Question 515 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}+2\text{x}&\text{b}+2\text{y}&\text{c}+2\text{z}\\\text{x}&\text{y}&\text{z}\\\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}+2\text{x}&\text{b}+2\text{y}&\text{c}+2\text{z}\\\text{x}&\text{y}&\text{z}\\\end{vmatrix}$
$=\begin{vmatrix}\text{a}+2\text{x}&\text{b}+2\text{y}&\text{c}+2\text{z}\\\text{a}+2\text{x}&\text{b}+2\text{y}&\text{c}+2\text{z}\\\text{x}&\text{y}&\text{z}\end{vmatrix} [$Applying $R_1 \rightarrow R_1 + 2R_3]$
$=\begin{vmatrix}0&0&0\\\text{a}+2\text{x}&\text{b}+2\text{y}&\text{c}+2\text{z}\\\text{x}&\text{y}&\text{z}\end{vmatrix}=0[$ Applying $R_1 \rightarrow R_1 + R_2]$
View full question & answer→Question 525 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 535 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: $R_1 \rightarrow R_1 + R_2 + R_3$
$=\begin{vmatrix}\text{ab}+\text{bc}+\text{ca}&\text{ab}+\text{bc}+\text{ca}&\text{ab}+\text{bc}+\text{ca}\\\text{a}\text{b}+\text{bc}&-\text{ac}&\text{bc}+\text{ab}\\\text{a}\text{c}+\text{bc}&\text{b}\text{c}+\text{ac}&-\text{ab}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})\begin{vmatrix}1&1&1\\\text{a}\text{b}+\text{bc}&-\text{ac}&\text{bc}+\text{ab}\\\text{a}\text{c}+\text{bc}&\text{b}\text{c}+\text{ac}&-\text{ab}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})\begin{vmatrix}0&1&0\\\text{a}\text{b}+\text{bc}+\text{ac}&-\text{ac}&\text{bc}+\text{ab}+\text{ac}\\0&\text{b}\text{c}+\text{ac}&-\text{ab}-\text{bc}-\text{ac}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})^3\begin{vmatrix}0&1&0\\0&-\text{ac}&1\\0&\text{b}\text{c}+\text{ac}&1\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})^3$
$=\text{R.H.S}$
View full question & answer→Question 545 Marks
$\begin{vmatrix}1+\text{a}&1&1\\1&1+\text{a}&\text{a}\\1&1&1+\text{a}\end{vmatrix}=\text{a}^3+3\text{a}^2$
Answer$\text{L.H.S}=\begin{vmatrix}1+\text{a}&1&1\\1&1+\text{a}&\text{a}\\1&1&1+\text{a}\end{vmatrix}$
$=1+\text{a}\begin{vmatrix}1+\text{a}&1\\1&1+\text{a}\end{vmatrix}-1\begin{vmatrix}1&1\\1&1+\text{a}\end{vmatrix}+1\begin{vmatrix}1&1+\text{a}\\1&1\end{vmatrix}$
$=(1+\text{a})[(1+\text{a})^2-1]-1(1+\text{a}-1)+(1-1-\text{a})$
$=(1+\text{a})[1+\text{a}^2+2\text{a}-1]-\text{a}-\text{a}$
$=1+\text{a}+\text{a}^2+\text{a}^3+2\text{a}+2\text{a}^2-2\text{a}$
$=\text{a}^3+3\text{a}^2$
$=\text{R.H.S}$
View full question & answer→Question 555 Marks
If $\triangle=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix},$ $\triangle_1=\begin{vmatrix}1&1&1\\\text{yz}&\text{zx}&\text{xy}\\\text{x}&\text{y}&\text{z}\end{vmatrix},$ then prove that $\triangle+\triangle_1=0$
Answer$\triangle+\triangle_1=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix}+\begin{vmatrix}1&1&1\\\text{yz}&\text{zx}&\text{xy}\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix}+\begin{vmatrix}1&\text{yz}&\text{x}\\1&\text{zx}&\text{y}\\1&\text{xy}&\text{z} \end{vmatrix}$
$[$Interchanging rows and coloumns in $\triangle_1]$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix}-\begin{vmatrix}1&\text{x}&\text{yz}\\1&\text{y}&\text{zx}\\1&\text{z}&\text{xy} \end{vmatrix}$
$[$Applying $\text{C}_2\leftrightarrow\text{C}_3$ in $\triangle_1]$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\0&\text{y}-\text{x}&\text{y}^2-\text{x}^2\\0&\text{z}-\text{x}&\text{z}^2-\text{x}^2\end{vmatrix}-\begin{vmatrix}1&\text{x}&\text{yz}\\0&\text{y}-\text{x}&\text{zx}-\text{yz}\\0&\text{z}-\text{x}&\text{xy}-\text{yz}\end{vmatrix}
[$ Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1]$
$=(\text{y}-\text{x})(\text{z}-\text{x})\begin{vmatrix}1&\text{x}&\text{x}^2\\0&1&\text{y}+\text{x}\\0&1&\text{z}+\text{x}\end{vmatrix}-(\text{y}-\text{x})(\text{z}-\text{x})\begin{vmatrix}1&\text{x}&\text{yz}\\0&1&-\text{z}\\0&1&-\text{y}\end{vmatrix}
[$Taking $(y - x)$ common from $R_2$ and $(z - x)$ common from $R_3]$
$=(\text{y}-\text{x})(\text{z}-\text{x})(\text{z}+\text{x}-\text{y}-\text{x})-(\text{y}-\text{x})(\text{z}-\text{x})(-\text{y}+\text{z})
[$Expanding along first column$]$
$=(\text{y}-\text{x})(\text{z}-\text{x})(\text{z}-\text{y})(1-1)$
$=0$
$\therefore\ \triangle+\triangle_1=0$
View full question & answer→Question 565 Marks
Evaluate the following determinant:
$\begin{vmatrix}1&3&9&27\\3&9&27&1\\9&27&1&3\\27&1&3&9 \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&3&9&27\\3&9&27&1\\9&27&1&3\\27&1&3&9 \end{vmatrix}$
$=1\begin{vmatrix}9&27&1\\27&1&3\\1&3&9 \end{vmatrix}-3\begin{vmatrix}3&27&1\\9&1&3\\27&3&6\end{vmatrix}\\+9\begin{vmatrix}3&9&1\\9&27&3\\27&1&9 \end{vmatrix}-27\begin{vmatrix}3&9&27\\9&27&1\\27&1&3 \end{vmatrix}$
$=1\{9(9-9)-27(243-3)+1(81-1)\}-3\{3(9-9)-27(81-81)+1(27-27)\}\\+9\{3(243-3)-9(81-81)+1(9-729)\}-27\{(81-1)-9(27-27)+27(9-729)\}$
$=1\{0-6480+80\}-3\{0-0+0\}+9\{720-0-720\}-27\{80-0-19440\}$
$=-6400+522720$
$=516320$
View full question & answer→Question 575 Marks
Show that:
$\begin{vmatrix}\text{y}+\text{z}&\text{x}&\text{y}\\\text{z}+\text{x}&\text{z}&\text{x}\\\text{x}+\text{y}&\text{y}&\text{z}\end{vmatrix}=(\text{x}+\text{y}+\text{z})(\text{x}-\text{z})^3$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{y}+\text{z}&\text{x}&\text{y}\\\text{z}+\text{x}&\text{z}&\text{x}\\\text{x}+\text{y}&\text{y}&\text{z}\end{vmatrix}$
$=\begin{vmatrix}2(\text{x}+\text{y}+\text{z})&\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}\\\text{z}+\text{x}&\text{z}&\text{x}\\\text{x}+\text{y}&\text{y}&\text{z}\end{vmatrix} [$Applying $R_1 \rightarrow R_1 + R_2 + R_3]$
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}2&1&1\\\text{z}+\text{x}&\text{z}&\text{x}\\\text{x}+\text{y}&\text{y}&\text{z}\end{vmatrix}$
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}0&1&1\\0&\text{z}&\text{x}\\\text{x}-\text{z}&\text{y}&\text{z}\end{vmatrix} [$Applying $C_1 \rightarrow C_1 - C_2 - C_3]$
$=(\text{x}+\text{y}+\text{z})\left\{(\text{x}-\text{z})\times\begin{vmatrix}1&1\\\text{z}&\text{x}\end{vmatrix}\right\} [$Expanding along $C_1]$
$=(\text{x}+\text{y}+\text{z})(\text{x}-\text{z})^3$
View full question & answer→Question 585 Marks
Find values of $k,$ if area of triangle is $4$ square units whose vertices are:
$(-2, 0), (0, 4), (0, k)$
Answer$4=\frac{1}{2}\begin{vmatrix}-2&0&1\\0&4&1\\0&\text{k}&1 \end{vmatrix}$
$\pm8=\begin{vmatrix}-2&0&1\\0&4&1\\0&\text{k}&1 \end{vmatrix}$
Expanding along $R_1$
$\pm8=-2(4-\text{k})-0(0-0)+1(0)$
$\pm=8=-8+2\text{k}$
Taking positive (+) sign
$\pm=8=-8+2\text{k}$
Taking positive $(-)$ sing
$-8 = -8 + 2k$ or $k = 0$
Hence $k = 0, 8$
View full question & answer→Question 595 Marks
Write the minors and cofactors of element of the first column of the following matrices and hence evaluate the determinant in case:
$\text{A}=\begin{vmatrix}2&-1&0&1\\-3&0&1&-2\\1&1&-1&1\\2&-1&5&0 \end{vmatrix}$
Answer$\text{M}_{11}=0(0-5)-1(0+1)-2(5-1)=-1-8=-9$
$\text{M}_{21}=-1(0-5)+1(5-1)=5+4=9$
$\text{M}_{31}=-1(0+10)+1(0+1)=-10+1=-9$
$\text{M}_{41}=-1(1-2)+1(0-1)=1-1=0$
$\text{C}_{11}=(-1)^{1+1}\text{M}_{11}=-9$
$\text{C}_{21}=(-1)^{2+1}\text{M}_{21}=(-1)\times9$
$\text{C}_{31}=(-1)^{3+1}\text{M}_{31}=-9$
$\text{C}_{41}=(-1)^{4+1}\text{M}_{41}=0$
$\text{D}=2\begin{vmatrix}0&1&-2\\1&-1&1\\-1&5&0 \end{vmatrix}+1\begin{vmatrix}-3&1&-2\\1&-1&1\\2&5&0\end{vmatrix}-1\begin{vmatrix}-3&0&1\\1&1&-1\\2&-1&5\end{vmatrix}$
$=-18-27+15=-30$
View full question & answer→Question 605 Marks
Prove that:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\end{vmatrix}=\text{a}^3+\text{b}^3+\text{c}^3-3\text{abc}$
Answer$\text{L.H.S}=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\end{vmatrix}$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\\text{a}+\text{b}+\text{c}&\text{c}+\text{a}+\text{b}&\text{a}+\text{b}+\text{c}\end{vmatrix} [$Applying $R_3 \rightarrow R_3 + R_2]$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\1&1&1\end{vmatrix} [$Taking $(a + b + c)$ common$]$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\1&1&1\end{vmatrix} [$Applying $R_2 \rightarrow R_1 - R_2]$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{a}-\text{b}&\text{b}-\text{c}&\text{c}\\\text{b}-\text{c}&\text{c}-\text{a}&\text{a}\\0&0&1\end{vmatrix} [C_1 \rightarrow C_1 - C_2$ and $C_2 \rightarrow C_2 - C_3]$
$=(\text{a}+\text{b}+\text{c})\big[-1\{(\text{a}-\text{b})(\text{c}-\text{a})-(\text{b}-\text{c})^2\}\big]$
$=(\text{a}+\text{b}+\text{c})\big[-\{\text{ac}-\text{bc}-\text{a}^2+\text{ab}-\text{b}^2-\text{c}^2+2\text{bc}\}\big]$
$=(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca})$
$=\text{a}^3+\text{b}^3+\text{c}^3-3\text{abc}=\text{ R.H.S}$
View full question & answer→Question 615 Marks
Prove that:
$\begin{vmatrix}\text{a}&\text{b}-\text{c}&\text{c}-\text{b}\\\text{a}-\text{c}&\text{b}&\text{c}-\text{a}\\\text{a}-\text{b}&\text{b}-\text{a}&\text{c}\end{vmatrix}$
$=(\text{a}+\text{b}-\text{c})(\text{b}+\text{c}-\text{a})(\text{c}+\text{a}-\text{b})$
Answer$\text{L.H.S}=\begin{vmatrix}\text{a}&\text{b}-\text{c}&\text{c}-\text{b}\\\text{a}-\text{c}&\text{b}&\text{c}-\text{a}\\\text{a}-\text{b}&\text{b}-\text{a}&\text{c}\end{vmatrix}$
$R_1 \rightarrow R_{1 }- R_2 - R_3$
$=\begin{vmatrix}-\text{a}+\text{c}+\text{b}&-\text{b}-\text{c}+\text{a}&-\text{c}-\text{b}+\text{a}\\\text{a}-\text{c}&\text{b}&\text{c}-\text{a}\\\text{a}-\text{b}&\text{b}-\text{a}&\text{c}\end{vmatrix}$
$=(\text{b}+\text{c}-\text{a})\begin{vmatrix}1&-1&1\\\text{a}-\text{c}&\text{b}&\text{c}-\text{a}\\\text{a}-\text{b}&\text{b}-\text{a}&\text{c}\end{vmatrix}$
$=(\text{b}+\text{c}-\text{a})\begin{vmatrix}1&0&0\\\text{a}-\text{c}&\text{b}+\text{a}-\text{c}&0\\\text{a}-\text{b}&0&\text{c}+\text{a}-\text{b}\end{vmatrix}$
$=(\text{a}+\text{b}-\text{c})(\text{b}+\text{c}-\text{a})(\text{c}+\text{a}-\text{b})$
$=\text{R.H.S}$
View full question & answer→Question 625 Marks
Prove that: $\begin{vmatrix} 1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}=\begin{vmatrix} 1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\\1&\text{c}&\text{c}^2\end{vmatrix}$
Answer$\begin{vmatrix} 1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
Apply $R_1 \rightarrow R_1a, R_2 \rightarrow R_2b, R_3 \rightarrow R_3c$
$=\frac{1}{\text{abc}}\begin{vmatrix} \text{a}&\text{a}^2&\text{abc}\\\text{b}&\text{b}^2&\text{cab}\\\text{c}&\text{c}^2&\text{abc}\end{vmatrix}$
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix} \text{a}&\text{a}^2&1\\\text{b}&\text{b}^2&1\\\text{c}&\text{c}^2&1\end{vmatrix}$
$=-\begin{vmatrix} \text{a}&1&\text{a}^2\\\text{b}&1&\text{b}^2\\\text{c}&1&\text{c}^2\end{vmatrix}$
$=\begin{vmatrix} 1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\\1&\text{c}&\text{c}^2\end{vmatrix}$
View full question & answer→Question 635 Marks
Without expanding, show that the values of the following determinant are zero: $\begin{vmatrix}1&43&6\\7&35&4\\3&17&2\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&43&6\\7&35&4\\3&17&2\end{vmatrix}$
$=\begin{vmatrix}1&1&6\\7&7&4\\3&3&2\end{vmatrix}=0 [$Appliying $C_2 \rightarrow C_2 - 7C_3]$
View full question & answer→Question 645 Marks
Solve the following systems of linear equations by cramer's rule:
9x + 5y = 10,
3x - 2y = 8
AnswerGiven, 9x + 5y = 10
3y - 2x = 8
Rearranging the second equation, the two equations can be written as
9x + 5y = 10
-2x + 3y = 8
Now,
$\text{D}=\begin{vmatrix}9&5\\-2&3\end{vmatrix}=27+10=37$
$\text{D}_1=\begin{vmatrix}10&5\\8&3\end{vmatrix}=30-40=-10$
$\text{D}_2=\begin{vmatrix}9&10\\-2&8\end{vmatrix}=72+20=-92$
Using Cramer's rule we get
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-10}{37}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{92}{37}$
$\therefore\text{x}=\frac{-10}{37}$ and $\text{y}=\frac{92}{37}$
View full question & answer→Question 655 Marks
Solve the following determinant equations:
$\begin{vmatrix}\text{x}+\text{a}&\text{x}&\text{x}\\\text{x}&\text{x}+\text{a}&\text{x}\\\text{x}&\text{x}&\text{x}+\text{a}\end{vmatrix}=0,\text{a}\neq0$
AnswerLet $\begin{vmatrix}\text{x}+\text{a}&\text{x}&\text{x}\\\text{x}&\text{x}+\text{a}&\text{x}\\\text{x}&\text{x}&\text{x}+\text{a}\end{vmatrix}$
$=\begin{vmatrix}3\text{x}+\text{a}&\text{x}&\text{x}\\3\text{x}+\text{a}&\text{x}+\text{a}&\text{x}\\3\text{x}+\text{a}&\text{x}&\text{x}+\text{a}\end{vmatrix} [$Applying $C_1 \rightarrow C_1 + C_2 + C_3]$
=$(3\text{x}+\text{a})\begin{vmatrix}1&\text{x}&\text{x}\\1&\text{x}+\text{a}&\text{x}\\1&\text{x}&\text{x}+\text{a}\end{vmatrix}$
=$(3\text{x}+\text{a})\begin{vmatrix}1&\text{x}&\text{x}\\0&\text{a}&0\\1&\text{x}&\text{x}+\text{a}\end{vmatrix}$ $[$Applying $R_2 \rightarrow R_2 - R_1]$
=($3\text{x}+\text{a})\begin{vmatrix}1&\text{x}&\text{x}\\0&\text{a}&0\\1&0&\text{a}\end{vmatrix} [$Applying $R_3 \rightarrow R_3 - R_1]$
$=(3\text{x}+\text{a})=(\text{a}^2-0)=0$
$\text{x}=\frac{-\text{a}}{3}$
View full question & answer→Question 665 Marks
Without expanding, show that the values of the following determinant are zero: $\begin{vmatrix}1&\text{a}&\text{a}^2-\text{bc}\\1&\text{b}&\text{b}^2-\text{ac}\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&\text{a}&\text{a}^2-\text{bc}\\1&\text{b}&\text{b}^2-\text{ac}\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$
$=\begin{vmatrix}0&\text{a}-\text{b}&\text{a}^2-\text{bc}-\text{b}^2+\text{ac}\\0&\text{b}-\text{c}&\text{b}^2-\text{ac}-\text{c}^2+\text{ab}\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix} [$applying $R_1 \rightarrow R_1 - R_2, R_2 \rightarrow R_2 - R_3]$
$=\begin{vmatrix}0&\text{a}-\text{b}&(\text{a}-\text{b})(\text{a}+\text{b})+\text{c}(\text{a}-\text{b})\\0&\text{b}-\text{c}&(\text{b}-\text{c})(\text{b}+\text{c})+\text{a}(\text{b}-\text{c})\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}0&1&(\text{a}+\text{b}+\text{c})\\0&1&(\text{a}+\text{b}+\text{c})\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$
$\Rightarrow\triangle=0$
View full question & answer→Question 675 Marks
If $\begin{vmatrix}\text{p}&\text{q}&\text{c}\\\text{a}&\text{q}&\text{c}\\\text{a}&\text{b}&\text{r} \end{vmatrix}=0,$ find the value of $\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}},$ $\text{p}\neq\text{a},\text{q}\neq\text{b},\text{r}\neq\text{c}.$
Answer$\text{L.H.S}=\begin{vmatrix}\text{p}&\text{q}&\text{c}\\\text{a}&\text{q}&\text{c}\\\text{a}&\text{b}&\text{r} \end{vmatrix}$
$=\begin{vmatrix}\text{p}&\text{q}&\text{c}\\0&\text{q}-\text{b}&\text{c}-\text{r}\\\text{a}&\text{b}&\text{r} \end{vmatrix} [$Applying $R_2 \rightarrow R_2 - R_3]$
$=\text{p}[\text{r}(\text{q}-\text{b})-\text{b}(\text{c}-\text{r})]+\text{a}[\text{b}(\text{c}-\text{r})-\text{c}(\text{q}-\text{b})]$
$=\text{pr}(\text{q}-\text{b})+\text{pb}(\text{r}-\text{c})-\text{ab}(\text{r}-\text{c})-\text{ac}(\text{q}-\text{b})$
$=(\text{pr}-\text{ac})(\text{q}-\text{b})+\text{b}(\text{p}-\text{a})(\text{r}-\text{c})$
Since, $\triangle=0$
$\therefore(\text{pr}-\text{ac})(\text{q}-\text{b})+\text{b}(\text{p}-\text{a})(\text{r}-\text{c})=0$
$\Rightarrow\frac{\text{pr}-\text{ac}}{(\text{p}-\text{a})(\text{r}-\text{c})}+\frac{\text{b}}{\text{q}-\text{b}}=0$
$\Rightarrow\frac{\text{pr}-\text{ar}+\text{ar}-\text{ac}}{(\text{p}-\text{a})(\text{r}-\text{c})}+\frac{\text{b}}{\text{q}-\text{b}}=0$
$\Rightarrow\frac{\text{r}(\text{p}-\text{a})+\text{a}(\text{r}-\text{c})}{(\text{p}-\text{a})(\text{r}-\text{c})}+\frac{\text{b}}{\text{q}-\text{b}}=0$
$\Rightarrow\frac{\text{r}}{\text{r}-\text{c}}+\frac{\text{a}}{\text{p}-\text{a}}+\frac{\text{b}}{\text{q}-\text{b}}=0$
$\Rightarrow\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}}\\=\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}-\frac{\text{a}}{\text{p}-\text{a}}-\frac{\text{b}}{\text{q}-\text{b}}$
$\Rightarrow\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}}=\frac{\text{p}-\text{a}}{\text{p}-\text{a}}+\frac{\text{q}-\text{b}}{\text{q}-\text{b}}$
$\Rightarrow\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}}=2$
View full question & answer→Question 685 Marks
Prove that:
$\begin{vmatrix}\frac{\text{a}^2+\text{b}^2}{\text{c}}&\text{c}&\text{c}\\\text{a}&\frac{\text{b}^2+\text{c}^2}{\text{a}}&\text{a}\\\text{b}&\text{b}&\frac{\text{c}^2+\text{a}^2}{\text{b}}\end{vmatrix}=4\text{abc}$
Answer$\text{L.H.S}=\begin{vmatrix}\frac{\text{a}^2+\text{b}^2}{\text{c}}&\text{c}&\text{c}\\\text{a}&\frac{\text{b}^2+\text{c}^2}{\text{a}}&\text{a}\\\text{b}&\text{b}&\frac{\text{c}^2+\text{a}^2}{\text{b}}\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}\text{a}^2+\text{b}^2&\text{c}^2&\text{c}^2\\\text{a}^2&\text{b}^2+\text{c}^2&\text{a}^2\\\text{b}^2&\text{b}^2&\text{c}^2+\text{a}^2\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}\text{a}^2+\text{b}^2&\text{c}^2-\text{a}^2-\text{b}^2&\text{c}^2-\text{a}^2-\text{b}^2\\\text{a}^2&\text{b}^2+\text{c}^2-\text{a}^2&0\\\text{b}^2&0&\text{c}^2+\text{a}^2-\text{b}^2\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}0&-2\text{b}^2&-2\text{a}^2\\\text{a}^2&\text{b}^2+\text{c}^2-\text{a}^2&0\\\text{b}^2&0&\text{c}^2+\text{a}^2-\text{b}^2\end{vmatrix}$
$=\frac{1}{\text{abc}}(-\text{a}^2)\begin{vmatrix}-2\text{b}^2&-2\text{a}^2\\0&\text{c}^2+\text{a}^2-\text{b}^2\end{vmatrix}+\text{b}^2\begin{vmatrix}-2\text{b}^2&-2\text{a}^2\\\text{b}^2+\text{c}^2-\text{a}^2&0\end{vmatrix}$
$=\frac{1}{\text{abc}}\big[-\text{a}^2\{-2\text{b}^2(\text{c}^2+\text{a}^2-\text{b}^2)\}+\text{b}^2\{0+2\text{a}^2(\text{b}^2+\text{c}^2-\text{a}^2)\}\big]$
$=\frac{1}{\text{abc}}\big[-\text{a}^2\{-2\text{b}^2\text{c}^2-2\text{b}^2\text{a}^2+2\text{b}^4\}+\text{b}^2\{2\text{a}^2\text{b}^2+2\text{a}^2\text{c}^2-2\text{a}^4\}\big]$
$=\frac{1}{\text{abc}}\big[2\text{a}^2\text{b}^2\text{c}^2+2\text{a}^4\text{b}^2-2\text{a}^4\text{b}^4+2\text{a}^2\text{b}^4+2\text{a}^2\text{b}^2\text{c}^2-2\text{a}^4\text{b}^2\big]$
$=\frac{1}{\text{abc}}4\text{a}^2\text{b}^2\text{c}^2$
$=4\text{abc}$
$=\text{R.H.S}$
View full question & answer→Question 695 Marks
Show that x = 2 is a root of the equation $\begin{vmatrix}\text{x}&-6&-1\\2&-3\text{x}&\text{x}-3\\-3&2\text{x}&\text{x}+2\end{vmatrix}=0$ and solve it completely.
AnswerLet us show that x = 2 is a root of the given equation:
Putting x = 2 in the L.H.S, we get
$\begin{vmatrix}2&-6&-1\\2&-6&-1\\-3&4&4 \end{vmatrix}=0$
$\because\text{R}_1=\text{R}_2$
Hence, x = 2 is a root of the given equation.
Now, we see if there are any other roots. For this we need to solve the equation:
$\begin{vmatrix}\text{x}&-6&-1\\2&-3\text{x}&\text{x}-3\\-3&2\text{x}&\text{x}+2\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-1&-6&-1\\\text{x}-1&-3\text{x}&\text{x}-3\\\text{x}-1&2\text{x}&\text{x}+2\end{vmatrix}=0$
$\Rightarrow(\text{x}-1)\begin{vmatrix}1&-6&-1\\1&-3\text{x}&\text{x}-3\\1&2\text{x}&\text{x}+2\end{vmatrix}=0$
$\Rightarrow(\text{x}-1)\begin{vmatrix}1&-6&-1\\0&-3\text{x}+6&\text{x}-3+1\\0&2\text{x}+6&\text{x}+2+1\end{vmatrix}=0$
$\Rightarrow(\text{x}-1)\begin{vmatrix}1&-6&-1\\0&-3(\text{x}-2)&\text{x}-2\\0&2(\text{x}+3)&\text{x}+3\end{vmatrix}=0$
$\Rightarrow(\text{x}-1)(\text{x}-2)(\text{x}+3)\begin{vmatrix}1&-6&-1\\0&-3&1\\0&2&1 \end{vmatrix}=0\\$
$\Rightarrow(\text{x}-1)(\text{x}-2)(\text{x}+3)=0$
$\Rightarrow(\text{x}-1)=0,(\text{x}-2)=0,(\text{x}+3)=0$
$\Rightarrow\text{x}=1,\text{x}=2,\text{x}=-3$
View full question & answer→Question 705 Marks
$\begin{vmatrix}\text{b}+\text{c}&\text{a}&\text{a}\\\text{b}&\text{c}+\text{a}&\text{b}\\\text{c}&\text{c}&\text{a}+\text{b}\end{vmatrix}=4\text{abc}$
Answer$\text{L.H.S}=\begin{vmatrix}\text{b}+\text{c}&\text{a}&\text{a}\\\text{b}&\text{c}+\text{a}&\text{b}\\\text{c}&\text{c}&\text{a}+\text{b}\end{vmatrix}$
$=\begin{vmatrix}0&-2\text{c}&-2\text{b}\\\text{b}&\text{c}+\text{a}&\text{b}\\\text{c}&\text{c}&\text{a}+\text{b}\end{vmatrix} [$Applying $R_1 \rightarrow R_1 - (R_2 + R_3)]$
$=\begin{vmatrix}0&-2\text{c}&-2\text{b}\\\text{b}&\text{c}+\text{a}-\text{b}&0\\\text{c}&0&\text{a}+\text{b}-\text{c}\end{vmatrix} [$Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=0\begin{vmatrix}\text{c}+\text{a}-\text{b}&0\\0&\text{a}+\text{b}-\text{c}\end{vmatrix}-(-2\text{c})\begin{vmatrix}\text{b}&0\\\text{c}&\text{a}+\text{b}-\text{c}\end{vmatrix}-2\text{b}\begin{vmatrix}\text{b}&\text{c}+\text{a}-\text{b}\\\text{c}&0\end{vmatrix}$
$=2\text{c}[\text{b}(\text{a}+\text{b}-\text{c})-0]-2\text{b}[0-\text{c}(\text{c}+\text{a}-\text{b})]$
$=2\text{bc}[\text{a}+\text{b}-\text{c}]-2\text{bc}[\text{b}-\text{c}-\text{a}]$
$=2\text{bc}[(\text{a}+\text{b}-\text{c})-(\text{b}-\text{c}-\text{a})]$
$=4\text{abc}$
$=\text{R.H.S}$
View full question & answer→Question 715 Marks
Without expanding, show that the values of the following determinant are zero: $\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2 \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}0&-3&2\\0&-1&2\\0&5&2 \end{vmatrix} [$Applying $C_1 \rightarrow C_1 + 2C_2]$
$\Rightarrow\triangle=0$
View full question & answer→Question 725 Marks
Find the real values of $\lambda$ for which the following system of linear equations has non-trivial solutions. Also, find the non-trivial solutions:
$2\lambda\text{x}-2\text{y}+3\text{z}=0,$
$\text{x}+\lambda\text{y}+2\text{z}=0,$
$2\text{x}+0\text{y}+\lambda\text{z}=0$
AnswerThe given system of equations can be written as,
$2\lambda\text{x}-2\text{y}+3\text{z}=0$
$\text{x}+\lambda\text{y}+2\text{z}=0$
$2\text{x}+0\text{y}+\lambda\text{z}=0$
The given system of equations will have non-trivial solutions if D = 0
$\Rightarrow\begin{vmatrix}2\lambda&-2&3\\1&\lambda&2\\2&0&\lambda\end{vmatrix}=0$
$\Rightarrow 2\lambda(\lambda^2)+2(\lambda-4)+3(-2\lambda)=0$
$\Rightarrow 2\lambda^3 - 4\lambda - 8 = 0$
$\Rightarrow \lambda = 2$
So, the given system of equations will have non-trivial solutions if $\lambda = 2$
Now, we shall find solutions for $\lambda = 2$
Replacing z by k in the first two equations, we get
$2\lambda\text{x}-2\text{y}=-3\text{k}$
$\text{x}+\lambda\text{y}=-2\text{k}$
$\text{x}=\frac{\begin{vmatrix}-3\text{k}&-2\\-2\text{k}&\lambda\end{vmatrix}}{\begin{vmatrix}2\lambda&-2\\1&\lambda\end{vmatrix}}=\frac{-3\text{k}\lambda-4\text{k}}{2\lambda^2+2} $
$=\frac{-3\text{k}(2)-4\text{k}}{2(2)^2+2}=\frac{-6\text{k}-4\text{k}}{10}=-\text{k}$
$\text{y}=\frac{\begin{vmatrix}2\lambda&-3\text{k}\\1&-2\text{k}\end{vmatrix}}{\begin{vmatrix}2\lambda&-2\\1&\lambda\end{vmatrix}}=\frac{-4\text{k}\lambda+3\text{k}}{2\lambda^2+2}$
$=\frac{-4\text{k}(2)+3\text{k}}{2(2)^2+2}=\frac{-5\text{k}}{10}=\frac{-\text{k}}{2} $
Substituting these value of x and y in the third equation, we get
$\text{L.H.S}= 2(-\text{k})+0\Big(-\frac{\text{k}}{2}\Big)+2\text{k}$
$=0=\text{R.H.S}$
Thus,
$\lambda=2, \text{x}=-\text{k},\text{y}=-\frac{\text{k}}{2}$ and $\text{z}=\text{k}\ [\text{k}\in\text{R}]$
View full question & answer→Question 735 Marks
Solve the following determinant equations: $\begin{vmatrix}3\text{x}-8&3&3\\3&3\text{x}-8&8\\3&3&3\text{x}-8\end{vmatrix}=0$
Answer$\begin{vmatrix}3\text{x}-8&3&3\\3&3\text{x}-8&8\\3&3&3\text{x}-8\end{vmatrix}=0$
Apply $C_1 \rightarrow C_1+ C_2 + C_3$
$\Rightarrow\begin{vmatrix}3\text{x}-2&3&3\\3\text{x}-2&3\text{x}-8&8\\3\text{x}-2&3&3\text{x}-8\end{vmatrix}=0$
$\Rightarrow(3\text{x}-2)\begin{vmatrix}1&3&3\\1&3\text{x}-8&8\\1&3&3\text{x}-8\end{vmatrix}=0$
$\Rightarrow(3\text{x}-2)\begin{vmatrix}1&3&3\\0&3\text{x}-11&0\\0&0&3\text{x}-11\end{vmatrix}=0$
$\Rightarrow(3\text{x}-2)(3\text{x}-11)^2=0$
$\Rightarrow(3\text{x}-2)=0$ or $(3\text{x}-11)^2=0$
$\Rightarrow\text{x}=\frac{2}{3}$ or $\text{x}=\pm\frac{11}{3}$
View full question & answer→Question 745 Marks
Solve the following systems of linear equations by cramer's rule:
2x - y = -2,
3x + 4y = 3
AnswerLet $\text{D}=\begin{vmatrix}2&-1\\3&4\end{vmatrix}=11$
$\text{D}_1=\begin{vmatrix}-2&-1\\3&4\end{vmatrix}=-5$
$\text{D}_2=\begin{vmatrix}2&-2\\3&3\end{vmatrix}=12$
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-5}{11}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{12}{11}$
View full question & answer→Question 755 Marks
Evaluate the following determinant:
$\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2 \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2 \end{vmatrix}$
$=6(-2-10)-(-3)(4+20)+(10-10)$
$=-72+72+0$
$=0$
View full question & answer→Question 765 Marks
If $\text{A}=\begin{vmatrix}2&5\\2&1\end{vmatrix}$ and $\text{B}=\begin{vmatrix}4&-3\\2&5\end{vmatrix},$ verify that |AB| = |A| |B|.
AnswerLet $\text{A}=\begin{vmatrix}2&5\\2&1\end{vmatrix}$
$\Rightarrow|\text{A}|=2-10=-8$
$\text{B}=\begin{vmatrix}4&-3\\2&5\end{vmatrix}$
$\Rightarrow|\text{B}|=20+6=26$
Now $\text{AB}=\begin{vmatrix}2&5\\2&1\end{vmatrix}\begin{vmatrix}4&-3\\2&5\end{vmatrix}$
$=\begin{vmatrix}2\times4+5\times2&2\times(-3)+5\times5\\2\times4+1\times2&2\times(-3)+1\times5\end{vmatrix}$
$=\begin{vmatrix}8+10&-6+25\\8+2&-6+5\end{vmatrix}$
$=\begin{vmatrix}18&19\\10&-1\end{vmatrix}$
$\Rightarrow|\text{AB}|=18\times(-1)-(10)(19)$
$=-18-190=-208$
Now $|\text{AB}|=|\text{A}|\times|\text{B}|$
$-208=(-8)\times(26)$
$-208=-208$
Hence verified.
View full question & answer→Question 775 Marks
Prove that: $\begin{vmatrix}\text{a}^2+1&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2+1&\text{bc}\\\text{ca}&\text{cb}&\text{c}^2+1 \end{vmatrix}=1+\text{a}^2+\text{b}^2+\text{c}^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^2+1&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2+1&\text{bc}\\\text{ca}&\text{cb}&\text{c}^2+1 \end{vmatrix}$
$=(\text{abc})\begin{vmatrix}\text{a}+\frac{1}{\text{a}}&\text{b}&\text{c}\\\text{a}&\text{b}+\frac{1}{\text{b}}&\text{c}\\\text{a}&\text{b}&\text{c}+\frac{1}{\text{c}} \end{vmatrix}$
$[$Taking out $a, b$ and $c$ common from $R_1, R_2$ and $R_3]$
$=(\text{abc})\begin{vmatrix}\text{a}+\frac{1}{\text{a}}&\text{b}&\text{c}\\-\frac{1}{\text{a}}&\frac{1}{\text{b}}&0\\-\frac{1}{\text{a}}&0&\frac{1}{\text{c}} \end{vmatrix}$
$[$Applying$ R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1]$
$=(\text{abc})\Big(\frac{1}{\text{abc}}\Big)\begin{vmatrix}\text{a}^2+1&\text{b}^2&\text{c}^2\\-1&1&0\\-1&0&1\end{vmatrix}$
$[$Applying $C_1 \rightarrow aC_1, C_2 \rightarrow bC_2$ and $C_3 \rightarrow cC_3]$
$=\begin{vmatrix}\text{a}^2+1&\text{b}^2&\text{c}^2\\-1&1&0\\-1&0&1\end{vmatrix}$
$=(-1)\begin{vmatrix}\text{b}^2&\text{c}^2\\1&0\end{vmatrix}+(1)\begin{vmatrix}\text{a}^2+1&\text{b}^2\\-1&1\end{vmatrix}$
$=(-1)(-\text{c}^2)+(\text{a}^2+1+\text{b}^2)$
$=(\text{a}^2+1+\text{b}^2+\text{c}^2)$
$=(\text{a}^2+\text{b}^2+\text{c}^2+1)$
$=\text{R.H.S}$
View full question & answer→Question 785 Marks
Evaluate the following: $\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}$
Applying $R_1 \rightarrow R_1 - R_2$ and $R_3 \rightarrow R_3 - R_2$
$\triangle=\begin{vmatrix}\text{a}&-\text{a}&0\\\text{x}&\text{a}+\text{y}&\text{z}\\0&-\text{a}&\text{a}\end{vmatrix}$
Applying $C_2 \rightarrow C_2 + C_1$
$\triangle=\begin{vmatrix}\text{a}&0&0\\\text{x}&\text{a}+\text{y}&\text{z}\\0&-\text{a}&\text{a}\end{vmatrix}$
$\triangle=\text{a}[\text{a}(\text{a}+\text{x}+\text{y})+\text{az}]+0+0$
$\triangle=\text{a}^2(\text{a}+\text{x}+\text{y}+\text{z})$
View full question & answer→Question 795 Marks
Prove that:
$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z}).$
Answer$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z})$
$\text{L.H.S}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}$
$=\text{xyz}\begin{vmatrix}1&1&1\\\text{x}&\text{y}&\text{z}\\\text{x}^3&\text{y}^3&\text{z}^3\end{vmatrix}$
$=\text{xyz}\begin{vmatrix}0&1&0\\\text{x}-\text{y}&\text{y}&\text{z}-\text{y}\\\text{x}^3-\text{y}^3&\text{y}^3&\text{z}^3-\text{y}^3\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})\begin{vmatrix}0&1&0\\1&\text{y}&1\\\text{x}^2+\text{y}^2+\text{xy}&\text{y}^3&\text{z}^2+\text{y}^2+\text{zy}\end{vmatrix}$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})[\text{z}^2+\text{y}^2+\text{zy}-\text{x}^2-\text{y}^2-\text{xy}]$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})[(\text{z}-\text{x})(\text{z}+\text{x})+\text{y}(\text{z}-\text{x})]$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})(\text{z}-\text{x})[\text{z}+\text{x}+\text{y}]$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z})$
$=\text{R.H.S}$
View full question & answer→Question 805 Marks
Prove that:
$\begin{vmatrix}1&\text{b}+\text{c}&\text{b}^2+\text{c}^2\\1&\text{c}+\text{a}&\text{c}^2+\text{a}^2\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2 \end{vmatrix}=(\text{a}+\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$
AnswerLet $\text{L.H.S}=\begin{vmatrix}1&\text{b}+\text{c}&\text{b}^2+\text{c}^2\\1&\text{c}+\text{a}&\text{c}^2+\text{a}^2\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2 \end{vmatrix}$
$=\begin{vmatrix}0&(\text{b}+\text{c})-(\text{c}+\text{a})&(\text{b}^2+\text{c}^2)-(\text{c}^2+\text{a}^2)\\0&(\text{c}+\text{a})-(\text{a}+\text{b})&(\text{c}^2+\text{a}^2)-(\text{a}^2+\text{b}^2)\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2\end{vmatrix} [$Applying $R_1 \rightarrow R_1 - R_2 $ and $R_2 \rightarrow R_2 - R_3]$
$=\begin{vmatrix}0&\text{b}-\text{a}&\text{b}^2-\text{a}^2\\0&\text{c}-\text{b}&\text{c}^2-\text{b}^2\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2\end{vmatrix}$
$=(-1)^2\begin{vmatrix}0&\text{a}-\text{b}&\text{a}^2-\text{b}^2\\0&\text{b}-\text{c}&\text{b}^2-\text{c}^2\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2\end{vmatrix} [$Taking out $(-1)$ common from $R_1$ and $R_2]$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}0&1&\text{a}+\text{b}\\0&1&\text{b}+\text{c}\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2\end{vmatrix}$
$=(\text{a}+\text{b})(\text{b}-\text{c})\left\{1\times\begin{vmatrix} 1 & \text{a}+\text{b} \\ 1 & \text{b}+\text{c} \end{vmatrix}\right\}$
$=(\text{a}+\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$
$=\text{R.H.S}$
View full question & answer→Question 815 Marks
Write the minors and cofactors of element of the first column of the following matrices and hence evaluate the determinant in case:
$\text{A}=\begin{vmatrix}\text{a}&\text{h}&\text{g}\\\text{h}&\text{b}&\text{f}\\\text{g}&\text{f}&\text{c} \end{vmatrix}$
Answer$\text{M}_{11}=\begin{vmatrix}\text{b}&\text{f}\\\text{f}&\text{c} \end{vmatrix}=\text{bc}-\text{f}^2$
$\text{M}_{21}=\begin{vmatrix}\text{h}&\text{g}\\\text{f}&\text{c} \end{vmatrix}=\text{hc}-\text{fg}$
$\text{M}_{31}=\begin{vmatrix}\text{h}&\text{g}\\\text{b}&\text{f} \end{vmatrix}=\text{hf}-\text{gb}$
$\text{C}_{11}=(-1)^{1+1}\text{M}_{11}=\text{bc}-\text{f}^2$
$\text{C}_{21}=(-1)^{2+1}\text{M}_{11}=-(\text{hc}-\text{fg})=\text{fg}-\text{hc}$
$\text{C}_{31}=(-1)^{3+1}\text{M}_{31}=\text{hf}-\text{gb}$
$\text{D}=\text{a}(\text{bc}-\text{f}^2)-\text{h}(\text{hc}-\text{fg})+\text{g}(\text{fh}-\text{bg})$
$=\text{abc}-\text{af}^2-\text{h}^2\text{c}+\text{fgh}+\text{fgh}-\text{bg}^2$
$=\text{abc}+2\text{hfg}-\text{af}^2-\text{bg}^2-\text{ch}^2$
View full question & answer→Question 825 Marks
Prove the following identities:
$\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}=\text{a}^2(\text{a}+\text{x}+\text{y}+\text{z})$
Answer$\text{L.H.S}=\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}$
$=\begin{vmatrix}\text{a}+\text{x}+\text{y}+\text{z}&\text{y}&\text{z}\\\text{a}+\text{x}+\text{y}+\text{z}&\text{a}+\text{y}&\text{z}\\\text{a}+\text{x}+\text{y}+\text{z}&\text{y}&\text{a}+\text{z}\end{vmatrix} [$Applying $C_1 \rightarrow C_1 + C_2 + C_3]$
=$(\text{a}+\text{x}+\text{y}+\text{z})\begin{vmatrix}1&\text{y}&\text{z}\\1&\text{a}+\text{y}&\text{z}\\1&\text{y}&\text{a}+\text{z}\end{vmatrix}$
=$(\text{a}+\text{x}+\text{y}+\text{z})\begin{vmatrix}1&\text{y}&\text{z}\\0&\text{a}&0\\0&0&\text{a}\end{vmatrix} [$Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_2 - R_1]$
$=(\text{a}+\text{x}+\text{y}+\text{z})\text{a}^2 [$Expanding along first column$]$
$=\text{a}^2(\text{a}+\text{x}+\text{y}+\text{z})$
$=\text{R.H.S}$
$\therefore\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}=\text{a}^2(\text{a}+\text{x}+\text{y}+\text{z})$
View full question & answer→Question 835 Marks
Prove that:
$\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\1&\text{b}^2+\text{ca}&\text{b}^3\\1&\text{c}^2+\text{ab}&\text{c}^3 \end{vmatrix}$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}^2+\text{b}^2+\text{c}^2)$
Answer$\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\1&\text{b}^2+\text{ca}&\text{b}^3\\1&\text{c}^2+\text{ab}&\text{c}^3 \end{vmatrix}$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}^2+\text{b}^2+\text{c}^2)$
$\text{L.H.S}=\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\1&\text{b}^2+\text{ca}&\text{b}^3\\1&\text{c}^2+\text{ab}&\text{c}^3 \end{vmatrix}$
Apply: $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$
$=\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\0&\text{b}^2+\text{ca}-\text{a}^2-\text{bc}&\text{b}^3-\text{a}^3\\0&\text{c}^2+\text{abb}+\text{ca}-\text{a}^2-\text{bc}&\text{c}^3-\text{a}^3 \end{vmatrix}$
$=\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\0&(\text{b}^2-\text{a}^2)-\text{c}(\text{b}-\text{a})&\text{b}^3-\text{a}^3\\0&(\text{c}^2-\text{a}^2)-\text{b}(\text{c}-\text{a})&\text{c}^3-\text{a}^3 \end{vmatrix}$
$=\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\0&(\text{b}-\text{a})(\text{b}+\text{a}-\text{c})&\text{b}^3-\text{a}^3\\0&(\text{c}-\text{a})(\text{c}+\text{a}-\text{b})&\text{c}^3-\text{a}^3 \end{vmatrix}$
$=(\text{b}-\text{a})(\text{c}-\text{a})\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\0&(\text{b}+\text{a}-\text{c})&\text{b}^2+\text{a}^2+\text{ab}\\0&(\text{c}+\text{a}-\text{b})&\text{c}^2+\text{a}^2+\text{ac} \end{vmatrix}$
$=(\text{b}-\text{a})(\text{c}-\text{a})\big[(\text{b}+\text{a}-\text{c})(\text{c}^2+\text{a}^2+\text{ac})\\-(\text{b}^2+\text{a}^2+\text{ab})(\text{c}^2+\text{a}^2+\text{ac})\big]$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}^2+\text{b}^2+\text{c}^2)$
$=\text{R.H.S}$
View full question & answer→Question 845 Marks
Solve the following determinant equations:
$\begin{vmatrix}1&1&\text{x}\\\text{p}+1&\text{p}+1&\text{p}+\text{x}\\3&\text{x}+1&\text{x}+2\end{vmatrix}=0$
AnswerLet $\begin{vmatrix}1&1&\text{x}\\\text{p}+1&\text{p}+1&\text{p}+\text{x}\\3&\text{x}+1&\text{x}+2\end{vmatrix}=0$
$=\begin{vmatrix}1&1&\text{x}\\\text{p}&\text{p}&\text{p}\\3&\text{x}+1&\text{x}+2\end{vmatrix} [$Applying $R_{2 }\rightarrow R_2 - R_1]$
$=\text{p}\begin{vmatrix}1&1&\text{x}\\1&1&1\\3&\text{x}+1&\text{x}+2\end{vmatrix}$
$=\text{p}\begin{vmatrix}1&1&\text{x}\\1&1&1\\2&\text{x}&2\end{vmatrix}$
$=\text{p}\begin{vmatrix}0&1&\text{x}\\0&1&1\\2-\text{x}&\text{x}&2\end{vmatrix} [$Applying $C_1 \rightarrow C_2 - C_1]$
$=\text{p}\left\{(2-\text{x})\times\begin{vmatrix}1&\text{x}\\1&1 \end{vmatrix}\right\} [$Expanding along $C_1]$
$=\text{p}(2-\text{x})(1-\text{x})=0$
$\text{x}=1,2$
View full question & answer→Question 855 Marks
Solve the following systems of linear equations by cramer's rule:
2x + 3y = 10,
x + 6y = 4
AnswerGiven, 2x + 3y = 10
x + 6y = 4
Using Cramer's Rule, we get
$\text{D}=\begin{vmatrix}2&3\\1&6\end{vmatrix}=12-3=9$
$\text{D}_1=\begin{vmatrix}10&3\\4&6\end{vmatrix}=60-12=48$
$\text{D}_2=\begin{vmatrix}2&10\\1&4\end{vmatrix}=8-10=-2$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{48}{9}=\frac{16}{3}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-2}{9}$
$\therefore\text{x}=\frac{16}{3}$ and $\text{y}=\frac{-2}{9}$
View full question & answer→Question 865 Marks
Without expanding, show that the values of the following determinant are zero: $\begin{vmatrix}\frac{1}{\text{a}}&\text{a}^2&\text{bc}\\\frac{1}{\text{b}}&\text{b}^2&\text{ac}\\\frac{1}{\text{c}}&\text{c}^2&\text{ab} \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}\frac{1}{\text{a}}&\text{a}^2&\text{bc}\\\frac{1}{\text{b}}&\text{b}^2&\text{ac}\\\frac{1}{\text{c}}&\text{c}^2&\text{ab} \end{vmatrix}$
$=\begin{vmatrix}1&\text{a}^3&\text{abc}\\1&\text{b}^3&\text{abc}\\1&\text{c}^3&\text{abc}\end{vmatrix} [$Applying $R_1 \rightarrow aR_1, R_2 \rightarrow bR_2$ and $R_3 \rightarrow cR_3]$
$=\text{abc}\begin{vmatrix}1&\text{a}^3&1\\1&\text{b}^3&1\\1&\text{c}^3&1 \end{vmatrix}$
$\Rightarrow\triangle=0$
View full question & answer→Question 875 Marks
Show that the following systems of linear equations is inconsistent:
3x - y + 2z = 6,
2x - y + z = 2,
3x + 6y + 5z = 20
Answer$\text{D}=\begin{vmatrix}3&-1&2\\2&-1&1\\3&6&5\end{vmatrix}$
$=3(-5-6)+1(10-3)+2(12+3)=4$
Since D is non zero,
$\text{D}_1=\begin{vmatrix}6&-1&2\\2&-1&1\\20&6&5\end{vmatrix}$
$=6(-5-6)+1(10-20)+2(12+20)$
$=-66-10+64=-12$
$\text{D}_2=\begin{vmatrix}3&6&2\\2&2&1\\3&20&5\end{vmatrix}$
$=3(10-20)-6(10-3)+2(40-6)$
$=-30+42+68=-4$
$\text{D}_3=\begin{vmatrix}3&-1&6\\2&-1&2\\3&6&20\end{vmatrix}$
$=3(-20-12)+1(40-6)+6(12+3)$
$=-96+34+90=28$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-12}{4}=-3$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-4}{4}=-1$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{28}{4}=7$
$\therefore\text{x}=-3,\text{ y}=-1$ and $\text{z}=7$
View full question & answer→Question 885 Marks
Without expanding, show that the values of the following determinant are zero: $\begin{vmatrix}1^2&2^2&3^2&4^2\\2^2&3^2&4^2&5^2\\3^2&4^2&5^2&6^2\\4^2&5^2&6^2&7^2\end{vmatrix}$
Answer$\begin{vmatrix}1^2&2^2&3^2&4^2\\2^2&3^2&4^2&5^2\\3^2&4^2&5^2&6^2\\4^2&5^2&6^2&7^2\end{vmatrix}$
Applying: $C_3 \rightarrow C_3 - C_2, C_4 \rightarrow C_4 - C_1$
$=\begin{vmatrix}1^1&2^2&3^2-2^2&4^2-1^2\\2^2&3^2&4^2-3^2&5^2-2^2\\3^3&4^2&5^2-4^2&6^2-3^2\\4^2&5^2&6^2-5^2&7^2-4^2 \end{vmatrix}$
$=\begin{vmatrix}1^1&2^2&5&15\\2^2&3^2&7&21\\3^3&4^2&9&27\\4^2&5^2&11&33 \end{vmatrix}$
Take $3$ common from $C_4$
$=0$
$\because\text{C}_3=\text{C}_4$
View full question & answer→Question 895 Marks
Evaluate the following:
$\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
$\triangle=\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
$=\text{x}^2\text{y}^2\text{z}^2\begin{vmatrix}0&\text{x}&\text{x}\\\text{y}&0&\text{y}\\\text{z}&\text{z}&0\end{vmatrix}$
$[$Taking $x^2$ common from from $C_1, y^2$ common from $C_2$ and $z^2$ common from $C_3]$
$=\text{x}^3\text{y}^3\text{z}^3\begin{vmatrix}0&0&1\\1&-1&1\\1&1&0\end{vmatrix}$
$[$Applying $C_2 \rightarrow C_2 - C_3]$
$=\text{x}^3\text{y}^3\text{z}^3(1+1) [$Expanding along first row$]$
$=2\text{x}^3\text{y}^3\text{z}^3$
$\therefore\ \triangle=2\text{x}^3\text{y}^3\text{z}^3$
View full question & answer→Question 905 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sqrt{23}+\sqrt{3}&\sqrt{5}&\sqrt{5}\\\sqrt{15}+\sqrt{46}&5&\sqrt{10}\\3+\sqrt{115}&\sqrt{15}&5\end{vmatrix}$
Answer$\begin{vmatrix}\sqrt{23}+\sqrt{3}&\sqrt{5}&\sqrt{5}\\\sqrt{15}+\sqrt{46}&5&\sqrt{10}\\3+\sqrt{115}&\sqrt{15}&5\end{vmatrix}$
$=\begin{vmatrix}\sqrt{3}&\sqrt{5}&\sqrt{5}\\\sqrt{15}&5&\sqrt{10}\\3&\sqrt{15}&5\end{vmatrix}+\begin{vmatrix}\sqrt{23}&\sqrt{5}&\sqrt{5}\\\sqrt{46}&5&\sqrt{10}\\\sqrt{115}&\sqrt{15}&5\end{vmatrix}$
$=\sqrt{3}\begin{vmatrix}1&\sqrt{5}&\sqrt{5}\\\sqrt{5}&5&\sqrt{10}\\\sqrt{3}&\sqrt{15}&5\end{vmatrix}+\sqrt{23}\begin{vmatrix}1&\sqrt{5}&\sqrt{5}\\\sqrt{2}&5&\sqrt{10}\\\sqrt{5}&\sqrt{15}&5\end{vmatrix}$
$=\sqrt{3}\times\sqrt{5}\begin{vmatrix}1&\sqrt{5}&\sqrt{5}\\\sqrt{5}&5&\sqrt{10}\\\sqrt{3}&\sqrt{3}&5\end{vmatrix}+\sqrt{23}\times\sqrt{5}\begin{vmatrix}1&\sqrt{5}&1\\\sqrt{2}&5&\sqrt{2}\\\sqrt{5}&\sqrt{15}&5\end{vmatrix}$
$=0+0$
$=0$
View full question & answer→Question 915 Marks
Prove that: $\begin{vmatrix}\text{a}&\text{a}+\text{b}&\text{a}+2\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}=9(\text{a}+\text{b})\text{b}^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}&\text{a}+\text{b}&\text{a}+2\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}$
$=\begin{vmatrix}3\text{a}+3\text{b}&3\text{a}+3\text{b}&3\text{a}+3\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix} [$Applying $R_1 \rightarrow R_2 + R_2 + R_3]$
$=3(\text{a}+\text{b})\begin{vmatrix}1&1&1\\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix} $[Taking out $3(a + b)$ common from $R_1]$
$=3(\text{a}+\text{b})\begin{vmatrix}0&0&1\\2\text{b}&-\text{b}&\text{a}+\text{b}\\-\text{b}&2\text{b}&\text{a} \end{vmatrix} [$Applying $C_1 \rightarrow C_1 - C_2$ and $C_2 \rightarrow C_2 - C_3]$
$=3(\text{a}+\text{b})\text{b}^2\begin{vmatrix}0&0&1\\2&-1&\text{a}+\text{b}\\-1&2&\text{a} \end{vmatrix} [$Taking out $b$ common from $C_1$ and $C_2]$
$=3(\text{a}+\text{b})\text{b}^2\times3$
$=9(\text{a}+\text{b})\text{b}^2$
$=\text{R.H.S}$
View full question & answer→Question 925 Marks
Show that $\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\\\text{x}-2&\text{x}-3&\text{x}-\beta\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}=0,$ where $\alpha,\beta,\gamma$ are in A.P.
AnswerSince, $\alpha,\beta,\gamma$ are in A.P, $2\beta=\alpha+\gamma$
$\text{L.H.S}=\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\\\text{x}-2&\text{x}-3&\text{x}-\beta\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}$
$\text{R}_2\rightarrow\text{R}_2-\frac{\text{R}_1}{2}-\frac{\text{R}_3}{2}$
$=\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\$\text{x}-2)-\frac{\text{x}-3}{2}-\frac{\text{x}-1}{2}&(\text{x}-3)-\frac{\text{x}-4}{2}-\frac{\text{x}-2}{2}&(\text{x}-\beta)-\frac{\text{x}-\alpha}{2}-\frac{\text{x}-\gamma}{2}\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}$
$=\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\\0&0&0\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}$ $[\because2\beta=\alpha+\gamma]$
$=0$
$=\text{R.H.S}$
View full question & answer→Question 935 Marks
Prove the following identities:
$\begin{vmatrix}\text{a}^3&2&\text{a}\\\text{b}^3&2&\text{b}\\\text{c}^3&2&\text{c}\end{vmatrix}$
$=2(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})(\text{a}+\text{b}+\text{c})$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^3&2&\text{a}\\\text{b}^3&2&\text{b}\\\text{c}^3&2&\text{c}\end{vmatrix}$
$=2\begin{vmatrix}\text{a}^3&1&\text{a}\\\text{b}^3&1&\text{b}\\\text{c}^3&1&\text{c}\end{vmatrix}$
$=2\{\text{a}^3(\text{c}-\text{d})-1(\text{b}^3\text{c}-\text{bc}^3)+\text{a}(\text{b}^3-\text{c}^3)\}$
$=2\{\text{a}^3(\text{c}-\text{b})-\text{bc}(\text{b}-\text{c})(\text{b}+\text{c})+\text{a}(\text{b}-\text{c})(\text{b}^2+\text{bc}+\text{c}^2)\}$
$=(\text{b}-\text{c})\{-\text{a}^3-\text{bc}(\text{b}+\text{c})+\text{a}(\text{b}^2+\text{bc}+\text{c}^2)\}$
$=2(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})(\text{a}+\text{b}+\text{c})$
$=\text{R.H.S}$
View full question & answer→Question 945 Marks
Prove that:
$\begin{vmatrix}\text{b}+\text{c}&\text{a}-\text{b}&\text{a}\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}\end{vmatrix}=3\text{abc}-\text{a}^3-\text{b}^3-\text{c}^3$
Answer$\begin{vmatrix}\text{b}+\text{c}&\text{a}-\text{b}&\text{a}\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}\end{vmatrix}=3\text{abc}-\text{a}^3-\text{b}^3-\text{c}^3$
$\text{L.H.S}=\begin{vmatrix}\text{b}+\text{c}&\text{a}-\text{b}&\text{a}\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}\end{vmatrix}$
$=\begin{vmatrix}\text{b}+\text{c}+\text{a}&-\text{b}&\text{a}\\\text{c}+\text{a}+\text{b}&-\text{c}&\text{b}\\\text{a}+\text{b}+\text{c}&-\text{a}&\text{c}\end{vmatrix}$
$=-(\text{b}+\text{c}+\text{a})\begin{vmatrix}1&\text{b}&\text{a}\\1&\text{c}&\text{b}\\1&\text{a}&\text{c}\end{vmatrix}$
$=-(\text{b}+\text{c}+\text{a})\begin{vmatrix}1&\text{b}&\text{a}\\0&\text{c}-\text{b}&\text{b}-\text{a}\\0&\text{a}-\text{b}&\text{c}-\text{a}\end{vmatrix}$
$=-(\text{b}+\text{c}+\text{a})[(\text{c}-\text{b})(\text{c}-\text{a})-(\text{b}-\text{a})(\text{a}-\text{b})]$
$=3\text{abc}-\text{a}^3-\text{b}^3-\text{c}^3$
$=\text{R.H.S}$
View full question & answer→Question 955 Marks
Prove the following identities:
$\begin{vmatrix}2\text{y}&\text{y}-\text{z}-\text{x}&2\text{y}\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$
$=(\text{x}+\text{y}+\text{z})^3$
Answer$\text{L.H.S}=\begin{vmatrix}2\text{y}&\text{y}-\text{z}-\text{x}&2\text{y}\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$
$=\begin{vmatrix}\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix} [R_1 = R_1 + R_2 + R_3]$
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}1&1&1\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}1&0&0\\2\text{z}&0&-\text{x}-\text{y}-\text{z}\\\text{x}-\text{y}-\text{z}&\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}\end{vmatrix}$$ [C_2 = C_2 - C_1, C_3 = C_3 - C_1]$
$=(\text{x}+\text{y}+\text{z})\big[1\{0+(\text{x}+\text{y}+\text{z})(\text{x}+\text{y}+\text{z})\}\big]$
$=(\text{x}+\text{y}+\text{z})^3$
$=\text{R.H.S}$
Hence proved.
View full question & answer→Question 965 Marks
Solve the following determinant equations: $\begin{vmatrix}15-2\text{x}&11-3\text{x}&7-\text{x}\\11&17&14\\10&16&13\end{vmatrix}=0$
AnswerLet $\begin{vmatrix}15-2\text{x}&11-3\text{x}&7-\text{x}\\11&17&14\\10&16&13\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}15-2\text{x}-14+2\text{x}&11-3\text{x}&7-\text{x}\\11-28&17&14\\10-26&16&13\end{vmatrix}=0 [$Applying $C_1 \rightarrow C_1 - 2C_3]$
$\Rightarrow\begin{vmatrix}1&11-3\text{x}&7-\text{x}\\-17&17&14\\-16&16&13\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}12-3\text{x}&4-2\text{x}&7-\text{x}\\0&3&14\\0&3&13\end{vmatrix}=0 [$Applying $C_1 \rightarrow C_1 + C_2$ and $C_2 \rightarrow C_2 - C_3]$
$\Rightarrow(12-3\text{x})((3)\times13-(3\times14))=0$
$\Rightarrow(12-3\text{x})(-3)=0$
$\Rightarrow12-3\text{x}=0$
$\Rightarrow3\text{x}=12$
$\Rightarrow\text{x}=4$
View full question & answer→Question 975 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 985 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 995 Marks
Evaluate: $\begin{vmatrix}\text{x}+\lambda&\text{x}&\text{x}\\\text{x}&\text{x}+\lambda&\text{x}\\\text{x}&\text{x}&\text{x}+\lambda\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}\text{x}+\lambda&\text{x}&\text{x}\\\text{x}&\text{x}+\lambda&\text{x}\\\text{x}&\text{x}&\text{x}+\lambda\end{vmatrix}$
$=\begin{vmatrix}\lambda&0&\text{x}\\-\lambda&\lambda&\text{x}\\0&-\lambda&\text{x}+\lambda\end{vmatrix} [$Applying $C_1 \rightarrow C_1 - C_2, C_2 \rightarrow C_2 - C_3]$
$=\begin{vmatrix}\lambda&0&\text{x}\\-\lambda&0&2\text{x}+\lambda\\0&-\lambda&\text{x}+\lambda\end{vmatrix} [$Applying $R_1 \rightarrow R_2 + R_3]$
$=\lambda\begin{vmatrix}0&2\text{x}+\lambda\\-\lambda&\text{x}+\lambda\end{vmatrix}+\text{x}\begin{vmatrix}-\lambda&0\\0&-\lambda\end{vmatrix}$
$=\lambda\big[\lambda(2\text{x}+\lambda)\big]+\text{x}\lambda^2$
$=\lambda^2(2\text{x}+\lambda+\lambda^2\text{x})$
$=3\lambda^2\text{x}+\lambda^3$
$=\lambda^2(3\text{x}+\lambda)$
View full question & answer→Question 1005 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 $R_3 \rightarrow R_1 - R_3]$
$=(\text{x}-\text{a})(\text{x}-\text{b})\begin{vmatrix}1&\text{x}&\text{x}^2\\0&1&\text{x}+\text{a}\\0&1&\text{x}+\text{b}\end{vmatrix}$
$=(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}+\text{b}-\text{x}-\text{a})=0$
$\text{x}=\text{a},\text{b}$
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