78977
Let \(M\) be a \(3 \times 3\) non-singular matrix with det \((M)=\alpha\). If \(\left[M^{-1} \operatorname{adj}(\operatorname{adj}(M)]=K I\right.\), then the value of \(K\) is
1 1
2 \(\alpha\)
3 \(\alpha^{2}\)
4 \(\alpha^{3}\)
Explanation:
(B) : We know that, \(M(\operatorname{adj} M)=|M| I\) Replacing \(M\) by adj \(M\), we get - \(\operatorname{adj} \mathbf{M}[\operatorname{adj}(\operatorname{adj} M)]=\operatorname{det}(\operatorname{adj} M) I\) \(\operatorname{det}(\mathrm{M}) \mathrm{M}^{-1}[\operatorname{adj}(\operatorname{adj} \mathrm{M})]=\alpha^{2} \mathrm{I}\) \(\Rightarrow \quad \alpha M^{-1}[\operatorname{adj}(\operatorname{adj} M)]=\alpha^{2} I\) \(\Rightarrow \quad \mathrm{M}^{-1}[\operatorname{adj}(\operatorname{adj} \mathrm{M})]=\alpha \mathrm{I}\) But \(\quad \mathrm{M}^{-1}[\operatorname{adj}(\operatorname{adj} \mathrm{M})]=\mathrm{KI}\) Hence, \(\mathrm{K}=\alpha\)
BITSAT-2015
Matrix and Determinant
78978
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 \(k\) is
1 7
2 -7
3 15
4 -11
Explanation:
(C) : It is given that, \(A=\left[\begin{array}{ccc} 3 & -2 & 4 \tag{i}\\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{array}\right]\) And \(\quad \mathrm{A}^{-1}=\frac{1}{\mathrm{k}} \operatorname{adj}(\mathrm{A})\) Also, we know that \(\mathrm{A}^{-1}=\frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|} \tag{ii}\) \(\therefore\) By comparing (i) and (ii), we get - \(|\mathrm{A}|=\mathrm{k}\) \(|\mathrm{A}|=\left|\begin{array}{ccc} 3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{array}\right|\) \(\mathrm{A}=3(2+1)+2(1+0)+4(1-0)\) \(|\mathrm{A}|=15\)
BITSAT-2018
Matrix and Determinant
78979
If \(A\) and \(B\) are matrices and \(B=A B A^{-1}\) then the value of \((A+B)(A-B)\) is
1 \(A^{2}+B^{2}\)
2 \(A^{2}-B^{2}\)
3 \(A+B\)
4 \(\mathrm{A}-\mathrm{B}\)
Explanation:
(B) : It is given that, \(\mathrm{B}=\mathrm{ABA}^{-1}\) But \(\quad \mathrm{B}=\mathrm{BAA}^{-1}\) \(\mathrm{ABA}^{-1}=\mathrm{BAA}^{-1}\) \(\mathrm{AB}=\mathrm{BA}\) Now, from the given expression - \((A+B)(A-B)\) \(=A^{2}-A B+B A-B^{2}\) \(=A^{2}-A B+A B-B^{2}\) \(=A^{2}-B^{2}\) \([\because \mathrm{AB}=\mathrm{BA}]\) So, \(\quad(A+B)(A-B)=A^{2}-B^{2}\)
78977
Let \(M\) be a \(3 \times 3\) non-singular matrix with det \((M)=\alpha\). If \(\left[M^{-1} \operatorname{adj}(\operatorname{adj}(M)]=K I\right.\), then the value of \(K\) is
1 1
2 \(\alpha\)
3 \(\alpha^{2}\)
4 \(\alpha^{3}\)
Explanation:
(B) : We know that, \(M(\operatorname{adj} M)=|M| I\) Replacing \(M\) by adj \(M\), we get - \(\operatorname{adj} \mathbf{M}[\operatorname{adj}(\operatorname{adj} M)]=\operatorname{det}(\operatorname{adj} M) I\) \(\operatorname{det}(\mathrm{M}) \mathrm{M}^{-1}[\operatorname{adj}(\operatorname{adj} \mathrm{M})]=\alpha^{2} \mathrm{I}\) \(\Rightarrow \quad \alpha M^{-1}[\operatorname{adj}(\operatorname{adj} M)]=\alpha^{2} I\) \(\Rightarrow \quad \mathrm{M}^{-1}[\operatorname{adj}(\operatorname{adj} \mathrm{M})]=\alpha \mathrm{I}\) But \(\quad \mathrm{M}^{-1}[\operatorname{adj}(\operatorname{adj} \mathrm{M})]=\mathrm{KI}\) Hence, \(\mathrm{K}=\alpha\)
BITSAT-2015
Matrix and Determinant
78978
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 \(k\) is
1 7
2 -7
3 15
4 -11
Explanation:
(C) : It is given that, \(A=\left[\begin{array}{ccc} 3 & -2 & 4 \tag{i}\\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{array}\right]\) And \(\quad \mathrm{A}^{-1}=\frac{1}{\mathrm{k}} \operatorname{adj}(\mathrm{A})\) Also, we know that \(\mathrm{A}^{-1}=\frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|} \tag{ii}\) \(\therefore\) By comparing (i) and (ii), we get - \(|\mathrm{A}|=\mathrm{k}\) \(|\mathrm{A}|=\left|\begin{array}{ccc} 3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{array}\right|\) \(\mathrm{A}=3(2+1)+2(1+0)+4(1-0)\) \(|\mathrm{A}|=15\)
BITSAT-2018
Matrix and Determinant
78979
If \(A\) and \(B\) are matrices and \(B=A B A^{-1}\) then the value of \((A+B)(A-B)\) is
1 \(A^{2}+B^{2}\)
2 \(A^{2}-B^{2}\)
3 \(A+B\)
4 \(\mathrm{A}-\mathrm{B}\)
Explanation:
(B) : It is given that, \(\mathrm{B}=\mathrm{ABA}^{-1}\) But \(\quad \mathrm{B}=\mathrm{BAA}^{-1}\) \(\mathrm{ABA}^{-1}=\mathrm{BAA}^{-1}\) \(\mathrm{AB}=\mathrm{BA}\) Now, from the given expression - \((A+B)(A-B)\) \(=A^{2}-A B+B A-B^{2}\) \(=A^{2}-A B+A B-B^{2}\) \(=A^{2}-B^{2}\) \([\because \mathrm{AB}=\mathrm{BA}]\) So, \(\quad(A+B)(A-B)=A^{2}-B^{2}\)
78977
Let \(M\) be a \(3 \times 3\) non-singular matrix with det \((M)=\alpha\). If \(\left[M^{-1} \operatorname{adj}(\operatorname{adj}(M)]=K I\right.\), then the value of \(K\) is
1 1
2 \(\alpha\)
3 \(\alpha^{2}\)
4 \(\alpha^{3}\)
Explanation:
(B) : We know that, \(M(\operatorname{adj} M)=|M| I\) Replacing \(M\) by adj \(M\), we get - \(\operatorname{adj} \mathbf{M}[\operatorname{adj}(\operatorname{adj} M)]=\operatorname{det}(\operatorname{adj} M) I\) \(\operatorname{det}(\mathrm{M}) \mathrm{M}^{-1}[\operatorname{adj}(\operatorname{adj} \mathrm{M})]=\alpha^{2} \mathrm{I}\) \(\Rightarrow \quad \alpha M^{-1}[\operatorname{adj}(\operatorname{adj} M)]=\alpha^{2} I\) \(\Rightarrow \quad \mathrm{M}^{-1}[\operatorname{adj}(\operatorname{adj} \mathrm{M})]=\alpha \mathrm{I}\) But \(\quad \mathrm{M}^{-1}[\operatorname{adj}(\operatorname{adj} \mathrm{M})]=\mathrm{KI}\) Hence, \(\mathrm{K}=\alpha\)
BITSAT-2015
Matrix and Determinant
78978
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 \(k\) is
1 7
2 -7
3 15
4 -11
Explanation:
(C) : It is given that, \(A=\left[\begin{array}{ccc} 3 & -2 & 4 \tag{i}\\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{array}\right]\) And \(\quad \mathrm{A}^{-1}=\frac{1}{\mathrm{k}} \operatorname{adj}(\mathrm{A})\) Also, we know that \(\mathrm{A}^{-1}=\frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|} \tag{ii}\) \(\therefore\) By comparing (i) and (ii), we get - \(|\mathrm{A}|=\mathrm{k}\) \(|\mathrm{A}|=\left|\begin{array}{ccc} 3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{array}\right|\) \(\mathrm{A}=3(2+1)+2(1+0)+4(1-0)\) \(|\mathrm{A}|=15\)
BITSAT-2018
Matrix and Determinant
78979
If \(A\) and \(B\) are matrices and \(B=A B A^{-1}\) then the value of \((A+B)(A-B)\) is
1 \(A^{2}+B^{2}\)
2 \(A^{2}-B^{2}\)
3 \(A+B\)
4 \(\mathrm{A}-\mathrm{B}\)
Explanation:
(B) : It is given that, \(\mathrm{B}=\mathrm{ABA}^{-1}\) But \(\quad \mathrm{B}=\mathrm{BAA}^{-1}\) \(\mathrm{ABA}^{-1}=\mathrm{BAA}^{-1}\) \(\mathrm{AB}=\mathrm{BA}\) Now, from the given expression - \((A+B)(A-B)\) \(=A^{2}-A B+B A-B^{2}\) \(=A^{2}-A B+A B-B^{2}\) \(=A^{2}-B^{2}\) \([\because \mathrm{AB}=\mathrm{BA}]\) So, \(\quad(A+B)(A-B)=A^{2}-B^{2}\)
78977
Let \(M\) be a \(3 \times 3\) non-singular matrix with det \((M)=\alpha\). If \(\left[M^{-1} \operatorname{adj}(\operatorname{adj}(M)]=K I\right.\), then the value of \(K\) is
1 1
2 \(\alpha\)
3 \(\alpha^{2}\)
4 \(\alpha^{3}\)
Explanation:
(B) : We know that, \(M(\operatorname{adj} M)=|M| I\) Replacing \(M\) by adj \(M\), we get - \(\operatorname{adj} \mathbf{M}[\operatorname{adj}(\operatorname{adj} M)]=\operatorname{det}(\operatorname{adj} M) I\) \(\operatorname{det}(\mathrm{M}) \mathrm{M}^{-1}[\operatorname{adj}(\operatorname{adj} \mathrm{M})]=\alpha^{2} \mathrm{I}\) \(\Rightarrow \quad \alpha M^{-1}[\operatorname{adj}(\operatorname{adj} M)]=\alpha^{2} I\) \(\Rightarrow \quad \mathrm{M}^{-1}[\operatorname{adj}(\operatorname{adj} \mathrm{M})]=\alpha \mathrm{I}\) But \(\quad \mathrm{M}^{-1}[\operatorname{adj}(\operatorname{adj} \mathrm{M})]=\mathrm{KI}\) Hence, \(\mathrm{K}=\alpha\)
BITSAT-2015
Matrix and Determinant
78978
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 \(k\) is
1 7
2 -7
3 15
4 -11
Explanation:
(C) : It is given that, \(A=\left[\begin{array}{ccc} 3 & -2 & 4 \tag{i}\\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{array}\right]\) And \(\quad \mathrm{A}^{-1}=\frac{1}{\mathrm{k}} \operatorname{adj}(\mathrm{A})\) Also, we know that \(\mathrm{A}^{-1}=\frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|} \tag{ii}\) \(\therefore\) By comparing (i) and (ii), we get - \(|\mathrm{A}|=\mathrm{k}\) \(|\mathrm{A}|=\left|\begin{array}{ccc} 3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{array}\right|\) \(\mathrm{A}=3(2+1)+2(1+0)+4(1-0)\) \(|\mathrm{A}|=15\)
BITSAT-2018
Matrix and Determinant
78979
If \(A\) and \(B\) are matrices and \(B=A B A^{-1}\) then the value of \((A+B)(A-B)\) is
1 \(A^{2}+B^{2}\)
2 \(A^{2}-B^{2}\)
3 \(A+B\)
4 \(\mathrm{A}-\mathrm{B}\)
Explanation:
(B) : It is given that, \(\mathrm{B}=\mathrm{ABA}^{-1}\) But \(\quad \mathrm{B}=\mathrm{BAA}^{-1}\) \(\mathrm{ABA}^{-1}=\mathrm{BAA}^{-1}\) \(\mathrm{AB}=\mathrm{BA}\) Now, from the given expression - \((A+B)(A-B)\) \(=A^{2}-A B+B A-B^{2}\) \(=A^{2}-A B+A B-B^{2}\) \(=A^{2}-B^{2}\) \([\because \mathrm{AB}=\mathrm{BA}]\) So, \(\quad(A+B)(A-B)=A^{2}-B^{2}\)