Class 12 Maths NCERT Solutions Chapter 4 Miscellaneous Exercise – Determinants

Expanding along R_1:

\Delta = x(-x^2-1)-\sin\theta(-x\sin\theta-\cos\theta)+\cos\theta(-\sin\theta+x\cos\theta)

= -x^3-x + x\sin^2\theta+\sin\theta\cos\theta -\sin\theta\cos\theta+x\cos^2\theta

The \sin\theta\cos\theta terms cancel, and x\sin^2\theta+x\cos^2\theta = x(\sin^2\theta+\cos^2\theta)=x:

\Delta = -x^3-x+x = -x^3

Call the left-hand determinant \Delta_1. Multiply R_1 by a, R_2 by b, and R_3 by c — this multiplies the whole determinant by abc:

abc\cdot\Delta_1 = \begin{vmatrix} a^2 & a^3 & abc \\ b^2 & b^3 & abc \\ c^2 & c^3 & abc \end{vmatrix}

Every entry in column 3 is now abc, so factoring it out of that column:

abc\cdot\Delta_1 = abc\begin{vmatrix} a^2 & a^3 & 1 \\ b^2 & b^3 & 1 \\ c^2 & c^3 & 1 \end{vmatrix} \Rightarrow \Delta_1 = \begin{vmatrix} a^2 & a^3 & 1 \\ b^2 & b^3 & 1 \\ c^2 & c^3 & 1 \end{vmatrix}

Cyclically shifting the columns — moving column 3 to the front — is an even permutation of three columns, so the determinant's value (and sign) is unchanged:

Expanding along R_1:

\Delta = \cos\alpha\cos\beta(\cos\alpha\cos\beta) - \cos\alpha\sin\beta(-\cos\alpha\sin\beta) + (-\sin\alpha)(-\sin\alpha)

(each 2\times2 minor here simplifies because the last column of the original matrix has a zero in row 2)

= \cos^2\alpha\cos^2\beta + \cos^2\alpha\sin^2\beta + \sin^2\alpha

= \cos^2\alpha(\cos^2\beta+\sin^2\beta) + \sin^2\alpha = \cos^2\alpha+\sin^2\alpha

Apply C_1\to C_1+C_2+C_3. Every row's three entries sum to 2(a+b+c), so the new first column is constant:

\Delta = 2(a+b+c)\begin{vmatrix} 1 & c+a & a+b \\ 1 & a+b & b+c \\ 1 & b+c & c+a \end{vmatrix}

Now apply R_2\to R_2-R_1 and R_3\to R_3-R_1:

\Delta = 2(a+b+c)\begin{vmatrix} 1 & c+a & a+b \\ 0 & b-c & c-a \\ 0 & b-a & c-b \end{vmatrix}

Expanding along column 1: \Delta = 2(a+b+c)\left[(b-c)(c-b)-(c-a)(b-a)\right]

Simplifying the bracket gives -\left[a^2+b^2+c^2-ab-bc-ca\right] = -\dfrac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]

So \Delta = -(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]

Setting \Delta=0: (a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]=0

Apply C_1\to C_1+C_2+C_3. Every row sums to 3x+a:

\Delta = (3x+a)\begin{vmatrix} 1 & x & x \\ 1 & x+a & x \\ 1 & x & x+a \end{vmatrix}

Apply R_2\to R_2-R_1 and R_3\to R_3-R_1:

\Delta = (3x+a)\begin{vmatrix} 1 & x & x \\ 0 & a & 0 \\ 0 & 0 & a \end{vmatrix} = (3x+a)\cdot1\cdot(a\cdot a - 0) = (3x+a)a^2

Setting \Delta=0: (3x+a)a^2=0. Since a\neq0, we need 3x+a=0.

Expanding directly along R_1, the three minors work out to:

M_{11}=b^2c^2-ac(b^2+bc)=b^2c^2-ab^2c-abc^2

M_{12}=(a^2+ab)c^2-a^2bc=a^2c^2+abc^2-a^2bc

M_{13}=(a^2+ab)(b^2+bc)-ab^3=a^2b^2+a^2bc+ab^2c

So \Delta = a^2M_{11}-bc\,M_{12}+(ac+c^2)M_{13}. Expanding each product and collecting like terms — the a^3b^2c,\ a^3bc^2,\ a^2bc^3,\ ab^2c^3 terms all cancel in pairs, leaving only the a^2b^2c^2 terms, of which there are four:

Using (AB)^{-1}=B^{-1}A^{-1}, we first need B^{-1}. Computing |B| by expanding along R_1: |B| = 1(3)-2(-1)-2(2) = 3+2-4 = 1

Computing all cofactors of B gives \text{adj }B = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}, so B^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} (since |B|=1).

Now (AB)^{-1} = B^{-1}A^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}

Multiplying row by column:

Expanding along R_1: |A| = 1(14)+2(-11)+1(-5) = 14-22-5 = -13, so A is invertible.

(ii) is immediate from the definition of an inverse: since A\cdot A^{-1}=A^{-1}\cdot A=I, matrix A itself satisfies the defining property of "the inverse of A^{-1}".

So (A^{-1})^{-1}=A for any invertible matrix, including this one.

(i) Using the identity \text{adj}(M) = |M|\,M^{-1} for any invertible matrix M:

\text{adj}(A^{-1}) = |A^{-1}|\,(A^{-1})^{-1} = \dfrac{1}{|A|}\cdot A (using part (ii) and |A^{-1}|=1/|A|)

Also, since \text{adj}\,A = |A|A^{-1}: [\text{adj }A]^{-1} = \left(|A|A^{-1}\right)^{-1} = \dfrac{1}{|A|}(A^{-1})^{-1} = \dfrac{1}{|A|}A

Apply C_1\to C_1+C_2+C_3. Every row sums to 2(x+y):

\Delta = 2(x+y)\begin{vmatrix} 1 & y & x+y \\ 1 & x+y & x \\ 1 & x & y \end{vmatrix}

Apply R_2\to R_2-R_1, R_3\to R_3-R_1:

\Delta = 2(x+y)\begin{vmatrix} 1 & y & x+y \\ 0 & x & -y \\ 0 & x-y & -x \end{vmatrix} = 2(x+y)\left[x(-x)-(-y)(x-y)\right]

= 2(x+y)\left[-x^2+xy-y^2\right] = -2(x+y)(x^2-xy+y^2)

Using the sum-of-cubes identity (x+y)(x^2-xy+y^2)=x^3+y^3:

Apply R_2\to R_2-R_1 and R_3\to R_3-R_1:

\Delta = \begin{vmatrix} 1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x \end{vmatrix}

Expanding along column 1: \Delta = 1\times\begin{vmatrix} y & 0 \\ 0 & x \end{vmatrix} = 1\times(xy-0)

Since a, b, c are in A.P., let the common difference be d, so a-b=b-c=-d.

Apply R_1\to R_1-R_2 and R_2\to R_2-R_3:

R_1-R_2 = \left[-1,\ -1,\ 2(a-b)\right] = \left[-1,-1,-2d\right]

R_2-R_3 = \left[-1,\ -1,\ 2(b-c)\right] = \left[-1,-1,-2d\right]

The new first and second rows are identical.

Expanding along R_1:

\text{Det}(A) = 1(1+\sin^2\theta) - \sin\theta(-\sin\theta+\sin\theta) + 1(\sin^2\theta+1)

The middle term vanishes since -\sin\theta+\sin\theta=0:

\text{Det}(A) = (1+\sin^2\theta) + (\sin^2\theta+1) = 2+2\sin^2\theta

Since 0\leq\sin^2\theta\leq1 for every real \theta, it follows that 2\leq2+2\sin^2\theta\leq4.