm{A}=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]\) and \(\mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 5 & 1\end{array}\right]\) \(\Rightarrow\left[\begin{array}{cc} \alpha & 0 \\ 1 & 1 \end{array}\right]\left[\begin{array}{ll} \alpha & 0 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 5 & 1 \end{array}\right]\) \(\Rightarrow\left[\begin{array}{cc} \alpha^{2} & 0 \\ \alpha+1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 5 & 1 \end{array}\right]\) On comparing corresponding elements on both the side we get - \(\Rightarrow \alpha^{2}=1, \alpha+1=5\) \(\Rightarrow \alpha= \pm 1, \alpha=4\) \(\because\) There is no common value \(\therefore\) There is no real value of \(\alpha\) for which \(\mathrm{A}^{2}=\mathrm{B}\)
BITSAT-2005
Matrix and Determinant
78829
Let \(A=\left(\begin{array}{ccc}1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6\end{array}\right)\), then the values of \(t\) for which inverse of \(A\) does not exist
1 \(-2,1\)
2 3,2
3 \(2,-3\)
4 \(3,-1\)
Explanation:
(C) : Given that, \(A=\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6 \end{array}\right]\) We know that inverse of A does not exist only when \(|\mathrm{A}|=0\) \(\left|\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & \mathrm{t} \\ 4 & 7-\mathrm{t} & -6 \end{array}\right|=0\) \(1(-30-t(7-t))-3(-12-4 t)+2(2(7-t)-20)=0\) \(-30-7 \mathrm{t}+\mathrm{t}^{2}+36+12 \mathrm{t}+2(14-2 \mathrm{t}-20)=0\) \(-30-7 \mathrm{t}+\mathrm{t}^{2}+36+12 \mathrm{t}+2(-2 \mathrm{t}-6)=0\) \(-30-7 \mathrm{t}+\mathrm{t}^{2}+36+12 \mathrm{t}-12-4 \mathrm{t}=0\) \(\mathrm{t}^{2}+\mathrm{t}-6=0\) \(\mathrm{t}^{2}+3 \mathrm{t}-2 \mathrm{t}-6=0\) \(\mathrm{t}(\mathrm{t}+3)-2(\mathrm{t}+3)=0\) \((\mathrm{t}+3)(\mathrm{t}-2)=0\) \(\mathrm{t}=2,-3\)
VITEEE-2006
Matrix and Determinant
78830
If matrix \(A=\left[\begin{array}{ccc}3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1\end{array}\right]\) and \(A^{-1}=\frac{1}{k} \operatorname{adj}(A)\), then \(\mathrm{k}\) is
1 7
2 -7
3 15
4 -11
Explanation:
(C) : We have, \(A=\left[\begin{array}{ccc} 3 & -2 & 4 \tag{i}\\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{array}\right]\) And \(\quad \mathrm{A}^{-1}=\frac{1}{\mathrm{k}} \operatorname{adj}(\mathrm{A})\) Also we know that, \(A^{-1}=\frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}\) On comparing (i) and (ii) we get- \(\frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}=\frac{1}{\mathrm{k}} \operatorname{adj}(\mathrm{A})\) \(|\mathrm{A}|=\mathrm{k}\) \(|\mathrm{A}|=\left|\begin{array}{ccc}3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1\end{array}\right|\) Hence, \(|\mathrm{A}|=3(2+1)+2(1+0)+4(1-0)\) \(|\mathrm{A}|=3(3)+2(1)+4(1)\) \(|\mathrm{A}|=9+2+4\) \(|\mathrm{~A}|=15\) \(|\mathrm{~A}|=\mathrm{k}=15\)
Matrix and Determinant
78831
If the matrix \(A=\left|\begin{array}{ccc}1 & 3 & 1 \\ -1 & 2 & -3 \\ 0 & 1 & 2\end{array}\right|\) then adj (adj \(\left.A\right)\) is equal to
78828
If \(A=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]\) and \(B=\left[\begin{array}{ll}1 & 0 \\ 5 & 1\end{array}\right]\), then value of \(\alpha\) for which \(\mathrm{A}^{2}=\mathrm{B}\), is
1 1
2 -1
3 4
4 No real values
Explanation:
(D) : Given that, \(\mathrm{A}=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]\) and \(\mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 5 & 1\end{array}\right]\) \(\Rightarrow\left[\begin{array}{cc} \alpha & 0 \\ 1 & 1 \end{array}\right]\left[\begin{array}{ll} \alpha & 0 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 5 & 1 \end{array}\right]\) \(\Rightarrow\left[\begin{array}{cc} \alpha^{2} & 0 \\ \alpha+1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 5 & 1 \end{array}\right]\) On comparing corresponding elements on both the side we get - \(\Rightarrow \alpha^{2}=1, \alpha+1=5\) \(\Rightarrow \alpha= \pm 1, \alpha=4\) \(\because\) There is no common value \(\therefore\) There is no real value of \(\alpha\) for which \(\mathrm{A}^{2}=\mathrm{B}\)
BITSAT-2005
Matrix and Determinant
78829
Let \(A=\left(\begin{array}{ccc}1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6\end{array}\right)\), then the values of \(t\) for which inverse of \(A\) does not exist
1 \(-2,1\)
2 3,2
3 \(2,-3\)
4 \(3,-1\)
Explanation:
(C) : Given that, \(A=\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6 \end{array}\right]\) We know that inverse of A does not exist only when \(|\mathrm{A}|=0\) \(\left|\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & \mathrm{t} \\ 4 & 7-\mathrm{t} & -6 \end{array}\right|=0\) \(1(-30-t(7-t))-3(-12-4 t)+2(2(7-t)-20)=0\) \(-30-7 \mathrm{t}+\mathrm{t}^{2}+36+12 \mathrm{t}+2(14-2 \mathrm{t}-20)=0\) \(-30-7 \mathrm{t}+\mathrm{t}^{2}+36+12 \mathrm{t}+2(-2 \mathrm{t}-6)=0\) \(-30-7 \mathrm{t}+\mathrm{t}^{2}+36+12 \mathrm{t}-12-4 \mathrm{t}=0\) \(\mathrm{t}^{2}+\mathrm{t}-6=0\) \(\mathrm{t}^{2}+3 \mathrm{t}-2 \mathrm{t}-6=0\) \(\mathrm{t}(\mathrm{t}+3)-2(\mathrm{t}+3)=0\) \((\mathrm{t}+3)(\mathrm{t}-2)=0\) \(\mathrm{t}=2,-3\)
VITEEE-2006
Matrix and Determinant
78830
If matrix \(A=\left[\begin{array}{ccc}3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1\end{array}\right]\) and \(A^{-1}=\frac{1}{k} \operatorname{adj}(A)\), then \(\mathrm{k}\) is
1 7
2 -7
3 15
4 -11
Explanation:
(C) : We have, \(A=\left[\begin{array}{ccc} 3 & -2 & 4 \tag{i}\\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{array}\right]\) And \(\quad \mathrm{A}^{-1}=\frac{1}{\mathrm{k}} \operatorname{adj}(\mathrm{A})\) Also we know that, \(A^{-1}=\frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}\) On comparing (i) and (ii) we get- \(\frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}=\frac{1}{\mathrm{k}} \operatorname{adj}(\mathrm{A})\) \(|\mathrm{A}|=\mathrm{k}\) \(|\mathrm{A}|=\left|\begin{array}{ccc}3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1\end{array}\right|\) Hence, \(|\mathrm{A}|=3(2+1)+2(1+0)+4(1-0)\) \(|\mathrm{A}|=3(3)+2(1)+4(1)\) \(|\mathrm{A}|=9+2+4\) \(|\mathrm{~A}|=15\) \(|\mathrm{~A}|=\mathrm{k}=15\)
Matrix and Determinant
78831
If the matrix \(A=\left|\begin{array}{ccc}1 & 3 & 1 \\ -1 & 2 & -3 \\ 0 & 1 & 2\end{array}\right|\) then adj (adj \(\left.A\right)\) is equal to
78828
If \(A=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]\) and \(B=\left[\begin{array}{ll}1 & 0 \\ 5 & 1\end{array}\right]\), then value of \(\alpha\) for which \(\mathrm{A}^{2}=\mathrm{B}\), is
1 1
2 -1
3 4
4 No real values
Explanation:
(D) : Given that, \(\mathrm{A}=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]\) and \(\mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 5 & 1\end{array}\right]\) \(\Rightarrow\left[\begin{array}{cc} \alpha & 0 \\ 1 & 1 \end{array}\right]\left[\begin{array}{ll} \alpha & 0 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 5 & 1 \end{array}\right]\) \(\Rightarrow\left[\begin{array}{cc} \alpha^{2} & 0 \\ \alpha+1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 5 & 1 \end{array}\right]\) On comparing corresponding elements on both the side we get - \(\Rightarrow \alpha^{2}=1, \alpha+1=5\) \(\Rightarrow \alpha= \pm 1, \alpha=4\) \(\because\) There is no common value \(\therefore\) There is no real value of \(\alpha\) for which \(\mathrm{A}^{2}=\mathrm{B}\)
BITSAT-2005
Matrix and Determinant
78829
Let \(A=\left(\begin{array}{ccc}1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6\end{array}\right)\), then the values of \(t\) for which inverse of \(A\) does not exist
1 \(-2,1\)
2 3,2
3 \(2,-3\)
4 \(3,-1\)
Explanation:
(C) : Given that, \(A=\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6 \end{array}\right]\) We know that inverse of A does not exist only when \(|\mathrm{A}|=0\) \(\left|\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & \mathrm{t} \\ 4 & 7-\mathrm{t} & -6 \end{array}\right|=0\) \(1(-30-t(7-t))-3(-12-4 t)+2(2(7-t)-20)=0\) \(-30-7 \mathrm{t}+\mathrm{t}^{2}+36+12 \mathrm{t}+2(14-2 \mathrm{t}-20)=0\) \(-30-7 \mathrm{t}+\mathrm{t}^{2}+36+12 \mathrm{t}+2(-2 \mathrm{t}-6)=0\) \(-30-7 \mathrm{t}+\mathrm{t}^{2}+36+12 \mathrm{t}-12-4 \mathrm{t}=0\) \(\mathrm{t}^{2}+\mathrm{t}-6=0\) \(\mathrm{t}^{2}+3 \mathrm{t}-2 \mathrm{t}-6=0\) \(\mathrm{t}(\mathrm{t}+3)-2(\mathrm{t}+3)=0\) \((\mathrm{t}+3)(\mathrm{t}-2)=0\) \(\mathrm{t}=2,-3\)
VITEEE-2006
Matrix and Determinant
78830
If matrix \(A=\left[\begin{array}{ccc}3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1\end{array}\right]\) and \(A^{-1}=\frac{1}{k} \operatorname{adj}(A)\), then \(\mathrm{k}\) is
1 7
2 -7
3 15
4 -11
Explanation:
(C) : We have, \(A=\left[\begin{array}{ccc} 3 & -2 & 4 \tag{i}\\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{array}\right]\) And \(\quad \mathrm{A}^{-1}=\frac{1}{\mathrm{k}} \operatorname{adj}(\mathrm{A})\) Also we know that, \(A^{-1}=\frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}\) On comparing (i) and (ii) we get- \(\frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}=\frac{1}{\mathrm{k}} \operatorname{adj}(\mathrm{A})\) \(|\mathrm{A}|=\mathrm{k}\) \(|\mathrm{A}|=\left|\begin{array}{ccc}3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1\end{array}\right|\) Hence, \(|\mathrm{A}|=3(2+1)+2(1+0)+4(1-0)\) \(|\mathrm{A}|=3(3)+2(1)+4(1)\) \(|\mathrm{A}|=9+2+4\) \(|\mathrm{~A}|=15\) \(|\mathrm{~A}|=\mathrm{k}=15\)
Matrix and Determinant
78831
If the matrix \(A=\left|\begin{array}{ccc}1 & 3 & 1 \\ -1 & 2 & -3 \\ 0 & 1 & 2\end{array}\right|\) then adj (adj \(\left.A\right)\) is equal to
78828
If \(A=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]\) and \(B=\left[\begin{array}{ll}1 & 0 \\ 5 & 1\end{array}\right]\), then value of \(\alpha\) for which \(\mathrm{A}^{2}=\mathrm{B}\), is
1 1
2 -1
3 4
4 No real values
Explanation:
(D) : Given that, \(\mathrm{A}=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]\) and \(\mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 5 & 1\end{array}\right]\) \(\Rightarrow\left[\begin{array}{cc} \alpha & 0 \\ 1 & 1 \end{array}\right]\left[\begin{array}{ll} \alpha & 0 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 5 & 1 \end{array}\right]\) \(\Rightarrow\left[\begin{array}{cc} \alpha^{2} & 0 \\ \alpha+1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 5 & 1 \end{array}\right]\) On comparing corresponding elements on both the side we get - \(\Rightarrow \alpha^{2}=1, \alpha+1=5\) \(\Rightarrow \alpha= \pm 1, \alpha=4\) \(\because\) There is no common value \(\therefore\) There is no real value of \(\alpha\) for which \(\mathrm{A}^{2}=\mathrm{B}\)
BITSAT-2005
Matrix and Determinant
78829
Let \(A=\left(\begin{array}{ccc}1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6\end{array}\right)\), then the values of \(t\) for which inverse of \(A\) does not exist
1 \(-2,1\)
2 3,2
3 \(2,-3\)
4 \(3,-1\)
Explanation:
(C) : Given that, \(A=\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6 \end{array}\right]\) We know that inverse of A does not exist only when \(|\mathrm{A}|=0\) \(\left|\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & \mathrm{t} \\ 4 & 7-\mathrm{t} & -6 \end{array}\right|=0\) \(1(-30-t(7-t))-3(-12-4 t)+2(2(7-t)-20)=0\) \(-30-7 \mathrm{t}+\mathrm{t}^{2}+36+12 \mathrm{t}+2(14-2 \mathrm{t}-20)=0\) \(-30-7 \mathrm{t}+\mathrm{t}^{2}+36+12 \mathrm{t}+2(-2 \mathrm{t}-6)=0\) \(-30-7 \mathrm{t}+\mathrm{t}^{2}+36+12 \mathrm{t}-12-4 \mathrm{t}=0\) \(\mathrm{t}^{2}+\mathrm{t}-6=0\) \(\mathrm{t}^{2}+3 \mathrm{t}-2 \mathrm{t}-6=0\) \(\mathrm{t}(\mathrm{t}+3)-2(\mathrm{t}+3)=0\) \((\mathrm{t}+3)(\mathrm{t}-2)=0\) \(\mathrm{t}=2,-3\)
VITEEE-2006
Matrix and Determinant
78830
If matrix \(A=\left[\begin{array}{ccc}3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1\end{array}\right]\) and \(A^{-1}=\frac{1}{k} \operatorname{adj}(A)\), then \(\mathrm{k}\) is
1 7
2 -7
3 15
4 -11
Explanation:
(C) : We have, \(A=\left[\begin{array}{ccc} 3 & -2 & 4 \tag{i}\\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{array}\right]\) And \(\quad \mathrm{A}^{-1}=\frac{1}{\mathrm{k}} \operatorname{adj}(\mathrm{A})\) Also we know that, \(A^{-1}=\frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}\) On comparing (i) and (ii) we get- \(\frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}=\frac{1}{\mathrm{k}} \operatorname{adj}(\mathrm{A})\) \(|\mathrm{A}|=\mathrm{k}\) \(|\mathrm{A}|=\left|\begin{array}{ccc}3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1\end{array}\right|\) Hence, \(|\mathrm{A}|=3(2+1)+2(1+0)+4(1-0)\) \(|\mathrm{A}|=3(3)+2(1)+4(1)\) \(|\mathrm{A}|=9+2+4\) \(|\mathrm{~A}|=15\) \(|\mathrm{~A}|=\mathrm{k}=15\)
Matrix and Determinant
78831
If the matrix \(A=\left|\begin{array}{ccc}1 & 3 & 1 \\ -1 & 2 & -3 \\ 0 & 1 & 2\end{array}\right|\) then adj (adj \(\left.A\right)\) is equal to