Adjoint and Inverse of Matrices
Matrix and Determinant

78787 If \(A=\left[\begin{array}{lll}1 & 2 & 1 \\ 2 & 1 & 0\end{array}\right], B=\left[\begin{array}{ll}1 & 2 \\ 2 & 1 \\ 0 & 1\end{array}\right]\), then \((A B)^{-1}\) is

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

78788 If \(A=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}4 & 1 \\ 3 & 1\end{array}\right]\), then \((A+B)^{-1}=\)

1 \(\frac{1}{7}\left[\begin{array}{ll}3 & 2 \\ 4 & 5\end{array}\right]\)
2 \(\frac{1}{7}\left[\begin{array}{cc}3 & -2 \\ -4 & 5\end{array}\right]\)
3 \(7\left[\begin{array}{ll}3 & 2 \\ 4 & 5\end{array}\right]\)
4 \(7\left[\begin{array}{cc}3 & -2 \\ -4 & 5\end{array}\right]\)
Matrix and Determinant

78789 If \(\mathrm{AX}=\mathrm{B}\), where \(\mathrm{A}=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4\end{array}\right], \mathrm{B}=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]\)
and \(X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]\), then \(x+y+z=\)

1 6
2 2
3 1
4 3,
Matrix and Determinant

78790 If \(A=\left[\begin{array}{ccc}0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]\), then

1 \(\mathrm{A}\) is not invertible
2 \(\mathrm{A}^{-1}=2 \mathrm{~A}\)
3 \(\mathrm{A}^{-1}=\mathrm{I}\)
4 \(\mathrm{A}=\mathrm{A}^{-1}\)
Matrix and Determinant

78787 If \(A=\left[\begin{array}{lll}1 & 2 & 1 \\ 2 & 1 & 0\end{array}\right], B=\left[\begin{array}{ll}1 & 2 \\ 2 & 1 \\ 0 & 1\end{array}\right]\), then \((A B)^{-1}\) is

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

78788 If \(A=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}4 & 1 \\ 3 & 1\end{array}\right]\), then \((A+B)^{-1}=\)

1 \(\frac{1}{7}\left[\begin{array}{ll}3 & 2 \\ 4 & 5\end{array}\right]\)
2 \(\frac{1}{7}\left[\begin{array}{cc}3 & -2 \\ -4 & 5\end{array}\right]\)
3 \(7\left[\begin{array}{ll}3 & 2 \\ 4 & 5\end{array}\right]\)
4 \(7\left[\begin{array}{cc}3 & -2 \\ -4 & 5\end{array}\right]\)
Matrix and Determinant

78789 If \(\mathrm{AX}=\mathrm{B}\), where \(\mathrm{A}=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4\end{array}\right], \mathrm{B}=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]\)
and \(X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]\), then \(x+y+z=\)

1 6
2 2
3 1
4 3,
Matrix and Determinant

78790 If \(A=\left[\begin{array}{ccc}0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]\), then

1 \(\mathrm{A}\) is not invertible
2 \(\mathrm{A}^{-1}=2 \mathrm{~A}\)
3 \(\mathrm{A}^{-1}=\mathrm{I}\)
4 \(\mathrm{A}=\mathrm{A}^{-1}\)
Matrix and Determinant

78787 If \(A=\left[\begin{array}{lll}1 & 2 & 1 \\ 2 & 1 & 0\end{array}\right], B=\left[\begin{array}{ll}1 & 2 \\ 2 & 1 \\ 0 & 1\end{array}\right]\), then \((A B)^{-1}\) is

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

78788 If \(A=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}4 & 1 \\ 3 & 1\end{array}\right]\), then \((A+B)^{-1}=\)

1 \(\frac{1}{7}\left[\begin{array}{ll}3 & 2 \\ 4 & 5\end{array}\right]\)
2 \(\frac{1}{7}\left[\begin{array}{cc}3 & -2 \\ -4 & 5\end{array}\right]\)
3 \(7\left[\begin{array}{ll}3 & 2 \\ 4 & 5\end{array}\right]\)
4 \(7\left[\begin{array}{cc}3 & -2 \\ -4 & 5\end{array}\right]\)
Matrix and Determinant

78789 If \(\mathrm{AX}=\mathrm{B}\), where \(\mathrm{A}=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4\end{array}\right], \mathrm{B}=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]\)
and \(X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]\), then \(x+y+z=\)

1 6
2 2
3 1
4 3,
Matrix and Determinant

78790 If \(A=\left[\begin{array}{ccc}0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]\), then

1 \(\mathrm{A}\) is not invertible
2 \(\mathrm{A}^{-1}=2 \mathrm{~A}\)
3 \(\mathrm{A}^{-1}=\mathrm{I}\)
4 \(\mathrm{A}=\mathrm{A}^{-1}\)
Matrix and Determinant

78787 If \(A=\left[\begin{array}{lll}1 & 2 & 1 \\ 2 & 1 & 0\end{array}\right], B=\left[\begin{array}{ll}1 & 2 \\ 2 & 1 \\ 0 & 1\end{array}\right]\), then \((A B)^{-1}\) is

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

78788 If \(A=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}4 & 1 \\ 3 & 1\end{array}\right]\), then \((A+B)^{-1}=\)

1 \(\frac{1}{7}\left[\begin{array}{ll}3 & 2 \\ 4 & 5\end{array}\right]\)
2 \(\frac{1}{7}\left[\begin{array}{cc}3 & -2 \\ -4 & 5\end{array}\right]\)
3 \(7\left[\begin{array}{ll}3 & 2 \\ 4 & 5\end{array}\right]\)
4 \(7\left[\begin{array}{cc}3 & -2 \\ -4 & 5\end{array}\right]\)
Matrix and Determinant

78789 If \(\mathrm{AX}=\mathrm{B}\), where \(\mathrm{A}=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4\end{array}\right], \mathrm{B}=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]\)
and \(X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]\), then \(x+y+z=\)

1 6
2 2
3 1
4 3,
Matrix and Determinant

78790 If \(A=\left[\begin{array}{ccc}0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]\), then

1 \(\mathrm{A}\) is not invertible
2 \(\mathrm{A}^{-1}=2 \mathrm{~A}\)
3 \(\mathrm{A}^{-1}=\mathrm{I}\)
4 \(\mathrm{A}=\mathrm{A}^{-1}\)