Adjoint and Inverse of Matrices
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Matrix and Determinant

78811 If \(A=\left[\begin{array}{ccc}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]\) then \(\left(A^{2}-5 A\right) A^{-1}=\)

1 \(\left[\begin{array}{ccc}4 & 2 & 3 \\ -1 & 4 & 2 \\ 1 & 2 & 1\end{array}\right]\)
2 \(\left[\begin{array}{ccc}-4 & 2 & 3 \\ -1 & -4 & 2 \\ 1 & 2 & -1\end{array}\right]\)
3 \(\left[\begin{array}{ccc}-4 & -1 & 1 \\ 2 & -4 & 2 \\ 3 & 2 & -1\end{array}\right]\)
4 \(\left[\begin{array}{ccc}-1 & -2 & 1 \\ 4 & -2 & -3 \\ 1 & 4 & -2\end{array}\right]\)
MHT CET-2018
Matrix and Determinant

78812 If \(A=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right], B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\) then
\(\left(\mathrm{B}^{-1} \mathrm{~A}^{-1}\right)^{-1}=\)

1 \(\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]\)
2 \(\left[\begin{array}{cc}2 & 2 \\ -2 & 3\end{array}\right]\)
3 \(\left[\begin{array}{cc}2 & -3 \\ 2 & 2\end{array}\right]\)
4 \(\left[\begin{array}{cc}1 & -1 \\ -2 & 3\end{array}\right]\)
MHT CET-2016
Matrix and Determinant

78813 If matrix \(A=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]\), such that, \(A X=I\) then \(\mathrm{X}=\) \(\qquad\)

1 \(\overline{\frac{1}{5}}\left[\begin{array}{cc}1 & 3 \\ 2 & -1\end{array}\right]\)
2 \(\frac{1}{5}\left[\begin{array}{cc}4 & 2 \\ 4 & -1\end{array}\right]\)
3 \(\frac{1}{5}\left[\begin{array}{cc}-3 & 2 \\ 4 & -1\end{array}\right]\)
4 \(\frac{1}{5}\left[\begin{array}{ll}-1 & 2 \\ -1 & 4\end{array}\right]\)
MHT CET-2016
Matrix and Determinant

78814 The inverse of matrix \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) is

1 \(\left[\begin{array}{cc}\mathrm{d} & -\mathrm{b} \\ -\mathrm{c} & \mathrm{a}\end{array}\right]\)
2 \(\frac{1}{(\mathrm{ad}-\mathrm{bc})}\left[\begin{array}{cc}\mathrm{d} & -\mathrm{b} \\ -\mathrm{c} & \mathrm{a}\end{array}\right]\)
3 \(\frac{1}{|\mathrm{~A}|}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\)
4 \(\left[\begin{array}{ll}\mathrm{b} & -\mathrm{a} \\ \mathrm{d} & -\mathrm{c}\end{array}\right]\)
MHT CET-2010
Matrix and Determinant

78816 If \(A=\left[\begin{array}{ll}5 & 4 \\ 3 & 2\end{array}\right]\), then \(A^{-1}\) is equal to

1 \(\frac{1}{2}\left[\begin{array}{cc}-2 & 4 \\ 3 & 5\end{array}\right]\)
2 \(\frac{1}{2}\left[\begin{array}{ll}5 & 3 \\ 4 & 2\end{array}\right]\)
3 \(\frac{1}{2}\left[\begin{array}{cc}5 & -4 \\ -3 & 2\end{array}\right]\)
4 \(-\frac{1}{2}\left[\begin{array}{cc}2 & -4 \\ -3 & 5\end{array}\right]\)
MHT CET-2007
NEET DIGITAL LIBRARY
Matrix and Determinant

78811 If \(A=\left[\begin{array}{ccc}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]\) then \(\left(A^{2}-5 A\right) A^{-1}=\)

1 \(\left[\begin{array}{ccc}4 & 2 & 3 \\ -1 & 4 & 2 \\ 1 & 2 & 1\end{array}\right]\)
2 \(\left[\begin{array}{ccc}-4 & 2 & 3 \\ -1 & -4 & 2 \\ 1 & 2 & -1\end{array}\right]\)
3 \(\left[\begin{array}{ccc}-4 & -1 & 1 \\ 2 & -4 & 2 \\ 3 & 2 & -1\end{array}\right]\)
4 \(\left[\begin{array}{ccc}-1 & -2 & 1 \\ 4 & -2 & -3 \\ 1 & 4 & -2\end{array}\right]\)
MHT CET-2018
Matrix and Determinant

78812 If \(A=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right], B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\) then
\(\left(\mathrm{B}^{-1} \mathrm{~A}^{-1}\right)^{-1}=\)

1 \(\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]\)
2 \(\left[\begin{array}{cc}2 & 2 \\ -2 & 3\end{array}\right]\)
3 \(\left[\begin{array}{cc}2 & -3 \\ 2 & 2\end{array}\right]\)
4 \(\left[\begin{array}{cc}1 & -1 \\ -2 & 3\end{array}\right]\)
MHT CET-2016
Matrix and Determinant

78813 If matrix \(A=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]\), such that, \(A X=I\) then \(\mathrm{X}=\) \(\qquad\)

1 \(\overline{\frac{1}{5}}\left[\begin{array}{cc}1 & 3 \\ 2 & -1\end{array}\right]\)
2 \(\frac{1}{5}\left[\begin{array}{cc}4 & 2 \\ 4 & -1\end{array}\right]\)
3 \(\frac{1}{5}\left[\begin{array}{cc}-3 & 2 \\ 4 & -1\end{array}\right]\)
4 \(\frac{1}{5}\left[\begin{array}{ll}-1 & 2 \\ -1 & 4\end{array}\right]\)
MHT CET-2016
Matrix and Determinant

78814 The inverse of matrix \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) is

1 \(\left[\begin{array}{cc}\mathrm{d} & -\mathrm{b} \\ -\mathrm{c} & \mathrm{a}\end{array}\right]\)
2 \(\frac{1}{(\mathrm{ad}-\mathrm{bc})}\left[\begin{array}{cc}\mathrm{d} & -\mathrm{b} \\ -\mathrm{c} & \mathrm{a}\end{array}\right]\)
3 \(\frac{1}{|\mathrm{~A}|}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\)
4 \(\left[\begin{array}{ll}\mathrm{b} & -\mathrm{a} \\ \mathrm{d} & -\mathrm{c}\end{array}\right]\)
MHT CET-2010
Matrix and Determinant

78816 If \(A=\left[\begin{array}{ll}5 & 4 \\ 3 & 2\end{array}\right]\), then \(A^{-1}\) is equal to

1 \(\frac{1}{2}\left[\begin{array}{cc}-2 & 4 \\ 3 & 5\end{array}\right]\)
2 \(\frac{1}{2}\left[\begin{array}{ll}5 & 3 \\ 4 & 2\end{array}\right]\)
3 \(\frac{1}{2}\left[\begin{array}{cc}5 & -4 \\ -3 & 2\end{array}\right]\)
4 \(-\frac{1}{2}\left[\begin{array}{cc}2 & -4 \\ -3 & 5\end{array}\right]\)
MHT CET-2007
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Matrix and Determinant

78811 If \(A=\left[\begin{array}{ccc}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]\) then \(\left(A^{2}-5 A\right) A^{-1}=\)

1 \(\left[\begin{array}{ccc}4 & 2 & 3 \\ -1 & 4 & 2 \\ 1 & 2 & 1\end{array}\right]\)
2 \(\left[\begin{array}{ccc}-4 & 2 & 3 \\ -1 & -4 & 2 \\ 1 & 2 & -1\end{array}\right]\)
3 \(\left[\begin{array}{ccc}-4 & -1 & 1 \\ 2 & -4 & 2 \\ 3 & 2 & -1\end{array}\right]\)
4 \(\left[\begin{array}{ccc}-1 & -2 & 1 \\ 4 & -2 & -3 \\ 1 & 4 & -2\end{array}\right]\)
MHT CET-2018
Matrix and Determinant

78812 If \(A=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right], B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\) then
\(\left(\mathrm{B}^{-1} \mathrm{~A}^{-1}\right)^{-1}=\)

1 \(\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]\)
2 \(\left[\begin{array}{cc}2 & 2 \\ -2 & 3\end{array}\right]\)
3 \(\left[\begin{array}{cc}2 & -3 \\ 2 & 2\end{array}\right]\)
4 \(\left[\begin{array}{cc}1 & -1 \\ -2 & 3\end{array}\right]\)
MHT CET-2016
Matrix and Determinant

78813 If matrix \(A=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]\), such that, \(A X=I\) then \(\mathrm{X}=\) \(\qquad\)

1 \(\overline{\frac{1}{5}}\left[\begin{array}{cc}1 & 3 \\ 2 & -1\end{array}\right]\)
2 \(\frac{1}{5}\left[\begin{array}{cc}4 & 2 \\ 4 & -1\end{array}\right]\)
3 \(\frac{1}{5}\left[\begin{array}{cc}-3 & 2 \\ 4 & -1\end{array}\right]\)
4 \(\frac{1}{5}\left[\begin{array}{ll}-1 & 2 \\ -1 & 4\end{array}\right]\)
MHT CET-2016
Matrix and Determinant

78814 The inverse of matrix \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) is

1 \(\left[\begin{array}{cc}\mathrm{d} & -\mathrm{b} \\ -\mathrm{c} & \mathrm{a}\end{array}\right]\)
2 \(\frac{1}{(\mathrm{ad}-\mathrm{bc})}\left[\begin{array}{cc}\mathrm{d} & -\mathrm{b} \\ -\mathrm{c} & \mathrm{a}\end{array}\right]\)
3 \(\frac{1}{|\mathrm{~A}|}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\)
4 \(\left[\begin{array}{ll}\mathrm{b} & -\mathrm{a} \\ \mathrm{d} & -\mathrm{c}\end{array}\right]\)
MHT CET-2010
Matrix and Determinant

78816 If \(A=\left[\begin{array}{ll}5 & 4 \\ 3 & 2\end{array}\right]\), then \(A^{-1}\) is equal to

1 \(\frac{1}{2}\left[\begin{array}{cc}-2 & 4 \\ 3 & 5\end{array}\right]\)
2 \(\frac{1}{2}\left[\begin{array}{ll}5 & 3 \\ 4 & 2\end{array}\right]\)
3 \(\frac{1}{2}\left[\begin{array}{cc}5 & -4 \\ -3 & 2\end{array}\right]\)
4 \(-\frac{1}{2}\left[\begin{array}{cc}2 & -4 \\ -3 & 5\end{array}\right]\)
MHT CET-2007
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Matrix and Determinant

78811 If \(A=\left[\begin{array}{ccc}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]\) then \(\left(A^{2}-5 A\right) A^{-1}=\)

1 \(\left[\begin{array}{ccc}4 & 2 & 3 \\ -1 & 4 & 2 \\ 1 & 2 & 1\end{array}\right]\)
2 \(\left[\begin{array}{ccc}-4 & 2 & 3 \\ -1 & -4 & 2 \\ 1 & 2 & -1\end{array}\right]\)
3 \(\left[\begin{array}{ccc}-4 & -1 & 1 \\ 2 & -4 & 2 \\ 3 & 2 & -1\end{array}\right]\)
4 \(\left[\begin{array}{ccc}-1 & -2 & 1 \\ 4 & -2 & -3 \\ 1 & 4 & -2\end{array}\right]\)
MHT CET-2018
Matrix and Determinant

78812 If \(A=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right], B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\) then
\(\left(\mathrm{B}^{-1} \mathrm{~A}^{-1}\right)^{-1}=\)

1 \(\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]\)
2 \(\left[\begin{array}{cc}2 & 2 \\ -2 & 3\end{array}\right]\)
3 \(\left[\begin{array}{cc}2 & -3 \\ 2 & 2\end{array}\right]\)
4 \(\left[\begin{array}{cc}1 & -1 \\ -2 & 3\end{array}\right]\)
MHT CET-2016
Matrix and Determinant

78813 If matrix \(A=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]\), such that, \(A X=I\) then \(\mathrm{X}=\) \(\qquad\)

1 \(\overline{\frac{1}{5}}\left[\begin{array}{cc}1 & 3 \\ 2 & -1\end{array}\right]\)
2 \(\frac{1}{5}\left[\begin{array}{cc}4 & 2 \\ 4 & -1\end{array}\right]\)
3 \(\frac{1}{5}\left[\begin{array}{cc}-3 & 2 \\ 4 & -1\end{array}\right]\)
4 \(\frac{1}{5}\left[\begin{array}{ll}-1 & 2 \\ -1 & 4\end{array}\right]\)
MHT CET-2016
Matrix and Determinant

78814 The inverse of matrix \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) is

1 \(\left[\begin{array}{cc}\mathrm{d} & -\mathrm{b} \\ -\mathrm{c} & \mathrm{a}\end{array}\right]\)
2 \(\frac{1}{(\mathrm{ad}-\mathrm{bc})}\left[\begin{array}{cc}\mathrm{d} & -\mathrm{b} \\ -\mathrm{c} & \mathrm{a}\end{array}\right]\)
3 \(\frac{1}{|\mathrm{~A}|}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\)
4 \(\left[\begin{array}{ll}\mathrm{b} & -\mathrm{a} \\ \mathrm{d} & -\mathrm{c}\end{array}\right]\)
MHT CET-2010
Matrix and Determinant

78816 If \(A=\left[\begin{array}{ll}5 & 4 \\ 3 & 2\end{array}\right]\), then \(A^{-1}\) is equal to

1 \(\frac{1}{2}\left[\begin{array}{cc}-2 & 4 \\ 3 & 5\end{array}\right]\)
2 \(\frac{1}{2}\left[\begin{array}{ll}5 & 3 \\ 4 & 2\end{array}\right]\)
3 \(\frac{1}{2}\left[\begin{array}{cc}5 & -4 \\ -3 & 2\end{array}\right]\)
4 \(-\frac{1}{2}\left[\begin{array}{cc}2 & -4 \\ -3 & 5\end{array}\right]\)
MHT CET-2007
NEET DIGITAL LIBRARY
Matrix and Determinant

78811 If \(A=\left[\begin{array}{ccc}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]\) then \(\left(A^{2}-5 A\right) A^{-1}=\)

1 \(\left[\begin{array}{ccc}4 & 2 & 3 \\ -1 & 4 & 2 \\ 1 & 2 & 1\end{array}\right]\)
2 \(\left[\begin{array}{ccc}-4 & 2 & 3 \\ -1 & -4 & 2 \\ 1 & 2 & -1\end{array}\right]\)
3 \(\left[\begin{array}{ccc}-4 & -1 & 1 \\ 2 & -4 & 2 \\ 3 & 2 & -1\end{array}\right]\)
4 \(\left[\begin{array}{ccc}-1 & -2 & 1 \\ 4 & -2 & -3 \\ 1 & 4 & -2\end{array}\right]\)
MHT CET-2018
Matrix and Determinant

78812 If \(A=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right], B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\) then
\(\left(\mathrm{B}^{-1} \mathrm{~A}^{-1}\right)^{-1}=\)

1 \(\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]\)
2 \(\left[\begin{array}{cc}2 & 2 \\ -2 & 3\end{array}\right]\)
3 \(\left[\begin{array}{cc}2 & -3 \\ 2 & 2\end{array}\right]\)
4 \(\left[\begin{array}{cc}1 & -1 \\ -2 & 3\end{array}\right]\)
MHT CET-2016
Matrix and Determinant

78813 If matrix \(A=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]\), such that, \(A X=I\) then \(\mathrm{X}=\) \(\qquad\)

1 \(\overline{\frac{1}{5}}\left[\begin{array}{cc}1 & 3 \\ 2 & -1\end{array}\right]\)
2 \(\frac{1}{5}\left[\begin{array}{cc}4 & 2 \\ 4 & -1\end{array}\right]\)
3 \(\frac{1}{5}\left[\begin{array}{cc}-3 & 2 \\ 4 & -1\end{array}\right]\)
4 \(\frac{1}{5}\left[\begin{array}{ll}-1 & 2 \\ -1 & 4\end{array}\right]\)
MHT CET-2016
Matrix and Determinant

78814 The inverse of matrix \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) is

1 \(\left[\begin{array}{cc}\mathrm{d} & -\mathrm{b} \\ -\mathrm{c} & \mathrm{a}\end{array}\right]\)
2 \(\frac{1}{(\mathrm{ad}-\mathrm{bc})}\left[\begin{array}{cc}\mathrm{d} & -\mathrm{b} \\ -\mathrm{c} & \mathrm{a}\end{array}\right]\)
3 \(\frac{1}{|\mathrm{~A}|}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\)
4 \(\left[\begin{array}{ll}\mathrm{b} & -\mathrm{a} \\ \mathrm{d} & -\mathrm{c}\end{array}\right]\)
MHT CET-2010
Matrix and Determinant

78816 If \(A=\left[\begin{array}{ll}5 & 4 \\ 3 & 2\end{array}\right]\), then \(A^{-1}\) is equal to

1 \(\frac{1}{2}\left[\begin{array}{cc}-2 & 4 \\ 3 & 5\end{array}\right]\)
2 \(\frac{1}{2}\left[\begin{array}{ll}5 & 3 \\ 4 & 2\end{array}\right]\)
3 \(\frac{1}{2}\left[\begin{array}{cc}5 & -4 \\ -3 & 2\end{array}\right]\)
4 \(-\frac{1}{2}\left[\begin{array}{cc}2 & -4 \\ -3 & 5\end{array}\right]\)
MHT CET-2007