Question 515 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}$
Answer$\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}$
$=\begin{vmatrix}\sin\alpha\sin\delta&\cos\alpha\cos\delta&\cos(\alpha+\delta)\\\sin\beta\sin\delta&\cos\beta\cos\delta&\cos(\beta+\delta)\\\sin\gamma\sin\delta&\cos\gamma\cos\delta&\cos(\gamma+\delta)\end{vmatrix}\ [$Applying $\text{ C}_1\rightarrow\sin\delta\text{ C}_1\text{ and}\text{ C}_2\rightarrow\cos\delta\text{ C}_2]$
$=\begin{vmatrix}\sin\alpha\sin\delta&\cos(\alpha+\delta)&\cos(\alpha+\delta)\\\sin\beta\sin\delta&\cos(\beta+\delta)&\cos(\beta+\delta)\\\sin\gamma\sin\delta&\cos(\gamma+\delta)&\cos(\gamma+\delta)\end{vmatrix}\ [$Applying $C_2 \rightarrow C_2 - C_1]$
$=0$
View full question & answer→Question 525 Marks
Prove the following identities:
$\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=(5\text{x}+\lambda)(\lambda-\text{x})^2$
Answer$\text{L.H.S}=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}-\text{x}-\lambda&\text{x}+\lambda-2\text{x}&0\\2\text{x}-\text{x}-\lambda&0&\text{x}+\lambda-2\text{x}\end{vmatrix}\ [$Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1]$
$=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\-(\lambda-\text{x})&\lambda-\text{x}&0\\-(\lambda-\text{x})&0&\lambda-\text{x}\end{vmatrix}$
$=(\lambda-\text{x})^2\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\-1&1&0\\-1&0&1\end{vmatrix} \ [$Taking $(\lambda-\text{x})$ common from $R_2$ and $(\lambda-\text{x})$ common from $R_3]$
$=(\lambda-\text{x})^2[-1(-2\text{x})+1(\text{x}+\lambda+2\text{x})] \ [$Expanding along last row$]$
$=(\lambda-\text{x})^2(\lambda+5\text{x})$
$=\text{R.H.S}$
$\because\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=(\lambda-\text{x})^2(\lambda+5\text{x})$
View full question & answer→Question 535 Marks
Prove the following identities:
$\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}=4\text{xyz}$
Answer$\text{L.H.S}=\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
Applying $R_1 \rightarrow R_1 - R_2$
$=\begin{vmatrix}\text{y}&-\text{x}&\text{y}-\text{x}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
Applying $R_1 \rightarrow R_1 - R_3$
$=\begin{vmatrix}0&-2\text{x}&-2\text{x}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
$=2\text{x}[\text{z}(\text{x}+\text{y})-\text{xy}]-2\text{x}[\text{zx}-\text{y}(\text{z}+\text{x})]$
$=2\text{x}[\text{zx}+\text{zy}-\text{xy}-\text{zx}+\text{yz}+\text{yx}]$
$=4\text{xyz}$
View full question & answer→Question 545 Marks
Prove that:
$\begin{vmatrix}\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\end{vmatrix}=2\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\end{vmatrix}$
Using the property of determinants that if each element of a row or column is expressed as the sum of two or more quantities, the determinant is expressed as the sum of two or more determinants, we get
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}+(-1)\begin{vmatrix}\text{a}&\text{c}&\text{b}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{b}&\text{a} \end{vmatrix}$
$[$Applying $\text{C}_1\leftrightarrow\text{C}_3$ in second determinant to get negative value of the deteminant$]$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}+(-1)(-1)\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}$ $[$Applying $\text{C}_2\leftrightarrow\text{C}_3]$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}=\text{R.H.S}$
View full question & answer→Question 555 Marks
Solve the following systems of linear equations by cramer's rule:
3x + ay = 4,
2x + ay = 2, $\text{a}\neq0$
AnswerGiven, 3x + ay = 4
2x + ay = 2
Using Cramer's rule, we get
$\text{D}=\begin{vmatrix}3&\text{a}\\2&\text{a}\end{vmatrix}=3\text{a}-2\text{a}=\text{a}$
$\text{D}_1=\begin{vmatrix}4&\text{a}\\2&\text{a}\end{vmatrix}=4\text{a}-2\text{a}=2\text{a}$
$\text{D}_2=\begin{vmatrix}3&4\\2&2\end{vmatrix}=6-8=-2$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{2\text{a}}{\text{a}}=2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-2}{\text{a}}=-\frac{2}{\text{a}}$
$\therefore\text{x}=2$ and $\text{y}=-\frac{2}{\text{a}}$
View full question & answer→Question 565 Marks
$\triangle=\begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha \end{vmatrix}$
AnswerGiven,
$\triangle=\begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha \end{vmatrix}$
$\Rightarrow\triangle=(-1)^{1+1}\cos\alpha\cos\beta(\cos\alpha\cos\beta-0)\\+(-1)^{1+2}\cos\alpha\sin\beta(-\sin\beta\cos\alpha-0)\\+(-1)^{1+3}(-\sin\alpha)(-\sin^2\beta\sin\alpha-\sin\alpha\cos^2\beta)$
$[$Expanding along $R_1]$ $=\cos\alpha\cos\beta(\cos\alpha\cos\beta-0)-\cos\alpha\sin\beta(-\sin\beta\cos\alpha-0)\\-\sin\alpha(\sin^2\beta\sin\alpha-\sin\alpha-\sin\alpha\cos^2\beta)$
$=\cos^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta+\sin^2\alpha\sin^2\beta+\sin^2\alpha\cos^2\beta$
$=\cos^2\alpha(\cos^2\beta+\sin^\beta)+\sin^2\alpha(\sin^2\beta+\cos^2\beta)$
$\Rightarrow\triangle=\cos^2\alpha+\sin^2\alpha$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$\Rightarrow\triangle=1$
View full question & answer→Question 575 Marks
If $\text{A}=\begin{bmatrix}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix},$ show that $A^{-1} = A^3.$
AnswerWe have, $\text{A}=\begin{bmatrix}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}\begin{bmatrix}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{bmatrix}$
$=\begin{bmatrix} 9-6+0 & -9+9-4 & 12-12+4 \\ 6-6+0 & -6+9-4 & 8-12+4 \\ 0-2+0 & 0+3-1 & 0-4+1 \end{bmatrix}$
$=\begin{bmatrix}3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix}$
Now, $\text{A}^3=\text{A}^2\times\text{A}=\begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix}\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$
$=\begin{bmatrix}9-8 & -9+12-4 & 12-16+4 \\ 0-2+0 & 0+3+0 & -4 \\ -6+4+0 & 6-6+3 & -8+8-3 \end{bmatrix}$
$= \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & 0 \\ -2 & 3 & -3 \end{bmatrix}$
Again, $\text{A}^3\times\text{A}=\begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & 0 \\ -2 & 3 & -3\end{bmatrix}\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$
$=\begin{bmatrix}-3-2+0 & -3+3+0 & 4-4+0 \\ -6+6 & 6-9+4 & -8+12-4 \\ 0 & 0 & 1 \end{bmatrix}$
$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=\text{I}_3 \ \big[$Identity matrix of order $3\big]$
$\Rightarrow\ \text{A}^3\times\text{A}=\text{I}_3$
$\Rightarrow\ \text{A}^3=\text{A}^{-1}$
View full question & answer→Question 585 Marks
Without expanding, prove that:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\\\text{a}&\text{b}&\text{c}\end{vmatrix}=\begin{vmatrix}\text{y}&\text{b}&\text{q}\\\text{x}&\text{a}&\text{p}\\\text{z}&\text{c}&\text{r}\end{vmatrix}$
Answer$=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\\\text{a}&\text{b}&\text{c}\end{vmatrix}$ $\text{R}_2\leftrightarrow\text{R}_3$
$=-\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{a}&\text{b}&\text{c}\\\text{p}&\text{q}&\text{r}\end{vmatrix}$ $\text{R}_1\leftrightarrow\text{R}_2$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\end{vmatrix}$
$\begin{vmatrix}\text{y}&\text{b}&\text{q}\\\text{x}&\text{a}&\text{p}\\\text{z}&\text{c}&\text{r}\end{vmatrix}$
$=\begin{vmatrix}\text{y}&\text{x}&\text{z}\\\text{b}&\text{a}&\text{c}\\\text{q}&\text{p}&\text{r}\end{vmatrix}$ $\text{C}_1\leftrightarrow\text{C}_2$
$=-\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{a}&\text{b}&\text{c}\\\text{p}&\text{q}&\text{r}\end{vmatrix}$ $\text{R}_1\leftrightarrow\text{R}_2$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\end{vmatrix}$
View full question & answer→Question 595 Marks
Show that the following systems of linear equations is inconsistent:
$3x - y + 2z = 3,$
$2x + y + 3z = 5,$
$x - 2y - z = 1$
AnswerGiven,
$3x - y + 2z = 3,$
$2x + y + 3z = 5,$
$x - 2y - z = 1$
$\text{D}=\begin{vmatrix}3&-1&2\\2&1&3\\1&-2&-1\end{vmatrix}$
$=3(-1+6)+1(-2-3)+2(-4-1)=0$
$\text{D}_1=\begin{vmatrix}3&-1&2\\5&1&3\\1&-2&-1\end{vmatrix}$
$=3(-1+6)+1(-5-3)+2(-10-1)=-15$
$\text{D}_2=\begin{vmatrix}3&3&2\\2&5&3\\1&1&-1\end{vmatrix}$
$=3(-5-3)-3(-2-3)+2(2-5)=-15$
$\text{D}_3=\begin{vmatrix}3&-1&3\\2&1&5\\1&-2&1\end{vmatrix}$
$=3(1+10)+1(2-5)+3(-4-1)=-15$
Here$, d$ is zero, but $D_1, D_2$ and $D_3$ are non$-$zero. Thus, the system of linear equations is inconsistent.
View full question & answer→Question 605 Marks
Solve the following determinant equations:
$\begin{vmatrix}3&-2&\sin(3\theta)\\-7&8&\cos(2\theta)\\-11&14&2\end{vmatrix}=0$
Answer$\begin{vmatrix}3&-2&\sin3\theta\\-7&8&\cos2\theta\\-11&14&2\end{vmatrix}=0$
$\Rightarrow3(16-14\cos2\theta)+2(-14+11\cos2\theta)\\+\sin3\theta(-98+88)=0$
$\Rightarrow20(1-\cos2\theta)+10\sin3\theta=0$
$\Rightarrow20(2\sin^2\theta)+10(3\sin\theta-4\sin^3\theta)=0$
$\Rightarrow4\sin^2\theta+3\sin\theta-4\sin^3\theta=0$
$\Rightarrow4\sin^2\theta+3-4\sin^2\theta=0$
$\Rightarrow4\sin^2\theta-4\sin\theta-3=0$
$\Rightarrow(2\sin\theta+1)(2\sin\theta-3)=0$
$\Rightarrow\sin\theta=-\frac{1}{2}$ or $\sin\theta=\frac{3}{2}=1.5$
As $\sin\theta\in[-1,1]$
$\therefore\sin\theta=-\frac{1}{2}$
$\Rightarrow\theta=\text{n}\pi+(-1)^{\text{n}}\frac{\pi}{6},\text{n }\epsilon\text{ z}$
View full question & answer→Question 615 Marks
Prove that $\begin{vmatrix}\text{bc}-\text{a}^2&\text{ca}-\text{b}^2&\text{ab}-\text{c}^2\\\text{ca}-\text{b}^2&\text{ab}-\text{c}^2&\text{bc}-\text{a}^2\\\text{ab}-\text{c}^2&\text{bc}-\text{a}^2&\text{ca}-\text{b}^2\end{vmatrix}$ is divisible by $(a + b + c)$ and find the quotient.
Answer$\Delta=\begin{vmatrix}\text{bc}-\text{a}^2&\text{ca}-\text{b}^2&\text{ab}-\text{c}^2\\\text{ca}-\text{b}^2&\text{ab}-\text{c}^2&\text{bc}-\text{a}^2\\\text{ab}-\text{c}^2&\text{bc}-\text{a}^2&\text{ca}-\text{b}^2\end{vmatrix}$
$\big[$Applying $\text{C}_1\rightarrow\text{C}_1-\text{C}_2$ and $\text{C}_2\rightarrow\text{C}_2-\text{C}_3\big]$
$\Delta=\begin{vmatrix}\text{bc}-\text{a}^2-\text{ca}+\text{b}^2&\text{ca}-\text{b}^2-\text{ab} +\text{c}^2&\text{ab}-\text{c}^2\\\text{ca}-\text{b}^2-\text{ab}+\text{c}^2&\text{ab}-\text{c}^2-\text{bc}+\text{a}^2&\text{bc}-\text{a}^2\\\text{ab}-\text{c}^2-\text{bc}+\text{a}^2&\text{bc}-\text{a}^2-\text{ca}+\text{b}^2&\text{ca}-\text{b}^2\\\end{vmatrix}$
$=\begin{vmatrix}(\text{b}-\text{a})(\text{a}+\text{b}+\text{c})&(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})&\text{ab}-\text{c}^2\text{c}-\text{b})(\text{a}+\text{b}+\text{c})&(\text{a}-\text{c})(\text{a}+\text{b}+\text{c})&\text{bc}-\text{a}^2\text{a}-\text{c})(\text{a}+\text{b}+\text{c})&(\text{b}-\text{a})(\text{a}+\text{b}+\text{c})&\text{ca}-\text{b}^2\end{vmatrix}$
$\big[$Taking $(\text{a}+\text{b}+\text{c})$ common from $\text{C}_1$ and $\text{C}_2$ each$\big]$
$\Delta=(\text{a}+\text{b}+\text{c})=\begin{vmatrix}\text{b}-\text{a}&\text{c}-\text{b} &\text{ab}-\text{c}^2\\\text{c}-\text{b}&\text{a}-\text{c}&\text{bc}-\text{a}^2\\\text{a}-\text{c}&\text{b}-\text{a}&\text{ca}-\text{b}^2\end{vmatrix}$
$\big[$Applying $\text{R}_1\rightarrow\text{R}_1+\text{R}_2+\text{R}_3\big]$
$\Delta=(\text{a}+\text{b}+\text{c})=\begin{vmatrix}0&0&\text{ab}+\text{bc}+\text{ca}-(\text{a}^2+\text{b}^2+\text{c}^2)\\\text{c}-\text{b}&\text{a}-\text{c}&\text{bc}-\text{a}^2\\\text{a}-\text{c}&\text{b}-\text{a}&\text{ca}-\text{b}^2\end{vmatrix}$
$[$Expanding along $R_1]$
$\Delta=(\text{a}+\text{b}+\text{c})^2\big[\text{ab}+\text{bc}+\text{ca}-(\text{a}^2+\text{b}^2+\text{c}^2)\big]\big[(\text{c}-\text{b})(\text{b}-\text{a})(\text{a}-\text{c})^2\big]$
$=(\text{a}+\text{b}+\text{c})^2(\text{ab}+\text{bc}+\text{ca}-\text{a}^2-\text{b}^2-\text{c}^2)\times(\text{bc}-\text{ac}-\text{b}^2+\text{ab}-\text{a}^2-\text{c}^2+2\text{ac})$
$=(\text{a}+\text{b}+\text{c})\big[(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca})^2\big]$
Hence, given deteminant is divisible by $(a + b + c)$ and quotient is
$(\text{a}+\text{b}+\text{c})\big(\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}\big)^2$
View full question & answer→Question 625 Marks
If $a, b$ and $c$ are all non-zero and $\begin{vmatrix}1+\text{a}&1&1\\1&1+\text{b}&1\\1&1&1+\text{c} \end{vmatrix}=0,$ then prove that $\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}+1=0.$
Answer$\begin{vmatrix}1+\text{a}&1&1\\1&1+\text{b}&1\\1&1&1+\text{c} \end{vmatrix}=0$
$C_1 \rightarrow C_1 - C_2$
$\begin{vmatrix}\text{a}&1&1\\-\text{b}&1+\text{b}&1\\1&1&1+\text{c} \end{vmatrix}=0$
$C_2 \rightarrow C_2 - C_3$
$\begin{vmatrix}\text{a}&0&1\\-\text{b}&\text{b}&1\\0&-\text{c}&1+\text{c} \end{vmatrix}=0$
Expanding along $R_1,$ we get
$\text{a}(\text{b}+\text{bc}+\text{c})+1(\text{bc})=0$
$\Rightarrow\text{ab}+\text{abc}+\text{ac}+\text{bc}=0$
Dividing by $abc,$ we get
$\frac{1}{\text{c}}+1+\frac{1}{\text{b}}+\frac{1}{\text{a}}=0$
$\therefore\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}+1=0$
View full question & answer→Question 635 Marks
Solve the following systems of linear equations by cramer's rule:
x + 2y = 1,
3x + y = 4
Answer$\text{D}=\begin{vmatrix}1&2\\3&1\end{vmatrix}=-5$
$\text{D}_1=\begin{vmatrix}1&2\\4&1\end{vmatrix}=-7$
$\text{D}_2=\begin{vmatrix}1&1\\3&4\end{vmatrix}=1$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{7}{5}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=-\frac{1}{5}$
$\therefore\text{x}=\frac{7}{5}$ and $\text{y}=-\frac{1}{5}$
View full question & answer→Question 645 Marks
If $\text{A}=\frac{1}{9}\begin{bmatrix}-8 & 1 & 4\\4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix},$ prove that $A^{-1} = A^3.$
Answer$\text{A}=\frac{1}{9}\begin{bmatrix}-8 & 1 & 4\\4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix}\text{ and A}^\text{T}=\frac{1}{9}\begin{bmatrix}-8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{bmatrix}$
$|\text{A}|=\frac{1}{9}\big[-8(16+56)-1(9)+4(-36)\big]=-81$
$\ce{C_{11}= 72'' C_{21} = -36'' C_{31} = -9}$
$\ce{C_{12} = -9'' C_{22} = -36'' C_{32} = +72}$
$\ce{C_{13} = -36'' C_{23} = -63'' C_{33} = -36}$
$\text{A}^{-1}=\frac{1}{-81}\begin{bmatrix} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & -63 & -36 \end{bmatrix}=\frac{1}{9}\begin{bmatrix}-8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{bmatrix}$
Hence proved.
View full question & answer→Question 655 Marks
Prove that:
$\begin{vmatrix}(\text{b}+\text{c})^2&\text{a}^2&\text{bc}\\\text{c}+\text{a})^2&\text{b}^2&\text{ca} \\\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$ $=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
Answer$\text{L.H.S}=\begin{vmatrix}(\text{b}+\text{c})^2&\text{a}^2&\text{bc}\\\text{c}+\text{a})^2&\text{b}^2&\text{ca}\\\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$
$=\begin{vmatrix}(\text{b}+\text{c})^2-(\text{c}+\text{a})^2&\text{a}^2-\text{b}^2&\text{bc}-\text{ca}\\\text{c}+\text{a})^2-(\text{a}+\text{b})^2&\text{b}^2-\text{c}^2&\text{ca}-\text{ab}\\\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix} [$Applying $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_1]$
$=\begin{vmatrix}(\text{b}+\text{c})(\text{b}+2\text{c}+\text{a})&(\text{a}+\text{b})(\text{a}-\text{b})&\text{c}(\text{b}-\text{a})\\\text{c}-\text{a})(\text{b}+2\text{a}+\text{c})&(\text{b}-\text{c})(\text{b}+\text{c})&\text{a}(\text{c}-\text{b})\\\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}-(\text{b}+2\text{c}+\text{a})&\text{a}+\text{b}&-\text{c}\\-(\text{b}+2\text{a}+\text{c})&\text{b}+\text{c}&-\text{a}\\\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$
$[$Applying $x_2 - y_2 = (x + y)(x - y)$ and taking out $(a - b)$ common from $R_1$ and $(b - c)$ from $R_2]$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}-2(\text{b}+\text{c}+\text{a})&\text{a}+\text{b}&-\text{c}\\-2(\text{b}+\text{a}+\text{c})&\text{b}+\text{c}&-\text{a}\\\text{a}+\text{b})^2-\text{c}^2&\text{c}^2&\text{ab}\end{vmatrix} [$Applying $C_1 \rightarrow C_1 - C_2]$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}-2(\text{b}+\text{c}+\text{a})&\text{a}+\text{b}&-\text{c}\\-2(\text{b}+\text{a}+\text{c})&\text{b}+\text{c}&-\text{a}\\\text{a}+\text{b}+\text{c})(\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$
$[$Applying $x^2 - y^2 = (x + y)(x - y)$ in $C_1]$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})\begin{vmatrix}-2&\text{a}+\text{b}&-\text{c}\\-2&\text{b}+\text{c}&-\text{a}\\\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$
$[$Taking out $(a + b + c)$ common from $C_1]$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})\begin{vmatrix}-2&\text{a}+\text{b}&-\text{c}\\0&\text{c}-\text{a}&\text{c}-\text{a}\\\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$
$[$ Applying $R_2 \rightarrow R_2 - R_1]$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})\begin{vmatrix}-2&\text{a}+\text{b}&-\text{c}\\0&1&1\\\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$
$[$Taking out $(c - a)$ common from $R_2]$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})\begin{vmatrix}-2&\text{a}+\text{b}+\text{c}&-\text{c}\\0&0&1\\\text{a}+\text{b}-\text{c})&\text{c}^2-\text{ab}&\text{ab}\end{vmatrix}$
$[$Applying $C_2 \rightarrow C_2 - C_3]$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a}) \left\{(-1)\begin{vmatrix}-2&\text{a}+\text{b}+\text{c}&\\\text{a}+\text{b}-\text{c})&\text{c}^2-\text{ab}\end{vmatrix}\right\} [$Expanding along $R_2]$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})\{-2\text{c}^2+2\text{ab}-\text{a}^2-\text{b}^2-2\text{ab}+\text{c}^2\}$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})(-\text{a}^2-\text{b}^2-\text{c}^2)$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
$=\text{R.H.S}$
View full question & answer→Question 665 Marks
By using properties of determinants, show that: $\begin{vmatrix}a^2+1&ab&ac\\ab&b^2+1&bc\\ca&cb&c^2+1\end{vmatrix}=1+a^2+b^2+c^2$
Answer$\text{L.H.S.}=\begin{vmatrix}a^2+1&ab&ac\\ab&b^2+1&bc\\ca&cb&c^2+1\end{vmatrix} $
Multiplying $C_1, C_2, C_3$ by $a, b, c$ respectively and then dividing the determinant by $abc,$
$=\frac{1}{abc}\begin{vmatrix}a(a^2+1)&ab^2&ac^2\\a^2b&b(b^2+1)&bc^2\\a^2c&b^2c&c(c^2+1)\end{vmatrix}$
$=\frac{abc}{abc}\begin{vmatrix}a^2+1&b^2&c^2\\a^2&b^2+1&c^2\\a^2&b^2&c^2+1\end{vmatrix}$
$=\frac{abc}{abc}\begin{vmatrix}1+a^2+b^2+c^2&b^2&c^2\\1+a^2+b^2+c^2&b^2+1&c^2\\1+a^2+b^2+c^2&b^2&c^2+1\end{vmatrix}\ \ \left[\text{C}_1\rightarrow\text{C}_1+\text{C}_2+\text{C}_3\right]$
$=(1+a^2+b^2+c^2)\begin{vmatrix}1&b^2&c^2\\1&b^2+1&c^2\\1&b^2&c^2+1\end{vmatrix}$
$=(1+a^2+b^2+c^2)\begin{vmatrix}1&b^2&c^2\\0&1&0\\0&0&1\end{vmatrix}\ \left[\text{R}_2\rightarrow\text{R}_2-\text{R}_1\text{and R}_3\rightarrow\text{R}_3-\text{R}_1\right]$
$=(1+a^2+b^2+c^2)(1)(1-0)=1+a^2+b^2+c^2=\text{R.H.S.}$ Proved.
View full question & answer→Question 675 Marks
Without expanding, show that the values of the following determinant are zero: $\begin{vmatrix}\cos(\text{x}+\text{y})& \sin(\text{x}+\text{y})&\cos2\text{y}\\\sin\text{x}&\cos\text{x}&\sin\text{y}\\-\cos\text{x}&\sin\text{x}&-\cos\text{y} \end{vmatrix}$
Answer$\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\sin\text{x}&\cos\text{x}&\sin\text{y}\\-\cos\text{x}&\sin\text{x}&-\cos\text{y} \end{vmatrix}$
$=\frac{1}{\sin\text{y}\cos\text{y}}\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y})\\\sin\text{x}\sin\text{y}&\cos\text{x}\sin\text{y}&\sin^2\text{y}\\-\cos\text{x}\cos\text{y}&\sin\text{x}\cos\text{y}&-\cos^2\text{y} \end{vmatrix}$
$[$Applying $R_2 \rightarrow$ siny $R_2$ and $R_3 \rightarrow$ cosy $R_3]$
$=\frac{1}{\sin\text{y}\cos\text{y}}\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\sin\text{x}\sin\text{y}-\cos\text{x}\cos\text{y}&\cos\text{x}\sin\text{y}+\sin\text{x}\sin\text{y}&\sin^2\text{y}-\cos^2\text{y}\\-\cos\text{x}\cos\text{y}&\sin\text{x}\cos\text{y}&-\cos^2\text{y} \end{vmatrix}$
$[$Applying $R_2 \rightarrow R_2 + R_3]$
$=\frac{1}{\sin\text{y}\cos\text{y}}\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\-\cos\text{x}\cos\text{y}&\sin\text{x}\cos\text{y}&-\cos^2\text{y} \end{vmatrix}$
$=0$
View full question & answer→Question 685 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sin^2\text{A}&\cot\text{A}&1\\\sin^2\text{B}&\cot\text{B}&1\\\sin^2\text{C}&\cot\text{C}&1\end{vmatrix}$
Answer$\begin{vmatrix}\sin^2\text{A}&\cot\text{A}&1\\\sin^2\text{B}&\cot\text{B}&1\\\sin^2\text{C}&\cot\text{C}&1\end{vmatrix}$
$=\begin{vmatrix}\sin^2\text{A}-\sin^2\text{B}&\cot\text{A}-\cot\text{B}&0\\\sin^2\text{B}&\cot\text{B}&1\\\sin^2\text{C}-\sin^2\text{B}&\cot\text{C}-\cot\text{B}&0\end{vmatrix} [$Applying $R_1 \rightarrow R_1 - R_2$ and $R_3 \rightarrow R_3 - R_2]$
$=\begin{vmatrix}\sin(\text{A}+\text{B})\sin(\text{A}-\text{B})&\frac{\cos\text{A}\sin\text{B}-\cos\text{B}\sin\text{A}}{\sin\text{A}\sin\text{B}}&0\\\sin^2\text{B}&\cot\text{B}&1\\\sin(\text{C}+\text{B})\sin(\text{C}-\text{B})&\frac{\cos\text{C}\sin\text{B}-\cos\text{B}\sin\text{C}}{\sin\text{B}\sin\text{C}}&0\end{vmatrix}$
$=\begin{vmatrix}\sin(\pi-\text{C})\sin(\text{A}-\text{B})&\frac{-\sin(\text{A}-\text{B})}{\sin\text{A}\sin\text{B}}&0\\\sin^2\text{B}&\cot\text{B}&1\\\sin(\pi-\text{A})\sin(\text{C}-\text{B})&\frac{-\sin(\text{C}-\text{B})}{\sin\text{B}\sin\text{C}}&0\end{vmatrix}$ $[\because\text{A}+\text{B}+\text{C}=\pi]$
$=\begin{vmatrix}\sin\text{C}\sin(\text{A}-\text{B})&\frac{-\sin(\text{A}-\text{B})}{\sin\text{A}\sin\text{B}}&0\\\sin^2\text{B}&\frac{\cos\text{B}}{\sin\text{B}}&1\\\sin\text{A}\sin(\text{C}-\text{B})&\frac{-\sin(\text{C}-\text{B})}{\sin\text{A}\sin\text{B}}&0\end{vmatrix}$
$=\frac{\sin(\text{A}-\text{B})\sin(\text{C}-\text{B})}{\sin\text{B}}\begin{vmatrix}\sin\text{C}&\frac{-1}{\sin\text{A}}&0\\\sin^2\text{B}&\cos\text{B}&1\\\sin\text{A}&\frac{-1}{\sin\text{C}}&0\end{vmatrix}$
$=\frac{\sin(\text{A}-\text{B})\sin(\text{C}-\text{B})}{\sin\text{B}\sin\text{A}\sin\text{C}}\begin{vmatrix}\sin\text{C}\sin\text{A}&-1&0\\\sin^2\text{B}&\cos\text{B}&1\\\sin\text{A}\sin\text{C}&-1&0\end{vmatrix} [$Applying $R_1 \rightarrow \sin A R_1 $ and $R_3 \rightarrow \sin C R_3]$
$=\frac{\sin(\text{A}-\text{B})\sin(\text{C}-\text{B})}{\sin\text{B}\sin\text{A}\sin\text{C}}\begin{vmatrix}0&0&0\\\sin^2\text{B}&\cos\text{B}&1\\\sin\text{A}\sin\text{C}&-1&0\end{vmatrix} [$Applying $R_1 \rightarrow R_1 - R_3]$
$=0$
View full question & answer→Question 695 Marks
Prove that:
$\begin{vmatrix}\text{a}^2&2\text{ab}&\text{b}^2\\\text{b}^2&\text{a}^2&2\text{ab}\\2\text{ab}&\text{b}^2&\text{a}^2\end{vmatrix}=(\text{a}^3+\text{b}^3)^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^2&2\text{ab}&\text{b}^2\\\text{b}^2&\text{a}^2&2\text{ab}\\2\text{ab}&\text{b}^2&\text{a}^2\end{vmatrix}$
$=\text{a}^2\begin{vmatrix}\text{a}^2&2\text{ab}\\\text{b}^2&\text{a}^2\end{vmatrix}-(2\text{ab})\begin{vmatrix}\text{b}^2&2\text{ab}\\2\text{ab}&\text{a}^2\end{vmatrix}+\text{b}^2\begin{vmatrix}\text{b}^2&\text{a}^2\\2\text{ab}&\text{b}^2\end{vmatrix}$ [Expanding]
$=\text{a}^2(\text{a}^4-2\text{ab}^3)-(2\text{ab})(\text{b}^2\text{a}^2-4\text{a}^2\text{b}^2)+\text{b}^2(\text{b}^4-2\text{a}^3\text{b})$
$=\text{a}^6-2\text{a}^3\text{b}^3-2\text{a}^3\text{b}^3+8\text{a}^3\text{b}^3+\text{b}^6-2\text{a}^3\text{b}^3$
$=\text{a}^6+2\text{a}^3\text{b}^3+(\text{b}^3)^2$
$=(\text{a}^3+\text{b}^3)^2$
$=\text{R.H.S}$
View full question & answer→Question 705 Marks
$$Find the inverse of each of the matrix:
$\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}$
Answer$\text{Let A}=\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}$ $\therefore\ |\text{A}|=\begin{vmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{vmatrix}$
$=1\left(-\cos^2\alpha-\sin^2\alpha\right)-0+0=-\left(\cos^2\alpha+\sin^2\alpha\right)=-1\neq0$
$\text{A}_{11}=+\begin{vmatrix}\cos\alpha&\sin\alpha\\\sin\alpha&-\cos\alpha\end{vmatrix}=+\left(-\cos^2\alpha-\sin^2\alpha\right)=-\left(\cos^2\alpha+\sin^2\alpha\right)=-1$
$\text{A}_{12}=-\begin{vmatrix}0&\sin\alpha\\0&-\cos\alpha\end{vmatrix}=-(0-0)=0,$
$\text{A}_{13}=+\begin{vmatrix}0&\cos\alpha\\0&\sin\alpha\end{vmatrix}=+(0-0)=0,$
$\text{A}_{21}=-\begin{vmatrix}0&0\\\sin\alpha&-\cos\alpha\end{vmatrix}=-(0-0)=0,$
$\text{A}_{22}=+\begin{vmatrix}1&0\\0&-\cos\alpha\end{vmatrix}=+(-\cos\alpha-0)=-\cos\alpha,$
$\text{A}_{23}=-\begin{vmatrix}1&0\\0&\sin\alpha\end{vmatrix}=(-\sin\alpha-0)=\sin\alpha,$
$\text{A}_{31}=+\begin{vmatrix}0&0\\\cos\alpha&\sin\alpha\end{vmatrix}=(0-0)=0,$
$\text{A}_{32}=-\begin{vmatrix}1&0\\0&\sin\alpha\end{vmatrix}=-(\sin\alpha-0)=-\sin\alpha,$
$\text{A}_{33}=+\begin{vmatrix}1&0\\0&\cos\alpha\end{vmatrix}=+(\cos\alpha-0)=\cos\alpha$
$\therefore\ \text{adj.A}=\begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix}=\begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix}$
$\therefore\ \text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj. A}=-\begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix}=\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}$
View full question & answer→Question 715 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}49&1&6\\39&7&4\\26&2&3 \end{vmatrix}$
Answer$\begin{vmatrix}49&1&6\\39&7&4\\26&2&3 \end{vmatrix}$
Applying: $\ce{C_1 \rightarrow C_1 + (-8)C_3}$
$=\begin{vmatrix}1&1&6\\7&7&4\\2&2&3 \end{vmatrix}=0$
$\because\text{C}_1=\text{C}_2$
View full question & answer→Question 725 Marks
Solve the following system of homogeneous linear equations:
3x + y + z = 0,
x - 4y + 3z = 0,
2x + 5y - 2z = 0
AnswerGiven,
3x + y + z = 0,
x - 4y + 3z = 0,
2x + 5y - 2z = 0
$\text{D}=\begin{vmatrix}3&1&1\\1&-4&3\\2&5&-2\end{vmatrix}=0$
The system has infinitely many solution Puting z = k in the first two equation we get
3x + y = -k
x - 4y = -3k
Solving these equations by Cramer's rule, we get
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{\begin{vmatrix}-\text{k}&1\\-3\text{k}&-4\end{vmatrix}}{\begin{vmatrix}3&1\\1&-4\end{vmatrix}}=-\frac{7\text{k}}{13}$
$\text{y}=\frac{\text{D}_2}{\text{D}} =\frac{\begin{vmatrix}3&-\text{k}\\1&-3\text{k}\end{vmatrix}}{\begin{vmatrix}3&1\\1&-4\end{vmatrix}}=\frac{8\text{k}}{13} \text{z}=\text{k}$
$\text{z}=\text{k}$
$\Rightarrow\text{x}=-\frac{7\text{k}}{13},\text{ y}=\frac{8\text{k}}{13}$ and $\text{z}=\text{x}$
Or x = -7k, y = 8k and z = 13k
Clearly, these value satisfy the thried equation.
Thus, x = -7k, y = 8k, z = 13k $[\text{k}\in\text{R}]$
View full question & answer→Question 735 Marks
If $a, b, c$ are real numbers such that $\begin{vmatrix}\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}=0,$ then show that either $a + b + c = 0$ or $a = b= c.$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}$
$=\begin{vmatrix}2(\text{a}+\text{b}+\text{c})&2(\text{a}+\text{b}+\text{c})&2(\text{a}+\text{b}+\text{c})\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix} [$Applying $R_1 \rightarrow R_1 + R_2 + R_3]$
$=2(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&1&1\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}$
$=2(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&0&0\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}-\text{a}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}-\text{b}\end{vmatrix} [$Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=2(\text{a}+\text{b}+\text{c})\left\{1\begin{vmatrix}\text{b}-\text{c}&\text{b}-\text{a}\\\text{c}-\text{a}&\text{c}-\text{b}\end{vmatrix}\right\}$
$=2(\text{a}+\text{b}+\text{c})\{(\text{b}-\text{c})(\text{c}-\text{b})-(\text{b}-\text{a})(\text{c}-\text{a})\}$
$=-2(\text{a}+\text{b}+\text{c})\{\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}\}$
$=-(\text{a}+\text{b}+\text{c})\{2\text{a}^2+2\text{b}^2+2\text{c}^2-2\text{ab}-2\text{bc}-2\text{ca}\}$
$=-(\text{a}+\text{b}+\text{c})\{(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2\}$
But $\triangle=0 [$Given$]$
$\Rightarrow-(\text{a}+\text{b}+\text{c})\{(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2\}=0$
Either,
$\Rightarrow(\text{a}+\text{b}+\text{c})=0$ or $(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2=0$
$\Rightarrow(\text{a}+\text{b}+\text{c})=0$ or $\text{a}=\text{b}=\text{c}$
Hence proved.
View full question & answer→Question 745 Marks
Find the adjoint of the following matrices: $\begin{bmatrix} \text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix}$Verify that $($adjoint $A) A = |A|I = A ($adjoint $A)$ for the above matrices.
AnswerHere, $\text{A}=\begin{bmatrix} \text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix}$
cofactors of $A$ are:
$C_{11} = d$
$C_{12} = -d$
$C_{21} = -b$
$C_{22} = a$
$\therefore\ \text{adjoint A}=\begin{bmatrix}\text{C}_{11} & \text{C}_{12} \\ \text{C}_{21} & \text{C}_{22} \end{bmatrix}^\text{T}$
$\therefore\ \text{adjoint A}=\begin{bmatrix}\text{d} & -\text{C} \\ -\text{b} & \text{a} \end{bmatrix}$
$=\begin{bmatrix}\text{d} & -\text{b} \\ -\text{C} & \text{a} \end{bmatrix}$
Now, $\text{adjoint A}=\begin{bmatrix}\text{d} & -\text{b} \\ -\text{C} & \text{a} \end{bmatrix}\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix}$
$=\begin{bmatrix}\text{ad}-\text{bc} & \text{bd}-\text{bd} \\ -\text{ac}+\text{ac} & \text{ad}-\text{bc} \end{bmatrix}$
$=\begin{bmatrix}\text{ad}-\text{bc} & 0 \\ 0 & \text{ad}-\text{bc} \end{bmatrix}$
And, $|\text{A}|\text{I}=\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$=(\text{ad}-\text{bc})\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$=\begin{bmatrix}\text{ad}-\text{bc} & 0 \\ 0 & \text{ad}-\text{bc} \end{bmatrix}$
Also, $\text{A(adjoint A)}=\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix}\begin{bmatrix} \text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
$=\begin{bmatrix}\text{ad}-\text{bc} & 0 \\ 0 & \text{ad}-\text{bc} \end{bmatrix}$
$\therefore\ \text{(adjoint A)A}=|\text{A}|\text{I}=\text{A(adjoint A)}$
View full question & answer→Question 755 Marks
Solve the following systems of linear equations by cramer's rule:
5x + 7y = -2
4x + 6y = -3
AnswerLet $\text{D}=\begin{vmatrix}5&7\\4&6\end{vmatrix}=-2$
$\text{D}_1=\begin{vmatrix}-2&7\\-3&6\end{vmatrix}=9$
$\text{D}_2=\begin{vmatrix}5&-2\\4&-3\end{vmatrix}=-7$
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{9}{2}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-7}{2}$
View full question & answer→Question 765 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}2&3&7\\13&17&5\\15&20&12 \end{vmatrix}$
Answer$\begin{vmatrix}2&3&7\\13&17&5\\15&20&12 \end{vmatrix}$
Use: $R_3 \rightarrow R_3 - R_2$
$=\begin{vmatrix}2&3&7\\13&17&5\\2&3&7 \end{vmatrix}$
$=0$
$\because\text{R}_3=\text{R}_1$
View full question & answer→Question 775 Marks
For what value of x the matrix A is singular?
$\text{A}=\begin{vmatrix}1+\text{x}&7\\3-\text{x}&8 \end{vmatrix}$
Answer$\text{A}=\begin{vmatrix}1+\text{x}&7\\3-\text{x}&8 \end{vmatrix}$
A matrix A is called singular if |A| = 0
Now expanding along the first row |A|
$=(\text{x}-1)\begin{vmatrix}\text{x}-1&1\\1&\text{x}-1 \end{vmatrix}-1\begin{vmatrix}1&1\\1&\text{x}-1 \end{vmatrix}+1\begin{vmatrix}1&1\\\text{x}-1&-1 \end{vmatrix}$
$=(\text{x}-1)\big[(\text{x}-1)^2-1\big]-1[\text{x}-1-1]+1[1-\text{x}+1]$
$=(\text{x}-1)(\text{x}^2+1-2\text{x}-1)-1(\text{x}-2)+1(2-\text{x})$
$=(\text{x}-1)(\text{x}^2-2\text{x})-\text{x}+2+2-\text{x}$
$=(\text{x}-1)\times(\text{x})\times(\text{x}-2)+(4-2\text{x})$
$=(\text{x}-1)\times(\text{x})\times(\text{x}-2)+2(2-\text{x})$
$=(\text{x}-1)\times(\text{x})\times(\text{x}-2)-2(\text{x}-2)$
$=(\text{x}-2)[\text{x}(\text{x}-1)-2]$ (Taking (x - 2) common)
Since A is a singular matrix, so |A| = 0
i.e., $(\text{x}-2)(\text{x}^2-\text{x}-2)=0$
either $(\text{x}-2)=0$ or $\text{x}^2-\text{x}-2=0$
$\text{x}=2$ or $\text{x}^2-2\text{x}+\text{x}-2=0$
$\text{x}(\text{x}-2)+1(\text{x}-2)=0$
$(\text{x}-2)(\text{x}+1)=0$
$\text{x}=2,-1$
$\text{x}=2\text{ or}-1$
View full question & answer→Question 785 Marks
Find the inverse of the matrix $\text{A}=\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \frac{1+\text{bc}}{\text{a}} \end{bmatrix}$ ans show that $aA^{-1} = (a^2 + bc + 1) I - aA.$
Answer$\text{A}=\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \frac{1+\text{bc}}{\text{a}} \end{bmatrix}$
$\Rightarrow\ |\text{A}|=(1+\text{bc})-\text{bc}=1\neq0$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\frac{1}{1}\text{A}=\begin{bmatrix}\frac{1+\text{bc}}{\text{a}} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
Now $aA^{-1} = (a^2 + bc + 1)I - aA$
$\text{L.H.S}:\text{aA}^{-1}=\text{a}\begin{bmatrix}\frac{1+\text{bc}}{\text{a}} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
$=\begin{bmatrix} 1+\text{bc} & -\text{ab} \\ -\text{ac} & \text{a}^2 \end{bmatrix}$
$\text{R.H.S}:(\text{a}^2+\text{bc}+1)\text{I}-\text{aA}$
$=\begin{bmatrix}\text{a}^2+\text{bc}+1 & 0 \\0 & \text{a}^2+\text{bc}+1 \end{bmatrix}-\begin{bmatrix}\text{a}^2 & \text{ab} \\ \text{ac} & 1+\text{bc} \end{bmatrix}$
$=\begin{bmatrix}1+\text{bc} & -\text{ab} \\ -\text{ac} & \text{a}^2 \end{bmatrix}$
Since, $L.H.S = R.H.S$
Hence proved.
View full question & answer→Question 795 Marks
Find the inverse of the following matrices and verify that $A^{-1} A = I_3.\begin{bmatrix}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}$
Answer$\text{A}=\begin{bmatrix}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}$
Now, $\text{C}_{11}=\begin{bmatrix}4 & 3 \\ 3 & 4 \end{bmatrix}=7,\ \text{C}_{12}=-\begin{bmatrix}1 & 3 \\ 1 & 4 \end{bmatrix}=-1$
$\text{and C}_{13}=\begin{bmatrix}1 & 4 \\ 1 & 3 \end{bmatrix}=-1$
$\text{C}_{21}=-\begin{bmatrix}3 & 3 \\ 3 & 4 \end{bmatrix}=-3,\ \text{C}_{22}=\begin{bmatrix}1 & 3 \\ 1 & 4 \end{bmatrix}=1$
$\text{and C}_{23}=-\begin{bmatrix}1 & 3 \\ 1 & 3 \end{bmatrix}=0$
$\text{C}_{31}=\begin{bmatrix}3 & 3 \\ 4 & 3 \end{bmatrix}=-3,\ \text{C}_{32}=-\begin{bmatrix}1 & 3 \\ 1 & 3 \end{bmatrix}=0$
$\text{and C}_{33}=\begin{bmatrix}1 & 3 \\ 1 & 4 \end{bmatrix}=1$
$\text{adj A}=\begin{bmatrix}7 & -3 & -3 \\ -1 & 1 & 0 \\ -3 & 0 & 1\end{bmatrix}^\text{T}=\begin{bmatrix}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{bmatrix}$
and $|A| = 1$
$\text{A}^{-1}=\begin{bmatrix}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{bmatrix}$
Now, $\text{A}^{-1}\text{A}=\begin{bmatrix}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}$
$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\text{I}_3$
View full question & answer→Question 805 Marks
Let $\text{F}(\alpha)=\begin{bmatrix}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}$ and
$\text{G}(\beta)=\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}$
Show that
$\big[\text{F}(\alpha)\big]^{-1}=\text{F}(-\alpha)$
Answer$\text{F}(\alpha)=\begin{bmatrix}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow\ \text{F}(-\alpha)=\begin{bmatrix}\cos(-\alpha) & -\sin(-\alpha) & 0 \\ \sin(-\alpha) & \cos(-\alpha) & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$=\begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\-\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Now, $\text{C}_{11}=\begin{vmatrix}\cos\alpha & 0 \\0 & 1 \end{vmatrix}=\cos\alpha,\ \text{C}_{12}=-\begin{vmatrix}\sin\alpha & 0 \\ 0 & 1 \end{vmatrix}=-\sin\alpha$
$\text{and C}_{13}=\begin{vmatrix}\sin\alpha & \cos\alpha \\0 & 0 \end{vmatrix}=0$
$\text{C}_{21}=-\begin{vmatrix}-\sin\alpha & 0 \\0 & 1 \end{vmatrix}=\sin\alpha,\ \text{C}_{22}=\begin{vmatrix}\cos\alpha & 0 \\0 & 1 \end{vmatrix}=\cos\alpha$
$\text{and C}_{23}=-\begin{vmatrix}\cos\alpha & -\sin\alpha \\0 & 0 \end{vmatrix}=0$
$\text{C}_{31}=\begin{bmatrix} -\sin\alpha & 0 \\ \cos\alpha & 0 \end{bmatrix}=0, \text{C}_{32}=-\begin{bmatrix}\cos\alpha & 0 \\ \sin\alpha & 0 \end{bmatrix}=0$
$\text{and C}_{33}=\begin{bmatrix}\cos\alpha & -\sin\alpha \\ 0 & 0 \end{bmatrix}=0$
$\Rightarrow\ \text{adj}\left\{\text{F}(\alpha)\right\}=\begin{bmatrix}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}^\text{T}$
$=\begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow\ |\text{F}(\alpha)|=1$
$\therefore\ \big[\text{F}(\alpha)\big]^{-1}=\begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}\ .....(\text{i})$
$\Rightarrow\ \big[\text{F}(\alpha)\big]^{-1}=\text{F}(-\alpha)$
View full question & answer→Question 815 Marks
Using properties of determinants, prove that:
$\begin{vmatrix}3\text{a}&-\text{a+b}&-\text{a+c}\\-\text{b+a}&3\text{b}&-\text{b+c}\\-\text{c+a}&\text{c+b}&3\text{c}\end{vmatrix}=3 (\text{a + b + c}) (\text{ab + bc + ca})$
Answer$\triangle=\begin{vmatrix}3\text{a}&-\text{a + b}&-\text{a + c}\\-\text{b + a}&3\text{b}&-\text{b + c}\\-\text{c + a}&\text{c + b}&3\text{c}\end{vmatrix}$
Applying $C_1 \rightarrow C_1 + C_2 + C_3,$ we have:
$\triangle=\begin{vmatrix}\text{a + b + c}&-\text{a + b}&-\text{a + c}\\\text{a + b + c}&3\text{b}&-\text{b + c}\\\text{a + b + c}&-\text{c + b}&3\text{c}\end{vmatrix}$
$= (\text{a + b + c})\begin{vmatrix}1&-\text{a + b}&-\text{a + c}\\1&3\text{b}&-\text{b + c}\\1&-\text{c + b}&3\text{c}\end{vmatrix}$
Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$, we have:
$\triangle=(\text{a + b + c})\begin{vmatrix}1&-\text{a + b}&-\text{a + c}\\0&2\text{b + a}&\text{a}-\text{b}\\0&\text{a}-\text{c}&2\text{c + a}\end{vmatrix}$
Expanding along $C_1,$ we have:
$\triangle= (a + b + c) [(2b + a) (2c + a) - (a - b) (a - c)]$
$= (a + b + c) [4bc + 2ab + 2ac + a^2 - a^2 + ac + ba - bc]$
$= (a + b + c) (3ab + 3bc + 3ac)$
$= 3(a + b + c) (ab + bc + ca)$
Hence, the given result is proved.
View full question & answer→Question 825 Marks
For the following pairs of matrices verify that $(AB)^{-1} = B^{-1} A^{-1}:$
$\text{A}=\begin{bmatrix}3 & 2 \\7 & 5 \end{bmatrix}\text{ and B}=\begin{bmatrix}4 & 6 \\3 & 2 \end{bmatrix}$
Answer$\text{A}=\begin{bmatrix}3 & 2 \\7 & 5 \end{bmatrix}\therefore|\text{A}|=1\neq0$
$\text{adj A}=\begin{bmatrix}5 & -2 \\-7 & 3 \end{bmatrix}\Rightarrow\ \text{A}^{-1}=\frac{\text{adj A}}{|\text{A}|}=\frac{1}{1}\begin{bmatrix}5 & -2 \\-7 & 3 \end{bmatrix}$
$\text{B}=\begin{bmatrix}4 & 6 \\3 & 2 \end{bmatrix}\therefore|\text{B}|=-10\neq0$
$\text{adj B}=\begin{bmatrix}2 & -6 \\-3 & 4 \end{bmatrix} \Rightarrow\ \text{B}^{-1}=\frac{\text{adj B}}{|\text{B}|}=\frac{1}{-10}\begin{bmatrix}2 & -6 \\-3 & 4 \end{bmatrix}$
Also, $\text{AB}=\begin{bmatrix}3 & 2 \\7 & 5 \end{bmatrix}\begin{bmatrix}4 & 6 \\3 & 2 \end{bmatrix}=\begin{bmatrix}18 & 22 \\43 & 52 \end{bmatrix}$
$|\text{AB}|=936-946=-10\neq0$
$\text{adj(AB)}=\begin{bmatrix}52 & -22 \\-43 & 18 \end{bmatrix}$
$(\text{AB})^{-1}=\frac{\text{adj(AB)}}{|\text{AB}|}=\frac{1}{|\text{Ab}|}\begin{bmatrix}52 & -22 \\-43 & 18 \end{bmatrix}$
$=\frac{1}{-10}\begin{bmatrix}+52 & -22 \\-43 & +18 \end{bmatrix}=\frac{1}{10}\begin{bmatrix}-52 & 22 \\43 & -18 \end{bmatrix}$
$\text{B}^{-1}\text{A}^{-1}=\frac{-1}{10}\begin{bmatrix}2 & -6 \\-3 & 4 \end{bmatrix}\begin{bmatrix}5 & -2 \\-7 & 3 \end{bmatrix}$
$=\frac{-1}{10}\begin{bmatrix}52 & -22 \\-43 & 18 \end{bmatrix}=\frac{1}{10}\begin{bmatrix}-52 & 22 \\43 & -18 \end{bmatrix}$
Hence, $(AB)^{-1} = B^{-1}A^{-1}$
View full question & answer→Question 835 Marks
If $\begin{vmatrix}\text{a}&\text{b}-\text{y}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0,$ then using properties of determinants, find the value of $\frac{\text{a}}{\text{x}}+\frac{\text{b}}{\text{y}}+\frac{\text{c}}{\text{z}},$ where $\text{x},\text{y},\text{z}\neq0.$
Answer$\Rightarrow\begin{vmatrix}\text{a}&\text{b}-\text{y}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0$
$R_1 \rightarrow R_1 - R_2$
$\Rightarrow\begin{vmatrix}\text{x}&-\text{y}&0\\\text{a}-\text{x}&\text{b}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0$
$R_{2 }\rightarrow R_2 - R_3$
$\Rightarrow\begin{vmatrix}\text{x}&-\text{y}&0\\0&\text{y}&-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0$
Expanding along first row, we get
$\text{x}(\text{yc}+\text{zb}-\text{zy})+\text{y}(0-\text{za}+\text{zx})=0$
$\Rightarrow\text{xyc}+\text{xzb}-2\text{xyz}+\text{zya}=0$
$\Rightarrow\text{xyc}+\text{xzb}-2\text{xyz}+\text{zya}=0$
Dividing by $xyz,$ we get
$\frac{\text{c}}{\text{z}}+\frac{\text{b}}{\text{y}}-2+\frac{\text{a}}{\text{x}}=0$
$\therefore\frac{\text{a}}{\text{x}}+\frac{\text{b}}{\text{y}}+\frac{\text{c}}{\text{z}}=2$
View full question & answer→Question 845 Marks
Solve the following systems of linear equations by cramer's rule:
2x - y = 17,
3x + 5y = 6
AnswerGiven, 2x - y = 17
3x + 5y = 6
Using cramers Rule, we get
$\text{D}=\begin{vmatrix}2&1\\3&5\end{vmatrix}=10+3=13$
$\text{D}_1=\begin{vmatrix}17&-1\\6&5\end{vmatrix}=85+6=91$
$\text{D}_2=\begin{vmatrix}2&17\\3&6\end{vmatrix}=12-51=-39$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{91}{13}=7$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-39}{13}=-3$
$\therefore\text{x}=7$ and $\text{y}=-3$
View full question & answer→Question 855 Marks
Solve the following systems of linear equations by cramer's rule:
2x - y = 1,
7x - 2y = -7
AnswerLet $\text{D}=\begin{vmatrix}2&-1\\7&-2 \end{vmatrix}=-4+7=3$
$\text{D}_1=\begin{vmatrix}1&-1\\-7&-2 \end{vmatrix}=-9$
$\text{D}_2=\begin{vmatrix}2&1\\7&-7\end{vmatrix}=-21$
Now, $\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-9}{3}=-3$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-21}{3}=-7$
Hence x = -3, y = -7
View full question & answer→Question 865 Marks
Show that the matrix $\text{A}=\begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}$ satisfies the equation,$ \ce{A^3 - A^2 - 3A - I_3 = 0.}$ Hence, find $A^{-1}.$
AnswerWe have, $\text{A}=\begin{bmatrix} 1 & 0 & -2 \\-2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}$
$\Rightarrow\ |\text{A}|=\begin{bmatrix} 1 & 0 & -2 \\-2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}=1(-9)+0-2(-8)=-9+16=7$
Since, $|\text{A}|\neq0$
Hence, $A^{-1}$ exists.
Now,
$\text{A}^2=\begin{bmatrix} 1 & 0 & -2 \\-2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & -2 \\-2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}$
$=\begin{bmatrix}1+0-6 & 0+0-8 & -2+0-2 \\ -2+2+6 & 4-2+2 & 4-2+2 \\ 3-8+3 & 0-4+4 & -6+8+1\end{bmatrix}$
$=\begin{bmatrix} -5 & -8 & -4 \\ -6 & 9 & 4 \\ -2 & 0 & 3 \end{bmatrix}$
$\text{A}^3=\text{A}^2\text{A}=\begin{bmatrix} -5 & -8 & 2 \\ 6 & 9 & 4 \\ -2 & 0 &3 \end{bmatrix}\begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ -2 & -1 & 2 \end{bmatrix}$
$=\begin{bmatrix} 5+16-12 & 0+8+16 & 10-16-4 \\ 6-18+12 & 0-9+16 & -12+18+4 \\ -2+0+9 & 0+0+12 & 4+0+3 \end{bmatrix}$
$=\begin{bmatrix}-1 & -8 & -10 \\0 & 7 & 10 \\ 7 & 12 & 7 \end{bmatrix}$
Now, $\text{A}^3-\text{A}^2-3\text{A}-\text{I}_3=\begin{bmatrix}-1 & -8 & -10 \\0 & 7 & 10 \\ 7 & 12 & 7 \end{bmatrix}-\begin{bmatrix} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3\end{bmatrix}$
$=-3\begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ -2 & -1 & 1 \end{bmatrix}-\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$=\begin{bmatrix}-1+5-3-1 & -8+8+0+0 & -10+4+6-0 \\ 0-6+6-0 & 7-9+3-1 & 10-4-6-0 \\ 7+2-9-0 & 12+0-12-0 & 7-3-3-1\end{bmatrix}$
$=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}=0$
Hence proved.
Now, $\ce{A^3 - A^2 - 3A - I_3 = 0}($Null matrix$)$
$\Rightarrow \ce{A^{-1}(A^3 - A^2- 3A - I_3) = A^{-1} 0}($Pre $-$ multiplying by $A-1)$
$\Rightarrow \ce{A^2 - A^1 - 3I_3 = A^{-1}}$
$\Rightarrow\ \begin{bmatrix}-5 & -8 & -4\\6 & 9 & 4\\ -2 & 0 & 4 \end{bmatrix}-\begin{bmatrix}1 & 0 & 2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}-3\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\text{A}^{-1}$
$\Rightarrow\ \begin{bmatrix}-5-1-3 & -8-0-0 & -4+2+0\\ 6+2+0 & 9+1-3 & 4-2 \\ -2-3-0 & 0-4-0 & 3-1-3 \end{bmatrix}$
$=\begin{bmatrix} -9 & -8 & -2\\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{bmatrix}=\text{A}^{-1}$
$\Rightarrow\ \text{A}^{-1}=\begin{bmatrix} -9 & -8 & -2\\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{bmatrix}$
View full question & answer→Question 875 Marks
Write the minors and cofactors of element of the first column of the following matrices and hence evaluate the determinant in case:
$\text{A}=\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab} \end{vmatrix}$
Answer$\text{M}_{11}=\begin{vmatrix}\text{b} &\text{ca}\\ \text{c}&\text{ab} \end{vmatrix}=\text{ab}^{2}-\text{c}^2\text{a}=\text{a}(\text{b}^2-\text{c}^2)$
$\text{M}_{21}=\begin{vmatrix}\text{a} &\text{bc}\\ \text{c}&\text{ab} \end{vmatrix}=\text{a}^{2}\text{b}-\text{c}^2\text{b}=\text{b}(\text{a}^2-\text{c}^2)$
$\text{M}_{31}=\begin{vmatrix}\text{a} &\text{bc}\\ \text{b}&\text{ca} \end{vmatrix}=\text{a}^{2}\text{c}-\text{b}^2\text{c}=\text{c}(\text{a}^2-\text{b}^2)$
$\text{C}_{11}=(-1)^{1+1}\text{M}_{11}=\text{a}(\text{b}^2-\text{c}^2)$
$\text{C}_{21}=(-1)^{2+1}\text{M}_{21}=-\text{b}(\text{a}^2-\text{c}^2)$
$\text{C}_{31}=(-1)^{3+1}\text{M}_{31}=\text{c}(\text{a}^2-\text{b}^2)$
$\text{D}=1\text{a}(\text{b}^2-\text{c}^2)-\text{a}(\text{ab}-\text{ca})+\text{b}(\text{c}-\text{b})$
$=\text{ab}^2-\text{ac}^2-\text{a}^2\text{b}+\text{a}^2\text{c}+\text{c}^2\text{b}-\text{b}^2\text{c}$
$=\text{a}^{2}(\text{c}-\text{b})+\text{b}^{2}(\text{a}-\text{c})+\text{c}^{2}(\text{b}-\text{a})$
View full question & answer→Question 885 Marks
Evaluate the following: $\begin{vmatrix}\text{x}&1&1\\1&\text{x}&1\\1&1&\text{x}\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}\text{x}&1&1\\1&\text{x}&1\\1&1&\text{x}\end{vmatrix}$
$\triangle=\begin{vmatrix}\text{x}&1&1\\1&\text{x}&1\\1&1&\text{x}\end{vmatrix}$
$=\begin{vmatrix}\text{x}-1&1-\text{x}&0\\1&\text{x}&1\\0&1-\text{x}&\text{x}-1\end{vmatrix} [$Applying $R_1 \rightarrow R_1 - R_2$ and $R_3 \rightarrow R_3 - R_2]$
$=(\text{x}-1)^2\begin{vmatrix}1&-1&0\\1&\text{x}&1\\0&-1&1\end{vmatrix}$
$=(\text{x}-1)^2\begin{vmatrix}1&-1&0\\1&\text{x}+1&1\\0&0&1\end{vmatrix} [$Applying $C_2 \rightarrow C_2 + C_3]$
$=(\text{x}-1)^2(\text{x}+1+1) [$Expanding along last row$]$
$=(\text{x}-1)^2(\text{x}+2)$
$\therefore\triangle=(\text{x}-1)^2(\text{x}+2)$
View full question & answer→Question 895 Marks
Solve the follwing system of equations by matrix method:
$3x + 4y - 5 = 0$
$x - y + 3 = 0$
AnswerThe given system of equations can be written in matrix form as follow:
$\begin{bmatrix}3&4\\ 1&-1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}5\\ -3\end{bmatrix}$
$AX = B$
Here,
$\text{A}=\begin{bmatrix}3&4\\ 1&-1\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{ and }\text{ B}=\begin{bmatrix}5\\ -3\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}3&4\\ 1&-1\end{vmatrix}$
$=-3-4$
$=-7\neq0$
So, the given system has a unique solution given by $X = A^{-1} B.$
Let $C_{ij}$ be the cofactors of the elements $a_{ij}$ in $A = [a_{ij}].$ Then,
$\text{C}_{11}=(-1)^{1+1}(-1)=-1,\text{C}_{12}=(-1)^{1+2}(1)=-1$
$\text{C}_{21}=(-1)^{2+1(4)}=-4,\text{C}_{22}=-{(-1)}^{2+2}{(3)}={(3)}$
$\text{adj A}=\begin{bmatrix}-1&-1\\ -4&3\end{bmatrix}^{T}=\begin{bmatrix}-1&-4\\ -1&3\end{bmatrix}$
$\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{-7}\begin{bmatrix}-1&-4\\ -1&3\end{bmatrix}$
$X = A^{-1} B$
$=\frac{1}{-7}\begin{bmatrix}-1&-4\\ -1&3\end{bmatrix}\begin{bmatrix}5\\ -3\end{bmatrix}$
$=\frac{1}{-7}\begin{bmatrix}-5+12\\ -5-9\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{7}{-7}\\ \frac{-14}{-7}\end{bmatrix}$
$\therefore x = -1$ and $y = 2$
View full question & answer→Question 905 Marks
Show that the following system of linear equation is inconsistent:
4x − 2y = 3
6x − 3y = 5
AnswerThis system can be written as:
$\begin{bmatrix}4&-2\\ 6&-3\end{bmatrix}\begin{bmatrix} \text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}3\\ 5\end{bmatrix}$
or $\text{AX = B}$
$\text{|A|}=-12+12=0$
So, A is singular, Now system will be inconsisent, if
$(\text{adj A})\times\text{B}\neq0$
$\text{C}_{11}=-3$
$\text{C}_{12}=-6$
$\text{C}_{21}=2$
$\text{C}_{22}=4$
$\text{adj A}=\begin{bmatrix}-3&-6\\ 2&4\end{bmatrix}^\text{T}\begin{bmatrix}-3&2\\ -6&4\end{bmatrix}$
$(\text{adj A})\times(\text{B})=\begin{bmatrix}-3&2\\ -6&4\end{bmatrix}\begin{bmatrix}3\\ 5\end{bmatrix}$
$=\begin{bmatrix}-9+10\\ -18+20\end{bmatrix}$
$=\begin{bmatrix}1\\ 2\end{bmatrix}$
$\neq0$
Hence, the above system is inconsisent.
View full question & answer→Question 915 Marks
Solve the following systems of homogeneous linear equations by matrix method:
2x + 3y - z = 0
x - y - 2z = 0
3x + y + 3z = 0
Answer2x + 3y - z = 0 x - y - 2z = 0 3x + y + 3z = 0 Hence, $\text{A}=\begin{bmatrix}2&3&-1\\1&-1&-2\\3&1&3\end{bmatrix}$ $|\text{A}|=\begin{bmatrix}2&3&-1\\1&-1&-2\\3&1&3\end{bmatrix}$ $=2(-3+2)-3(3+6)-1(4)$$=-2-27-4$
$\neq0$
Hence, the system has only trivial solutions given by x = y = z = 0
View full question & answer→Question 925 Marks
Solve the follwing system of equations by matrix method:
$5x + 2y = 3$
$3x + 2y = 5$
AnswerThe above system can be written in matrix form as: $\begin{bmatrix}5&2\\ 3&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}3\\ 2\end{bmatrix}$ Or $\text{AX = B}$ Where, $\text{A}=\begin{bmatrix}5&2\\ 3&2\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix},\text{B}=\begin{bmatrix}3\\ 5\end{bmatrix}$
Now, $\text{|A|}=10-6=4\neq0$
So, the above system has a unique solution, given by $\text{X}=\text{A}^{-1}\text{B}$
Let $C_{ij}$ be the co factor of $a_{ij}$ in $A,$ then$\text{C}_{11} = 2,\text{C}_{12} = -3$
$\text{C}_{21} = -2,\text{C}_{22} = 5$
Also, $\text{Adj A}=\begin{bmatrix}2&-3\\ -2&5\end{bmatrix}^\text{T}=\begin{bmatrix}2&-2\\ -3&5\end{bmatrix}$ $\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj A}=\frac{1}{4}\begin{bmatrix}2&-2\\ -3&5\end{bmatrix}$
Now, $X = A^{-1 }B$
$=\frac{1}{4}\begin{bmatrix}2&-2\\ -3&5\end{bmatrix}\begin{bmatrix}3\\ 5\end{bmatrix}$ $=\frac{1}{4}\begin{bmatrix}-4\\ 16\end{bmatrix}$ $\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}-1\\ 4\end{bmatrix}$
Hence, $x = -1,y = 4$
View full question & answer→Question 935 Marks
Show that the following system of linear equations is consistent and also find solution: $6x + 4y = 2 , 9x + 6y =3$
AnswerHere,
$6\text{x}+4\text{y}=2\ \dots(1)$
$9\text{x}+6\text{y}\ \dots(2)$
$AX = B$
Here,
$\text{A}=\begin{bmatrix}6&4\\ 9&6\end{bmatrix},\text{X}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2\\ 3\end{bmatrix}$
$\begin{bmatrix}6&4\\ 9&6\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}2\\ 3\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}6&4\\ 9&6\end{vmatrix}$
$= 36 -36$
$= 0$
So, A is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because $($adj $A)B \neq 0$ or $($adj $A) = 0.$
Let $C_{ij}$ be the co$-$factors of the elements $a_{ij}$ in $A [a_{ij}].$ Then,
$\text{C}_{11}=6,\\ \text{C}_{12}=-9,\\ \text{C}_{21}=-4,\\ \text{C}_{22}=6$
$\text{adj A}=\begin{bmatrix}6&-9\\ -4&6\end{bmatrix}^\text{T}$
$=\begin{bmatrix}6&-4\\ -9&6\end{bmatrix}$
$\text{(adj A) B}=\begin{bmatrix}6&-4\\ -9&6\end{bmatrix}\begin{bmatrix}2\\ 3\end{bmatrix}$
$=\begin{bmatrix}12-12\\ -18+18\end{bmatrix}$
$=\begin{bmatrix}0\\ 0\end{bmatrix}$
If $|A| = 0$ and $($adj $A) B = 0,$ then the system is consistent and has infinitely many solutions.
Thus, $AX = B$ has infinitely many solutions.
Substituting $y = k$ in the eq. $(1),$ we get
$6\text{x}+4\text{k}=2$
$\Rightarrow6\text{x}=2-4\text{k}$
$\Rightarrow\text{x}=\frac{2-\text{4k}}{6}$
$\Rightarrow\text{x}=\frac{1-\text{2k}}{3}$
$\therefore \text{x}=\frac{1-2\text{k}}{3}\text{ and }\text{y}=\text{k}$
These values of $x$ and $y$ satisfy the third equation.
Thus, $\text{x}=\frac{1-2\text{k}}{3}$ and $y = k ($ where $k$ is a real number $)$ satisfy the given system of equations.
View full question & answer→Question 945 Marks
If $\text{a}+\text{b}+\text{c}\neq0\text{ and}\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b}\end{vmatrix}=0,$ then prove that $a = b = c.$
AnswerLet $\Delta=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b}\end{vmatrix}$
$\big[\text{Applying R}_1\rightarrow\text{R}_1+\text{R}_2+\text{R}_3\big]$
$\Delta=\begin{vmatrix}\text{a}+\text{b}+\text{c}&\text{a}+\text{b}+\text{c}&\text{a}+\text{b}+\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b}\end{vmatrix}$ $=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&1&1\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b}\end{vmatrix}$
$\big[\text{Applying C}_1\rightarrow\text{C}_1-\text{C}_3\text{ and C}_2\rightarrow\text{C}_2-\text{C}_1\big]$
$\Delta=(\text{a}+\text{b}+\text{c})\begin{vmatrix}0&0&1\\\text{b}-\text{a}&\text{c}-\text{a}&\text{a}\\\text{c}-\text{b}&\text{a}-\text{b}&\text{b}\end{vmatrix}$
$[$Expanding along $R_1]$
$=(\text{a}+\text{b}+\text{c})[(\text{b}-\text{a})(\text{a}-\text{b})-(\text{c}-\text{a})(\text{c}-\text{b})]$
$=(\text{a}+\text{b}+\text{c})(\text{ba}-\text{b}^2-\text{a}^2+\text{ab}-\text{c}^2+\text{cb}+\text{ac}-\text{ab})$
$=-(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca})$
$=\frac{-1}{2}(\text{a}+\text{b}+\text{c})\big[2\text{a}^2+2\text{b}^2-2\text{c}^2-2\text{ab}-2\text{bc}-2\text{ca}\big]$
$=-\frac{1}{2}(\text{a}+\text{b}+\text{c})\big[(\text{a}^2+\text{b}^2-2\text{ab})+(\text{b}^2+\text{c}^2-2\text{bc})+(\text{c}^2+\text{a}^2-2\text{ac})\big]$
$-\frac{-1}{2}(\text{a}+\text{b}+\text{c})\big[(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2\big]$
Given $\Delta=0$
$\Rightarrow\ \frac{-1}{2}(\text{a}+\text{b}+\text{c})\big[(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2\big]=0$
$\Rightarrow\ (\text{a}-\text{b})^2+(\text{b}-\text{c}^2)+(\text{c}-\text{a})^2=0$ $[\because\ \text{a}+\text{b}+\text{c}\neq0,\text{ given}]$
$\Rightarrow\ \text{a}-\text{b}=\text{b}-\text{c}=\text{c}-\text{a}=0$
$\Rightarrow\ \text{a}=\text{b}=\text{c}$
View full question & answer→Question 955 Marks
Solve the following system of equations by matrix method:
$3x + y = 7$
$5x + 3y = 12$
AnswerThe given system of equations can be written in matrix form as follows:
$\begin{bmatrix}3&1\\ 5&3\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}7\\ 12\end{bmatrix}$
$AX = B$
Here,
$\text{A}=\begin{bmatrix}3&1\\ 5&3\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{ and }\text{B}=\begin{bmatrix}7\\ 12\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}3&1\\ 5&3\end{vmatrix}$
$=9-5$
$=4\neq0$
So, the given system has a unique solution given by $X = A^{-1}\ B.$
Let $C_{ij}$ be the co-factors of the elements $a_{ij}$ in $A = [a_{ij}].$ then,
$\text{C}_{11}=-{(-1)}^{1+1}{(3)}=3,\text{C}_{12}={(-1)}^{1+2}{(5)}=-5$
$\text{C}_{21}={(-1)}^{2+1}{(1)}=-1,\text{C}_{22}=-{(-1)}^{2+2}{(3)}=3$
$\text{adj A}=\begin{bmatrix}3&-5\\ -1&3\end{bmatrix}^\text{T}$
$=\begin{bmatrix}3&-1\\-5&3\end{bmatrix}$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj A}$
$=\frac{1}{4}\begin{bmatrix}3&-1\\ -5&3\end{bmatrix}$
$X = A^{-1}B$
$=\frac{1}{4}\begin{bmatrix}3&-1\\ -5&3\end{bmatrix}\begin{bmatrix}7\\ 12\end{bmatrix}$
$=\frac{1}{4}\begin{bmatrix}21-12\\ -35+36\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{9}{4}\\ \frac{1}{4}\end{bmatrix}$
$\therefore\ \text{x}=\frac{9}{4}\text{ and }\text{y}=\frac{1}{4}$
View full question & answer→Question 965 Marks
Solve the following determinant equations:
$\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\end{vmatrix}=0,\text{a}\neq\text{b}$
AnswerLet $\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\end{vmatrix}$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\0&\text{x}-\text{a}&\text{x}^2-\text{a}^2\\1&\text{b}&\text{b}^2\end{vmatrix}\ [$Applying $R_2 \rightarrow R_1 - R_2]$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\0&\text{x}-\text{a}&\text{x}^2-\text{a}^2\\0&\text{x}-\text{b}&\text{x}^2-\text{b}^2\end{vmatrix} \ [$Applying $R_3 \rightarrow R_1 - R_3]$
$=(\text{x}-\text{a})(\text{x}-\text{b})\begin{vmatrix}1&\text{x}&\text{x}^2\\0&1&\text{x}+\text{a}\\0&1&\text{x}+\text{b}\end{vmatrix}$
$=(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}+\text{b}-\text{x}-\text{a})=0$
$\text{x}=\text{a},\text{b}$
View full question & answer→Question 975 Marks
If $\text{A}=\begin{bmatrix}4 & 3 \\ 2 & 5 \end{bmatrix},$ find $x$ and $y$ such that $\ce{A^2 = zA + yI = 0}. $Hence, evaluate $A^{-1}.$
Answer$\text{A}=\begin{bmatrix}4 & 3 \\2 & 5 \end{bmatrix}$
$\therefore\ \text{A}^2=\begin{bmatrix}22 & 27 \\ 18 & 31 \end{bmatrix}$
Now, $\ce{A^2 - xA + yI = 0}$
$\Rightarrow\ \begin{bmatrix}22 & 27 \\ 18 & 31 \end{bmatrix}-\begin{bmatrix}4\text{x} & 3\text{x} \\2\text{x} & 5\text{x} \end{bmatrix}+\begin{bmatrix}\text{y} & 0 \\0 & \text{y} \end{bmatrix}=\begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}22-4\text{x}-\text{y} & 27-3\text{x} \\ 18-2\text{x} & 31-5\text{x}-\text{y} \end{bmatrix}=\begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}$
Thus, we have
$22 - 4x + y = 0, 37 - 3x = 0, 18 - 2x = 0$ and $31 - 5x + y = 0$
$\Rightarrow -3x = -27$
$\Rightarrow x = 9$
On putting $x = 9$ in $22 - 4x + y = 0,$ we get
$22 - 36 + y = 0$
$\Rightarrow -14 = -y$
$\Rightarrow y = 14$
Now,
$\ce{A^2 - 9A + 14I = 0}$
$\ce{\Rightarrow A^2 - 9A = -14I}$
$\Rightarrow \ce{A^{-1}A^2 - 9AA^{-1} = -14IA^{-1}}[$Pre$-$multiplying both sides by $A^{-1}]$
$\Rightarrow \ce{A - 9I = -14A^{-1}}$
$\Rightarrow\ \text{A}^{-1}=-\frac{1}{14}(\text{A}-9\text{I})$
$\Rightarrow\ \text{A}^{-1}=-\frac{1}{14}\left\{\begin{bmatrix}4 & 3 \\2 & 5 \end{bmatrix}-\begin{bmatrix}9 & 0 \\0 & 9 \end{bmatrix}\right\}$
$=\frac{1}{14}\begin{bmatrix}5 & -3 \\-2 & 4 \end{bmatrix}$
View full question & answer→Question 985 Marks
$\begin{vmatrix}-\text{a}(\text{b}^2+\text{c}^2-\text{a}^2)&2\text{b}^3&2\text{c}^3\\2\text{a}^3&-\text{b}(\text{c}^2+\text{a}^2-\text{b}^2)&2\text{c}^3\\2\text{a}^3&2\text{b}^3&-\text{c}(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}(\text{a}^2+\text{b}^2+\text{c}^2)$
Answer$\begin{vmatrix}-\text{a}(\text{b}^2+\text{c}^2-\text{a}^2)&2\text{b}^3&2\text{c}^3\\2\text{a}^3&-\text{b}(\text{c}^2+\text{a}^2-\text{b}^2)&2\text{c}^3\\2\text{a}^3&2\text{b}^3&-\text{c}(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}(\text{a}^2+\text{b}^2+\text{c}^2)$
$\text{L.H.S}=\begin{vmatrix}-\text{a}(\text{b}^2+\text{c}^2-\text{a}^2)&2\text{b}^3&2\text{c}^3\\2\text{a}^3&-\text{b}(\text{c}^2+\text{a}^2-\text{b}^2)&2\text{c}^3\\2\text{a}^3&2\text{b}^3&-\text{c}(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
Tkae $a, b$ and $c$ common from $C_1, C_2$ and $C_3$ respectively.
$=\text{abc}\begin{vmatrix}-(\text{b}^2+\text{c}^2-\text{a}^2)&2\text{b}^2&2\text{c}^2\\2\text{a}^2&-(\text{c}^2+\text{a}^2-\text{b}^2)&2\text{c}^2\\2\text{a}^2&2\text{b}^2&-(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
Apply: $R_1 \rightarrow R_1 - R_3, R_2 \rightarrow R_2 - R_3$
$=\text{abc}\begin{vmatrix}-(\text{b}^2+\text{c}^2-\text{a}^2)-2\text{a}^2&0&2\text{c}^2+(\text{a}^2+\text{b}^2-\text{c}^2)\\0&-(\text{c}^2+\text{a}^2-\text{b}^2)-2\text{b}^2&2\text{c}^2+(\text{a}^2+\text{b}^2-\text{c}^2)\\2\text{a}^2&2\text{b}^2&-(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}\begin{vmatrix}-(\text{b}^2+\text{c}^2-\text{a}^2)-2\text{a}^2&0&(\text{a}^2+\text{b}^2-\text{c}^2)\\0&-(\text{c}^2+\text{a}^2-\text{b}^2)-2\text{b}^2&2\text{c}^2+(\text{a}^2+\text{b}^2-\text{c}^2)\\2\text{a}^2&2\text{b}^2&-(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}(\text{b}^2+\text{c}^2+\text{a}^2)^2\begin{vmatrix}-1&0&1\\0&-1&1\\2\text{a}^2&2\text{b}^2&-(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}(\text{b}^2+\text{c}^2+\text{a}^2)^2\begin{vmatrix}-1&0&1\\0&-1&1\\2\text{a}^2&2\text{b}^2&-(\text{a}^2+\text{b}^2-\text{c}^2)+2\text{a}^2\end{vmatrix}$
$=\text{abc}(\text{b}^2+\text{c}^2+\text{a}^2)^2\begin{vmatrix}-1&0&1\\0&-1&1\\2\text{a}^2&2\text{b}^2&-\text{b}^2+\text{c}^2+\text{a}^2\end{vmatrix}$
$=-\text{abc}(\text{b}^2+\text{c}^2+\text{a}^2)[(-1)(-\text{b}^2+\text{c}^2+\text{a}^2)-(1)(2\text{b}^2)]$
$=\text{abc}(\text{a}^2+\text{b}^2+\text{c}^2)^3$
$=\text{R.H.S}$
View full question & answer→Question 995 Marks
If the points $(x, -2), (5, 2), (8, 8)$ are collinear, find $x$ using determinants.
AnswerThe points $(k, -2), (5, 2), (8, 8)$ are collinear.
$\begin{vmatrix}\text{x}&-2&1\\5&2&1\\8&8&1\end{vmatrix}=0$
$\triangle=\begin{vmatrix}\text{x}&-2&1\\5&2&1\\8&8&1\end{vmatrix}$
$=\begin{vmatrix}\text{x}&-2&1\\5-\text{x}&4&0\\8&8&1\end{vmatrix} \ [$Applying $R_2 \rightarrow R_2 - R_1]$
$=\begin{vmatrix}\text{x}&-2&1\\5-\text{x}&4&0\\8-\text{x}&10&0\end{vmatrix}\ [$Applying $R_3 \rightarrow R_3 - R_1]$
$=\begin{vmatrix}5-\text{x}&4\\8-\text{x}&10\end{vmatrix}$
$=50-10\text{x}-32+4\text{x}$
$=18-6\text{x}=0$
$\Rightarrow\text{x}=3$
View full question & answer→Question 1005 Marks
Prove that:
$\begin{vmatrix}-\text{bc}&\text{b}^2+\text{bc}&\text{c}^2+\text{bc}\\\text{a}^2+\text{ac}&-\text{ac}&\text{c}^2+\text{ac}\\\text{a}^2+\text{ab}&\text{b}^2+\text{ab}&-\text{ab}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})^3$
Answer$\text{L.H.S}=\begin{vmatrix}-\text{bc}&\text{b}^2+\text{bc}&\text{c}^2+\text{bc}\\\text{a}^2+\text{ac}&-\text{ac}&\text{c}^2+\text{ac}\\\text{a}^2+\text{ab}&\text{b}^2+\text{ab}&-\text{ab}\end{vmatrix}$
Multiply $R_1, R_2$ and $R_3$ by $a, b$ and $c$ respectively.
$=\frac{1}{\text{abc}}\begin{vmatrix}-\text{abc}&\text{ab}^2+\text{abc}&\text{ac}^2+\text{abc}\\\text{a}^2\text{b}+\text{abc}&-\text{abc}&\text{bc}^2+\text{abc}\\\text{a}^2\text{c}+\text{abc}&\text{b}^2\text{c}+\text{abc}&-\text{abc}\end{vmatrix}$
Take $a, b$ and $c$ common from $C_1, C_2$ and $C_3$ respectively.
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}-\text{bc}&\text{ab}+\text{ac}&\text{ac}+\text{ab}\\\text{a}\text{b}+\text{bc}&-\text{ac}&\text{bc}+\text{ab}\\\text{a}\text{c}+\text{bc}&\text{b}\text{c}+\text{ac}&-\text{ab}\end{vmatrix}$
Apply: $R_1 \rightarrow R_1 + R_2 + R_3$
$=\begin{vmatrix}\text{ab}+\text{bc}+\text{ca}&\text{ab}+\text{bc}+\text{ca}&\text{ab}+\text{bc}+\text{ca}\\\text{a}\text{b}+\text{bc}&-\text{ac}&\text{bc}+\text{ab}\\\text{a}\text{c}+\text{bc}&\text{b}\text{c}+\text{ac}&-\text{ab}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})\begin{vmatrix}1&1&1\\\text{a}\text{b}+\text{bc}&-\text{ac}&\text{bc}+\text{ab}\\\text{a}\text{c}+\text{bc}&\text{b}\text{c}+\text{ac}&-\text{ab}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})\begin{vmatrix}0&1&0\\\text{a}\text{b}+\text{bc}+\text{ac}&-\text{ac}&\text{bc}+\text{ab}+\text{ac}\\0&\text{b}\text{c}+\text{ac}&-\text{ab}-\text{bc}-\text{ac}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})^3\begin{vmatrix}0&1&0\\0&-\text{ac}&1\\0&\text{b}\text{c}+\text{ac}&1\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})^3$
$=\text{R.H.S}$
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