NCERT Solutions · Class 12 Mathematics Determinants

The adjoint of a square matrix is the transpose of its matrix of cofactors: \(\displaystyle \mathrm{adj}\,A=[A_{ij}]^{T}\), where \(\displaystyle A_{ij}=(-1)^{i+j}M_{ij}\).For \(\displaystyle A=\left[\begin{array}{rr}1 & 2 \\ 3 & 4\end{array}\right]\), deleting the row and column of each entry leaves a single number, so \[A_{11}=+4,\qquad A_{12}=-3,\qquad A_{21}=-2,\qquad A_{22}=+1.\] The cofactor matrix is \(\displaystyle \left[\begin{array}{rr}4 & -3 \\ -2 & 1\end{array}\right]\); transposing it, \[\mathrm{adj}\,A=\left[\begin{array}{rr}4 & -2 \\ -3 & 1\end{array}\right].\] This is the standard \(\displaystyle 2\times 2\) pattern: interchange the two diagonal entries and reverse the signs of the other two.

\(\displaystyle \mathrm{adj}\,A\) is the transpose of the cofactor matrix, with \(\displaystyle A_{ij}=(-1)^{i+j}M_{ij}\) and \(\displaystyle M_{ij}\) the \(\displaystyle 2\times 2\) minor left after deleting row \(\displaystyle i\) and column \(\displaystyle j\) of \[A=\left[\begin{array}{rrr}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right].\]Row $\displaystyle 1$ cofactors: \[A_{11}=+\left|\begin{array}{rr}3 & 5 \\ 0 & 1\end{array}\right|=3,\qquad A_{12}=-\left|\begin{array}{rr}2 & 5 \\ -2 & 1\end{array}\right|=-(2+10)=-12,\qquad A_{13}=+\left|\begin{array}{rr}2 & 3 \\ -2 & 0\end{array}\right|=0+6=6.\] Row $\displaystyle 2$ cofactors: \[A_{21}=-\left|\begin{array}{rr}-1 & 2 \\ 0 & 1\end{array}\right|=-(-1-0)=1,\qquad A_{22}=+\left|\begin{array}{rr}1 & 2 \\ -2 & 1\end{array}\right|=1+4=5,\qquad A_{23}=-\left|\begin{array}{rr}1 & -1 \\ -2 & 0\end{array}\right|=-(0-2)=2.\] Row $\displaystyle 3$ cofactors: \[A_{31}=+\left|\begin{array}{rr}-1 & 2 \\ 3 & 5\end{array}\right|=-5-6=-11,\qquad A_{32}=-\left|\begin{array}{rr}1 & 2 \\ 2 & 5\end{array}\right|=-(5-4)=-1,\qquad A_{33}=+\left|\begin{array}{rr}1 & -1 \\ 2 & 3\end{array}\right|=3+2=5.\]Now transpose the array of cofactors -- this is the step most often dropped, and it sends \(\displaystyle A_{12}=-12\) into row $\displaystyle 2$, column $\displaystyle 1$: \[\mathrm{adj}\,A=\left[\begin{array}{rrr}3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5\end{array}\right].\]

For \(\displaystyle A=\left[\begin{array}{rr}2 & 3 \\ -4 & -6\end{array}\right]\), \[|A|=(2)(-6)-(3)(-4)=-12+12=0,\] so \(\displaystyle A\) is singular. The identity \(\displaystyle A(\mathrm{adj}\,A)=(\mathrm{adj}\,A)A=|A|I\) holds for every square matrix, singular or not, so it is still to be checked.By the \(\displaystyle 2\times 2\) rule (swap the diagonal entries, change the sign of the others), \[\mathrm{adj}\,A=\left[\begin{array}{rr}-6 & -3 \\ 4 & 2\end{array}\right].\] Then \[A(\mathrm{adj}\,A)=\left[\begin{array}{rr}2 & 3 \\ -4 & -6\end{array}\right]\left[\begin{array}{rr}-6 & -3 \\ 4 & 2\end{array}\right]=\left[\begin{array}{rr}-12+12 & -6+6 \\ 24-24 & 12-12\end{array}\right]=\left[\begin{array}{rr}0 & 0 \\ 0 & 0\end{array}\right],\] \[(\mathrm{adj}\,A)A=\left[\begin{array}{rr}-6 & -3 \\ 4 & 2\end{array}\right]\left[\begin{array}{rr}2 & 3 \\ -4 & -6\end{array}\right]=\left[\begin{array}{rr}-12+12 & -18+18 \\ 8-8 & 12-12\end{array}\right]=\left[\begin{array}{rr}0 & 0 \\ 0 & 0\end{array}\right].\] Also \(\displaystyle |A|\,I=0\cdot I=O\).Hence \(\displaystyle A(\mathrm{adj}\,A)=(\mathrm{adj}\,A)A=|A|I\), all three being the zero matrix -- which is exactly why a matrix with \(\displaystyle |A|=0\) can have no inverse.

Let \(\displaystyle A=\left[\begin{array}{rrr}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]\). Expand \(\displaystyle |A|\) along the second column, which carries two zeros: \[|A|=(-1)A_{12}+0\cdot A_{22}+0\cdot A_{32},\qquad A_{12}=-\left|\begin{array}{rr}3 & -2 \\ 1 & 3\end{array}\right|=-(9+2)=-11,\] so \(\displaystyle |A|=(-1)(-11)=11\).The remaining cofactors: \[A_{11}=+\left|\begin{array}{rr}0 & -2 \\ 0 & 3\end{array}\right|=0,\qquad A_{13}=+\left|\begin{array}{rr}3 & 0 \\ 1 & 0\end{array}\right|=0,\] \[A_{21}=-\left|\begin{array}{rr}-1 & 2 \\ 0 & 3\end{array}\right|=-(-3)=3,\qquad A_{22}=+\left|\begin{array}{rr}1 & 2 \\ 1 & 3\end{array}\right|=1,\qquad A_{23}=-\left|\begin{array}{rr}1 & -1 \\ 1 & 0\end{array}\right|=-(0+1)=-1,\] \[A_{31}=+\left|\begin{array}{rr}-1 & 2 \\ 0 & -2\end{array}\right|=2,\qquad A_{32}=-\left|\begin{array}{rr}1 & 2 \\ 3 & -2\end{array}\right|=-(-2-6)=8,\qquad A_{33}=+\left|\begin{array}{rr}1 & -1 \\ 3 & 0\end{array}\right|=3.\] Transposing the cofactor matrix, \[\mathrm{adj}\,A=\left[\begin{array}{rrr}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right].\]Now multiply, in both orders: \[A(\mathrm{adj}\,A)=\left[\begin{array}{rrr}0+11+0 & 3-1-2 & 2-8+6 \\ 0+0+0 & 9+0+2 & 6+0-6 \\ 0+0+0 & 3+0-3 & 2+0+9\end{array}\right]=\left[\begin{array}{rrr}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right],\] \[(\mathrm{adj}\,A)A=\left[\begin{array}{rrr}0+9+2 & 0+0+0 & 0-6+6 \\ -11+3+8 & 11+0+0 & -22-2+24 \\ 0-3+3 & 0+0+0 & 0+2+9\end{array}\right]=\left[\begin{array}{rrr}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right].\] Both products equal \(\displaystyle 11I=|A|\,I\). Hence \(\displaystyle A(\mathrm{adj}\,A)=(\mathrm{adj}\,A)A=|A|I\) is verified.

A square matrix is invertible precisely when \(\displaystyle |A|\neq 0\), and then \[A^{-1}=\frac{1}{|A|}\,\mathrm{adj}\,A.\]For \(\displaystyle A=\left[\begin{array}{rr}2 & -2 \\ 4 & 3\end{array}\right]\), \[|A|=(2)(3)-(-2)(4)=6+8=14\neq 0,\] so the inverse exists. (Mind the sign: \(\displaystyle -(-2)(4)=+8\), not \(\displaystyle -8\).)Using the \(\displaystyle 2\times 2\) adjoint rule, \[\mathrm{adj}\,A=\left[\begin{array}{rr}3 & 2 \\ -4 & 2\end{array}\right],\] and therefore \[A^{-1}=\frac{1}{14}\left[\begin{array}{rr}3 & 2 \\ -4 & 2\end{array}\right].\]

Use \(\displaystyle A^{-1}=\dfrac{1}{|A|}\mathrm{adj}\,A\), valid when \(\displaystyle |A|\neq 0\).For \(\displaystyle A=\left[\begin{array}{rr}-1 & 5 \\ -3 & 2\end{array}\right]\), \[|A|=(-1)(2)-(5)(-3)=-2+15=13\neq 0,\] so \(\displaystyle A^{-1}\) exists.Interchanging the diagonal entries and reversing the signs of the off-diagonal ones, \[\mathrm{adj}\,A=\left[\begin{array}{rr}2 & -5 \\ 3 & -1\end{array}\right].\] Hence \[A^{-1}=\frac{1}{13}\left[\begin{array}{rr}2 & -5 \\ 3 & -1\end{array}\right].\]

Let \(\displaystyle A=\left[\begin{array}{rrr}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]\). Expanding along the first column (only one non-zero entry) and again along the first column of the resulting minor, \[|A|=1\left|\begin{array}{rr}2 & 4 \\ 0 & 5\end{array}\right|=1\cdot 2\cdot 5=10\neq 0,\] the familiar fact that the determinant of a triangular matrix is the product of its diagonal entries. So \(\displaystyle A^{-1}\) exists.Cofactors: \[A_{11}=\left|\begin{array}{rr}2 & 4 \\ 0 & 5\end{array}\right|=10,\qquad A_{12}=-\left|\begin{array}{rr}0 & 4 \\ 0 & 5\end{array}\right|=0,\qquad A_{13}=\left|\begin{array}{rr}0 & 2 \\ 0 & 0\end{array}\right|=0,\] \[A_{21}=-\left|\begin{array}{rr}2 & 3 \\ 0 & 5\end{array}\right|=-10,\qquad A_{22}=\left|\begin{array}{rr}1 & 3 \\ 0 & 5\end{array}\right|=5,\qquad A_{23}=-\left|\begin{array}{rr}1 & 2 \\ 0 & 0\end{array}\right|=0,\] \[A_{31}=\left|\begin{array}{rr}2 & 3 \\ 2 & 4\end{array}\right|=8-6=2,\qquad A_{32}=-\left|\begin{array}{rr}1 & 3 \\ 0 & 4\end{array}\right|=-4,\qquad A_{33}=\left|\begin{array}{rr}1 & 2 \\ 0 & 2\end{array}\right|=2.\] Transposing, \[\mathrm{adj}\,A=\left[\begin{array}{rrr}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right].\] Therefore \[A^{-1}=\frac{1}{10}\left[\begin{array}{rrr}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right].\] (The inverse of an upper triangular matrix is again upper triangular, a useful check.)

Let \(\displaystyle A=\left[\begin{array}{rrr}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]\). Expanding along the first row, \[|A|=1\left|\begin{array}{rr}3 & 0 \\ 2 & -1\end{array}\right|-0+0=1(-3-0)=-3\neq 0,\] so \(\displaystyle A^{-1}\) exists.Cofactors: \[A_{11}=\left|\begin{array}{rr}3 & 0 \\ 2 & -1\end{array}\right|=-3,\qquad A_{12}=-\left|\begin{array}{rr}3 & 0 \\ 5 & -1\end{array}\right|=-(-3)=3,\qquad A_{13}=\left|\begin{array}{rr}3 & 3 \\ 5 & 2\end{array}\right|=6-15=-9,\] \[A_{21}=-\left|\begin{array}{rr}0 & 0 \\ 2 & -1\end{array}\right|=0,\qquad A_{22}=\left|\begin{array}{rr}1 & 0 \\ 5 & -1\end{array}\right|=-1,\qquad A_{23}=-\left|\begin{array}{rr}1 & 0 \\ 5 & 2\end{array}\right|=-2,\] \[A_{31}=\left|\begin{array}{rr}0 & 0 \\ 3 & 0\end{array}\right|=0,\qquad A_{32}=-\left|\begin{array}{rr}1 & 0 \\ 3 & 0\end{array}\right|=0,\qquad A_{33}=\left|\begin{array}{rr}1 & 0 \\ 3 & 3\end{array}\right|=3.\] Transposing, \[\mathrm{adj}\,A=\left[\begin{array}{rrr}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right].\] Dividing by \(\displaystyle |A|=-3\) (every sign flips -- the step to be careful with): \[A^{-1}=\frac{1}{-3}\left[\begin{array}{rrr}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]=\frac{1}{3}\left[\begin{array}{rrr}3 & 0 & 0 \\ -3 & 1 & 0 \\ 9 & 2 & -3\end{array}\right]=\left[\begin{array}{rrr}1 & 0 & 0 \\ -1 & \tfrac{1}{3} & 0 \\ 3 & \tfrac{2}{3} & -1\end{array}\right].\]

Let \(\displaystyle A=\left[\begin{array}{rrr}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right]\). Expanding along the first row, \[|A|=2\left|\begin{array}{rr}-1 & 0 \\ 2 & 1\end{array}\right|-1\left|\begin{array}{rr}4 & 0 \\ -7 & 1\end{array}\right|+3\left|\begin{array}{rr}4 & -1 \\ -7 & 2\end{array}\right|=2(-1)-1(4)+3(8-7)=-2-4+3=-3\neq 0,\] so \(\displaystyle A^{-1}\) exists.Cofactors: \[A_{11}=-1,\qquad A_{12}=-\left|\begin{array}{rr}4 & 0 \\ -7 & 1\end{array}\right|=-4,\qquad A_{13}=\left|\begin{array}{rr}4 & -1 \\ -7 & 2\end{array}\right|=8-7=1,\] \[A_{21}=-\left|\begin{array}{rr}1 & 3 \\ 2 & 1\end{array}\right|=-(1-6)=5,\qquad A_{22}=\left|\begin{array}{rr}2 & 3 \\ -7 & 1\end{array}\right|=2+21=23,\qquad A_{23}=-\left|\begin{array}{rr}2 & 1 \\ -7 & 2\end{array}\right|=-(4+7)=-11,\] \[A_{31}=\left|\begin{array}{rr}1 & 3 \\ -1 & 0\end{array}\right|=0+3=3,\qquad A_{32}=-\left|\begin{array}{rr}2 & 3 \\ 4 & 0\end{array}\right|=-(0-12)=12,\qquad A_{33}=\left|\begin{array}{rr}2 & 1 \\ 4 & -1\end{array}\right|=-2-4=-6.\] Transposing, \[\mathrm{adj}\,A=\left[\begin{array}{rrr}-1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6\end{array}\right].\] Hence \[A^{-1}=\frac{1}{-3}\left[\begin{array}{rrr}-1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6\end{array}\right]=\frac{1}{3}\left[\begin{array}{rrr}1 & -5 & -3 \\ 4 & -23 & -12 \\ -1 & 11 & 6\end{array}\right].\]

Let \(\displaystyle A=\left[\begin{array}{rrr}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\). Expanding along the first row, \[|A|=1\left|\begin{array}{rr}2 & -3 \\ -2 & 4\end{array}\right|-(-1)\left|\begin{array}{rr}0 & -3 \\ 3 & 4\end{array}\right|+2\left|\begin{array}{rr}0 & 2 \\ 3 & -2\end{array}\right|=(8-6)+(0+9)+2(0-6)=2+9-12=-1\neq 0,\] so \(\displaystyle A^{-1}\) exists.Cofactors: \[A_{11}=2,\qquad A_{12}=-\left|\begin{array}{rr}0 & -3 \\ 3 & 4\end{array}\right|=-9,\qquad A_{13}=\left|\begin{array}{rr}0 & 2 \\ 3 & -2\end{array}\right|=-6,\] \[A_{21}=-\left|\begin{array}{rr}-1 & 2 \\ -2 & 4\end{array}\right|=-(-4+4)=0,\qquad A_{22}=\left|\begin{array}{rr}1 & 2 \\ 3 & 4\end{array}\right|=-2,\qquad A_{23}=-\left|\begin{array}{rr}1 & -1 \\ 3 & -2\end{array}\right|=-(-2+3)=-1,\] \[A_{31}=\left|\begin{array}{rr}-1 & 2 \\ 2 & -3\end{array}\right|=3-4=-1,\qquad A_{32}=-\left|\begin{array}{rr}1 & 2 \\ 0 & -3\end{array}\right|=3,\qquad A_{33}=\left|\begin{array}{rr}1 & -1 \\ 0 & 2\end{array}\right|=2.\] Transposing, \[\mathrm{adj}\,A=\left[\begin{array}{rrr}2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2\end{array}\right].\] Since \(\displaystyle |A|=-1\), dividing simply reverses every sign: \[A^{-1}=\frac{1}{-1}\left[\begin{array}{rrr}2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2\end{array}\right]=\left[\begin{array}{rrr}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right].\]