NCERT Solutions · Class 12 Mathematics Determinants

Let \(\displaystyle A=\left[\begin{array}{rrr}1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha\end{array}\right]\). Expanding along the first row and using \(\displaystyle \sin^{2}\alpha+\cos^{2}\alpha=1\), \[|A|=1\left|\begin{array}{rr}\cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha\end{array}\right|=-\cos^{2}\alpha-\sin^{2}\alpha=-1.\] This is \(\displaystyle \neq 0\) for every value of \(\displaystyle \alpha\), so the inverse always exists.Cofactors: \[A_{11}=-1,\qquad A_{12}=-\left|\begin{array}{rr}0 & \sin\alpha \\ 0 & -\cos\alpha\end{array}\right|=0,\qquad A_{13}=\left|\begin{array}{rr}0 & \cos\alpha \\ 0 & \sin\alpha\end{array}\right|=0,\] \[A_{21}=-\left|\begin{array}{rr}0 & 0 \\ \sin\alpha & -\cos\alpha\end{array}\right|=0,\qquad A_{22}=\left|\begin{array}{rr}1 & 0 \\ 0 & -\cos\alpha\end{array}\right|=-\cos\alpha,\qquad A_{23}=-\left|\begin{array}{rr}1 & 0 \\ 0 & \sin\alpha\end{array}\right|=-\sin\alpha,\] \[A_{31}=0,\qquad A_{32}=-\left|\begin{array}{rr}1 & 0 \\ 0 & \sin\alpha\end{array}\right|=-\sin\alpha,\qquad A_{33}=\left|\begin{array}{rr}1 & 0 \\ 0 & \cos\alpha\end{array}\right|=\cos\alpha.\] The cofactor matrix is symmetric here, so transposing changes nothing: \[\mathrm{adj}\,A=\left[\begin{array}{rrr}-1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha\end{array}\right].\] Dividing by \(\displaystyle |A|=-1\), \[A^{-1}=\left[\begin{array}{rrr}1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha\end{array}\right]=A.\] So \(\displaystyle A\) is its own inverse, i.e. \(\displaystyle A^{2}=I\), for every \(\displaystyle \alpha\).

Each inverse is found from \(\displaystyle A^{-1}=\dfrac{1}{|A|}\mathrm{adj}\,A\).The two inverses. \(\displaystyle |A|=3(5)-7(2)=15-14=1\), so \[A^{-1}=\frac{1}{1}\left[\begin{array}{rr}5 & -7 \\ -2 & 3\end{array}\right]=\left[\begin{array}{rr}5 & -7 \\ -2 & 3\end{array}\right].\] \(\displaystyle |B|=6(9)-8(7)=54-56=-2\), so \[B^{-1}=\frac{1}{-2}\left[\begin{array}{rr}9 & -8 \\ -7 & 6\end{array}\right].\]Left-hand side. First \[AB=\left[\begin{array}{rr}3 & 7 \\ 2 & 5\end{array}\right]\left[\begin{array}{rr}6 & 8 \\ 7 & 9\end{array}\right]=\left[\begin{array}{rr}18+49 & 24+63 \\ 12+35 & 16+45\end{array}\right]=\left[\begin{array}{rr}67 & 87 \\ 47 & 61\end{array}\right],\] \[|AB|=67(61)-87(47)=4087-4089=-2\quad(=|A||B|,\text{ as expected}),\] \[(AB)^{-1}=\frac{1}{-2}\left[\begin{array}{rr}61 & -87 \\ -47 & 67\end{array}\right].\]Right-hand side. Note the order: \(\displaystyle B^{-1}\) comes first. \[B^{-1}A^{-1}=\frac{1}{-2}\left[\begin{array}{rr}9 & -8 \\ -7 & 6\end{array}\right]\left[\begin{array}{rr}5 & -7 \\ -2 & 3\end{array}\right]=\frac{1}{-2}\left[\begin{array}{rr}45+16 & -63-24 \\ -35-12 & 49+18\end{array}\right]=\frac{1}{-2}\left[\begin{array}{rr}61 & -87 \\ -47 & 67\end{array}\right].\]The two results agree, so \(\displaystyle (AB)^{-1}=B^{-1}A^{-1}\) is verified. (Multiplying in the other order, \(\displaystyle A^{-1}B^{-1}\), gives a different matrix -- the reversal of order is the whole content of the result.)

With \(\displaystyle A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]\), \[A^{2}=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]=\left[\begin{array}{rr}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]=\left[\begin{array}{rr}8 & 5 \\ -5 & 3\end{array}\right].\] Then \[A^{2}-5A+7I=\left[\begin{array}{rr}8 & 5 \\ -5 & 3\end{array}\right]-\left[\begin{array}{rr}15 & 5 \\ -5 & 10\end{array}\right]+\left[\begin{array}{rr}7 & 0 \\ 0 & 7\end{array}\right]=\left[\begin{array}{rr}8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7\end{array}\right]=\left[\begin{array}{rr}0 & 0 \\ 0 & 0\end{array}\right]=O.\]Hence \(\displaystyle A^{-1}\). Rearrange the relation so that \(\displaystyle A\) is a factor -- do not divide by \(\displaystyle A\), and do not assume in advance that \(\displaystyle A^{-1}\) exists: \[7I=5A-A^{2}=A(5I-A)\quad\Longrightarrow\quad A\left[\tfrac{1}{7}(5I-A)\right]=I.\] So \(\displaystyle A\) is invertible and \[A^{-1}=\frac{1}{7}(5I-A)=\frac{1}{7}\left(\left[\begin{array}{rr}5 & 0 \\ 0 & 5\end{array}\right]-\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]\right)=\frac{1}{7}\left[\begin{array}{rr}2 & -1 \\ 1 & 3\end{array}\right].\] (Check: \(\displaystyle |A|=6+1=7\) and \(\displaystyle \mathrm{adj}\,A=\left[\begin{array}{rr}2 & -1 \\ 1 & 3\end{array}\right]\), giving the same answer.)

With \(\displaystyle A=\left[\begin{array}{rr}3 & 2 \\ 1 & 1\end{array}\right]\), \[A^{2}=\left[\begin{array}{rr}3 & 2 \\ 1 & 1\end{array}\right]\left[\begin{array}{rr}3 & 2 \\ 1 & 1\end{array}\right]=\left[\begin{array}{rr}9+2 & 6+2 \\ 3+1 & 2+1\end{array}\right]=\left[\begin{array}{rr}11 & 8 \\ 4 & 3\end{array}\right].\] Substituting into \(\displaystyle A^{2}+aA+bI=O\): \[\left[\begin{array}{rr}11 & 8 \\ 4 & 3\end{array}\right]+\left[\begin{array}{rr}3a & 2a \\ a & a\end{array}\right]+\left[\begin{array}{rr}b & 0 \\ 0 & b\end{array}\right]=\left[\begin{array}{rr}11+3a+b & 8+2a \\ 4+a & 3+a+b\end{array}\right]=\left[\begin{array}{rr}0 & 0 \\ 0 & 0\end{array}\right].\] Two matrices are equal only if corresponding entries are equal, which gives four equations: \[11+3a+b=0,\qquad 8+2a=0,\qquad 4+a=0,\qquad 3+a+b=0.\] The \(\displaystyle (1,2)\) and \(\displaystyle (2,1)\) entries both give \(\displaystyle a=-4\) (consistent). Putting \(\displaystyle a=-4\) into the \(\displaystyle (1,1)\) entry: \[11-12+b=0\ \Rightarrow\ b=1,\] and the \(\displaystyle (2,2)\) entry checks out: \(\displaystyle 3-4+1=0\).\[\boxed{a=-4,\quad b=1}\] (These are \(\displaystyle -\)trace\(\displaystyle (A)\) and \(\displaystyle \det(A)\) respectively, as the Cayley--Hamilton theorem predicts.)

Let \(\displaystyle A=\left[\begin{array}{rrr}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]\).Step $\displaystyle 1$: \(\displaystyle A^{2}\). \[A^{2}=\left[\begin{array}{rrr}1+1+2 & 1+2-1 & 1-3+3 \\ 1+2-6 & 1+4+3 & 1-6-9 \\ 2-1+6 & 2-2-3 & 2+3+9\end{array}\right]=\left[\begin{array}{rrr}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right].\]Step $\displaystyle 2$: \(\displaystyle A^{3}=A^{2}A\). \[A^{3}=\left[\begin{array}{rrr}4+2+2 & 4+4-1 & 4-6+3 \\ -3+8-28 & -3+16+14 & -3-24-42 \\ 7-3+28 & 7-6-14 & 7+9+42\end{array}\right]=\left[\begin{array}{rrr}8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{array}\right].\]Step $\displaystyle 3$: combine. \[A^{3}-6A^{2}+5A+11I=\left[\begin{array}{rrr}8-24+5+11 & 7-12+5+0 & 1-6+5+0 \\ -23+18+5+0 & 27-48+10+11 & -69+84-15+0 \\ 32-42+10+0 & -13+18-5+0 & 58-84+15+11\end{array}\right]=\left[\begin{array}{rrr}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=O.\]Step $\displaystyle 4$: hence \(\displaystyle A^{-1}\). Factor \(\displaystyle A\) out of the first three terms rather than "dividing by \(\displaystyle A\)": \[11I=-\left(A^{3}-6A^{2}+5A\right)=A\left(-A^{2}+6A-5I\right)\ \Longrightarrow\ A\left[\tfrac{1}{11}\left(-A^{2}+6A-5I\right)\right]=I,\] so \(\displaystyle A\) is invertible with \(\displaystyle A^{-1}=\dfrac{1}{11}\left(-A^{2}+6A-5I\right)\).Now \[-A^{2}+6A-5I=\left[\begin{array}{rrr}-4+6-5 & -2+6 & -1+6 \\ 3+6 & -8+12-5 & 14-18 \\ -7+12 & 3-6 & -14+18-5\end{array}\right]=\left[\begin{array}{rrr}-3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1\end{array}\right],\] and therefore \[A^{-1}=\frac{1}{11}\left[\begin{array}{rrr}-3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1\end{array}\right].\]

Let \(\displaystyle A=\left[\begin{array}{rrr}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]\).Step $\displaystyle 1$: \(\displaystyle A^{2}\). \[A^{2}=\left[\begin{array}{rrr}4+1+1 & -2-2-1 & 2+1+2 \\ -2-2-1 & 1+4+1 & -1-2-2 \\ 2+1+2 & -1-2-2 & 1+1+4\end{array}\right]=\left[\begin{array}{rrr}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{array}\right].\]Step $\displaystyle 2$: \(\displaystyle A^{3}=A^{2}A\). \[A^{3}=\left[\begin{array}{rrr}12+5+5 & -6-10-5 & 6+5+10 \\ -10-6-5 & 5+12+5 & -5-6-10 \\ 10+5+6 & -5-10-6 & 5+5+12\end{array}\right]=\left[\begin{array}{rrr}22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22\end{array}\right].\]Step $\displaystyle 3$: verify the relation. \[A^{3}-6A^{2}+9A-4I=\left[\begin{array}{rrr}22-36+18-4 & -21+30-9 & 21-30+9 \\ -21+30-9 & 22-36+18-4 & -21+30-9 \\ 21-30+9 & -21+30-9 & 22-36+18-4\end{array}\right]=O,\] since every diagonal entry is \(\displaystyle 22-36+18-4=0\) and every off-diagonal entry is \(\displaystyle \pm(21-30+9)=0\).Step $\displaystyle 4$: hence \(\displaystyle A^{-1}\). Rewrite the identity with \(\displaystyle A\) factored out: \[4I=A^{3}-6A^{2}+9A=A\left(A^{2}-6A+9I\right)\ \Longrightarrow\ A\left[\tfrac{1}{4}\left(A^{2}-6A+9I\right)\right]=I,\] so \(\displaystyle A\) is invertible and \(\displaystyle A^{-1}=\dfrac{1}{4}\left(A^{2}-6A+9I\right)\).Computing that bracket, \[A^{2}-6A+9I=\left[\begin{array}{rrr}6-12+9 & -5+6 & 5-6 \\ -5+6 & 6-12+9 & -5+6 \\ 5-6 & -5+6 & 6-12+9\end{array}\right]=\left[\begin{array}{rrr}3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3\end{array}\right],\] so \[A^{-1}=\frac{1}{4}\left[\begin{array}{rrr}3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3\end{array}\right].\]

Start from the theorem that holds for every square matrix of order \(\displaystyle n\): \[A(\mathrm{adj}\,A)=|A|\,I_{n}.\] Here \(\displaystyle n=3\). Take determinants of both sides and use \(\displaystyle |XY|=|X||Y|\) together with \(\displaystyle |kI_{3}|=k^{3}\) (a scalar \(\displaystyle k\) multiplies each of the three rows): \[|A|\,|\mathrm{adj}\,A|=\left||A|\,I_{3}\right|=|A|^{3}.\] Because \(\displaystyle A\) is nonsingular, \(\displaystyle |A|\neq 0\), so we may cancel one factor of \(\displaystyle |A|\): \[|\mathrm{adj}\,A|=|A|^{2}.\] (The nonsingularity is exactly what licenses the cancellation -- without it the step is invalid.)Hence the correct option is (B) \(\displaystyle |A|^{2}\). In general \(\displaystyle |\mathrm{adj}\,A|=|A|^{\,n-1}\).

By the definition of the inverse, \(\displaystyle AA^{-1}=I\). Take determinants of both sides and use the product rule \(\displaystyle |XY|=|X||Y|\) with \(\displaystyle |I|=1\): \[\det(A)\cdot\det\!\left(A^{-1}\right)=\det(I)=1.\] Since \(\displaystyle A\) is invertible, \(\displaystyle \det(A)\neq 0\), so dividing is legitimate: \[\det\!\left(A^{-1}\right)=\frac{1}{\det(A)}.\]Hence the correct option is (B) \(\displaystyle \dfrac{1}{\det(A)}\). (The order $\displaystyle 2$ plays no part -- the result holds for an invertible matrix of any order.)