NEET Test Series from KOTA - 10 Papers In MS WORD
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Matrix and Determinant
78963
If \(A\) and \(B\) are square matrices of the same order such that \((A+B)(A-B)=A^{2}-B^{2}\), then \(\left(\mathbf{A B A}^{-1}\right)^{2}=\)
1 \(\mathrm{A}^{2} \mathrm{~B}^{2}\)
2 \(A^{2}\)
3 \(\mathrm{B}^{2}\)
4 1
Explanation:
(C) : According to given summation, \((A+B)(A-B)=A^{2}-B^{2}\) \(A A+B A-A B-B B=A^{2}-B^{2}\) \(A^{2}+B A-A B-B^{2}=A^{2}-B^{2}\) \(B A-A B=0 \Rightarrow B A=A B\) Now, \(\left(\mathrm{ABA}^{-1}\right)^{2}=\left(\mathrm{BAA}^{-1}\right)^{2}\) \(=(\mathrm{B} \cdot 1)^{2} \quad\left[\because \mathrm{AA}^{-1}=1\right]\) \(=\mathrm{B}^{2}\)
Karnataka CET-2009
Matrix and Determinant
78964
If \(A(\operatorname{adj} A)=5 I\) where \(I\) is the identity matrix of order 3 , then \(|\operatorname{adj} A|\) is equal to
1 5
2 10
3 125
4 25
Explanation:
(D) : According to given summation, \(\mathrm{AA}^{-1}=\mathrm{I}\) \(A \cdot \frac{1}{|A|}(\operatorname{adj} A)=I\) \(A \cdot(\operatorname{adj} A)=|A| I\) After comparing the above equation expression with given expression, we get - \(|A|=5\) \(\operatorname{adj} A=|A| A^{-1}\) \(|\operatorname{adj} A|=|| A\left|A^{-1}\right|=\left|A^{2}\right|=25\)
Karnataka CET-2008
Matrix and Determinant
78965
If \(A=\left[\begin{array}{ccc}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\), then \(A \cdot \operatorname{adj}(A)\) is equal to=
78963
If \(A\) and \(B\) are square matrices of the same order such that \((A+B)(A-B)=A^{2}-B^{2}\), then \(\left(\mathbf{A B A}^{-1}\right)^{2}=\)
1 \(\mathrm{A}^{2} \mathrm{~B}^{2}\)
2 \(A^{2}\)
3 \(\mathrm{B}^{2}\)
4 1
Explanation:
(C) : According to given summation, \((A+B)(A-B)=A^{2}-B^{2}\) \(A A+B A-A B-B B=A^{2}-B^{2}\) \(A^{2}+B A-A B-B^{2}=A^{2}-B^{2}\) \(B A-A B=0 \Rightarrow B A=A B\) Now, \(\left(\mathrm{ABA}^{-1}\right)^{2}=\left(\mathrm{BAA}^{-1}\right)^{2}\) \(=(\mathrm{B} \cdot 1)^{2} \quad\left[\because \mathrm{AA}^{-1}=1\right]\) \(=\mathrm{B}^{2}\)
Karnataka CET-2009
Matrix and Determinant
78964
If \(A(\operatorname{adj} A)=5 I\) where \(I\) is the identity matrix of order 3 , then \(|\operatorname{adj} A|\) is equal to
1 5
2 10
3 125
4 25
Explanation:
(D) : According to given summation, \(\mathrm{AA}^{-1}=\mathrm{I}\) \(A \cdot \frac{1}{|A|}(\operatorname{adj} A)=I\) \(A \cdot(\operatorname{adj} A)=|A| I\) After comparing the above equation expression with given expression, we get - \(|A|=5\) \(\operatorname{adj} A=|A| A^{-1}\) \(|\operatorname{adj} A|=|| A\left|A^{-1}\right|=\left|A^{2}\right|=25\)
Karnataka CET-2008
Matrix and Determinant
78965
If \(A=\left[\begin{array}{ccc}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\), then \(A \cdot \operatorname{adj}(A)\) is equal to=
78963
If \(A\) and \(B\) are square matrices of the same order such that \((A+B)(A-B)=A^{2}-B^{2}\), then \(\left(\mathbf{A B A}^{-1}\right)^{2}=\)
1 \(\mathrm{A}^{2} \mathrm{~B}^{2}\)
2 \(A^{2}\)
3 \(\mathrm{B}^{2}\)
4 1
Explanation:
(C) : According to given summation, \((A+B)(A-B)=A^{2}-B^{2}\) \(A A+B A-A B-B B=A^{2}-B^{2}\) \(A^{2}+B A-A B-B^{2}=A^{2}-B^{2}\) \(B A-A B=0 \Rightarrow B A=A B\) Now, \(\left(\mathrm{ABA}^{-1}\right)^{2}=\left(\mathrm{BAA}^{-1}\right)^{2}\) \(=(\mathrm{B} \cdot 1)^{2} \quad\left[\because \mathrm{AA}^{-1}=1\right]\) \(=\mathrm{B}^{2}\)
Karnataka CET-2009
Matrix and Determinant
78964
If \(A(\operatorname{adj} A)=5 I\) where \(I\) is the identity matrix of order 3 , then \(|\operatorname{adj} A|\) is equal to
1 5
2 10
3 125
4 25
Explanation:
(D) : According to given summation, \(\mathrm{AA}^{-1}=\mathrm{I}\) \(A \cdot \frac{1}{|A|}(\operatorname{adj} A)=I\) \(A \cdot(\operatorname{adj} A)=|A| I\) After comparing the above equation expression with given expression, we get - \(|A|=5\) \(\operatorname{adj} A=|A| A^{-1}\) \(|\operatorname{adj} A|=|| A\left|A^{-1}\right|=\left|A^{2}\right|=25\)
Karnataka CET-2008
Matrix and Determinant
78965
If \(A=\left[\begin{array}{ccc}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\), then \(A \cdot \operatorname{adj}(A)\) is equal to=
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Matrix and Determinant
78963
If \(A\) and \(B\) are square matrices of the same order such that \((A+B)(A-B)=A^{2}-B^{2}\), then \(\left(\mathbf{A B A}^{-1}\right)^{2}=\)
1 \(\mathrm{A}^{2} \mathrm{~B}^{2}\)
2 \(A^{2}\)
3 \(\mathrm{B}^{2}\)
4 1
Explanation:
(C) : According to given summation, \((A+B)(A-B)=A^{2}-B^{2}\) \(A A+B A-A B-B B=A^{2}-B^{2}\) \(A^{2}+B A-A B-B^{2}=A^{2}-B^{2}\) \(B A-A B=0 \Rightarrow B A=A B\) Now, \(\left(\mathrm{ABA}^{-1}\right)^{2}=\left(\mathrm{BAA}^{-1}\right)^{2}\) \(=(\mathrm{B} \cdot 1)^{2} \quad\left[\because \mathrm{AA}^{-1}=1\right]\) \(=\mathrm{B}^{2}\)
Karnataka CET-2009
Matrix and Determinant
78964
If \(A(\operatorname{adj} A)=5 I\) where \(I\) is the identity matrix of order 3 , then \(|\operatorname{adj} A|\) is equal to
1 5
2 10
3 125
4 25
Explanation:
(D) : According to given summation, \(\mathrm{AA}^{-1}=\mathrm{I}\) \(A \cdot \frac{1}{|A|}(\operatorname{adj} A)=I\) \(A \cdot(\operatorname{adj} A)=|A| I\) After comparing the above equation expression with given expression, we get - \(|A|=5\) \(\operatorname{adj} A=|A| A^{-1}\) \(|\operatorname{adj} A|=|| A\left|A^{-1}\right|=\left|A^{2}\right|=25\)
Karnataka CET-2008
Matrix and Determinant
78965
If \(A=\left[\begin{array}{ccc}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\), then \(A \cdot \operatorname{adj}(A)\) is equal to=