Question 12 Marks
Evaluate : $\left| {\begin{array}{*{20}{c}} 1&x&y \\ 1&{x + y}&y \\ 1&x&{x + y} \end{array}} \right|$
AnswerLet $\Delta = \left| {\begin{array}{*{20}{c}} 1&x&y \\ 1&{x + y}&y \\ 1&x&{x + y} \end{array}} \right|$ $\left[ {{R_2} \to {R_2} - {R_1}\,\,and\,\,{R_3} \to {R_3} - {R_1}} \right]$
$= \left| {\begin{array}{*{20}{c}} 1&x&y \\ 0&{x + y - y}&0 \\ 0&0&{x + y - y} \end{array}} \right|$
$ = \left| {\begin{array}{*{20}{c}} 1&x&y \\ 0&y&0 \\ 0&0&x \end{array}} \right|$
Expanding along Ist column
$= 1\left| {\begin{array}{*{20}{c}} y&0 \\ 0&x \end{array}} \right|$
= xy
View full question & answer→Question 22 Marks
Let $A =\left[\begin{array}{ccc} {1} & {2} & {1} \\ {2} & {3} & {1} \\ {1} & {1} & {5} \end{array}\right]$. verify that $(A^{–1})^{–1} = A$
AnswerLet $A=\left[\begin{array}{ccc} {1} & {-2} & {1} \\ {-2} & {3} & {1} \\ {1} & {1} & {5} \end{array}\right]$
$\therefore |A|=1(15-1) + 2(-10-1) + 1(-2-3) = 14-22-5 = -13$
Now, $\begin{aligned} &A_{11}=14, A_{12}=11, A_{13}=-5\\ &A_{21}=11, A_{22}=4, A_{23}=-3\\ &A_{31}=-5, A_{12}=-3, A_{13}=-1 \end{aligned}$
$\therefore $ $a d j A=\left[\begin{array}{ccc} {14} & {11} & {-5} \\ {11} & {4} & {-3} \\ {-5} & {-3} & {-1} \end{array}\right]$
$\therefore $ $A^{-1}=\frac{1}{|A|}(a d j A)$
= $-\frac{1}{13}\left[\begin{array}{ccc} {14} & {11} & {-5} \\ {11} & {4} & {-3} \\ {-5} & {-3} & {-1} \end{array}\right]=\frac{1}{13}\left[\begin{array}{ccc} {-14} & {-11} & {5} \\ {-11} & {-4} & {3} \\ {5} & {3} & {1} \end{array}\right]$
We have shown that
$A^{-1}=\frac{1}{13}\left[\begin{array}{ccc} {-14} & {-11} & {5} \\ {-11} & {-4} & {3} \\ {5} & {3} & {1} \end{array}\right]$
and, Adj $A^{-1} = \frac{1}{13}\left[\begin{array}{ccc} {-1} & {2} & {-1} \\ {2} & {-3} & {-1} \\ {-1} & {-1} & {-5} \end{array}\right]$
Now
$\left|A^{-1}\right|=\left(\frac{1}{13}\right)^{3}[-14 \times(-13)+11 \times(-26)+5 \times(-13)]=\left(\frac{1}{13}\right)^{3} \times(-169)=-\frac{1}{13}$
$\therefore $ $\left(A^{-1}\right)^{-1}=\frac{a d j A^{-1}}{\left|A^{-1}\right|}=\frac{1}{\left(-\frac{1}{13}\right)} \times \frac{1}{13}$$\left[\begin{array}{ccc} {-1} & {2} & {-1} \\ {2} & {-3} & {-1} \\ {-1} & {-1} & {-5} \end{array}\right]=\left[\begin{array}{ccc} {1} & {-2} & {1} \\ {-2} & {3} & {1} \\ {1} & {1} & {5} \end{array}\right]=A$
$\Rightarrow (A^{-1})^{-1} = A$
View full question & answer→Question 32 Marks
Let $A = \left[\begin{array}{ccc} {1} & {2} & {1} \\ {2} & {3} & {1} \\ {1} & {1} & {5} \end{array}\right]$. verify that $[adj\ A]^{–1} = adj (A^{–1})$
AnswerLet $A=\left[\begin{array}{ccc} {1} & {-2} & {1} \\ {-2} & {3} & {1} \\ {1} & {1} & {5} \end{array}\right]$
$\therefore |A|=1(15-1) + 2(-10-1) + 1(-2-3) = 14-22-5 = -13$
Now, $\begin{aligned} &A_{11}=14, A_{12}=11, A_{13}=-5\\ &A_{21}=11, A_{22}=4, A_{23}=-3\\ &A_{31}=-5, A_{12}=-3, A_{13}=-1 \end{aligned}$
$\therefore $ $a d j A=\left[\begin{array}{ccc} {14} & {11} & {-5} \\ {11} & {4} & {-3} \\ {-5} & {-3} & {-1} \end{array}\right]$
So, we have $A^{-1}=\frac{1}{|A|}(a d j A)$
= $-\frac{1}{13}\left[\begin{array}{ccc} {14} & {11} & {-5} \\ {11} & {4} & {-3} \\ {-5} & {-3} & {-1} \end{array}\right]=\frac{1}{13}\left[\begin{array}{ccc} {-14} & {-11} & {5} \\ {-11} & {-4} & {3} \\ {5} & {3} & {1} \end{array}\right]$
Clearly, $a d j A=$$\left[\begin{array}{ccc} {-14} & {-11} & {5} \\ {-11} & {-4} & {3} \\ {5} & {3} & {1} \end{array}\right]$
$\therefore |adj A| = 14(-4-9)-11(-11-15)-5(-33+20)$
$= 14(-13)-11(-26)-5(-13)$
$= -182 + 286 + 65 = 169$
we have,
adj (adj A) = $\left[\begin{array}{ccc} {-13} & {26} & {-13} \\ {26} & {-39} & {-13} \\ {-13} & {-13} & {-65} \end{array}\right]$
$\therefore $ $[\operatorname{adj} A]^{\prime}=\frac{1}{|a d j A|}(\operatorname{adj}(\operatorname{adj} A))$
= $\frac{1}{169}\left[\begin{array}{ccc} {-13} & {26} & {-13} \\ {26} & {-39} & {-13} \\ {-13} & {-13} & {-65} \end{array}\right]$
= $\frac{1}{13}\left[\begin{array}{ccc} {-1} & {2} & {-1} \\ {2} & {-3} & {-1} \\ {-1} & {-1} & {-5} \end{array}\right]$
Now, $A^{-1}=\frac{1}{13}\left[\begin{array}{ccc} {-14} & {-11} & {5} \\ {-11} & {-4} & {3} \\ {5} & {3} & {1} \end{array}\right]$= $\left[\begin{array}{ccc} {-\frac{14}{13}} & {-\frac{11}{13}} & {\frac{5}{13}} \\ {-\frac{11}{13}} & {-\frac{4}{13}} & {\frac{3}{13}} \\ {\frac{5}{13}} & {\frac{3}{13}} & {\frac{1}{13}} \end{array}\right]$
$\therefore adj\ (A^{-1}) = \left[\begin{array}{ccc} {-\frac{4}{169}-\frac{9}{169}} & {-\left(-\frac{11}{169}-\frac{15}{169}\right)} & {-\frac{33}{169}+\frac{20}{169}} \\ {-\left(-\frac{11}{169}-\frac{15}{169}\right)} & {-\frac{14}{169}-\frac{25}{169}} & {-\left(-\frac{42}{169}+\frac{55}{169}\right)} \\ {-\frac{33}{169}+\frac{20}{169}} & {-\left(-\frac{42}{169}+\frac{55}{169}\right)} & {\frac{56}{169}-\frac{121}{169}} \end{array}\right]$
= $\frac{1}{169}\left[\begin{array}{ccc} {-13} & {26} & {-13} \\ {26} & {-39} & {-13} \\ {-13} & {-13} & {-65} \end{array}\right]=\frac{1}{13}\left[\begin{array}{ccc} {-1} & {2} & {-1} \\ {2} & {-3} & {-1} \\ {-1} & {-1} & {-5} \end{array}\right]$
Hence, $[adj\ A]^{-1} = adj(A^{-1})$
View full question & answer→Question 42 Marks
Prove that the determinant $\left| {\begin{array}{*{20}{c}} x&{\sin \theta }&{\cos \theta } \\ { - \sin \theta }&{ - x}&1 \\ {\cos \theta }&1&x \end{array}} \right|$ is independent of $\theta $.
AnswerLet $\Delta = \left[ {\begin{array}{*{20}{c}} x&{\sin \theta }&{\cos \theta } \\ { - \sin \theta }&{ - x}&1 \\ {\cos \theta }&1&x \end{array}} \right]$Expanding along first row,
$\Delta = x\left| {\begin{array}{*{20}{c}} { - x}&1 \\ 1&x \end{array}} \right| - \sin \theta \left| {\begin{array}{*{20}{c}} { - \sin \theta }&1 \\ {\cos \theta }&x \end{array}} \right| $ $+ \cos \theta \left| {\begin{array}{*{20}{c}} { - \sin \theta }&{ - x} \\ {\cos \theta }&1 \end{array}} \right|$
$\Rightarrow \Delta = x\left( { - {x^2} - 1} \right) - \sin \theta \left( { - x\sin \theta - \cos \theta } \right) $ $+ \cos \theta \left( { - \sin \theta + x\cos \theta } \right)$
$\Rightarrow \Delta = - {x^3} - x + x{\sin ^2}\theta $ $+ \sin \theta \cos \theta - \sin \theta \cos \theta + x{\cos ^2}\theta$
$\Rightarrow \Delta = - {x^3} - x + x\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) $ $= -x^3 - x + x = - x^3$ which is independent of $\theta $
View full question & answer→Question 52 Marks
Examine the consistency of the system of equation $x + 3y = 5;\,\,2x + 6y = 8\,$
AnswerMatrix form of given equations is AX = B $\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&3 \\ 2&6 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5 \\ 6 \end{array}} \right]$
$\therefore A = \left[ {\begin{array}{*{20}{c}} 1&3 \\ 2&6 \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}} 5 \\ 8 \end{array}} \right]$
$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 1&3 \\ 2&6 \end{array}} \right|$ = 6 – 6 = 0
Now $\left( {adj.A} \right)B = \left[ {\begin{array}{*{20}{c}} 6&{ - 3} \\ { - 2}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5 \\ 8 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {30 - 24} \\ { - 10 + 8} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 6 \\ { - 2} \end{array}} \right] \ne 0$
Therefore, given equations are inconsistent, i.e., have no common solution.
View full question & answer→Question 62 Marks
Examine the consistency of the system of equation $2x - y = 5;\,\,x + y = 4$
AnswerMatrix form of given equations is AX = B $\Rightarrow \left[ {\begin{array}{*{20}{c}} 2&{ - 1} \\ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5 \\ 4 \end{array}} \right]$
$\therefore A = \left[ {\begin{array}{*{20}{c}} 2&{ - 1} \\ 1&1 \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}} 5 \\ 4 \end{array}} \right]$
$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 2&{ - 1} \\ 1&1 \end{array}} \right| = 2 - \left( { - 1} \right) = 3 \ne 0$
Therefore, Unique solution and hence, equations are consistent.
View full question & answer→Question 72 Marks
Examine the consistency of the system of equation x + 2y = 2; 2x + 3y = 3
AnswerMatrix form of given equations is AX = B $\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&2 \\ 2&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \\ 3 \end{array}} \right]$
$\therefore A = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 2&3 \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}} 2 \\ 3 \end{array}} \right]$
$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 1&2 \\ 2&3 \end{array}} \right| = 3 - 4 = - 1 \ne 0$
Therefore, Unique solution and hence equations are consistent.
View full question & answer→Question 82 Marks
Using cofactors of elements of second row, evaluate $\Delta = \left| {\begin{array}{*{20}{c}} 5&3&8 \\ 2&0&1 \\ 1&2&3 \end{array}} \right|$
Answer$\Delta = {a_{21}}{A_{21}} + {a_{22}}{A_{22}} + {a_{23}}{A_{23}}$
$ = - 2\left( {9 - 16} \right) + 0\left( {15 - 8} \right) - 1\left( {10 - 3} \right)$
$ = 14 + 0 - 7$
= 7.
Which is the required solution.
View full question & answer→Question 92 Marks
Write minors and cofactors of the elements of $\left| {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right|$.
AnswerLet $\Delta = \left| {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right|$$M_{11} =$ Minor of ${a_{11}} = \left| d \right| = d$ and ${A_{11}} = {\left( { - 1} \right)^{1 + 1}}{M_{11}} = {\left( { - 1} \right)^2}\left( d \right) = d$
$M_{12} =$ Minor of ${a_{12}} = \left| b \right| = b$ and ${A_{12}} = {\left( { - 1} \right)^{1 + 2}}{M_{12}} = {\left( { - 1} \right)^3}\left( b \right) = - b$
$M_{21} =$ Minor of ${a_{21}} = \left| c \right| = c$ and ${A_{21}} = {\left( { - 1} \right)^{2 + 1}}{M_{21}} = {\left( { - 1} \right)^3}\left( c \right) = - c$
$M_{22} =$ Minor of ${a_{22}} = \left| a \right| = a$ and ${A_{22}} = {\left( { - 1} \right)^{2 + 2}}{M_{22}} = {\left( { - 1} \right)^4}\left( a \right) = a$
View full question & answer→Question 102 Marks
Write minors and cofactors of the elements of $\left| {\begin{array}{*{20}{c}} 2&{ - 4} \\ 0&3 \end{array}} \right|$.
AnswerLet $\Delta = \left| {\begin{array}{*{20}{c}} 2&{ - 4} \\ 0&3 \end{array}} \right|$$M_{11} =$ Minor of ${a_{11}} = \left| 3 \right| = 3$ and ${A_{11}} = {\left( { - 1} \right)^{1 + 1}}{M_{11}} = {\left( { - 1} \right)^2}\left( 3 \right) = 3$
$M_{12} =$ Minor of ${a_{12}} = \left| 0 \right| = 0$ and ${A_{12}} = {\left( { - 1} \right)^{1 + 2}}{M_{12}} = {\left( { - 1} \right)^3}\left( 0 \right) = 0$
$M_{21} =$ Minor of ${a_{21}} = \left| { - 4} \right| = - 4$ and ${A_{21}} = {\left( { - 1} \right)^{1 + 2}}{M_{21}} = {\left( { - 1} \right)^3}\left( { - 4} \right) = 4$
$M_{22} =$ Minor of ${a_{22}} = \left| 2 \right| = 2$ and ${A_{22}} = {\left( { - 1} \right)^{2 + 2}}{M_{22}} = {\left( { - 1} \right)^4}\left( 2 \right) = 2$
View full question & answer→Question 112 Marks
Find equation of line joining $(3, 1)$ and $(9, 3)$ using determinants.
AnswerLet $(x, y)$ be any point on the line containing $(3, 1)$ and $(9, 3)$,then the required equation is,$\left| {\begin{array}{*{20}{c}} \begin{gathered} x \hfill \\ 3 \hfill \\ 9 \hfill \\ \end{gathered} &\begin{gathered} y \hfill \\ 1 \hfill \\ 3 \hfill \\ \end{gathered} &\begin{gathered} 1 \hfill \\ 1 \hfill \\ 1 \hfill \\ \end{gathered} \end{array}} \right| = 0$
Expanding along $R_1$ , we get,
$x[1-3]-y[3-9]+1[9-9]=0$
$\Rightarrow -2x+6y=0$
$ x=3y$ which is the required equation of the line.
View full question & answer→Question 122 Marks
Find the equation of line joining (1, 2) and (3, 6) using determinants.
AnswerLet $p\left( {x,y} \right)$ be any point on the line joining the points (1, 2) and (3, 6).
Then, Area of triangle that could be formed by these points is zero. $\therefore$ Area of triangle = $\frac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\ {{x_2}}&{{y_2}}&1 \\ {{x_3}}&{{y_3}}&1 \end{array}} \right| = 0$
$\Rightarrow$ $\frac{1}{2}\left| {\begin{array}{*{20}{c}} x&y&1 \\ 1&2&1 \\ 3&6&1 \end{array}} \right| = 0$
$\Rightarrow \frac{1}{2}\left[ {x\left( {2 - 6} \right) - y\left( {1 - 3} \right) + 1\left( {6 - 6} \right)} \right] = 0$
$\Rightarrow - 4x + 2y = 0$
$\Rightarrow - 2x + y = 0$
$ \Rightarrow y = 2x$ which is required line.
View full question & answer→Question 132 Marks
Find values of k if area of triangle is 4 sq. units and vertices are: (-2, 0),(0, 4),(0, k).
AnswerGiven: Area of triangle = $\frac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\ {{x_2}}&{{y_2}}&1 \\ {{x_3}}&{{y_3}}&1 \end{array}} \right| = 4$ sq.units $\Rightarrow \frac{1}{2}\left| {\begin{array}{*{20}{c}} { - 2}&0&1 \\ 0&4&1 \\ 0&k&1 \end{array}} \right| =\pm 4$
$\Rightarrow {\frac{1}{2}\left[ { - 2\left( {4 - k} \right) - 0 + 1\left( {0 - 0} \right)} \right]} = \pm4$
$\Rightarrow {\frac{1}{2}\left( { - 8 + 2k} \right)} =\pm 4$
$ \Rightarrow { - k + 4} = \pm4$
Taking positive sign, $- k + 4 = 4$
$ \Rightarrow k = 0$
Taking negative sign, $- k + 4 = - 4$
$\Rightarrow k = 8$
View full question & answer→Question 142 Marks
Find value of k if area of triangle is 4 sq. units and vertices are :(k ,0), (4, 0), (0,2).
AnswerGiven: Area of triangle =$\frac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\ {{x_2}}&{{y_2}}&1 \\ {{x_3}}&{{y_3}}&1 \end{array}} \right| = 4$ sq. units $\Rightarrow$ $\frac{1}{2}\left| {\begin{array}{*{20}{c}} k&0&1 \\ 4&0&1 \\ 0&2&1 \end{array}} \right| =\pm 4$
$\Rightarrow {\frac{1}{2}\left[ {k\left( {0 - 2} \right) - 0 + 1\left( {8 - 0} \right)} \right]} =\pm 4$
$\Rightarrow {\frac{1}{2}\left( { - 2k + 8} \right)} =\pm 4$
$ \Rightarrow { - k + 4} =\pm 4$
$ \Rightarrow - k + 4 = \pm 4$
Taking positive sign, $ - k + 4 = 4$
$ \Rightarrow k = 0$
Taking negative sign, $ - k + 4 = - 4$
$ \Rightarrow k = 8$
View full question & answer→Question 152 Marks
Show that points A( a, b + c), B(b, c + a), c(c, a + b) are collinear.
AnswerArea of triangle ABC = $\frac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\ {{x_2}}&{{y_2}}&1 \\ {{x_3}}&{{y_3}}&1 \end{array}} \right| = {\frac{1}{2}\left[ {\begin{array}{*{20}{c}} a&{b + c}&1 \\ b&{c + a}&1 \\ c&{a + b}&1 \end{array}} \right]} $ $= {\frac{1}{2}\left[ {a\left( {c + a - a} \right) - \left( {b + c} \right)\left( {b - c} \right) + 1\left\{ {b\left( {a + b} \right) - c\left( {c + a} \right)} \right\}} \right]} $
$= {\frac{1}{2}\left[ {a\left( {c - b} \right) - \left( {{b^2} - {c^2}} \right) + \left( {ab + {b^2} - {c^2}- ac} \right)} \right]} $
$ = {\frac{1}{2}\left( {ac - ab - {b^2} + {c^2} + ab + {b^2} - {c^2} - ac} \right)} $
$= {\frac{1}{2} \times 0} = 0$
Therefore, points A, B and C are collinear.
View full question & answer→Question 162 Marks
Find the area of the triangle with vertices at the points given (-2, -3), (3, 2) and (-1 , -8).
AnswerArea of triangle = $\frac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\ {{x_2}}&{{y_2}}&1 \\ {{x_3}}&{{y_3}}&1 \end{array}} \right|$ $= {\frac{1}{2}\left[ {\begin{array}{*{20}{c}} { - 2}&{ - 3}&1 \\ 3&2&1 \\ { - 1}&8&1 \end{array}} \right]} $
$= {\frac{1}{2}\left[ { - 2\left( {2 + 8} \right) - \left( { - 3} \right)\left( {3 + 1} \right) + 1\left( { - 24 + 2} \right)} \right]} $
$= {\frac{1}{2}\left[ { - 2\left( {10} \right) + 3\left( 4 \right) - 22} \right]} $
$= {\frac{1}{2}\left( { - 20 + 12 - 22} \right)}$
$ = {\frac{1}{2} \times \left( { - 30} \right)} $
$ = {\frac{1}{2} \times -30} = -15$
So, the area is 15 sq. unit
View full question & answer→Question 172 Marks
Find area of the triangle with vertices at the point given (2, 7), (1, 1), (10, 8)
AnswerArea of triangle $= \frac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\ {{x_2}}&{{y_2}}&1 \\ {{x_3}}&{{y_3}}&1 \end{array}} \right|$ $={\frac{1}{2}\left[ {\begin{array}{*{20}{c}} 2&7&1 \\ 1&1&1 \\ {10}&8&1 \end{array}} \right]} $
$= {\frac{1}{2}\left[ {2\left( {1 - 8} \right) - 7\left( {1 - 10} \right) + 1\left( {8 - 10} \right)} \right]} $
$= {\frac{1}{2}\left[ {2\left( { - 7} \right) - 7\left( { - 9} \right) - 2} \right]} $
$= {\frac{1}{2}\left( { - 14 + 63 - 2} \right)} $
$= {\frac{1}{2}\left( {63 - 16} \right)} $
$={\frac{{47}}{2}} $ sq. units
View full question & answer→Question 182 Marks
Find area of the triangle with vertices at the point given (1, 0), (6, 0), (4, 3).
AnswerArea of triangle = $\frac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\ {{x_2}}&{{y_2}}&1 \\ {{x_3}}&{{y_3}}&1 \end{array}} \right|$ $=\frac{1}{2} \left| {\begin{array}{*{20}{c}} 1&0&1 \\ 6&0&1 \\ 4&3&1 \end{array}} \right|$
$= \frac{1}{2}$[-3(1-6)]=$\frac{1}{2}$(-3)(-5)=$\frac{15}{2}$ sq.units
View full question & answer→Question 192 Marks
If $A = \left| {\begin{array}{*{20}{c}} 1&0&1 \\ 0&1&2 \\ 0&0&4 \end{array}} \right|$ then show that |3A| = 27|A|
AnswerGiven: $A = \left[ {\begin{array}{*{20}{c}} 1&0&1 \\ 0&1&2 \\ 0&0&4 \end{array}} \right],$ then $3A = 3\left[ {\begin{array}{*{20}{c}} 1&0&1 \\ 0&1&2 \\ 0&0&4 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3&0&3 \\ 0&3&6 \\ 0&0&{12} \end{array}} \right]$ L.H.S. = $\left| {3A} \right| = \left[ {\begin{array}{*{20}{c}} 3&0&3 \\ 0&3&6 \\ 0&0&{12} \end{array}} \right]$
= 3(36 - 0) $ = 3 \times 36 = 108$
R.H.S. $ = 27\left| A \right| = 27\left| {\begin{array}{*{20}{c}} 1&0&1 \\ 0&1&2 \\ 0&0&4 \end{array}} \right|$
= 27[1(4 - 0)] $ = 27 \times 4 = 108$
Since L.H.S. = R.H.S.
Hence, proved.
View full question & answer→Question 202 Marks
If $A = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 4&2 \end{array}} \right]$, then show that |2A| = 4|A|
Answer$2A = 2\left[ {\begin{array}{*{20}{c}} 1&2 \\ 4&2 \end{array}} \right]$ $= \left[ {\begin{array}{*{20}{c}} 2&4 \\ 8&4 \end{array}} \right] $
|2A| = 8 - 32 = -24
|A| = 2 - 8 = -6
4|A| = -24
Hence Proved.
View full question & answer→Question 212 Marks
Find minors and cofactors of all the elements of the determinant $\left| {\begin{array}{*{20}{c}} 1&{ - 2} \\ 4&3 \end{array}} \right|$
Answer${M_{11}} = 3,{A_{11}} = 3$ ${M_{12}} = 4,{A_{12}} = - 4\left[ {\because Aij = {{\left( { - 1} \right)}^{i + j}}.Mij} \right]$
${M_{21}} = - 2,{A_{21}} = 2$
${M_{22}} = 1,{A_{22}} = 1$
View full question & answer→Question 222 Marks
Find the minor of element 6 in the determinant $\Delta=\left|\begin{array}{lll} {1} & {2} & {3} \\ {4} & {5} & {6} \\ {7} & {8} & {9} \end{array}\right|$
AnswerSince 6 lies in the second row and third column, its minor $M_{23}$ is given by
$\mathrm{M}_{23}=\left|\begin{array}{ll} {1} & {2} \\ {7} & {8} \end{array}\right| = 8 – 14 = – 6$ (obtained by deleting $R_2$ and $C_3$ in $\Delta)$
View full question & answer→Question 232 Marks
Find the equation of the line joining A (1, 3) and B (0, 0) using determinants and find k if D (k, 0) is a point such that area of $\Delta ABD$ is 3 square units.
AnswerLet P (x, y) be any point on AB. Then the equation of line AB is, $$ $\frac{1}{2}\left| {\begin{array}{*{20}{c}} 0&0&1 \\ 1&3&1 \\ x&y&1 \end{array}} \right| = 0$
y = 3x
Area $\Delta ABD = 3$ square unit
$\frac{1}{2}\left| {\begin{array}{*{20}{c}} 1&3&1 \\ 0&0&1 \\ K&0&1 \end{array}} \right| = \pm 3$
$k = \pm 2$
View full question & answer→Question 242 Marks
Find the area of $\Delta $ whose vertices are (3, 8) (-4, 2) and (5, 1).
Answer$\Delta = \frac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\ {{x_2}}&{{y_2}}&1 \\ {{x_3}}&{{y_3}}&1 \end{array}} \right|$
$ = \frac{1}{2}\left| {\begin{array}{*{20}{c}} 3&8&1 \\ { - 4}&2&1 \\ 5&1&1 \end{array}} \right|$ $ = \frac{1}{2}\left[ {3\left( {2 - 1} \right) - 8\left( { - 4 - 5} \right) + 1\left( { - 4 - 10} \right)} \right]$
$ = \frac{1}{2}\left[ {3 + 72 - 14} \right] = \frac{{61}}{2}$
View full question & answer→Question 252 Marks
Find values of x for which $\left| {\begin{array}{*{20}{c}} 3&x \\ x&1 \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 3&2 \\ 4&1 \end{array}} \right|$.
Answer${\left( {3 - x} \right)^2} = 3 - 8$ $3 - {x^2} = 3 - 8$
$ - {x^2} = - 8$
$x = \pm \sqrt 8 $
$x = \pm 2\sqrt 2 $
View full question & answer→Question 262 Marks
Evaluate $\Delta = \left| {\begin{array}{*{20}{c}} 0&{\sin \alpha }&{ - \cos \alpha } \\ { - \sin \alpha }&0&{\sin \beta } \\ {\cos \alpha }&{ - \sin \beta }&0 \end{array}} \right|$
Answer$\Delta =0 \left|{\begin{array}{*{20}{c}} 0&{\sin \beta } \\ { - \sin \beta }&0 \end{array}} \right| - \sin \alpha \left|{\begin{array}{*{20}{c}} { - \sin \alpha }&{\sin \beta } \\ {\cos \alpha }&0 \end{array}} \right|$$ - \cos \alpha \left| {\begin{array}{*{20}{c}} { - \sin \alpha }&0 \\ {\cos \alpha }&{ - \sin \beta } \end{array}} \right|$
$=0-sin\alpha(0-\cos\alpha\sin\beta)-\cos\alpha(\sin\alpha\sin\beta)$
$=\sin\alpha\cos\alpha\sin\beta-\cos\alpha\sin\alpha\sin\beta =0 $
View full question & answer→Question 272 Marks
Evaluate the determinant $\Delta=\left|\begin{array}{rrr} {1} & {2} & {4} \\ {-1} & {3} & {0} \\ {4} & {1} & {0} \end{array}\right|$
AnswerNote that in the third column, two entries are zero. So expanding along the third column $(C_3),$ we get
$\Delta=4\left|\begin{array}{cc} {-1} & {3} \\ {4} & {1} \end{array}\right|-0\left|\begin{array}{cc} {1} & {2} \\ {4} & {1} \end{array}\right|+0\left|\begin{array}{cc} {1} & {2} \\ {-1} & {3} \end{array}\right|$
$= 4 (–1 – 12) – 0 + 0 = – 52$
View full question & answer→Question 282 Marks
Evaluate $\left|\begin{array}{cc} {x} & {x+1} \\ {x-1} & {x} \end{array}\right|$
AnswerWe have $\left|\begin{array}{cc} {x} & {x+1} \\ {x-1} & {x} \end{array}\right|$ $= x (x) – (x + 1) (x – 1) = x^2 – (x^2 – 1) = x^2 – x^2 + 1 = 1$
View full question & answer→Question 292 Marks
Use product $\left[\begin{array}{ccc} {1} & {1} & {2} \\ {0} & {2} & {3} \\ {3} & {2} & {4} \end{array}\right]\left[\begin{array}{ccc} {2} & {0} & {1} \\ {9} & {2} & {3} \\ {6} & {1} & {2} \end{array}\right]$to solve the system of equations $x – y + 2z = 1, 2y – 3z = 1, 3x – 2y + 4z = 2$
AnswerConsider the product $\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 0&2&{ - 3} \\ 3&{ - 2}&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 2}&0&1 \\ 9&2&{ - 3} \\ 6&1&{ - 2} \end{array}} \right] $
$= \left[\begin{array}{ccc} {-2-9+12} & {0-2+2} & {1+3-4} \\ {0+18-18} & {0+4-3} & {0-6+6} \\ {-6-18+24} & {0-4+4} & {3+6-8} \end{array}\right]$
$= \left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right]$
Hence ${\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 0&2&{ - 3} \\ 3&{ - 2}&4 \end{array}} \right]^{ - 1}} $
$= \left[ {\begin{array}{*{20}{c}} { - 2}&0&1 \\ 9&2&{ - 3} \\ 6&1&{ - 2} \end{array}} \right]$
Now, given system of equations can be written, in matrix form, as follows
$\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 0&2&{ - 3} \\ 3&{ - 2}&4 \end{array}} \right] \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ 1 \\ 2 \end{array}} \right]$
or $\begin{array}{l} {x} \\ {y} \\ {z} \end{array}=\left[\begin{array}{rrr} {1} & {-1} & {2} \\ {0} & {2} & {-3} \\ {3} & {-2} & {4} \end{array}\right]^{-1}\left[\begin{array}{l} {1} \\ {1} \\ {2} \end{array}\right]$
$= \left[\begin{array}{rrr} {2} & {0} & {1} \\ {9} & {2} & {3} \\ {6} & {1} & {2} \end{array}\right]\left[\begin{array}{l} {1} \\ {1} \\ {2} \end{array}\right]$
$= \left[\begin{array}{c} {-2+0+2} \\ {9+2-6} \\ {6+1-4} \end{array}\right]=\left[\begin{array}{c} {0} \\ {5} \\ {3} \end{array}\right]$
Hence $, x = 0, y = 5, z = 3$
View full question & answer→Question 302 Marks
The sum of three numbers is $6$. If we multiply third number by $3$ and add second number to it, we get $11$. By adding first and third numbers, we get double of the second number. Represent it algebraically and find the numbers using matrix method.
AnswerLet first ,second and third number be denoted by $x,y$ and $z,$ respectively.
Then, according to given conditions,we have,
$x + y + z = 6$
$y + 3z = 11$
$x + z = 2y$ or $x-2y+z=0$
This system can be written as $AX = B$ whose $A$
$= \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 0&1&3 \\ 1&{ - 2}&1 \end{array}} \right]X $
$= \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right]B = \left[ {\begin{array}{*{20}{c}} 6 \\ {11} \\ 0 \end{array}} \right]$
$\left| A \right|=1(1+6)-(0-3)+(0-1) = 9 \ne 0$
${A_{11}} = 7,{A_{12}} = 3,{A_{13}} = - 1$
${A_{21}} = - 3,{A_{22}} = 0,{A_{23}} = 3$
${A_{31}} = 2,{A_{32}} = - 3,{A_{33}} = 1$
$\text{adj } A = \left[ {\begin{array}{*{20}{c}} 7&{ - 3}&2 \\ 3&0&{ - 3} \\ { - 1}&3&1 \end{array}} \right]$
${A^{ - 1}}=\frac{1}{{\left| A \right|}}$
$\text{adj } {A} = \frac{1}{9}\left[ {\begin{array}{*{20}{c}} 7&{ - 3}&2 \\ 3&0&{ - 3} \\ { - 1}&3&1 \end{array}} \right]$
$X = {A^{ - 1}}B$
$\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \frac{1}{9}\left[ {\begin{array}{*{20}{c}} 7&{ - 3}&2 \\ 3&0&{ - 3} \\ { - 1}&3&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 6 \\ {11} \\ 0 \end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right]= \frac{1}{9}\left[ {\begin{array}{*{20}{c}} 42-33+0 \\ 18+0+0 \\ -6+33+0 \end{array}} \right]$
$=\frac19\begin{bmatrix}9\\18\\27\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}$
Hence $, x=1, y=2, z=3$
View full question & answer→Question 312 Marks
Solve the following system of equations by matrix method: $3x - 2y + 3z = 8, 2x + y - z = 1, 4x - 3y + 2z = 4$
AnswerThe system of equations can be written in the form $AX = B$, where
$A=\left[\begin{array}{ccc} {3} & {-2} & {3} \\ {2} & {1} & {-1} \\ {4} & {-3} & {2} \end{array}\right], X=\left[\begin{array}{c} {x} \\ {y} \\ {z} \end{array}\right] \text { and } B=\left[\begin{array}{l} {8} \\ {1} \\ {4} \end{array}\right]$
We see that
$|A|=3(2-3)+2(4+4)+3(-6-4)=-17 \neq 0$
Hence,$ \mathrm{A}$ is non $-$ singular and so its inverse exists.
Now
$A_{11}=-1, A_{12}=-8, A_{13}=-10 $
$A_{21}=-5, A_{22}=-6, A_{23}=1$
$ A_{31}=-1, A_{32}=9, A_{33}=7$
Therefore $A^{-1}=-\frac{1}{17}\left[\begin{array}{ccc} {-1} & {-5} & {-1} \\ {-8} & {-6} & {9} \\ {-10} & {1} & {7} \end{array}\right]$
So, $X=A^{-1} B=-\frac{1}{17}\left[\begin{array}{ccc} {-1} & {-5} & {-1} \\ {-8} & {-6} & {9} \\ {-10} & {1} & {7} \end{array}\right]\left[\begin{array}{l} {8} \\ {1} \\ {4} \end{array}\right]$
i.e. $\left[\begin{array}{l} {x} \\ {y} \\ {z} \end{array}\right]=-\frac{1}{17}\left[\begin{array}{c} {-17} \\ {-34} \\ {-51} \end{array}\right]=\left[\begin{array}{l} {1} \\ {2} \\ {3} \end{array}\right]$
Hence $x = 1, y = 2$ and $z = 3.$
View full question & answer→Question 322 Marks
Solve the system of equations $\begin{aligned} &2 x+5 y=1\\ &3 x+2 y=7 \end{aligned}$
AnswerThe system of equations can be written in the form $AX = B,$ where
$A = \left[\begin{array}{ll} {2} & {5} \\ {3} & {2} \end{array}\right], X=\left[\begin{array}{l} {x} \\ {y} \end{array}\right]$ and $B = \left[\begin{array}{l} {1} \\ {7} \end{array}\right]$
Now, $A = –11 \neq 0,$
Hence $,A$ is nonsingular matrix and so has a unique solution.
Note that $A^{-1}=-\frac{1}{11}\left[\begin{array}{cc} {2} & {-5} \\ {-3} & {2} \end{array}\right]$
Therefore $X = A^{–1}B = –\frac{1}{11}\left[\begin{array}{cc} {2} & {-5} \\ {-3} & {2} \end{array}\right]\left[\begin{array}{l} {1} \\ {7} \end{array}\right]$
i.e., $\left[\begin{array}{l} {x} \\ {y} \end{array}\right]=-\frac{1}{11}\left[\begin{array}{c} {-33} \\ {11} \end{array}\right]=\left[\begin{array}{c} {3} \\ {-1} \end{array}\right]$
Hence, $x = 3, y = – 1$
View full question & answer→Question 332 Marks
Show that the matrix $A = \left[ {\begin{array}{*{20}{c}} 2&3 \\ 1&2 \end{array}} \right]$ satisfies the equation $A^2 – 4A + I = 0.$ where I is $2 \times 2$ identity matrix and $O$ is $2 \times 2$ zero matrix. Using this equation, find $A^{-1}$
Answer${A^2} = \left[ {\begin{array}{*{20}{c}} 2&3 \\ 1&2 \end{array}} \right].\left[ {\begin{array}{*{20}{c}} 2&3 \\ 1&2 \end{array}} \right]$$= \left[ {\begin{array}{*{20}{c}} 7&{12} \\ 1&7 \end{array}} \right]$
${A^2} - 4A + I = \left[ {\begin{array}{*{20}{c}} 7&{12} \\ 1&7 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 8&{12} \\ 4&8 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$
$= 0$
${A^2} - 4A + I = 0$
${A^2} - 4A = - I$
$AA{A^{ - 1}} - 4A{A^{ - 1}} = I{A^{ - 1}}$
$AI - 4I = - I{A^{ - 1}}\left[ {\because A{A^{ - 1}} = I} \right]$
${A^{ - 1}} = 4I - A$
$= \left[ {\begin{array}{*{20}{c}} 2&{ - 3} \\ { - 1}&2 \end{array}} \right]$
View full question & answer→Question 342 Marks
If $A = \left[ {\begin{array}{*{20}{c}} 2&3 \\ 1&{ - 4} \end{array}} \right]$ and $ B = \left[ {\begin{array}{*{20}{c}} 1&{ - 2} \\ { - 1}&3 \end{array}} \right]$ then verify that $(AB)^{-1} = B^{-1} A^{-1}$
AnswerAB = $\left[ {\begin{array}{*{20}{c}} 2&3 \\ 1&{ - 4} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ - 2} \\ { - 1}&3 \end{array}} \right]$= $\begin{bmatrix}-1&5\\5&-14\end{bmatrix}|AB| = 14 -25 = -11$
$\Rightarrow (AB)$ is non singular and hence, $(AB)^{-1}$ exists
$\therefore (AB)^{-1}= \frac{1}{{|AB|}}adj(AB) = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}} { - 14}&{ - 5} \\ { - 5}&{ - 1} \end{array}} \right]$
Now, $|A| = - 8 - 3 = - 11$
$\Rightarrow A$ is non singular and hence $A^{-1}$ exists
Also, $|B| = 3 - 2 = 1$
$\Rightarrow B$ is non- singular and hence $B^{-1}$ exists
$A^{-1} = \frac{{ - 1}}{{11}}\left[ {\begin{array}{*{20}{c}} { - 4}&{ - 3} \\ { - 1}&2 \end{array}} \right]$
$B^{-1}= \frac{1}{1}\left[ {\begin{array}{*{20}{c}} 3&2 \\ 1&1 \end{array}} \right]$
$B^{-1}A^{-1} = \frac{{ - 1}}{{11}}\left[ {\begin{array}{*{20}{c}} 3&2 \\ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 4}&{ - 3} \\ { - 1}&2 \end{array}} \right]$
= $\frac{{ - 1}}{{11}}\left[ {\begin{array}{*{20}{c}} { - 14}&{ - 5} \\ { - 5}&{ - 1} \end{array}} \right]$
Hence, $(AB)^{-1}= B^{-1}A^{-1}$
View full question & answer→Question 352 Marks
If A = $\left[\begin{array}{lll} {1} & {3} & {3} \\ {1} & {4} & {3} \\ {1} & {3} & {4} \end{array}\right]$then verify that $A$ adj $A = | A|\ I$. Also find $A^{–1}.$
AnswerWe have $A = 1 (16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1 \neq 0$
Now $A_{11} = 7, A_{12} = –1, A_{13} = –1, A_{21} = –3, A_{22} = 1,A_{23} = 0, A_{31} = –3, A_{32} = 0, A_{33} = 1$
Therefore $\text { adj } \mathrm{A}=\left[\begin{array}{rrr} {7} & {-3} & {-3} \\ {-1} & {1} & {0} \\ {-1} & {0} & {1} \end{array}\right]$
Now A (adj A) = $\left[\begin{array}{ccc} {1} & {3} & {3} \\ {1} & {4} & {3} \\ {1} & {3} & {4} \end{array}\right]\left[\begin{array}{ccc} {7} & {-3} & {-3} \\ {-1} & {1} & {0} \\ {-1} & {0} & {1} \end{array}\right]$
= $\left[\begin{array}{ccc} {7-3-3} & {-3+3+0} & {-3+0+3} \\ {7-4-3} & {-3+4+0} & {-3+0+3} \\ {7-3-4} & {-3+3+0} & {-3+0+4} \end{array}\right]$
= $\left[\begin{array}{lll} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right]=(1) \cdot\left[\begin{array}{lll} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right]=|\mathrm{A}| . \mathrm{I}$
Also $\mathrm{A}^{-1}=\frac{1}{|\mathrm{A}|} \text { adj } \mathrm{A}$ = $\frac{1}{1}\left[\begin{array}{rrr} {7} & {-3} & {-3} \\ {-1} & {1} & {0} \\ {-1} & {0} & {1} \end{array}\right]=\left[\begin{array}{rrr} {7} & {-3} & {-3} \\ {-1} & {1} & {0} \\ {-1} & {0} & {1} \end{array}\right]$
View full question & answer→Question 362 Marks
Find adj A for $A = \left[ {\begin{array}{*{20}{c}} 2&3 \\ 1&4 \end{array}} \right].$
Answer$adjA = \left[ {\begin{array}{*{20}{c}} 4&{ - 3} \\ { - 1}&2 \end{array}} \right]$ 
View full question & answer→Question 372 Marks
Find minors and cofactors of the elements of the determinant $\left| {\begin{array}{*{20}{c}} 2&{ - 3}&5 \\ 6&0&4 \\ 1&5&{ - 7} \end{array}} \right|$ and verify that $a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33} = 0.$
AnswerWe have, $\mathrm{M}_{11}=\left|\begin{array}{cc} {0} & {4} \\ {5} & {-7} \end{array}\right|=0-20=-20 ; \mathrm{A}_{11}=(-1)^{1+1}(-20)=-20$
$\mathrm{M}_{12}=\left|\begin{array}{cc} {6} & {4} \\ {1} & {-7} \end{array}\right|=-42-4=-46 ; \quad \mathrm{A}_{12}=(-1)^{1+2}(-46)=46$
$\mathrm{M}_{13}=\left|\begin{array}{cc} {6} & {0} \\ {1} & {5} \end{array}\right|=30-0=30 ; \quad \mathrm{A}_{13}=(-1)^{1+3}(30)=30$
$\mathrm{M}_{21}=\left|\begin{array}{cc} {-3} & {5} \\ {5} & {-7} \end{array}\right|=21-25=-4 ; \quad \mathrm{A}_{21}=(-1)^{2+1}(-4)=4$
$\mathrm{M}_{22}=\left|\begin{array}{cc} {2} & {5} \\ {1} & {-7} \end{array}\right|=-14-5=-19 ; \quad \mathrm{A}_{22}=(-1)^{2+2}(-19)=-19$
$\mathrm{M}_{23}=\left|\begin{array}{cc} {2} & {-3} \\ {1} & {5} \end{array}\right|=10+3=13 ; \quad \mathrm{A}_{23}=(-1)^{2+3}(13)=-13$
$\mathbf{M}_{31}=\left|\begin{array}{cc} {-3} & {5} \\ {0} & {4} \end{array}\right|=-12-0=-12 ; \quad \mathrm{A}_{31}=(-1)^{3+1}(-12)=-12$
$\mathrm{M}_{32}=\left|\begin{array}{cc} {2} & {5} \\ {6} & {4} \end{array}\right|=8-30=-22 ;$$ \quad \mathrm{A}_{32}=(-1)^{3+2}(-22)=22$
and $\mathrm{M}_{33}=\left|\begin{array}{cc} {2} & {-3} \\ {6} & {0} \end{array}\right|=0+18=18;$ $A_{33}=(-1)^{3+3}(18)=18$
Now $a_{11} = 2, a_{12} = -3, a_{13} = 5; A_{31} = -12, A_{32} = 22, A_{33} = 18$
So, $a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33}$
$= 2 (-12) + (-3) (22) + 5 (18) = -24 - 66 + 90 = 0$
View full question & answer→Question 382 Marks
Find minors and cofactors of the elements $a_{11}, a_{21}$ in the determinant $\Delta=\left|\begin{array}{lll} {a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}} \end{array}\right|$
AnswerBy definition of minors and cofactors, we have
Minor of $a_{11} = M_{11} = \left|\begin{array}{ll} {a_{22}} & {a_{23}} \\ {a_{32}} & {a_{33}} \end{array}\right| = a_{22}\ a_{33} – a_{23}\ a_{32}$
Cofactor of $a_{11} = A_{11} = (–1)^{1+1} M_{11} = a_{22}\ a_{33} – a_{23}\ a_{32}$
Minor of $a_{21} = M_{21} = \left|\begin{array}{ll} {a_{12}} & {a_{13}} \\ {a_{32}} & {a_{33}} \end{array}\right| = a_{12} a_{33} – a_{13} a_{32}$
Cofactor of $a_{21} = A_{21} = (–1)^{2+1}\ M_{21} = (–1) (a_{12}\ a_{33} – a_{13}\ a_{32}) = – a_{12}\ a_{33} + a_{13}\ a_{32}$
View full question & answer→Question 392 Marks
Evaluate $\left|\begin{array}{rr} {2} & {4} \\ {-1} & {2} \end{array}\right|$
AnswerWe have $\left|\begin{array}{cc} {2} & {4} \\ {-1} & {2} \end{array}\right|$ = 2 (2) – 4(–1) = 4 + 4 = 8.
View full question & answer→