Solution of System of Linear Equation Using Matrix
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Matrix and Determinant

78972 If \(A=\left[\begin{array}{ccc}6 & -7 & 8 \\ 1 & -3 & 1 \\ 2 & 1 & -4\end{array}\right]\), then \(A^{-1}=\)

1 \(\frac{1}{80}\left[\begin{array}{ccc}11 & -20 & 17 \\ 6 & -40 & 2 \\ 7 & -20 & -11\end{array}\right]\)
2 \(\frac{1}{80}\left[\begin{array}{ccc}11 & 6 & 7 \\ -20 & -40 & -20 \\ 17 & 2 & -11\end{array}\right]\)
3 \(\frac{1}{80}\left[\begin{array}{ccc}-11 & 20 & -17 \\ -6 & 40 & -2 \\ -7 & 20 & 11\end{array}\right]\)
4 \(\frac{1}{80}\left[\begin{array}{ccc}11 & 20 & 17 \\ 6 & 40 & -2 \\ 7 & 20 & -11\end{array}\right]\)
COMEDK-2020
Matrix and Determinant

78973 If \(A=\left[\begin{array}{lll}0 & -1 & 2 \\ 2 & -2 & 0\end{array}\right], \quad B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0 \\ 1 & 1\end{array}\right]\) and \(M=A B\), then find \(\mathbf{M}^{-1}\).

1 \(\left[\begin{array}{cc}\frac{1}{6} & \frac{-1}{3} \\ \frac{1}{3} & \frac{1}{3}\end{array}\right]\)
2 \(\left[\begin{array}{cc}\frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{6}\end{array}\right]\)
3 \(\left[\begin{array}{cc}\frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{3}\end{array}\right]\)
4 \(\left[\begin{array}{cc}\frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6}\end{array}\right]\)
BITSAT-2006
Matrix and Determinant

78974 Inverse matrix of \(\left[\begin{array}{cc}2 & -3 \\ -4 & 2\end{array}\right]\)

1 \(-\frac{1}{8}\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]\)
2 \(-\frac{1}{8}\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right]\)
3 \(\frac{1}{8}\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]\)
4 \(\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]\)
BITSAT-2005
Matrix and Determinant

78975 If \(M(\alpha)=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]\);
\(M(\beta)=\left[\begin{array}{ccc}\cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta\end{array}\right]\) then \([M(\alpha) M(\beta)]^{-1}\)
is equal to-

1 \(\mathrm{M}(\beta) \mathrm{M}(\alpha)\)
2 \(\mathrm{M}(-\alpha) \mathrm{M}(-\beta)\)
3 \(\mathrm{M}(-\beta) \mathrm{M}(-\alpha)\)
4 \(-\mathrm{M}(\beta) \mathrm{M}(\alpha)\)
BITSAT-2009
NEET DIGITAL LIBRARY
Matrix and Determinant

78972 If \(A=\left[\begin{array}{ccc}6 & -7 & 8 \\ 1 & -3 & 1 \\ 2 & 1 & -4\end{array}\right]\), then \(A^{-1}=\)

1 \(\frac{1}{80}\left[\begin{array}{ccc}11 & -20 & 17 \\ 6 & -40 & 2 \\ 7 & -20 & -11\end{array}\right]\)
2 \(\frac{1}{80}\left[\begin{array}{ccc}11 & 6 & 7 \\ -20 & -40 & -20 \\ 17 & 2 & -11\end{array}\right]\)
3 \(\frac{1}{80}\left[\begin{array}{ccc}-11 & 20 & -17 \\ -6 & 40 & -2 \\ -7 & 20 & 11\end{array}\right]\)
4 \(\frac{1}{80}\left[\begin{array}{ccc}11 & 20 & 17 \\ 6 & 40 & -2 \\ 7 & 20 & -11\end{array}\right]\)
COMEDK-2020
Matrix and Determinant

78973 If \(A=\left[\begin{array}{lll}0 & -1 & 2 \\ 2 & -2 & 0\end{array}\right], \quad B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0 \\ 1 & 1\end{array}\right]\) and \(M=A B\), then find \(\mathbf{M}^{-1}\).

1 \(\left[\begin{array}{cc}\frac{1}{6} & \frac{-1}{3} \\ \frac{1}{3} & \frac{1}{3}\end{array}\right]\)
2 \(\left[\begin{array}{cc}\frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{6}\end{array}\right]\)
3 \(\left[\begin{array}{cc}\frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{3}\end{array}\right]\)
4 \(\left[\begin{array}{cc}\frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6}\end{array}\right]\)
BITSAT-2006
Matrix and Determinant

78974 Inverse matrix of \(\left[\begin{array}{cc}2 & -3 \\ -4 & 2\end{array}\right]\)

1 \(-\frac{1}{8}\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]\)
2 \(-\frac{1}{8}\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right]\)
3 \(\frac{1}{8}\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]\)
4 \(\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]\)
BITSAT-2005
Matrix and Determinant

78975 If \(M(\alpha)=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]\);
\(M(\beta)=\left[\begin{array}{ccc}\cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta\end{array}\right]\) then \([M(\alpha) M(\beta)]^{-1}\)
is equal to-

1 \(\mathrm{M}(\beta) \mathrm{M}(\alpha)\)
2 \(\mathrm{M}(-\alpha) \mathrm{M}(-\beta)\)
3 \(\mathrm{M}(-\beta) \mathrm{M}(-\alpha)\)
4 \(-\mathrm{M}(\beta) \mathrm{M}(\alpha)\)
BITSAT-2009
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Matrix and Determinant

78972 If \(A=\left[\begin{array}{ccc}6 & -7 & 8 \\ 1 & -3 & 1 \\ 2 & 1 & -4\end{array}\right]\), then \(A^{-1}=\)

1 \(\frac{1}{80}\left[\begin{array}{ccc}11 & -20 & 17 \\ 6 & -40 & 2 \\ 7 & -20 & -11\end{array}\right]\)
2 \(\frac{1}{80}\left[\begin{array}{ccc}11 & 6 & 7 \\ -20 & -40 & -20 \\ 17 & 2 & -11\end{array}\right]\)
3 \(\frac{1}{80}\left[\begin{array}{ccc}-11 & 20 & -17 \\ -6 & 40 & -2 \\ -7 & 20 & 11\end{array}\right]\)
4 \(\frac{1}{80}\left[\begin{array}{ccc}11 & 20 & 17 \\ 6 & 40 & -2 \\ 7 & 20 & -11\end{array}\right]\)
COMEDK-2020
Matrix and Determinant

78973 If \(A=\left[\begin{array}{lll}0 & -1 & 2 \\ 2 & -2 & 0\end{array}\right], \quad B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0 \\ 1 & 1\end{array}\right]\) and \(M=A B\), then find \(\mathbf{M}^{-1}\).

1 \(\left[\begin{array}{cc}\frac{1}{6} & \frac{-1}{3} \\ \frac{1}{3} & \frac{1}{3}\end{array}\right]\)
2 \(\left[\begin{array}{cc}\frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{6}\end{array}\right]\)
3 \(\left[\begin{array}{cc}\frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{3}\end{array}\right]\)
4 \(\left[\begin{array}{cc}\frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6}\end{array}\right]\)
BITSAT-2006
Matrix and Determinant

78974 Inverse matrix of \(\left[\begin{array}{cc}2 & -3 \\ -4 & 2\end{array}\right]\)

1 \(-\frac{1}{8}\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]\)
2 \(-\frac{1}{8}\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right]\)
3 \(\frac{1}{8}\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]\)
4 \(\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]\)
BITSAT-2005
Matrix and Determinant

78975 If \(M(\alpha)=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]\);
\(M(\beta)=\left[\begin{array}{ccc}\cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta\end{array}\right]\) then \([M(\alpha) M(\beta)]^{-1}\)
is equal to-

1 \(\mathrm{M}(\beta) \mathrm{M}(\alpha)\)
2 \(\mathrm{M}(-\alpha) \mathrm{M}(-\beta)\)
3 \(\mathrm{M}(-\beta) \mathrm{M}(-\alpha)\)
4 \(-\mathrm{M}(\beta) \mathrm{M}(\alpha)\)
BITSAT-2009
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NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
Matrix and Determinant

78972 If \(A=\left[\begin{array}{ccc}6 & -7 & 8 \\ 1 & -3 & 1 \\ 2 & 1 & -4\end{array}\right]\), then \(A^{-1}=\)

1 \(\frac{1}{80}\left[\begin{array}{ccc}11 & -20 & 17 \\ 6 & -40 & 2 \\ 7 & -20 & -11\end{array}\right]\)
2 \(\frac{1}{80}\left[\begin{array}{ccc}11 & 6 & 7 \\ -20 & -40 & -20 \\ 17 & 2 & -11\end{array}\right]\)
3 \(\frac{1}{80}\left[\begin{array}{ccc}-11 & 20 & -17 \\ -6 & 40 & -2 \\ -7 & 20 & 11\end{array}\right]\)
4 \(\frac{1}{80}\left[\begin{array}{ccc}11 & 20 & 17 \\ 6 & 40 & -2 \\ 7 & 20 & -11\end{array}\right]\)
COMEDK-2020
Matrix and Determinant

78973 If \(A=\left[\begin{array}{lll}0 & -1 & 2 \\ 2 & -2 & 0\end{array}\right], \quad B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0 \\ 1 & 1\end{array}\right]\) and \(M=A B\), then find \(\mathbf{M}^{-1}\).

1 \(\left[\begin{array}{cc}\frac{1}{6} & \frac{-1}{3} \\ \frac{1}{3} & \frac{1}{3}\end{array}\right]\)
2 \(\left[\begin{array}{cc}\frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{6}\end{array}\right]\)
3 \(\left[\begin{array}{cc}\frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{3}\end{array}\right]\)
4 \(\left[\begin{array}{cc}\frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6}\end{array}\right]\)
BITSAT-2006
Matrix and Determinant

78974 Inverse matrix of \(\left[\begin{array}{cc}2 & -3 \\ -4 & 2\end{array}\right]\)

1 \(-\frac{1}{8}\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]\)
2 \(-\frac{1}{8}\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right]\)
3 \(\frac{1}{8}\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]\)
4 \(\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]\)
BITSAT-2005
Matrix and Determinant

78975 If \(M(\alpha)=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]\);
\(M(\beta)=\left[\begin{array}{ccc}\cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta\end{array}\right]\) then \([M(\alpha) M(\beta)]^{-1}\)
is equal to-

1 \(\mathrm{M}(\beta) \mathrm{M}(\alpha)\)
2 \(\mathrm{M}(-\alpha) \mathrm{M}(-\beta)\)
3 \(\mathrm{M}(-\beta) \mathrm{M}(-\alpha)\)
4 \(-\mathrm{M}(\beta) \mathrm{M}(\alpha)\)
BITSAT-2009