NCERT Solutions · Class 12 Mathematics Determinants

Expand the determinant along the first row. \[\Delta=\left|\begin{array}{ccc}x & \sin\theta & \cos\theta\\ -\sin\theta & -x & 1\\ \cos\theta & 1 & x\end{array}\right|\] \[=x\left|\begin{array}{rr}-x & 1\\ 1 & x\end{array}\right|-\sin\theta\left|\begin{array}{rr}-\sin\theta & 1\\ \cos\theta & x\end{array}\right|+\cos\theta\left|\begin{array}{rr}-\sin\theta & -x\\ \cos\theta & 1\end{array}\right|\] Evaluating the three \(\displaystyle 2\times 2\) determinants, \[\Delta=x\left(-x^{2}-1\right)-\sin\theta\left(-x\sin\theta-\cos\theta\right)+\cos\theta\left(-\sin\theta+x\cos\theta\right)\] \[=-x^{3}-x+x\sin^{2}\theta+\sin\theta\cos\theta-\sin\theta\cos\theta+x\cos^{2}\theta\] The step that decides the question: the two \(\displaystyle \sin\theta\cos\theta\) terms come out with opposite signs and cancel, and the two remaining \(\displaystyle \theta\)-terms have the common factor \(\displaystyle x\), so the Pythagorean identity \(\displaystyle \sin^{2}\theta+\cos^{2}\theta=1\) applies: \[\Delta=-x^{3}-x+x\left(\sin^{2}\theta+\cos^{2}\theta\right)=-x^{3}-x+x=-x^{3}\] The value \(\displaystyle \Delta=-x^{3}\) contains no \(\displaystyle \theta\). Hence the determinant is independent of \(\displaystyle \theta\).

\[\Delta=\left|\begin{array}{ccc}\cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha\\ -\sin\beta & \cos\beta & 0\\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha\end{array}\right|\] Expand along \(\displaystyle R_{2}\); it carries a zero, so only two cofactors are needed. With \(\displaystyle C_{ij}=(-1)^{i+j}M_{ij}\), \[\Delta=(-\sin\beta)C_{21}+(\cos\beta)C_{22}+0\cdot C_{23}\] The minors are \[M_{21}=\left|\begin{array}{rr}\cos\alpha\sin\beta & -\sin\alpha\\ \sin\alpha\sin\beta & \cos\alpha\end{array}\right|=\cos^{2}\alpha\sin\beta+\sin^{2}\alpha\sin\beta=\sin\beta\left(\cos^{2}\alpha+\sin^{2}\alpha\right)=\sin\beta\] \[M_{22}=\left|\begin{array}{rr}\cos\alpha\cos\beta & -\sin\alpha\\ \sin\alpha\cos\beta & \cos\alpha\end{array}\right|=\cos^{2}\alpha\cos\beta+\sin^{2}\alpha\cos\beta=\cos\beta\] The sign is where this goes wrong: \(\displaystyle C_{21}=(-1)^{2+1}M_{21}=-\sin\beta\), while \(\displaystyle C_{22}=(-1)^{2+2}M_{22}=+\cos\beta\). Hence \[\Delta=(-\sin\beta)(-\sin\beta)+(\cos\beta)(\cos\beta)=\sin^{2}\beta+\cos^{2}\beta\] \[\Delta=1\] The determinant equals \(\displaystyle 1\) for every \(\displaystyle \alpha\) and \(\displaystyle \beta\).

Use the reversal law for inverses of a product: \[(\mathrm{AB})^{-1}=\mathrm{B}^{-1}\mathrm{A}^{-1}\] Note the order: \(\displaystyle \mathrm{B}^{-1}\) comes first. Since \(\displaystyle \mathrm{A}^{-1}\) is already given, only \(\displaystyle \mathrm{B}^{-1}\) has to be found; \(\displaystyle \mathrm{A}\) itself is never needed.For \(\displaystyle \mathrm{B}=\left[\begin{array}{rrr}1 & 2 & -2\\ -1 & 3 & 0\\ 0 & -2 & 1\end{array}\right]\), expanding \(\displaystyle |\mathrm{B}|\) along \(\displaystyle R_{1}\), \[|\mathrm{B}|=1(3\cdot 1-0\cdot(-2))-2((-1)\cdot 1-0\cdot 0)+(-2)((-1)(-2)-3\cdot 0)=3+2-4=1\] So \(\displaystyle |\mathrm{B}|=1\neq 0\) and \(\displaystyle \mathrm{B}^{-1}\) exists. The cofactors of \(\displaystyle \mathrm{B}\) are \[B_{11}=3,\; B_{12}=1,\; B_{13}=2,\quad B_{21}=2,\; B_{22}=1,\; B_{23}=2,\quad B_{31}=6,\; B_{32}=2,\; B_{33}=5\] The adjoint is the transpose of the cofactor matrix: \[\mathrm{adj}\,\mathrm{B}=\left[\begin{array}{rrr}3 & 2 & 6\\ 1 & 1 & 2\\ 2 & 2 & 5\end{array}\right],\qquad \mathrm{B}^{-1}=\frac{1}{|\mathrm{B}|}\,\mathrm{adj}\,\mathrm{B}=\left[\begin{array}{rrr}3 & 2 & 6\\ 1 & 1 & 2\\ 2 & 2 & 5\end{array}\right]\] Now multiply, keeping the order \(\displaystyle \mathrm{B}^{-1}\mathrm{A}^{-1}\): \[(\mathrm{AB})^{-1}=\left[\begin{array}{rrr}3 & 2 & 6\\ 1 & 1 & 2\\ 2 & 2 & 5\end{array}\right]\left[\begin{array}{rrr}3 & -1 & 1\\ -15 & 6 & -5\\ 5 & -2 & 2\end{array}\right]\] Row $\displaystyle 1$: \(\displaystyle 9-30+30=9\), \(\displaystyle -3+12-12=-3\), \(\displaystyle 3-10+12=5\). Row $\displaystyle 2$: \(\displaystyle 3-15+10=-2\), \(\displaystyle -1+6-4=1\), \(\displaystyle 1-5+4=0\). Row $\displaystyle 3$: \(\displaystyle 6-30+25=1\), \(\displaystyle -2+12-10=0\), \(\displaystyle 2-10+10=2\). \[(\mathrm{AB})^{-1}=\left[\begin{array}{rrr}9 & -3 & 5\\ -2 & 1 & 0\\ 1 & 0 & 2\end{array}\right]\]

Here \(\displaystyle \mathrm{A}=\left[\begin{array}{rrr}1 & 2 & 1\\ 2 & 3 & 1\\ 1 & 1 & 5\end{array}\right]\).Expanding along \(\displaystyle R_{1}\), \[|\mathrm{A}|=1(15-1)-2(10-1)+1(2-3)=14-18-1=-5\neq 0\] The cofactors are \(\displaystyle A_{11}=14,\;A_{12}=-9,\;A_{13}=-1,\;A_{21}=-9,\;A_{22}=4,\;A_{23}=1,\;A_{31}=-1,\;A_{32}=1,\;A_{33}=-1\), so (taking the transpose) \[\mathrm{adj}\,\mathrm{A}=\left[\begin{array}{rrr}14 & -9 & -1\\ -9 & 4 & 1\\ -1 & 1 & -1\end{array}\right],\qquad \mathrm{A}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{adj}\,\mathrm{A}=-\frac{1}{5}\left[\begin{array}{rrr}14 & -9 & -1\\ -9 & 4 & 1\\ -1 & 1 & -1\end{array}\right]\](i) \(\displaystyle [\mathrm{adj}\,\mathrm{A}]^{-1}=\mathrm{adj}\left(\mathrm{A}^{-1}\right)\).The key relation is \(\displaystyle \mathrm{M}\,(\mathrm{adj}\,\mathrm{M})=|\mathrm{M}|\,\mathrm{I}\), i.e. \(\displaystyle \mathrm{adj}\,\mathrm{M}=|\mathrm{M}|\,\mathrm{M}^{-1}\).Left side: \(\displaystyle \mathrm{adj}\,\mathrm{A}=|\mathrm{A}|\,\mathrm{A}^{-1}=-5\mathrm{A}^{-1}\), hence \[[\mathrm{adj}\,\mathrm{A}]^{-1}=\left(-5\mathrm{A}^{-1}\right)^{-1}=-\frac{1}{5}\mathrm{A}\] Right side: applying the same relation to \(\displaystyle \mathrm{A}^{-1}\), and using \(\displaystyle \left|\mathrm{A}^{-1}\right|=\dfrac{1}{|\mathrm{A}|}=-\dfrac{1}{5}\), \[\mathrm{adj}\left(\mathrm{A}^{-1}\right)=\left|\mathrm{A}^{-1}\right|\left(\mathrm{A}^{-1}\right)^{-1}=-\frac{1}{5}\mathrm{A}\] Both sides equal \[-\frac{1}{5}\mathrm{A}=\left[\begin{array}{rrr}-\tfrac{1}{5} & -\tfrac{2}{5} & -\tfrac{1}{5}\\ -\tfrac{2}{5} & -\tfrac{3}{5} & -\tfrac{1}{5}\\ -\tfrac{1}{5} & -\tfrac{1}{5} & -1\end{array}\right]\] Check of the left side by direct multiplication: the first row of \(\displaystyle \mathrm{adj}\,\mathrm{A}\) times the first column of \(\displaystyle -\tfrac{1}{5}\mathrm{A}\) gives \(\displaystyle -\tfrac{14}{5}+\tfrac{18}{5}+\tfrac{1}{5}=1\), and times the second and third columns gives \(\displaystyle -\tfrac{28}{5}+\tfrac{27}{5}+\tfrac{1}{5}=0\) and \(\displaystyle -\tfrac{14}{5}+\tfrac{9}{5}+1=0\); continuing likewise the product is \(\displaystyle \mathrm{I}\). Hence \(\displaystyle [\mathrm{adj}\,\mathrm{A}]^{-1}=\mathrm{adj}\left(\mathrm{A}^{-1}\right)\), verified.(ii) \(\displaystyle \left(\mathrm{A}^{-1}\right)^{-1}=\mathrm{A}\).Apply the inverse formula to the matrix \(\displaystyle \mathrm{A}^{-1}\): \[\left(\mathrm{A}^{-1}\right)^{-1}=\frac{1}{\left|\mathrm{A}^{-1}\right|}\,\mathrm{adj}\left(\mathrm{A}^{-1}\right)=\frac{1}{-\tfrac{1}{5}}\left(-\frac{1}{5}\mathrm{A}\right)=\mathrm{A}\] (Equivalently, \(\displaystyle \mathrm{A}\mathrm{A}^{-1}=\mathrm{A}^{-1}\mathrm{A}=\mathrm{I}\) already says \(\displaystyle \mathrm{A}\) is the inverse of \(\displaystyle \mathrm{A}^{-1}\), and the inverse of a matrix is unique.) Verified.

\[\Delta=\left|\begin{array}{ccc}x & y & x+y\\ y & x+y & x\\ x+y & x & y\end{array}\right|\] Every row has the same sum \(\displaystyle 2(x+y)\), so use the property that adding columns to a column leaves the determinant unchanged. Apply \(\displaystyle C_{1}\to C_{1}+C_{2}+C_{3}\): \[\Delta=\left|\begin{array}{ccc}2(x+y) & y & x+y\\ 2(x+y) & x+y & x\\ 2(x+y) & x & y\end{array}\right|=2(x+y)\left|\begin{array}{ccc}1 & y & x+y\\ 1 & x+y & x\\ 1 & x & y\end{array}\right|\] (a common factor of a column may be taken outside the determinant).Now \(\displaystyle R_{2}\to R_{2}-R_{1}\) and \(\displaystyle R_{3}\to R_{3}-R_{1}\): \[\Delta=2(x+y)\left|\begin{array}{ccc}1 & y & x+y\\ 0 & x & -y\\ 0 & x-y & -x\end{array}\right|\] Expanding along \(\displaystyle C_{1}\), \[\Delta=2(x+y)\left|\begin{array}{rr}x & -y\\ x-y & -x\end{array}\right|=2(x+y)\left[-x^{2}+y(x-y)\right]\] \[=2(x+y)\left(-x^{2}+xy-y^{2}\right)=-2(x+y)\left(x^{2}-xy+y^{2}\right)\] Recognising \(\displaystyle (x+y)\left(x^{2}-xy+y^{2}\right)=x^{3}+y^{3}\), \[\Delta=-2\left(x^{3}+y^{3}\right)\]

\[\Delta=\left|\begin{array}{ccc}1 & x & y\\ 1 & x+y & y\\ 1 & x & x+y\end{array}\right|\] The first column is constant, so subtracting \(\displaystyle R_{1}\) from the other rows (a row operation that does not change the value of a determinant) clears it. Apply \(\displaystyle R_{2}\to R_{2}-R_{1}\) and \(\displaystyle R_{3}\to R_{3}-R_{1}\): \[\Delta=\left|\begin{array}{ccc}1 & x & y\\ 0 & y & 0\\ 0 & 0 & x\end{array}\right|\] This is upper triangular, or expand along \(\displaystyle C_{1}\): \[\Delta=1\cdot\left|\begin{array}{rr}y & 0\\ 0 & x\end{array}\right|=1\cdot(yx-0)\] \[\Delta=xy\]

The equations are not linear in \(\displaystyle x,y,z\), but they are linear in the reciprocals. Since no denominator may vanish, \(\displaystyle x,y,z\neq 0\); put \[a=\frac{1}{x},\qquad b=\frac{1}{y},\qquad c=\frac{1}{z}\] The system becomes \[2a+3b+10c=4,\qquad 4a-6b+5c=1,\qquad 6a+9b-20c=2\] that is \(\displaystyle \mathrm{A}X=\mathrm{B}\) with \[\mathrm{A}=\left[\begin{array}{rrr}2 & 3 & 10\\ 4 & -6 & 5\\ 6 & 9 & -20\end{array}\right],\quad X=\left[\begin{array}{r}a\\ b\\ c\end{array}\right],\quad \mathrm{B}=\left[\begin{array}{r}4\\ 1\\ 2\end{array}\right]\] Expanding along \(\displaystyle R_{1}\), \[|\mathrm{A}|=2(120-45)-3(-80-30)+10(36+36)=150+330+720=1200\neq 0\] so the system has the unique solution \(\displaystyle X=\mathrm{A}^{-1}\mathrm{B}\). The cofactors of \(\displaystyle \mathrm{A}\) are \[A_{11}=75,\;A_{12}=110,\;A_{13}=72,\quad A_{21}=150,\;A_{22}=-100,\;A_{23}=0,\quad A_{31}=75,\;A_{32}=30,\;A_{33}=-24\] Taking the transpose, \[\mathrm{adj}\,\mathrm{A}=\left[\begin{array}{rrr}75 & 150 & 75\\ 110 & -100 & 30\\ 72 & 0 & -24\end{array}\right],\qquad \mathrm{A}^{-1}=\frac{1}{1200}\,\mathrm{adj}\,\mathrm{A}\] Hence \[\left[\begin{array}{r}a\\ b\\ c\end{array}\right]=\frac{1}{1200}\left[\begin{array}{rrr}75 & 150 & 75\\ 110 & -100 & 30\\ 72 & 0 & -24\end{array}\right]\left[\begin{array}{r}4\\ 1\\ 2\end{array}\right]=\frac{1}{1200}\left[\begin{array}{r}300+150+150\\ 440-100+60\\ 288+0-48\end{array}\right]=\frac{1}{1200}\left[\begin{array}{r}600\\ 400\\ 240\end{array}\right]\] So \(\displaystyle a=\dfrac{1}{2},\; b=\dfrac{1}{3},\; c=\dfrac{1}{5}\).The step most often forgotten is the last one: these are the reciprocals, so substitute back. \[x=\frac{1}{a}=2,\qquad y=\frac{1}{b}=3,\qquad z=\frac{1}{c}=5\] Check: \(\displaystyle \tfrac{2}{2}+\tfrac{3}{3}+\tfrac{10}{5}=1+1+2=4\); \(\displaystyle \tfrac{4}{2}-\tfrac{6}{3}+\tfrac{5}{5}=2-2+1=1\); \(\displaystyle \tfrac{6}{2}+\tfrac{9}{3}-\tfrac{20}{5}=3+3-4=2\). All three hold, and \(\displaystyle x,y,z\neq 0\) as required.

\(\displaystyle \mathrm{A}\) is diagonal, so expanding gives \(\displaystyle |\mathrm{A}|=xyz\), which is non-zero because \(\displaystyle x,y,z\) are non-zero. Hence \(\displaystyle \mathrm{A}^{-1}\) exists.Rather than testing options, construct the inverse from the definition \(\displaystyle \mathrm{A}\mathrm{A}^{-1}=\mathrm{I}\). Try \(\displaystyle \mathrm{D}=\left[\begin{array}{ccc}x^{-1} & 0 & 0\\ 0 & y^{-1} & 0\\ 0 & 0 & z^{-1}\end{array}\right]\). Multiplying two diagonal matrices multiplies corresponding diagonal entries: \[\mathrm{A}\mathrm{D}=\left[\begin{array}{ccc}x\cdot x^{-1} & 0 & 0\\ 0 & y\cdot y^{-1} & 0\\ 0 & 0 & z\cdot z^{-1}\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right]=\mathrm{I}\] and likewise \(\displaystyle \mathrm{D}\mathrm{A}=\mathrm{I}\). Since the inverse is unique, \(\displaystyle \mathrm{A}^{-1}=\mathrm{D}\).(The same result follows from \(\displaystyle \mathrm{A}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{adj}\,\mathrm{A}\): here \(\displaystyle \mathrm{adj}\,\mathrm{A}=\left[\begin{array}{ccc}yz & 0 & 0\\ 0 & xz & 0\\ 0 & 0 & xy\end{array}\right]\), and dividing by \(\displaystyle xyz\) gives \(\displaystyle \mathrm{diag}\left(x^{-1},y^{-1},z^{-1}\right)\).)Option (B) is \(\displaystyle xyz\) times this and option (D) is \(\displaystyle \frac{1}{xyz}\mathrm{I}\); neither is diagonal-reciprocal, so both fail. The correct answer is (A).

Expand \(\displaystyle \operatorname{Det}(\mathrm{A})\) along \(\displaystyle R_{1}\). \[\operatorname{Det}(\mathrm{A})=1\left|\begin{array}{cc}1 & \sin\theta\\ -\sin\theta & 1\end{array}\right|-\sin\theta\left|\begin{array}{rr}-\sin\theta & \sin\theta\\ -1 & 1\end{array}\right|+1\left|\begin{array}{rr}-\sin\theta & 1\\ -1 & -\sin\theta\end{array}\right|\] \[=1\left(1+\sin^{2}\theta\right)-\sin\theta\left(-\sin\theta+\sin\theta\right)+1\left(\sin^{2}\theta+1\right)\] The middle \(\displaystyle 2\times 2\) determinant is exactly \(\displaystyle 0\), so \[\operatorname{Det}(\mathrm{A})=\left(1+\sin^{2}\theta\right)+\left(1+\sin^{2}\theta\right)=2+2\sin^{2}\theta\] Now use the range. For \(\displaystyle 0\le\theta\le 2\pi\), \(\displaystyle \sin\theta\) takes every value in \(\displaystyle [-1,1]\), so \(\displaystyle \sin^{2}\theta\) takes every value in \(\displaystyle [0,1]\): \[0\le\sin^{2}\theta\le 1\;\Longrightarrow\;2\le 2+2\sin^{2}\theta\le 4\] Both endpoints are actually attained — \(\displaystyle \theta=0\) gives \(\displaystyle 2\) and \(\displaystyle \theta=\frac{\pi}{2}\) gives \(\displaystyle 4\) — so the interval is closed, which rules out the open intervals in (B) and (C), and \(\displaystyle \operatorname{Det}(\mathrm{A})\ge 2>0\) rules out (A). \[\operatorname{Det}(\mathrm{A})\in[2,4]\] The correct answer is (D).