Determinants

By using properties of determinants prove that the determinant $ \begin{vmatrix} a & \sin x & \cos x \\ -\sin x & -a & 1 \\ \cos x & 1 & a \end{vmatrix} $ is independent of x.

Let $\Delta = $ $ \begin{vmatrix} a & \sin x & \cos x \\ -\sin x & -a & 1 \\ \cos x & 1 & a \end{vmatrix} $

$C_1 \rightarrow C_1 + C_2 \implies \Delta =$ $ \begin{vmatrix} a+\sin x & \sin x & \cos x \\ -a-\sin x & -a & 1 \\ 1+\cos x & 1 & a \end{vmatrix} $

$R_1 \rightarrow R_1 + R_2, R_2 \rightarrow R_2+aR_3 \implies$

$\Delta =$ $ \begin{vmatrix} 0 & -a+\sin x & 1+\cos x \\ a\cos x-\sin x & 0 & 1+a^2 \\ 1+\cos x & 1 & a \end{vmatrix} $

Expanding along $R_1$ gives $\begin{aligned} \Delta = & -\left(-a+\sin x\right) \left[a\left(a\cos x -\sin x\right)-\left(1+a^2\right)\left(1+\cos x\right)\right] + \\ & \left(1+\cos x\right)\left(a\cos x - \sin x\right) \\ \text{i.e. } \Delta = & -\left(-a+\sin x\right)\left(a^2\cos x -a\sin x -1-\cos x -a^2 -a^2 \cos x\right) + \\ & a\cos x - \sin x + a \cos^2 x - \sin x \cos x \\ \text{i.e. } \Delta = & \left(-a+\sin x\right) \left(a\sin x + \cos x +1 +a^2\right) \\ & a\cos x - \sin x + a \cos^2 x - \sin x \cos x \\ \text{i.e. } \Delta = & -a^2 \sin x - a\cos x -a - a^3 +a \sin^2 x +\sin x \cos x +\sin x + \\ & a^2 \sin x + a\cos x - \sin x + a \cos^2 x - \sin x \cos x \\ \text{i.e. } \Delta = & -a^3 \text{ which is independent of x.} \end{aligned}$

Determinants

If $A+B+C=\pi$, then find the value of $\begin{vmatrix} \sin \left(A+B+C\right) & \sin B & \cos C \\ - \sin B & 0 & \tan A \\ \cos \left(A+B\right) & -\tan A & 0 \end{vmatrix}$

Since $A+B+C = \pi$, $\sin \left(A+B+C\right) = \sin \pi = 0$ and $\cos \left(A+B\right) = \cos \left(\pi-C\right) = -\cos C$

$ \begin{aligned} \text{Let } \Delta & = \begin{vmatrix} \sin \left(A+B+C\right) & \sin B & \cos C \\ - \sin B & 0 & \tan A \\ \cos \left(A+B\right) & -\tan A & 0 \end{vmatrix} \\ & \\ & = \begin{vmatrix} 0 & \sin B & \cos C \\ -\sin B & 0 & \tan A \\ -\cos C & -\tan A & 0 \end{vmatrix} \\ & \\ & = -\sin B \times \left[0- \tan A \times \left(-\cos C\right)\right] + \cos C \sin B \tan A \\ & \\ & = -\tan A \sin B \cos C + \tan A \sin B \cos C = 0 \end{aligned} $

Determinants

Using properties of determinants, show that $p \alpha^2 + 2q \alpha + r = 0$ given that p, q and r are not in G.P and $ \begin{vmatrix} 1 & \dfrac{q}{p} & \alpha + \dfrac{q}{p} \\ 1 & \dfrac{r}{q} & \alpha + \dfrac{r}{q} \\ p \alpha + q & q \alpha + r & 0 \end{vmatrix} = 0 $

$R_1 \rightarrow R_1-R_2 \implies$ $ \begin{vmatrix} 0 & \dfrac{q}{p}-\dfrac{r}{q} & \dfrac{q}{p}-\dfrac{r}{q} \\ 1 & \dfrac{r}{q} & \alpha+\dfrac{r}{q} \\ p\alpha + q & q \alpha + r & 0 \end{vmatrix} = 0 $

i.e. $\left(\dfrac{q}{p}-\dfrac{r}{q}\right)$ $ \begin{vmatrix} 0 & 1 & 1 \\ 1 & \dfrac{r}{q} & \alpha+\dfrac{r}{q} \\ p\alpha+q & q\alpha+r & 0 \end{vmatrix} = 0 $

$C_2 \rightarrow C_2 - C_3 \implies \left(\dfrac{q^2-pr}{pq}\right)$ $ \begin{vmatrix} 0 & 0 & 1 \\ 1 & -\alpha & \alpha+\dfrac{r}{q} \\ p\alpha+q & q\alpha+r & 0 \end{vmatrix} = 0 $

Expanding along $R_1$ gives

$\dfrac{\left(q^2-pr\right) \left[q\alpha+r +\alpha \left(p\alpha+q\right)\right]}{pq}= 0$

i.e. $\left(q^2-pr\right)\left(p\alpha^2 + 2q\alpha + r\right) = 0$ $\cdots$ (1)

Since p, q and r are not in G.P $\implies$ $q^2 \neq pr$

i.e. $q^2 - pr \neq 0$ $\cdots$ (2)

Therefore, from equations (1) and (2) we have

$p\alpha^2 + 2q \alpha + r = 0$

Determinants

If $\Delta = \begin{vmatrix} 1 & a & a^2 \\ a & a^2 & 1 \\ a^2 & 1 & a \end{vmatrix} = -4$, then find the value of $\begin{vmatrix} a^3 - 1 & 0 & a-a^4 \\ 0 & a-a^4 & a^3-1 \\ a-a^4 & a^3-1 & 0 \end{vmatrix}$

$\left[\text{Note: If } C_{ij} \text{ are the cofactors of } a_{ij} \text{ in } \left|A\right| = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \\ \text{ then } \begin{vmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_23 \\ C_{31} & C_{32} & C_{33} \end{vmatrix} = \left|A\right|^2\right] $
Cofactors of determinant $\Delta$ are
$ C_{11} = \left(-1\right)^{1+1} \begin{vmatrix} a^2 & 1 \\ 1 & a \end{vmatrix} = a^3 - 1 $

$ C_{12} = \left(-1\right)^{1+2} \begin{vmatrix} a & 1 \\ a^2 & a \end{vmatrix} = a^2-a^2=0$

$C_{13} = \left(-1\right)^{1+3} \begin{vmatrix} a & a^2 \\ a^2 & 1 \end{vmatrix} = a - a^4$

$C_{21} = \left(-1\right)^{2+1} \begin{vmatrix} a & a^2 \\ 1 & a \end{vmatrix} = a^2 - a^2 = 0$

$C_{22} = \left(-1\right)^{2+2} \begin{vmatrix} 1 & a^2 \\ a^2 & a \end{vmatrix} = a - a^4$

$C_{23} = \left(-1\right)^{2+3} \begin{vmatrix} 1 & a \\ a^2 & 1 \end{vmatrix} = -\left(1-a^3\right) = a^3 - 1$

$C_{31} = \left(-1\right)^{3+1} \begin{vmatrix} a & a^2 \\ a^2 & 1 \end{vmatrix} = a - a^4$

$C_{32} = \left(-1\right)^{3+2} \begin{vmatrix} 1 & a^2 \\ a & 1 \end{vmatrix} = -\left(1-a^3\right) = a^3 - 1$

$C_{33} = \left(-1\right)^{3+3} \begin{vmatrix} 1 & a \\ a & a^2 \end{vmatrix} = a^2 - a^2 = 0$

Now, $\begin{vmatrix} a^3 - 1 & 0 & a-a^4 \\ 0 & a-a^4 & a^3-1 \\ a-a^4 & a^3-1 & 0 \end{vmatrix} = \begin{vmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{vmatrix}$

$\therefore$ $\begin{vmatrix} a^3 - 1 & 0 & a-a^4 \\ 0 & a-a^4 & a^3-1 \\ a-a^4 & a^3-1 & 0 \end{vmatrix} = \left|\Delta\right|^2 = \left(-4\right)^2 = 16$

Determinants

Using properties of determinants prove that $\begin{vmatrix} a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c \end{vmatrix} = a^3$

Let $\Delta =$ $ \begin{vmatrix} a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c \end{vmatrix} $ $R_2 \rightarrow R_2 - 2R_1, R_3 \rightarrow R_3 - 3R_1 \implies \Delta =$ $ \begin{vmatrix} a & a+b & a+b+c \\ 0 & a & 2a+b \\ 0 & 3a & 7a+3b \end{vmatrix} $ $R_3 \rightarrow R_3 - 3R_2 \implies \Delta =$ $ \begin{vmatrix} a & a+b & a+b+c \\ 0 & a & 2a+b \\ 0 & 0 & a \end{vmatrix} $
Expanding along $R_1$ gives
$\Delta = a \times a \times a = a^3$

Determinants

Using properties of determinants prove that $ \begin{vmatrix} a & b-c & c+b \\ a+c & b & c-a \\ a-b & b+a & c \end{vmatrix} = \left(a+b+c\right) \left(a^2 + b^2 + c^2\right) $

$ \begin{aligned} \text{Let } \Delta & = \begin{vmatrix} a & b-c & c+b \\ a+c & b & c-a \\ a-b & b+a & c \end{vmatrix} \\ & \\ C_1 \rightarrow aC_1 \implies \Delta & = \dfrac{1}{a} \begin{vmatrix} a^2 & b-c & c+b \\ a^2 + ac & b & c-a \\ a^2 - ab & b+a & c \end{vmatrix} \\ & \\ C_1 \rightarrow C_1 + bC_2 + cC_3 \implies \Delta & = \dfrac{1}{a} \begin{vmatrix} a^2+b^2-bc+c^2+bc & b-c & c+b \\ a^2+ac+b^2+c^2-ac & b & c-a \\ a^2-ab+b^2+ab+c^2 & b+a & c \end{vmatrix} \\ & \\ & = \dfrac{1}{a} \begin{vmatrix} a^2+b^2+c^2 & b-c & c+b \\ a^2+b^2+c^2 & b & c-a \\ a^2+b^2+c^2 & b+a & c \end{vmatrix} \\ & \\ & = \dfrac{a^2+b^2+c^2}{a} \begin{vmatrix} 1 & b-c & c+b \\ 1 & b & c-a \\ 1 & b+a & c \end{vmatrix} \\ & \\ R_2 \rightarrow R_2-R_1 \implies \Delta & = \dfrac{a^2+b^2+c^2}{a} \begin{vmatrix} 1 & b-c & c+b \\ 0 & c & -a-b \\ 1 & b+a & c \end{vmatrix} \\ & \\ R_3 \rightarrow R_3-R_1 \implies \Delta & = \dfrac{a^2+b^2+c^2}{a} \begin{vmatrix} 1 & b-c & c+b \\ 0 & c & -a-b \\ 0 & a+c & -b \end{vmatrix} \\ & \\ & = \dfrac{\left(a^2+b^2+c^2\right) \left[-bc+\left(a+b\right)\left(a+c\right)\right]}{a} \left[\text{Note: Expanding along }C_1\right] \\ & = \dfrac{\left(a^2+b^2+c^2\right)\left(a^2+ab+ac+bc-bc\right)}{a} \\ & = \dfrac{\left(a^2+b^2+c^2\right) \left(a^2+ab+ac\right)}{a} \\ & = \left(a^2+b^2+c^2\right) \left(a+b+c\right) \end{aligned} $