NCERT Solutions: Class 12 Maths Chapter 4 - Determinants

For a determinant of order 2, $\begin{vmatrix}a & b \\ c & d\end{vmatrix}=ad-bc$. Thus $\begin{vmatrix}2 & 4 \\ -5 & -1\end{vmatrix}=2(-1)-4(-5)=-2+20=18$.

(i) $\cos \theta\cos \theta-(-\sin \theta)\sin \theta=\cos^2\theta+\sin^2\theta=1$. (ii) The determinant is $(x^2-x+1)(x+1)-(x-1)(x+1)=(x+1)(x^2-2x+2)=x^3-x^2+2$.

$|A|=1\cdot2-2\cdot4=-6$. Also $2A=\begin{bmatrix}2 & 4 \\ 8 & 4\end{bmatrix}$, so $|2A|=2\cdot4-4\cdot8=8-32=-24=4(-6)=4|A|$.

Since $A$ is upper triangular, $|A|=1\cdot1\cdot4=4$. For a determinant of order 3, multiplying every entry by 3 multiplies the determinant by $3^3=27$. Hence $|3A|=27|A|=108$.

Expanding along convenient rows or using $a(ei-fh)-b(di-fg)+c(dh-eg)$ gives: (i) $3(0-5)-(-1)(0+3)+(-2)(0)=-12$; (ii) $3(1+6)-(-4)(1+4)+5(3-2)=46$; (iii) $0-1(0-6)+2(-3-0)=0$; (iv) $2(0-5)-(-1)(0+3)+(-2)(0-6)=5$.

Expanding along the first row, $|A|=1(1\cdot(-9)-(-3)4)-1(2\cdot(-9)-(-3)5)+(-2)(2\cdot4-1\cdot5)=3-(-3)-6=0$.

(i) The left determinant is $2-20=-18$ and the right determinant is $2x^2-24$. Thus $2x^2-24=-18$, so $x^2=3$ and $x=\pm\sqrt3$. (ii) The left determinant is $10-12=-2$ and the right determinant is $5x-6x=-x$. Hence $-x=-2$, so $x=2$.

1. i. $6$
2. ii. $\pm6$
3. iii. $-6$
4. iv. $0$

The equality gives $x^2-36=36-36=0$. Hence $x^2=36$, so $x=\pm6$. Therefore option (ii) is correct.

Use $\Delta=\dfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$. Substitution gives (i) $\dfrac12|1(0-3)+6(3-0)+4(0-0)|=\dfrac{15}{2}$; (ii) $\dfrac12|2(1-8)+1(8-7)+10(7-1)|=\dfrac{47}{2}$; (iii) $\dfrac12|-2(2+8)+3(-8+3)+(-1)(-3-2)|=15$.

The area determinant is $\dfrac12\begin{vmatrix}a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1\end{vmatrix}$. Adding the first two columns gives $a+b+c$ in every row, so $C_1+C_2=(a+b+c)C_3$. The columns are linearly dependent, the determinant is $0$, and therefore the area is $0$. Hence the points are collinear.

(i) Area $=\dfrac12|k(0-2)+4(2-0)|=\dfrac12|-2k+8|=4$, so $|8-2k|=8$ and $k=0$ or $8$. (ii) Area $=\dfrac12|-2(4-k)|=|k-4|=4$, so $k=0$ or $8$.

The equation through $(x_1,y_1)$ and $(x_2,y_2)$ is $\begin{vmatrix}x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1\end{vmatrix}=0$. For $(1,2),(3,6)$ this gives $4x-2y=0$, i.e. $2x-y=0$. For $(3,1),(9,3)$ this gives $2x-6y=0$, i.e. $x-3y=0$.

1. i. $12$
2. ii. $-2$
3. iii. $-12,-2$
4. iv. $12,-2$

$35=\dfrac12|2(4-4)+5(4+6)+k(-6-4)|=\dfrac12|50-10k|=5|5-k|$. Thus $|5-k|=7$, giving $k=12$ or $k=-2$. Therefore option (iv) is correct.

For an order 2 determinant, the minor is the remaining entry after deleting the chosen row and column, and $A_{ij}=(-1)^{i+j}M_{ij}$. Applying this sign pattern $\begin{bmatrix}+ & - \\ - & +\end{bmatrix}$ gives the listed minors and cofactors.

Delete the row and column of each element to get each minor, then multiply by $(-1)^{i+j}$ for the cofactor. For (i), every off-diagonal minor is $0$ and each diagonal minor is $1$. For (ii), the computed minors form $\begin{bmatrix}11 & 6 & 3 \\ -4 & 2 & 1 \\ -20 & -13 & 5\end{bmatrix}$; applying the checkerboard signs gives $\begin{bmatrix}11 & -6 & 3 \\ 4 & 2 & -1 \\ -20 & 13 & 5\end{bmatrix}$.

The cofactors of the second row are $A_{21}=-(3\cdot3-8\cdot2)=7$, $A_{22}=5\cdot3-8\cdot1=7$, and $A_{23}=-(5\cdot2-3\cdot1)=-7$. Therefore $\Delta=2A_{21}+0A_{22}+1A_{23}=14-7=7$.

Expanding along the third column, $\Delta=yz(z-y)-zx(z-x)+xy(y-x)$. Factoring this symmetric expression gives $\Delta=(x-y)(y-z)(z-x)$.

1. i. $a_{11}A_{31}+a_{12}A_{32}+a_{13}A_{33}$
2. ii. $a_{11}A_{11}+a_{12}A_{21}+a_{13}A_{31}$
3. iii. $a_{21}A_{11}+a_{22}A_{12}+a_{23}A_{13}$
4. iv. $a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}$

Expanding a determinant down the first column gives $\Delta=a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}$. Hence option (iv) is correct.

For $A=\begin{bmatrix}a & b \\ c & d\end{bmatrix}$, $\operatorname{adj}A=\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}$. Therefore $\operatorname{adj}\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}=\begin{bmatrix}4 & -2 \\ -3 & 1\end{bmatrix}$.

The cofactor matrix of $A$ is $\begin{bmatrix}3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5\end{bmatrix}$. Transposing it gives $\operatorname{adj}A=\begin{bmatrix}3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5\end{bmatrix}$.

$|A|=2(-6)-3(-4)=0$. Also $\operatorname{adj}A=\begin{bmatrix}-6 & -3 \\ 4 & 2\end{bmatrix}$. Direct multiplication gives $A(\operatorname{adj}A)=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$ and $(\operatorname{adj}A)A=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$. Since $|A|I=0I=O$, the identity is verified.

Computing cofactors and transposing gives $\operatorname{adj}A=\begin{bmatrix}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{bmatrix}$. Also $|A|=11$. Multiplication gives $A(\operatorname{adj}A)=(\operatorname{adj}A)A=\begin{bmatrix}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{bmatrix}=|A|I$.

$|A|=2\cdot3-(-2)4=14\ne0$. Hence $A^{-1}=\dfrac1{14}\begin{bmatrix}3 & 2 \\ -4 & 2\end{bmatrix}=\begin{bmatrix}\dfrac{3}{14} & \dfrac{1}{7} \\ -\dfrac{2}{7} & \dfrac{1}{7}\end{bmatrix}$.

$|A|=(-1)2-5(-3)=13\ne0$. Thus $A^{-1}=\dfrac1{13}\begin{bmatrix}2 & -5 \\ 3 & -1\end{bmatrix}=\begin{bmatrix}\dfrac{2}{13} & -\dfrac{5}{13} \\ \dfrac{3}{13} & -\dfrac{1}{13}\end{bmatrix}$.

$A$ is upper triangular with determinant $1\cdot2\cdot5=10\ne0$. Solving $AX=I$ column by column gives $A^{-1}=\begin{bmatrix}1 & -1 & \dfrac{1}{5} \\ 0 & \dfrac{1}{2} & -\dfrac{2}{5} \\ 0 & 0 & \dfrac{1}{5}\end{bmatrix}$.

$|A|=1\cdot3\cdot(-1)=-3\ne0$. Solving $AX=I$ by forward substitution gives the inverse $\begin{bmatrix}1 & 0 & 0 \\ -1 & \dfrac{1}{3} & 0 \\ 3 & \dfrac{2}{3} & -1\end{bmatrix}$.

The determinant is $3\ne0$. Row-reducing $[A\mid I]$ to $[I\mid A^{-1}]$ gives $A^{-1}=\begin{bmatrix}\dfrac{1}{3} & -\dfrac{5}{3} & -1 \\ \dfrac{4}{3} & -\dfrac{23}{3} & -4 \\ -\dfrac{1}{3} & \dfrac{11}{3} & 2\end{bmatrix}$.

Here $|A|=1\ne0$, so the inverse exists. Computing cofactors and transposing gives $\operatorname{adj}A=\begin{bmatrix}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{bmatrix}$, hence $A^{-1}=\operatorname{adj}A$.

The lower $2\times2$ block $B=\begin{bmatrix}\cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha\end{bmatrix}$ satisfies $B^2=I$. Hence $A^2=I$, so $A^{-1}=A$.

$A^{-1}=\begin{bmatrix}5 & -7 \\ -2 & 3\end{bmatrix}$ and $B^{-1}=\dfrac1{-2}\begin{bmatrix}9 & -8 \\ -7 & 6\end{bmatrix}=\begin{bmatrix}-\dfrac92 & 4 \\ \dfrac72 & -3\end{bmatrix}$. Multiplying, $B^{-1}A^{-1}=\begin{bmatrix}-\dfrac{61}{2} & \dfrac{87}{2} \\ \dfrac{47}{2} & -\dfrac{67}{2}\end{bmatrix}$. Also $AB=\begin{bmatrix}67 & 87 \\ 47 & 61\end{bmatrix}$ with determinant $-2$, so $(AB)^{-1}=\dfrac1{-2}\begin{bmatrix}61 & -87 \\ -47 & 67\end{bmatrix}$, the same matrix.

$A^2=\begin{bmatrix}8 & 5 \\ -5 & 3\end{bmatrix}$. Then $A^2-5A+7I=\begin{bmatrix}8 & 5 \\ -5 & 3\end{bmatrix}-\begin{bmatrix}15 & 5 \\ -5 & 10\end{bmatrix}+\begin{bmatrix}7 & 0 \\ 0 & 7\end{bmatrix}=O$. Multiplying $A^2-5A+7I=O$ by $A^{-1}$ gives $A-5I+7A^{-1}=O$, so $A^{-1}=\dfrac{5I-A}{7}=\begin{bmatrix}\dfrac{2}{7} & -\dfrac{1}{7} \\ \dfrac{1}{7} & \dfrac{3}{7}\end{bmatrix}$.

$A^2=\begin{bmatrix}11 & 8 \\ 4 & 3\end{bmatrix}$. Hence $A^2+aA+bI=\begin{bmatrix}11+3a+b & 8+2a \\ 4+a & 3+a+b\end{bmatrix}$. Equating entries to zero gives $4+a=0$, so $a=-4$, and then $3+a+b=0$ gives $b=1$.

Direct multiplication gives $A^2=\begin{bmatrix}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{bmatrix}$ and $A^3=\begin{bmatrix}8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{bmatrix}$. Substitution gives $A^3-6A^2+5A+11I=O$. Multiplying this identity by $A^{-1}$ gives $A^{-1}=\dfrac{-A^2+6A-5I}{11}=\begin{bmatrix}-\dfrac{3}{11} & \dfrac{4}{11} & \dfrac{5}{11} \\ \dfrac{9}{11} & -\dfrac{1}{11} & -\dfrac{4}{11} \\ \dfrac{5}{11} & -\dfrac{3}{11} & -\dfrac{1}{11}\end{bmatrix}$.

$A^2=\begin{bmatrix}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{bmatrix}$ and $A^3=\begin{bmatrix}22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22\end{bmatrix}$. Substitution verifies $A^3-6A^2+9A-4I=O$. Multiplying by $A^{-1}$ gives $A^{-1}=\dfrac{A^2-6A+9I}{4}=\begin{bmatrix}\dfrac{3}{4} & \dfrac{1}{4} & -\dfrac{1}{4} \\ \dfrac{1}{4} & \dfrac{3}{4} & \dfrac{1}{4} \\ -\dfrac{1}{4} & \dfrac{1}{4} & \dfrac{3}{4}\end{bmatrix}$.

1. i. $|A|$
2. ii. $|A|^2$
3. iii. $|A|^3$
4. iv. $3|A|$

For a nonsingular matrix of order $n$, $|\operatorname{adj}A|=|A|^{n-1}$. Here $n=3$, so $|\operatorname{adj}A|=|A|^2$. Therefore option (ii) is correct.

1. i. $\det(A)$
2. ii. $\dfrac{1}{\det(A)}$
3. iii. $1$
4. iv. $0$

Since $AA^{-1}=I$, taking determinants gives $\det(A)\det(A^{-1})=1$. Hence $\det(A^{-1})=\dfrac1{\det(A)}$. Therefore option (ii) is correct.

The determinant of coefficients is $\begin{vmatrix}1 & 2 \\ 2 & 3\end{vmatrix}=3-4=-1\ne0$. Hence the system is consistent and has a unique solution. Solving gives $x=0$, $y=1$.

The coefficient determinant is $\begin{vmatrix}2 & -1 \\ 1 & 1\end{vmatrix}=3\ne0$. Therefore the system is consistent and has a unique solution. Solving gives $x=3$, $y=1$.

The left side of the second equation is twice the left side of the first equation, but twice the first right side would be $10$, not $8$. Thus the equations represent parallel distinct lines, so the system is inconsistent.

The coefficient determinant is $\begin{vmatrix}1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2a\end{vmatrix}=a$. If $a\ne0$, the system has a unique solution; solving gives $x=2-\dfrac4a$, $y=0$, $z=\dfrac4a-1$. If $a=0$, the third equation becomes $0=4$, impossible. Hence the system is inconsistent for $a=0$.

From $2y-z=-1$, $z=2y+1$. Substituting in $3x-y-2z=2$ gives $3x-5y=4$. But the third equation says $3x-5y=3$. This contradiction shows that the system is inconsistent.

The coefficient determinant is non-zero. Solving the equations gives $x=3$, $y=2$, $z=-2$. Substitution checks: $15-2-8=5$, $6+6-10=2$, and $15-4-12=-1$. Hence the system is consistent.

With $A=\begin{bmatrix}5 & 2 \\ 7 & 3\end{bmatrix}$, $|A|=15-14=1$. Hence $A^{-1}=\begin{bmatrix}3 & -2 \\ -7 & 5\end{bmatrix}$. Therefore $\begin{bmatrix}x \\ y\end{bmatrix}=A^{-1}\begin{bmatrix}4 \\ 5\end{bmatrix}=\begin{bmatrix}2 \\ -3\end{bmatrix}$.

$A=\begin{bmatrix}2 & -1 \\ 3 & 4\end{bmatrix}$ has determinant $11$. Thus $A^{-1}=\dfrac1{11}\begin{bmatrix}4 & 1 \\ -3 & 2\end{bmatrix}$. Multiplying by $\begin{bmatrix}-2 \\ 3\end{bmatrix}$ gives $x=-\dfrac5{11}$ and $y=\dfrac{12}{11}$.

$A=\begin{bmatrix}4 & -3 \\ 3 & -5\end{bmatrix}$ has determinant $-11$. Solving by $X=A^{-1}B$ gives $x=-\dfrac6{11}$ and $y=-\dfrac{19}{11}$. These values satisfy both equations.

Subtracting the second equation from the first gives $2x=-2$, so $x=-1$. Then $3(-1)+2y=5$ gives $2y=8$, so $y=4$. Thus the matrix-method solution is $(x,y)=(-1,4)$.

From $3y-5z=9$ and the first two equations, solving $AX=B$ gives $x=1$, $y=\dfrac12$, $z=-\dfrac32$. Substitution checks: $2+\dfrac12-\dfrac32=1$, $1-1+\dfrac32=\dfrac32$, and $\dfrac32+\dfrac{15}{2}=9$.

Writing the system as $AX=B$ and solving gives $x=2$, $y=-1$, $z=1$. Substitution verifies: $2-(-1)+1=4$, $4-1-3=0$, and $2-1+1=2$.

The coefficient determinant is non-zero, so the system has a unique solution. Solving $AX=B$ gives $x=1$, $y=2$, $z=-1$. These values give $2+6-3=5$, $1-4-1=-4$, and $3-2+2=3$.

Solving the matrix equation $AX=B$ gives $x=2$, $y=1$, $z=3$. Substituting: $2-1+6=7$, $6+4-15=-5$, and $4-1+9=12$, so the solution is correct.

Expanding along the first row, the determinant equals $x(-x^2-1)-\sin\theta(-x\sin\theta-\cos\theta)+\cos\theta(-\sin\theta+x\cos\theta)$. This simplifies to $-x^3-x+x\sin^2\theta+\sin\theta\cos\theta-\sin\theta\cos\theta+x\cos^2\theta=-x^3$, independent of $\theta$.

Expanding along the second row or recognizing the rows as an orthonormal transformation, the determinant simplifies to $\cos^2\alpha(\cos^2\beta+\sin^2\beta)+\sin^2\alpha(\cos^2\beta+\sin^2\beta)=\cos^2\alpha+\sin^2\alpha=1$.

$(AB)^{-1}=B^{-1}A^{-1}$. Computing $B^{-1}$ and multiplying by the given $A^{-1}$ gives $B^{-1}A^{-1}=\begin{bmatrix}9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2\end{bmatrix}$.

For the given matrix, $|A|=-5\ne0$ and $\operatorname{adj}A=\begin{bmatrix}14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1\end{bmatrix}$. Therefore $A^{-1}=\dfrac1{|A|}\operatorname{adj}A$. Since $\operatorname{adj}A=|A|A^{-1}$, $[\operatorname{adj}A]^{-1}=\dfrac1{|A|}A=\operatorname{adj}(A^{-1})$. This common matrix is $-\dfrac15A$, as shown in the answer. Also the inverse operation reverses itself, so $(A^{-1})^{-1}=A$, verified by multiplying $A^{-1}A=I$.

Expanding the determinant gives $x\{y(x+y)-x^2\}-y\{y^2-x(x+y)\}+(x+y)\{xy-(x+y)^2\}$. Simplifying yields $-2x^3-2y^3=-2(x^3+y^3)$.

Apply $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$ to get $\begin{vmatrix}1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x\end{vmatrix}$. The determinant is the product of the diagonal entries, $1\cdot y\cdot x=xy$.

Put $u=\dfrac1x$, $v=\dfrac1y$, $w=\dfrac1z$. Then $2u+3v+10w=4$, $4u-6v+5w=1$, and $6u+9v-20w=2$. Solving this linear system gives $u=\dfrac12$, $v=\dfrac13$, $w=\dfrac15$. Therefore $x=2$, $y=3$, and $z=5$.

1. i. $\begin{bmatrix}x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1}\end{bmatrix}$
2. ii. $\begin{bmatrix}x^{-1} & y^{-1} & z^{-1} \\ x^{-1} & y^{-1} & z^{-1} \\ x^{-1} & y^{-1} & z^{-1}\end{bmatrix}$
3. iii. $\begin{bmatrix}0 & x^{-1} & 0 \\ 0 & 0 & y^{-1} \\ z^{-1} & 0 & 0\end{bmatrix}$
4. iv. $\begin{bmatrix}0 & 0 & x^{-1} \\ 0 & y^{-1} & 0 \\ z^{-1} & 0 & 0\end{bmatrix}$

For a diagonal matrix with non-zero diagonal entries, the inverse is obtained by taking reciprocals of the diagonal entries. Hence $A^{-1}=\operatorname{diag}(x^{-1},y^{-1},z^{-1})$. Therefore option (i) is correct.

1. i. $\det(A)=0$
2. ii. $\det(A)\in(2,\infty)$
3. iii. $\det(A)\in(2,4)$
4. iv. $\det(A)\in[2,4]$

Let $s=\sin\theta$. Expanding gives $\det(A)=2(1+s^2)$. Since $0\le s^2\le1$, $2\le2(1+s^2)\le4$. Therefore $\det(A)\in[2,4]$, so option (iv) is correct.