78441
For how many values of \(x\) in the closed interval \([-4,-1]\) the matrix \(\left[\begin{array}{ccc}3 & x-1 & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2\end{array}\right]\) is singular :
1 zero
2 2
3 1
4 3
Explanation:
(C) : Given that a singular matrix, \(A=\left[\begin{array}{ccc} 3 & x-1 & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{array}\right]\) \(|A|=0\) Then, \(\quad|\mathrm{A}|=0\) \(\Rightarrow \quad\left|\begin{array}{ccc} 3 & -x+1 & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{array}\right|=0\) On applying \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\) and \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\) we get - \(\left|\begin{array}{ccc} 3 & \mathrm{x}-1 & 2 \\ 0 & -\mathrm{x} & \mathrm{x} \\ \mathrm{x} & -\mathrm{x} & 0 \end{array}\right|=0\) On applying \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\) we get - \(\left|\begin{array}{ccc} x+4 & -x+1 & 2 \\ 0 & -x & x \\ 0 & -x & 0 \end{array}\right|=0\) \(x+4\left(0+x^{2}\right)=0\) \(4 x^{2}+x^{3}=0\) \(x^{2}(x+4)=0\) \(x^{2}=0 \text { and } x+4=0\) \(x=0 \text { and } x=-4\) Given interval \(x \in[-4,-1]\) Hence, \(x=-4\) So, \(x\) have only one value.
Karnataka CET-2002
Matrix and Determinant
78443
Let \(M\) be \(2 \times 2\) symmetric matrix with integer entries, then \(M\) is invertible if
1 the first column of \(M\) is the transpose of second row of \(\mathrm{M}\)
2 the second row of \(\mathrm{M}\) is the transpose of first column of \(\mathrm{M}\)
3 \(\mathrm{M}\) is diagonal matrix with non-zero entries in the principal diagonal
4 the product of entries in the principal diagonal of \(M\) is the product of entries in the other diagonal
Explanation:
, d) Exp:(c, d) If \(\mathrm{M}\) is a diagonal matrix then, \(M=\left[\begin{array}{ll} \mathrm{a} & 0 \\ 0 & \mathrm{~d} \end{array}\right]\) \(|\mathrm{M}|=\mathrm{ad}-0\) \(|\mathrm{M}|=\mathrm{ad}\) Hence, \(|\mathrm{M}| \neq 0\) Therefore \(\mathrm{M}\) is invertible matrix. And \(M=\left[\begin{array}{ll} a & b \\ b & d \end{array}\right]\) \(|\mathrm{M}|=\mathrm{ad}-\mathrm{b}^{2}\) Hence, \(|\mathrm{M}| \neq 0\) Therefore, \(M\) is an invertible matrix. Hence, option (c) and (d) both are correct.
Karnataka CET-2021
Matrix and Determinant
78444
If \(\left(\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right) \mathbf{A}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\), then the matrix \(A\) is
(A) : Given that, \(\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] \mathrm{A}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\) Let, \(A=\left[\begin{array}{ll} a & b \\ x & y \end{array}\right]\) \({\left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{cc} a & b \\ x & y \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]}\) \({\left[\begin{array}{cc} 2 a+x & 2 b+y \\ 3 a+2 x & 3 b+2 y \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]}\) On comparing corresponding elements on both the side we get - \(2 \mathrm{a}+\mathrm{x}=1 \Rightarrow 2 \mathrm{a}=1-\mathrm{x}\) \(\mathrm{a}=\frac{1-\mathrm{x}}{2}\) And \(2 b+y=0\) \(2 b=-y \tag{ii}\) \(b=-\frac{y}{2}\) \(3 \mathrm{a}+2 \mathrm{x}=0\) From equation (i), \(3 \times \frac{1-\mathrm{x}}{2}+2 \mathrm{x}=0\) \(\frac{3-3 x+4 x}{2}=0\) \(\mathrm{x}=-3\) And \(\quad 3 b+2 y=1\) from equation(ii), \(3 \times-\frac{y}{2}+2 y=1\) \(\frac{-3 y+4 y}{2}=1\) \(y=2\) Putting the value of \(x\) and \(y\) in equation (i) and (ii) respectively- \(\mathrm{a}=2 \text { and } \mathrm{b}=-1\) \(\therefore \quad \mathrm{A}=\left[\begin{array}{ll} \mathrm{a} & \mathrm{b} \\ \mathrm{x} & \mathrm{y} \end{array}\right]=\left[\begin{array}{cc} 2 & -1 \\ -3 & 2 \end{array}\right]\)
Karnataka CET-2020
Matrix and Determinant
78446
If \(f(x)=\left[\begin{array}{ccc}x^{3}-x & a+x & b+x \\ x-a & x^{2}-x & c+x \\ x-b & x-c & 0\end{array}\right]\), then
1 \(\mathrm{f}(2)=0\)
2 \(\mathrm{f}(0)=0\)
3 \(\mathrm{f}(-1)=0\)
4 \(\mathrm{f}(1)=0\)
Explanation:
(B) : Given that, \(f(x)=\left[\begin{array}{ccc} x^{3}-x & a+x & b+x \\ x-a & x^{2}-x & c+x \\ x-b & x-c & 0 \end{array}\right]\) \(f(0)=\left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]\) If all the diagonal of the matrix is zero then the matrix is zero matrix. \(\mathrm{f}(0)=0\)
Karnataka CET-2020
Matrix and Determinant
78448
If \(A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]\), then \(A^{n}=2^{k} A\), where \(k=\)
78441
For how many values of \(x\) in the closed interval \([-4,-1]\) the matrix \(\left[\begin{array}{ccc}3 & x-1 & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2\end{array}\right]\) is singular :
1 zero
2 2
3 1
4 3
Explanation:
(C) : Given that a singular matrix, \(A=\left[\begin{array}{ccc} 3 & x-1 & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{array}\right]\) \(|A|=0\) Then, \(\quad|\mathrm{A}|=0\) \(\Rightarrow \quad\left|\begin{array}{ccc} 3 & -x+1 & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{array}\right|=0\) On applying \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\) and \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\) we get - \(\left|\begin{array}{ccc} 3 & \mathrm{x}-1 & 2 \\ 0 & -\mathrm{x} & \mathrm{x} \\ \mathrm{x} & -\mathrm{x} & 0 \end{array}\right|=0\) On applying \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\) we get - \(\left|\begin{array}{ccc} x+4 & -x+1 & 2 \\ 0 & -x & x \\ 0 & -x & 0 \end{array}\right|=0\) \(x+4\left(0+x^{2}\right)=0\) \(4 x^{2}+x^{3}=0\) \(x^{2}(x+4)=0\) \(x^{2}=0 \text { and } x+4=0\) \(x=0 \text { and } x=-4\) Given interval \(x \in[-4,-1]\) Hence, \(x=-4\) So, \(x\) have only one value.
Karnataka CET-2002
Matrix and Determinant
78443
Let \(M\) be \(2 \times 2\) symmetric matrix with integer entries, then \(M\) is invertible if
1 the first column of \(M\) is the transpose of second row of \(\mathrm{M}\)
2 the second row of \(\mathrm{M}\) is the transpose of first column of \(\mathrm{M}\)
3 \(\mathrm{M}\) is diagonal matrix with non-zero entries in the principal diagonal
4 the product of entries in the principal diagonal of \(M\) is the product of entries in the other diagonal
Explanation:
, d) Exp:(c, d) If \(\mathrm{M}\) is a diagonal matrix then, \(M=\left[\begin{array}{ll} \mathrm{a} & 0 \\ 0 & \mathrm{~d} \end{array}\right]\) \(|\mathrm{M}|=\mathrm{ad}-0\) \(|\mathrm{M}|=\mathrm{ad}\) Hence, \(|\mathrm{M}| \neq 0\) Therefore \(\mathrm{M}\) is invertible matrix. And \(M=\left[\begin{array}{ll} a & b \\ b & d \end{array}\right]\) \(|\mathrm{M}|=\mathrm{ad}-\mathrm{b}^{2}\) Hence, \(|\mathrm{M}| \neq 0\) Therefore, \(M\) is an invertible matrix. Hence, option (c) and (d) both are correct.
Karnataka CET-2021
Matrix and Determinant
78444
If \(\left(\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right) \mathbf{A}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\), then the matrix \(A\) is
(A) : Given that, \(\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] \mathrm{A}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\) Let, \(A=\left[\begin{array}{ll} a & b \\ x & y \end{array}\right]\) \({\left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{cc} a & b \\ x & y \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]}\) \({\left[\begin{array}{cc} 2 a+x & 2 b+y \\ 3 a+2 x & 3 b+2 y \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]}\) On comparing corresponding elements on both the side we get - \(2 \mathrm{a}+\mathrm{x}=1 \Rightarrow 2 \mathrm{a}=1-\mathrm{x}\) \(\mathrm{a}=\frac{1-\mathrm{x}}{2}\) And \(2 b+y=0\) \(2 b=-y \tag{ii}\) \(b=-\frac{y}{2}\) \(3 \mathrm{a}+2 \mathrm{x}=0\) From equation (i), \(3 \times \frac{1-\mathrm{x}}{2}+2 \mathrm{x}=0\) \(\frac{3-3 x+4 x}{2}=0\) \(\mathrm{x}=-3\) And \(\quad 3 b+2 y=1\) from equation(ii), \(3 \times-\frac{y}{2}+2 y=1\) \(\frac{-3 y+4 y}{2}=1\) \(y=2\) Putting the value of \(x\) and \(y\) in equation (i) and (ii) respectively- \(\mathrm{a}=2 \text { and } \mathrm{b}=-1\) \(\therefore \quad \mathrm{A}=\left[\begin{array}{ll} \mathrm{a} & \mathrm{b} \\ \mathrm{x} & \mathrm{y} \end{array}\right]=\left[\begin{array}{cc} 2 & -1 \\ -3 & 2 \end{array}\right]\)
Karnataka CET-2020
Matrix and Determinant
78446
If \(f(x)=\left[\begin{array}{ccc}x^{3}-x & a+x & b+x \\ x-a & x^{2}-x & c+x \\ x-b & x-c & 0\end{array}\right]\), then
1 \(\mathrm{f}(2)=0\)
2 \(\mathrm{f}(0)=0\)
3 \(\mathrm{f}(-1)=0\)
4 \(\mathrm{f}(1)=0\)
Explanation:
(B) : Given that, \(f(x)=\left[\begin{array}{ccc} x^{3}-x & a+x & b+x \\ x-a & x^{2}-x & c+x \\ x-b & x-c & 0 \end{array}\right]\) \(f(0)=\left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]\) If all the diagonal of the matrix is zero then the matrix is zero matrix. \(\mathrm{f}(0)=0\)
Karnataka CET-2020
Matrix and Determinant
78448
If \(A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]\), then \(A^{n}=2^{k} A\), where \(k=\)
78441
For how many values of \(x\) in the closed interval \([-4,-1]\) the matrix \(\left[\begin{array}{ccc}3 & x-1 & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2\end{array}\right]\) is singular :
1 zero
2 2
3 1
4 3
Explanation:
(C) : Given that a singular matrix, \(A=\left[\begin{array}{ccc} 3 & x-1 & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{array}\right]\) \(|A|=0\) Then, \(\quad|\mathrm{A}|=0\) \(\Rightarrow \quad\left|\begin{array}{ccc} 3 & -x+1 & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{array}\right|=0\) On applying \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\) and \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\) we get - \(\left|\begin{array}{ccc} 3 & \mathrm{x}-1 & 2 \\ 0 & -\mathrm{x} & \mathrm{x} \\ \mathrm{x} & -\mathrm{x} & 0 \end{array}\right|=0\) On applying \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\) we get - \(\left|\begin{array}{ccc} x+4 & -x+1 & 2 \\ 0 & -x & x \\ 0 & -x & 0 \end{array}\right|=0\) \(x+4\left(0+x^{2}\right)=0\) \(4 x^{2}+x^{3}=0\) \(x^{2}(x+4)=0\) \(x^{2}=0 \text { and } x+4=0\) \(x=0 \text { and } x=-4\) Given interval \(x \in[-4,-1]\) Hence, \(x=-4\) So, \(x\) have only one value.
Karnataka CET-2002
Matrix and Determinant
78443
Let \(M\) be \(2 \times 2\) symmetric matrix with integer entries, then \(M\) is invertible if
1 the first column of \(M\) is the transpose of second row of \(\mathrm{M}\)
2 the second row of \(\mathrm{M}\) is the transpose of first column of \(\mathrm{M}\)
3 \(\mathrm{M}\) is diagonal matrix with non-zero entries in the principal diagonal
4 the product of entries in the principal diagonal of \(M\) is the product of entries in the other diagonal
Explanation:
, d) Exp:(c, d) If \(\mathrm{M}\) is a diagonal matrix then, \(M=\left[\begin{array}{ll} \mathrm{a} & 0 \\ 0 & \mathrm{~d} \end{array}\right]\) \(|\mathrm{M}|=\mathrm{ad}-0\) \(|\mathrm{M}|=\mathrm{ad}\) Hence, \(|\mathrm{M}| \neq 0\) Therefore \(\mathrm{M}\) is invertible matrix. And \(M=\left[\begin{array}{ll} a & b \\ b & d \end{array}\right]\) \(|\mathrm{M}|=\mathrm{ad}-\mathrm{b}^{2}\) Hence, \(|\mathrm{M}| \neq 0\) Therefore, \(M\) is an invertible matrix. Hence, option (c) and (d) both are correct.
Karnataka CET-2021
Matrix and Determinant
78444
If \(\left(\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right) \mathbf{A}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\), then the matrix \(A\) is
(A) : Given that, \(\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] \mathrm{A}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\) Let, \(A=\left[\begin{array}{ll} a & b \\ x & y \end{array}\right]\) \({\left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{cc} a & b \\ x & y \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]}\) \({\left[\begin{array}{cc} 2 a+x & 2 b+y \\ 3 a+2 x & 3 b+2 y \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]}\) On comparing corresponding elements on both the side we get - \(2 \mathrm{a}+\mathrm{x}=1 \Rightarrow 2 \mathrm{a}=1-\mathrm{x}\) \(\mathrm{a}=\frac{1-\mathrm{x}}{2}\) And \(2 b+y=0\) \(2 b=-y \tag{ii}\) \(b=-\frac{y}{2}\) \(3 \mathrm{a}+2 \mathrm{x}=0\) From equation (i), \(3 \times \frac{1-\mathrm{x}}{2}+2 \mathrm{x}=0\) \(\frac{3-3 x+4 x}{2}=0\) \(\mathrm{x}=-3\) And \(\quad 3 b+2 y=1\) from equation(ii), \(3 \times-\frac{y}{2}+2 y=1\) \(\frac{-3 y+4 y}{2}=1\) \(y=2\) Putting the value of \(x\) and \(y\) in equation (i) and (ii) respectively- \(\mathrm{a}=2 \text { and } \mathrm{b}=-1\) \(\therefore \quad \mathrm{A}=\left[\begin{array}{ll} \mathrm{a} & \mathrm{b} \\ \mathrm{x} & \mathrm{y} \end{array}\right]=\left[\begin{array}{cc} 2 & -1 \\ -3 & 2 \end{array}\right]\)
Karnataka CET-2020
Matrix and Determinant
78446
If \(f(x)=\left[\begin{array}{ccc}x^{3}-x & a+x & b+x \\ x-a & x^{2}-x & c+x \\ x-b & x-c & 0\end{array}\right]\), then
1 \(\mathrm{f}(2)=0\)
2 \(\mathrm{f}(0)=0\)
3 \(\mathrm{f}(-1)=0\)
4 \(\mathrm{f}(1)=0\)
Explanation:
(B) : Given that, \(f(x)=\left[\begin{array}{ccc} x^{3}-x & a+x & b+x \\ x-a & x^{2}-x & c+x \\ x-b & x-c & 0 \end{array}\right]\) \(f(0)=\left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]\) If all the diagonal of the matrix is zero then the matrix is zero matrix. \(\mathrm{f}(0)=0\)
Karnataka CET-2020
Matrix and Determinant
78448
If \(A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]\), then \(A^{n}=2^{k} A\), where \(k=\)
78441
For how many values of \(x\) in the closed interval \([-4,-1]\) the matrix \(\left[\begin{array}{ccc}3 & x-1 & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2\end{array}\right]\) is singular :
1 zero
2 2
3 1
4 3
Explanation:
(C) : Given that a singular matrix, \(A=\left[\begin{array}{ccc} 3 & x-1 & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{array}\right]\) \(|A|=0\) Then, \(\quad|\mathrm{A}|=0\) \(\Rightarrow \quad\left|\begin{array}{ccc} 3 & -x+1 & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{array}\right|=0\) On applying \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\) and \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\) we get - \(\left|\begin{array}{ccc} 3 & \mathrm{x}-1 & 2 \\ 0 & -\mathrm{x} & \mathrm{x} \\ \mathrm{x} & -\mathrm{x} & 0 \end{array}\right|=0\) On applying \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\) we get - \(\left|\begin{array}{ccc} x+4 & -x+1 & 2 \\ 0 & -x & x \\ 0 & -x & 0 \end{array}\right|=0\) \(x+4\left(0+x^{2}\right)=0\) \(4 x^{2}+x^{3}=0\) \(x^{2}(x+4)=0\) \(x^{2}=0 \text { and } x+4=0\) \(x=0 \text { and } x=-4\) Given interval \(x \in[-4,-1]\) Hence, \(x=-4\) So, \(x\) have only one value.
Karnataka CET-2002
Matrix and Determinant
78443
Let \(M\) be \(2 \times 2\) symmetric matrix with integer entries, then \(M\) is invertible if
1 the first column of \(M\) is the transpose of second row of \(\mathrm{M}\)
2 the second row of \(\mathrm{M}\) is the transpose of first column of \(\mathrm{M}\)
3 \(\mathrm{M}\) is diagonal matrix with non-zero entries in the principal diagonal
4 the product of entries in the principal diagonal of \(M\) is the product of entries in the other diagonal
Explanation:
, d) Exp:(c, d) If \(\mathrm{M}\) is a diagonal matrix then, \(M=\left[\begin{array}{ll} \mathrm{a} & 0 \\ 0 & \mathrm{~d} \end{array}\right]\) \(|\mathrm{M}|=\mathrm{ad}-0\) \(|\mathrm{M}|=\mathrm{ad}\) Hence, \(|\mathrm{M}| \neq 0\) Therefore \(\mathrm{M}\) is invertible matrix. And \(M=\left[\begin{array}{ll} a & b \\ b & d \end{array}\right]\) \(|\mathrm{M}|=\mathrm{ad}-\mathrm{b}^{2}\) Hence, \(|\mathrm{M}| \neq 0\) Therefore, \(M\) is an invertible matrix. Hence, option (c) and (d) both are correct.
Karnataka CET-2021
Matrix and Determinant
78444
If \(\left(\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right) \mathbf{A}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\), then the matrix \(A\) is
(A) : Given that, \(\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] \mathrm{A}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\) Let, \(A=\left[\begin{array}{ll} a & b \\ x & y \end{array}\right]\) \({\left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{cc} a & b \\ x & y \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]}\) \({\left[\begin{array}{cc} 2 a+x & 2 b+y \\ 3 a+2 x & 3 b+2 y \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]}\) On comparing corresponding elements on both the side we get - \(2 \mathrm{a}+\mathrm{x}=1 \Rightarrow 2 \mathrm{a}=1-\mathrm{x}\) \(\mathrm{a}=\frac{1-\mathrm{x}}{2}\) And \(2 b+y=0\) \(2 b=-y \tag{ii}\) \(b=-\frac{y}{2}\) \(3 \mathrm{a}+2 \mathrm{x}=0\) From equation (i), \(3 \times \frac{1-\mathrm{x}}{2}+2 \mathrm{x}=0\) \(\frac{3-3 x+4 x}{2}=0\) \(\mathrm{x}=-3\) And \(\quad 3 b+2 y=1\) from equation(ii), \(3 \times-\frac{y}{2}+2 y=1\) \(\frac{-3 y+4 y}{2}=1\) \(y=2\) Putting the value of \(x\) and \(y\) in equation (i) and (ii) respectively- \(\mathrm{a}=2 \text { and } \mathrm{b}=-1\) \(\therefore \quad \mathrm{A}=\left[\begin{array}{ll} \mathrm{a} & \mathrm{b} \\ \mathrm{x} & \mathrm{y} \end{array}\right]=\left[\begin{array}{cc} 2 & -1 \\ -3 & 2 \end{array}\right]\)
Karnataka CET-2020
Matrix and Determinant
78446
If \(f(x)=\left[\begin{array}{ccc}x^{3}-x & a+x & b+x \\ x-a & x^{2}-x & c+x \\ x-b & x-c & 0\end{array}\right]\), then
1 \(\mathrm{f}(2)=0\)
2 \(\mathrm{f}(0)=0\)
3 \(\mathrm{f}(-1)=0\)
4 \(\mathrm{f}(1)=0\)
Explanation:
(B) : Given that, \(f(x)=\left[\begin{array}{ccc} x^{3}-x & a+x & b+x \\ x-a & x^{2}-x & c+x \\ x-b & x-c & 0 \end{array}\right]\) \(f(0)=\left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]\) If all the diagonal of the matrix is zero then the matrix is zero matrix. \(\mathrm{f}(0)=0\)
Karnataka CET-2020
Matrix and Determinant
78448
If \(A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]\), then \(A^{n}=2^{k} A\), where \(k=\)
78441
For how many values of \(x\) in the closed interval \([-4,-1]\) the matrix \(\left[\begin{array}{ccc}3 & x-1 & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2\end{array}\right]\) is singular :
1 zero
2 2
3 1
4 3
Explanation:
(C) : Given that a singular matrix, \(A=\left[\begin{array}{ccc} 3 & x-1 & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{array}\right]\) \(|A|=0\) Then, \(\quad|\mathrm{A}|=0\) \(\Rightarrow \quad\left|\begin{array}{ccc} 3 & -x+1 & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{array}\right|=0\) On applying \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\) and \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\) we get - \(\left|\begin{array}{ccc} 3 & \mathrm{x}-1 & 2 \\ 0 & -\mathrm{x} & \mathrm{x} \\ \mathrm{x} & -\mathrm{x} & 0 \end{array}\right|=0\) On applying \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\) we get - \(\left|\begin{array}{ccc} x+4 & -x+1 & 2 \\ 0 & -x & x \\ 0 & -x & 0 \end{array}\right|=0\) \(x+4\left(0+x^{2}\right)=0\) \(4 x^{2}+x^{3}=0\) \(x^{2}(x+4)=0\) \(x^{2}=0 \text { and } x+4=0\) \(x=0 \text { and } x=-4\) Given interval \(x \in[-4,-1]\) Hence, \(x=-4\) So, \(x\) have only one value.
Karnataka CET-2002
Matrix and Determinant
78443
Let \(M\) be \(2 \times 2\) symmetric matrix with integer entries, then \(M\) is invertible if
1 the first column of \(M\) is the transpose of second row of \(\mathrm{M}\)
2 the second row of \(\mathrm{M}\) is the transpose of first column of \(\mathrm{M}\)
3 \(\mathrm{M}\) is diagonal matrix with non-zero entries in the principal diagonal
4 the product of entries in the principal diagonal of \(M\) is the product of entries in the other diagonal
Explanation:
, d) Exp:(c, d) If \(\mathrm{M}\) is a diagonal matrix then, \(M=\left[\begin{array}{ll} \mathrm{a} & 0 \\ 0 & \mathrm{~d} \end{array}\right]\) \(|\mathrm{M}|=\mathrm{ad}-0\) \(|\mathrm{M}|=\mathrm{ad}\) Hence, \(|\mathrm{M}| \neq 0\) Therefore \(\mathrm{M}\) is invertible matrix. And \(M=\left[\begin{array}{ll} a & b \\ b & d \end{array}\right]\) \(|\mathrm{M}|=\mathrm{ad}-\mathrm{b}^{2}\) Hence, \(|\mathrm{M}| \neq 0\) Therefore, \(M\) is an invertible matrix. Hence, option (c) and (d) both are correct.
Karnataka CET-2021
Matrix and Determinant
78444
If \(\left(\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right) \mathbf{A}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\), then the matrix \(A\) is
(A) : Given that, \(\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] \mathrm{A}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\) Let, \(A=\left[\begin{array}{ll} a & b \\ x & y \end{array}\right]\) \({\left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{cc} a & b \\ x & y \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]}\) \({\left[\begin{array}{cc} 2 a+x & 2 b+y \\ 3 a+2 x & 3 b+2 y \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]}\) On comparing corresponding elements on both the side we get - \(2 \mathrm{a}+\mathrm{x}=1 \Rightarrow 2 \mathrm{a}=1-\mathrm{x}\) \(\mathrm{a}=\frac{1-\mathrm{x}}{2}\) And \(2 b+y=0\) \(2 b=-y \tag{ii}\) \(b=-\frac{y}{2}\) \(3 \mathrm{a}+2 \mathrm{x}=0\) From equation (i), \(3 \times \frac{1-\mathrm{x}}{2}+2 \mathrm{x}=0\) \(\frac{3-3 x+4 x}{2}=0\) \(\mathrm{x}=-3\) And \(\quad 3 b+2 y=1\) from equation(ii), \(3 \times-\frac{y}{2}+2 y=1\) \(\frac{-3 y+4 y}{2}=1\) \(y=2\) Putting the value of \(x\) and \(y\) in equation (i) and (ii) respectively- \(\mathrm{a}=2 \text { and } \mathrm{b}=-1\) \(\therefore \quad \mathrm{A}=\left[\begin{array}{ll} \mathrm{a} & \mathrm{b} \\ \mathrm{x} & \mathrm{y} \end{array}\right]=\left[\begin{array}{cc} 2 & -1 \\ -3 & 2 \end{array}\right]\)
Karnataka CET-2020
Matrix and Determinant
78446
If \(f(x)=\left[\begin{array}{ccc}x^{3}-x & a+x & b+x \\ x-a & x^{2}-x & c+x \\ x-b & x-c & 0\end{array}\right]\), then
1 \(\mathrm{f}(2)=0\)
2 \(\mathrm{f}(0)=0\)
3 \(\mathrm{f}(-1)=0\)
4 \(\mathrm{f}(1)=0\)
Explanation:
(B) : Given that, \(f(x)=\left[\begin{array}{ccc} x^{3}-x & a+x & b+x \\ x-a & x^{2}-x & c+x \\ x-b & x-c & 0 \end{array}\right]\) \(f(0)=\left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]\) If all the diagonal of the matrix is zero then the matrix is zero matrix. \(\mathrm{f}(0)=0\)
Karnataka CET-2020
Matrix and Determinant
78448
If \(A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]\), then \(A^{n}=2^{k} A\), where \(k=\)
