Simplify: $\sin A$ $\begin{bmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{bmatrix} + \cos A$ $\begin{bmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{bmatrix}$
$\begin{bmatrix}
\sin A & -\cos A \\
\cos A & \sin A
\end{bmatrix} = \sin^2 A + \cos^2 A = 1$
$\begin{bmatrix}
\cos A & \sin A \\
-\sin A & \cos A
\end{bmatrix} = \cos^2 A + \sin^2 A = 1$
Therefore, the given expression becomes:
$\sin A \times 1 + \cos A \times 1 = \sin A + \cos A$
If $A =
\begin{bmatrix}
3 & 1 & 2 \\
3 & 2 & -3 \\
2 & 0 & -1
\end{bmatrix}$, find $A^{-1}$.
Hence solve the system of equations
$3x+3y+2z=1$, $x+2y=4$, $2x-3y-z=5$
$\left|A\right| =$
$
\begin{vmatrix}
3 & 1 & 2 \\
3 & 2 & -3 \\
2 & 0 & -1
\end{vmatrix} = 3 \times \left(-2\right) -1 \times \left(-3+6\right) + 2 \times \left(-4\right) = -17 \neq 0
$
Therefore, $A^{-1}$ exists.
Cofactors of A are:
$A_{11} = \left(-1\right)^{1+1}$
$\begin{vmatrix}
2 & -3 \\
0 & -1
\end{vmatrix}=-2$
$A_{12} = \left(-1\right)^{1+2}$
$\begin{vmatrix}
3 & -3 \\
2 & -1
\end{vmatrix}=- \left(-3+6\right)=-3$
$A_{13} = \left(-1\right)^{1+3}$
$\begin{vmatrix}
3 & 2 \\
2 & 0
\end{vmatrix}=-4$
$A_{21} = \left(-1\right)^{2+1}$
$\begin{vmatrix}
1 & 2 \\
0 & -1
\end{vmatrix}=-\left(-1\right)=1$
$A_{22} = \left(-1\right)^{2+2}$
$\begin{vmatrix}
3 & 2 \\
2 & -1
\end{vmatrix}=-3-4=-7$
$A_{23} = \left(-1\right)^{2+3}$
$\begin{vmatrix}
3 & 1 \\
2 & 0
\end{vmatrix}=- \left(-2\right)=2$
$A_{31} = \left(-1\right)^{3+1}$
$\begin{vmatrix}
1 & 2 \\
2 & -3
\end{vmatrix}=-3-4=-7$
$A_{32} = \left(-1\right)^{3+2}$
$\begin{vmatrix}
3 & 2 \\
3 & -3
\end{vmatrix}=-\left(-9-6\right)=15$
$A_{33} = \left(-1\right)^{3+3}$
$\begin{vmatrix}
3 & 1 \\
3 & 2
\end{vmatrix}=6-3=3$
$\text{adj A}=$
$\begin{bmatrix}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
\end{bmatrix}^t = $
$\begin{bmatrix}
-2 & -3 & -4 \\
1 & -7 & 2 \\
-7 & 15 & 3
\end{bmatrix}^t=$
$
\begin{bmatrix}
-2 & 1 & -7 \\
-3 & -7 & 15 \\
-4 & 2 & 3
\end{bmatrix}
$
$A^{-1} = \dfrac{\text{adj A}}{\left|A\right|} = \dfrac{-1}{17}$
$
\begin{bmatrix}
-2 & 1 & -7 \\
-3 & -7 & 15 \\
-4 & 2 & 3
\end{bmatrix}
$
Let $A_1 =$
$\begin{bmatrix}
3 & 3 & 2 \\
1 & 2 & 0 \\
2 & -3 & -1
\end{bmatrix}$,
$B=$
$\begin{bmatrix}
1 \\
4 \\
5
\end{bmatrix}$ and $X=$
$\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}$
Then $A_1 X = B \implies X = A_1^{-1}B$
But $A_1 = A^t$ (Comparing A and $A_1$)
Therefore, $X = \left(A^t\right)^{-1}B = \left(A^{-1}\right)^t B$ $\left[\text{Note: } \left(A^t\right)^{-1} = \left(A^{-1}\right)^t\right]$
$X= \dfrac{-1}{17}$
$\begin{bmatrix}
-2 & 1 & -7 \\
-3 & -7 & 15 \\
-4 & 2 & 3
\end{bmatrix}^t$
$\begin{bmatrix}
1 \\
4 \\
5
\end{bmatrix} = \dfrac{-1}{17}$
$\begin{bmatrix}
-2 & -3 & -4 \\
1 & -7 & 2 \\
-7 & 15 & 3
\end{bmatrix}$
$\begin{bmatrix}
1 \\
4 \\
5
\end{bmatrix}$
$= \dfrac{-1}{17}$
$\begin{bmatrix}
-2-12-20 \\
1-28+10 \\
-7+60+15
\end{bmatrix}$
$= \dfrac{-1}{17}$
$\begin{bmatrix}
-34 \\
-17 \\
68
\end{bmatrix}$
i.e.
$\begin{bmatrix}
x \\
y \\
z
\end{bmatrix} = $
$\begin{bmatrix}
2 \\
1 \\
-4
\end{bmatrix}$
$\implies x=2 \,, y =1 \,, z=-4$
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
$\begin{aligned}
\Delta = & -\left(-a+\sin x\right) \left[a\left(a\cos x -\sin x\right)-\left(1+a^2\right)\left(1+\cos x\right)\right] + \\
& \left(1+\cos x\right)\left(a\cos x - \sin x\right) \\
\text{i.e. } \Delta = & -\left(-a+\sin x\right)\left(a^2\cos x -a\sin x -1-\cos x -a^2 -a^2 \cos x\right) + \\
& a\cos x - \sin x + a \cos^2 x - \sin x \cos x \\
\text{i.e. } \Delta = & \left(-a+\sin x\right) \left(a\sin x + \cos x +1 +a^2\right) \\
& a\cos x - \sin x + a \cos^2 x - \sin x \cos x \\
\text{i.e. } \Delta = & -a^2 \sin x - a\cos x -a - a^3 +a \sin^2 x +\sin x \cos x +\sin x + \\
& a^2 \sin x + a\cos x - \sin x + a \cos^2 x - \sin x \cos x \\
\text{i.e. } \Delta = & -a^3 \text{ which is independent of x.}
\end{aligned}$
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}
$
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
$p\alpha^2 + 2q \alpha + r = 0$
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, $\begin{vmatrix}
a^3 - 1 & 0 & a-a^4 \\
0 & a-a^4 & a^3-1 \\
a-a^4 & a^3-1 & 0
\end{vmatrix} = \begin{vmatrix}
C_{11} & C_{12} & C_{13} \\
C_{21} & C_{22} & C_{23} \\
C_{31} & C_{32} & C_{33}
\end{vmatrix}$
$\therefore$
$\begin{vmatrix}
a^3 - 1 & 0 & a-a^4 \\
0 & a-a^4 & a^3-1 \\
a-a^4 & a^3-1 & 0
\end{vmatrix} = \left|\Delta\right|^2 = \left(-4\right)^2 = 16$
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
$\Delta = a \times a \times a = a^3$