Question 12 Marks
Given $\left[\begin{array}{cc}2 & 1 \\ -3 & 4\end{array}\right] X=\left[\begin{array}{l}7 \\ 6\end{array}\right]$ write the order of matrix $x$
AnswerLet the order of matrix $x$ be $a \times b$
$\therefore\left[\begin{array}{cc}2 & 1 \\ -3 & 4\end{array}\right]_{2 \times 2} \times X_{a \times b}=\left[\begin{array}{l}7 \\ 6\end{array}\right]_{2 \times 1}$
$\Rightarrow a = 2$ and $b =1$
$\therefore$ The order of the matrix $x = ax , b = 2x \ 1$
View full question & answer→Question 22 Marks
If $[x, y]\left[\begin{array}{l}x \\ y\end{array}\right]=[25]$ and $\left[\begin{array}{ll}-x & y\end{array}\right]\left[\begin{array}{c}2 x \\ y\end{array}\right]=[-2]$ find $x$ and $y$ if $x, y \in Z ($integer$)$
Answer$[x, y]\left[\begin{array}{l}x \\ y\end{array}\right]=[25]$
$x^2+y^2=25$
and
$-2 x^2+y^2=-2$
$x, y \in Z ($interger$)$
If can observed that that above two equation are satisfied when $x= \pm 3$ and $\mathrm{y}=\pm 4$
View full question & answer→Question 32 Marks
Given $A=\left[\begin{array}{ccc}0 & 4 & 6 \\ 3 & 0 & -1\end{array}\right]$ and $B=\left[\begin{array}{cc}0 & 1 \\ -1 & 2 \\ -5 & -6\end{array}\right]$ find if possible $BA$
Answer$\begin{array}{l}B A=\left[\begin{array}{cc}0 & 1 \\ -1 & 2 \\ -5 & -6\end{array}\right]\left[\begin{array}{ccc}0 & 4 & 6 \\ 3 & 0 & -1\end{array}\right] \end{array} $
$ =\left[\begin{array}{ccc}0+3 & 0=0 & 0-1 \\ 0+6 & -4+0 & -6-2 \\ 0-18 & -20-0 & -30+6\end{array}\right] $
$ =\left[\begin{array}{ccc}3 & 0 & -1 \\ 6 & -4 & -8 \\ -18 & -20 & -24\end{array}\right]$
View full question & answer→Question 42 Marks
Given $A=\left[\begin{array}{ccc}0 & 4 & 6 \\ 3 & 0 & -1\end{array}\right]$ and $B=\left[\begin{array}{cc}0 & 1 \\ -1 & 2 \\ -5 & -6\end{array}\right]$ find if possible $AB$
Answer$\begin{array}{l}A B=\left[\begin{array}{ccc}0 & 4 & 6 \\ 3 & 0 & -1\end{array}\right]\left[\begin{array}{cc}0 & 1 \\ -1 & 2 \\ -5 & -6\end{array}\right] \end{array}$
$ =\left[\begin{array}{cc}0-4-30 & 0+8-36 \\ 0-0+5 & 3+0+6\end{array}\right] $
$ =\left[\begin{array}{cc}-34 & -28 \\ 5 & 9\end{array}\right]$
View full question & answer→Question 52 Marks
Find $x$ and $y$ if $\left[\begin{array}{cc}x & 0 \\ -3 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & y\end{array}\right]=\left[\begin{array}{cc}2 & 2 \\ -3 & -2\end{array}\right]$
Answer$\begin{array}{l}{\left[\begin{array}{cc}x & 0 \\ -3 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & y\end{array}\right]=\left[\begin{array}{cc}2 & 2 \\ -3 & -2\end{array}\right]}\end{array} $
$ {\left[\begin{array}{cc}x+0 & x+0 \\ -3+0 & -3+y\end{array}\right]=\left[\begin{array}{cc}2 & 2 \\ -3 & -2\end{array}\right]} $
$ {\left[\begin{array}{cc}x & x \\ -3 & -3+y\end{array}\right]=\left[\begin{array}{cc}2 & 2 \\ -3 & -2\end{array}\right]}$
Comparing the corresponding elements we get
$x = 2$
$-3+y=-2 $
$\Rightarrow y=1$
View full question & answer→Question 62 Marks
Find $x$ and $y$ if $\left[\begin{array}{cc}4 & 3 x \\ x & -2\end{array}\right]\left[\begin{array}{l}5 \\ 1\end{array}\right]=\left[\begin{array}{l}y \\ 8\end{array}\right]$
Answer$\begin{array}{l}{\left[\begin{array}{cc}4 & 3 x \\ x & -2\end{array}\right]\left[\begin{array}{l}5 \\ 1\end{array}\right]=\left[\begin{array}{l}y \\ 8\end{array}\right]} \end{array} $
$ {\left[\begin{array}{c}20+3 x \\ 5 x-2\end{array}\right]=\left[\begin{array}{l}y \\ 8\end{array}\right]}$
Comparing the corresponding elements we get
$5 x-2=8 $
$\Rightarrow x=2$
$20 + 3x = y $
$\Rightarrow y = 20 + 6 $
$= 26$
View full question & answer→Question 72 Marks
If $P=\left[\begin{array}{ll}2 & 6 \\ 3 & 9\end{array}\right]$ and $Q=\left[\begin{array}{ll}3 & x \\ y & 2\end{array}\right]$ find $x$ and $y$ such that $P Q=$ null matrix
Answer$P Q=[(2,6),(3,9)][(3, x),(y, 2)]=[(6+6 y, 2 x+12),(9+9 y, 3 x+18)]^{\prime}$
PQ = Null matrix
∴ [(6 + 6y,2x + 12),(9 + 9y, 3x + 18)] = [(0,0),(0,0)]`
Comparing the corresponding elements we get
2x + 12 = 0
Therefore x = -6
6 + 6y = 0
Therefore y = -1
View full question & answer→Question 82 Marks
If $A=\left[\begin{array}{cc}0 & 2 \\ 5 & -2\end{array}\right], B=\left[\begin{array}{cc}1 & -1 \\ 3 & 2\end{array}\right]$ and $I$ is a unit matrix of order $2 \times 2$ find $A^2$
Answer$\begin{array}{l}A^2=\left[\begin{array}{cc}0 & 2 \\ 5 & -2\end{array}\right]\left[\begin{array}{cc}0 & 2 \\ 5 & -2\end{array}\right] \end{array} $
$ =\left[\begin{array}{cc}0+10 & 0-4 \\ 0-10 & 10+4\end{array}\right] $
$ =\left[\begin{array}{cc}10 & -4 \\ -10 & 14\end{array}\right]$
View full question & answer→Question 92 Marks
If $A=\left[\begin{array}{cc}0 & 2 \\ 5 & -2\end{array}\right], B=\left[\begin{array}{cc}1 & -1 \\ 3 & 2\end{array}\right]$ and $I$ is unit matrix of order $2 \times 2$ Find $IB$
Answer$\begin{array}{l}I B=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 3 & 2\end{array}\right] \end{array} $
$ =\left[\begin{array}{cc}1+0 & -1+0 \\ 0+3 & 0+2\end{array}\right] $
$=\left[\begin{array}{cc}1 & -1 \\ 3 & 2\end{array}\right]=B$
View full question & answer→Question 102 Marks
If $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right], B=\left[\begin{array}{ll}6 & 1 \\ 1 & 1\end{array}\right]$ and $C=\left[\begin{array}{cc}-2 & -3 \\ 0 & 1\end{array}\right]$
find of the following and state if they are equal A + CB
Answer$C B=\left[\begin{array}{cc}-2 & -3 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}6 & 1 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}-12-3 & -2-3 \\ 0+1 & 0+1\end{array}\right]=\left[\begin{array}{cc}-15 & -5 \\ 1 & 1\end{array}\right]$
$A+C B=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]+\left[\begin{array}{cc}-15 & -5 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}-14 & -3 \\ 4 & 5\end{array}\right]$
Thus $C A+B \neq A+C B$
View full question & answer→Question 112 Marks
If $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right], B=\left[\begin{array}{ll}6 & 1 \\ 1 & 1\end{array}\right]$ and $C=\left[\begin{array}{cc}-2 & -3 \\ 0 & 1\end{array}\right]$
Find the following and state of they are equal CA + B
Answer$C A=\left[\begin{array}{cc}-2 & -3 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]=\left[\begin{array}{cc}-2-9 & -4-12 \\ 0+3 & 0+4\end{array}\right]=\left[\begin{array}{cc}-11 & -16 \\ 3 & 4\end{array}\right]$
$C A+B=\left[\begin{array}{cc}-11 & -16 \\ 3 & 4\end{array}\right]+\left[\begin{array}{cc}6 & 1 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}-5 & -15 \\ 4 & 5\end{array}\right]$
View full question & answer→Question 122 Marks
If $A=\left[\begin{array}{cc}0 & 2 \\ 5 & -2\end{array}\right], B=\left[\begin{array}{cc}1 & -1 \\ 3 & 2\end{array}\right]$ and $I$ is a unit matrix of order $2 \times 2.$ Find $AI.$
Answer$\begin{array}{l}A I=\left[\begin{array}{cc}0 & 2 \\ 5 & -2\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \end{array} $
$ =\left[\begin{array}{cc}0(1)+2(0) & 0(0)+(2)(1) \\ 5(1)+(-2)(0) & (5)(0)+(-2)(1)\end{array}\right] $
$ =\left[\begin{array}{cc}0+0 & 0+2 \\ 5+0 & 0-2\end{array}\right] $
$ =\left[\begin{array}{cc}0 & 2 \\ 5 & -2\end{array}\right]$
View full question & answer→Question 132 Marks
If A and B are any two 2 x 2 matrices such that AB = BA = B and B is not a zero matrix, what can you say about the matrix A?
AnswerAB = BA = B
We know that I x B = B x I = B, where I is the identity matrix.
Hence, A is the identity matrix.
View full question & answer→Question 142 Marks
Given $A=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right]$ Find $A^2-A B+2 B$
Answer$A^2-A B+2 B$
$\begin{array}{l}=\left[\begin{array}{cc}18 & 7 \\ 14 & 11\end{array}\right]-\left[\begin{array}{cc}2 & 1 \\ -4 & 3\end{array}\right]+2\left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right] \end{array} $
$ =\left[\begin{array}{cc}16 & 6 \\ 18 & 8\end{array}\right]+\left[\begin{array}{cc}2 & 0 \\ -4 & 2\end{array}\right] $
$ =\left[\begin{array}{cc}18 & 6 \\ 14 & 10\end{array}\right]$
View full question & answer→Question 152 Marks
If $M=\left[\begin{array}{l}0 \\ 1\end{array}\right]$ and $N=\left[\begin{array}{l}1 \\ 0\end{array}\right]$ show that $3 M+5 N=\left[\frac{5}{3}\right]$
Answer$3M + 5N$
$\begin{array}{l}=3\left[\begin{array}{l}0 \\ 1\end{array}\right]+5\left[\begin{array}{l}1 \\ 0\end{array}\right] \end{array} $
$ =\left[\begin{array}{l}0 \\ 3\end{array}\right]+\left[\begin{array}{l}5 \\ 0\end{array}\right] $
$ =\left[\begin{array}{l}5 \\ 3\end{array}\right]$
View full question & answer→Question 162 Marks
Given $A=\left[\begin{array}{cc}1 & 1 \\ -2 & 0\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]$ Solve for matrix $X: X + 2A = B$
Answer$X+2 A=B$
$X=B-2 A$
$\begin{array}{l}X=\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]-2\left[\begin{array}{cc}1 & 1 \\ -2 & 0\end{array}\right] \end{array} $
$ X=\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]-\left[\begin{array}{cc}2 & 2 \\ -4 & 0\end{array}\right] $
$ X=\left[\begin{array}{cc}0 & -3 \\ 5 & 1\end{array}\right]$
View full question & answer→Question 172 Marks
Given $A=\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right]$ and $A^t$ is its transpose matrix Find $A^t-\frac{1}{3} A$
Answer$A^t-\frac{1}{3} A $
$ =\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right]-\frac{1}{3}\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right] $
$=\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right]-\left[\begin{array}{cc}-1 & 2 \\ 0 & -3\end{array}\right] $
$ =\left[\begin{array}{cc}-2 & -2 \\ 6 & -6\end{array}\right]$
View full question & answer→Question 182 Marks
Given $A=\left[\begin{array}{ll}1 & 4 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{l}-4-1 \\ -3-2\end{array}\right]$ Find the matrix $C$ such that $C + B = \left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
Answer$C+B=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$C=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]-B$
$=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]-\left[\begin{array}{l}-4-1 \\ -3-2\end{array}\right]$
$=\left[\begin{array}{ll}0+4 & 0+1 \\ 0+3 & 0+2\end{array}\right]$
$=\left[\begin{array}{ll}4 & 1 \\ 3 & 2\end{array}\right]$
View full question & answer→Question 192 Marks
Given $A=\left[\begin{array}{ll}1 & 4 \\ 2 & 3\end{array}\right]$ and $B=\left|\begin{array}{l}-4-1 \\ -3-2\end{array}\right|$ Find the matrix $2A + B$
Answer$2\left[\begin{array}{ll}1 & 4 \\ 2 & 3\end{array}\right]+\left[\begin{array}{l}-4-1 \\ -3-2\end{array}\right]$
$=\left[\begin{array}{ll}2 & 8 \\ 4 & 6\end{array}\right]+\left[\begin{array}{l}-4-1 \\ -3-2\end{array}\right]$
$=\left[\begin{array}{ll}2-4 & 8-1 \\ 4-3 & 6-2\end{array}\right]$
$=\left[\begin{array}{rr}-2 & 7 \\ 1 & 4\end{array}\right]$
View full question & answer→Question 202 Marks
Given $A=2\left[\begin{array}{ll}2 & 1 \\ 3 & 0\end{array}\right]-3\left[\begin{array}{ll}1 & 1 \\ 5 & 2\end{array}\right]+\left[\begin{array}{cc}-3 & -1 \\ 0 & 0\end{array}\right] :$ find $A + 2C – B$
Answer$A + 2C – B$
$\begin{array}{l}=\left[\begin{array}{ll}2 & 1 \\ 3 & 0\end{array}\right]+2\left[\begin{array}{cc}-3 & -1 \\ 0 & 0\end{array}\right]+\left[\begin{array}{ll}1 & 1 \\ 5 & 2\end{array}\right] \end{array} $
$ =\left[\begin{array}{ll}2 & 1 \\ 3 & 0\end{array}\right]+\left[\begin{array}{cc}-6 & -2 \\ 0 & 0\end{array}\right]-\left[\begin{array}{ll}1 & 1 \\ 5 & 2\end{array}\right] $
$ =\left[\begin{array}{ll}2-6-1 & 1-2-1 \\ 3+0-5 & 0+0-2\end{array}\right] $
$ =[-5-2] $
$ =[-2-2]$
View full question & answer→Question 212 Marks
Given $A=2\left[\begin{array}{ll}2 & 1 \\ 3 & 0\end{array}\right],-3\left[\begin{array}{ll}1 & 1 \\ 5 & 2\end{array}\right]+\left[\begin{array}{cc}-3 & -1 \\ 0 & 0\end{array}\right]$ Find $2A - 3B + C$
Answer$2A - 3B + C$
$\begin{array}{l}=2\left[\begin{array}{ll}2 & 1 \\ 3 & 0\end{array}\right]-3\left[\begin{array}{ll}1 & 1 \\ 5 & 2\end{array}\right]+\left[\begin{array}{cc}-3 & -1 \\ 0 & 0\end{array}\right] \end{array} $
$ =\left[\begin{array}{ll}4 & 2 \\ 6 & 0\end{array}\right]-\left[\begin{array}{cc}3 & 3 \\ 15 & 6\end{array}\right]+\left[\begin{array}{cc}-3 & -1 \\ 0 & 0\end{array}\right] $
$=\left[\begin{array}{cc}4-3-3 & 2-3-1 \\ 6-15+0 & 0-6+0\end{array}\right] $
$=\left[\begin{array}{l}-2-2 \\ -9-6\end{array}\right]$
View full question & answer→Question 222 Marks
Find $x$ and $y$ if $x\left[\begin{array}{c}-1 \\ 2\end{array}\right]-4\left[\begin{array}{c}-2 \\ y\end{array}\right]=\left[\begin{array}{c}7 \\ -8\end{array}\right]$
Answer$\begin{array}{l}x\left[\begin{array}{c}-1 \\ 2\end{array}\right]-4\left[\begin{array}{c}-2 \\ y\end{array}\right]=\left[\begin{array}{c}7 -8\end{array}\right] \end{array} $
$ {\left[\begin{array}{c}-x \\ 2 x\end{array}\right]-\left[\begin{array}{c}-8 \\ 4 y\end{array}\right]=\left[\begin{array}{c}7 \\ -8\end{array}\right]} $
$ {\left[\begin{array}{c}-x+8 \\ 2 x-4 y\end{array}\right]=\left[\begin{array}{c}7 \\ -8\end{array}\right]}$
Comparing corresponding the elements we get
$-x + 8 = 7$ and $2x - 4y = -8$
Simplifying we get
$x=1$ and $y=\frac{5}{2}=2.5$
View full question & answer→Question 232 Marks
Find x and y if 3[4 x] + 2[y -3] = [10 0]
Answer3[4 x] + 2[y -3] = [10 0]
[12 3x] + [2y -6] = [10 0]
[12 + 2y 3x - 6] = [10 0]
Comparing the corrresponding elements we get
12 + 2y = 10 and 3x - 6 = 0
Simplifying we get y = -1 and x = 2
View full question & answer→Question 242 Marks
Evaluate $6\left[\begin{array}{c}3 \\ -2\end{array}\right]-2\left[\begin{array}{c}-8 \\ 1\end{array}\right]$
Answer$6\left[\begin{array}{c}3 \\ -2\end{array}\right]-2\left[\begin{array}{c}-8 \\ 1\end{array}\right]=\left[\begin{array}{c}18 \\ -12\end{array}\right]-\left[\begin{array}{c}-16 \\ 2\end{array}\right]=\left[\begin{array}{c}18+16 \\ -12-2\end{array}\right]=\left[\begin{array}{c}34 \\ -14\end{array}\right]$
View full question & answer→Question 252 Marks
Evaluate: $\quad 2\left[\begin{array}{ll}-1 & 0 \\ 2-3\end{array}\right]+\left[\begin{array}{ll}3 & 3 \\ \hline 5 & 0\end{array}\right]$
Answer$2\left[\begin{array}{cc}-1 & 0 \\ \hline2-3\end{array}\right]+\left[\begin{array}{ll}3 & 3 \\ 5 & \hline0\end{array}\right]$$=\left[\begin{array}{ll}-2 & 0 \\ \hline 4 & -6\end{array}\right]+\left[\begin{array}{ll}3 & 3 \\ \hline5 & 0\end{array}\right]$$=\left[\frac{-2+3}{4+5} \frac{0+3}{-6+0}]\right.$=$\left[\begin{array}{cc}1 & 3 \\ \hline9 &-6\end{array}\right]$
View full question & answer→Question 262 Marks
If $\left[\begin{array}{cc}1 & 4 \\ -2 & 3\end{array}\right]+2 M=3\left[\begin{array}{cc}3 & 2 \\ 0 & -3\end{array}\right]$ find the matrix $M$
Answer$\left[\begin{array}{cc}1 & 4 \\ -2 & 3\end{array}\right]+2 M=3\left[\begin{array}{cc}3 & 2 \\ 0 & -3\end{array}\right]$
$\begin{array}{l}\Rightarrow 2 M=\left[\begin{array}{cc}9 & 6 \\ 0 & -9\end{array}\right]-\left[\begin{array}{cc}1 & 4 \\ -2 & 3\end{array}\right]=\left[\begin{array}{cc}8 & 2 \\ 2 & -12\end{array}\right] \end{array} $
$ \Rightarrow M=\left[\begin{array}{cc}4 & 1 \\ 1 & -6\end{array}\right]$
View full question & answer→Question 272 Marks
Write the additive inverse of matrices A, B and C Where A = [6, -5]; B =
$\left[\begin{array}{cc}-2 & 0 \\ 4 & -1\end{array}\right]$ and $C=\left[\begin{array}{c}-7 \\ 4\end{array}\right]$
AnswerWe know additive inverse of a matrix is its negative
The additive inverse of A = -A = -[6 -5] = [-6 5]
Additive inverse of B = -B =$-\left[\begin{array}{cc}-2 & 0 \\ 4 & -1\end{array}\right]=\left[\begin{array}{cc}2 & 0 \\ -4 & 1\end{array}\right]$
Additive inverse of C = -C =$-C=-\left[\begin{array}{c}-7 \\ 4\end{array}\right]=\left[\begin{array}{c}7 \\ -4\end{array}\right]$
View full question & answer→Question 282 Marks
Given $M=\left[\begin{array}{cc}5 & -3 \\ -2 & 4\end{array}\right]$ Find its transpose matrix $M^t$. If possible find $M+M^t$
Answer$M=\left[\begin{array}{cc}5 & -3 \\ -2 & 4\end{array}\right] $
$ M^t=\left[\begin{array}{cc}5 & -2 \\ -3 & 4\end{array}\right]$
$M+M^t=\left[\begin{array}{cc}5 & -3 \\ -2 & 4\end{array}\right]+\left[\begin{array}{cc}5 & -2 \\ -3 & 4\end{array}\right]$
$=\left[\begin{array}{cc}5+5 & -3-2 \\ -2-3 & 4+4\end{array}\right]$
$=\left[\begin{array}{cc}10 & -5 \\ -5 & 8\end{array}\right]$
View full question & answer→Question 292 Marks
Find x and y from the given equations:
[-8 x] + [y -2] = [-3 2]
Answer[-8 x] + [y -2] = [-3 2]
$\Rightarrow[-8+y x-2]=[-3,2]$
Equation the corresponding elements we get
-8 + y = -3 and x - 2 = 2
Thus we get x = 4 and y = 5
View full question & answer→Question 302 Marks
Find $x$ and $y$ from the given equations $:\left[\begin{array}{cc}5 & 2 \\ -1 & y-1\end{array}\right]-\left[\begin{array}{cc}1 & x-1 \\ 2 & -3\end{array}\right]=\left[\begin{array}{cc}4 & 7 \\ -3 & 2\end{array}\right]$
Answer$\begin{array}{l}{\left[\begin{array}{cc}5 & 2 \\ -1 & y-1\end{array}\right]-\left[\begin{array}{cc}1 & x-1 \\ 2 & -3\end{array}\right]=\left[\begin{array}{cc}4 & 7 \\ -3 & 2\end{array}\right]} \end{array} $
$ \Rightarrow\left[\begin{array}{cc}5-1 & 2-x+1 \\ -1-2 & y-1+3\end{array}\right]=\left[\begin{array}{cc}4 & 7 \\ -3 & 2\end{array}\right] $
$ \Rightarrow\left[\begin{array}{cc}4 & 3-x \\ -3 & y+2\end{array}\right]=\left[\begin{array}{cc}4 & 7 \\ -3 & 2\end{array}\right]$
Equating the corresponding elements we get
$3 - x = 7$ and $y + 2 = 2$
Thus we get $x = -4$ and $y = 0$
View full question & answer→Question 312 Marks
$A+B-C=\left[\begin{array}{l}2 \\ 5\end{array}\right]-\left[\begin{array}{l}1 \\ 4\end{array}\right]-\left[\begin{array}{c}6 \\ -2\end{array}\right]$$=\left[\begin{array}{l}2+1-6 \\ 5+4+2\end{array}\right]=\left[\begin{array}{c}-3 \\ 11\end{array}\right]$, find
$:A – B +C$
Answer$A-B+C$
$=\left[\frac{2}{5}\right]-\left[\frac{1}{4}\right]+\left[\frac{6}{-2}\right]$
$\left[\frac{2-1+6}{5-4-2}\right]$
$=\left[\frac{7}{-1}\right]$
View full question & answer→Question 322 Marks
If $A=\left[\begin{array}{l}2 \\ 5\end{array}\right], B=\left[\begin{array}{l}1 \\ 4\end{array}\right]$ and $C=\left[\begin{array}{c}6 \\ -2\end{array}\right]$ Find $A + B - C$
Answer$A+B-C$
$=\left[\begin{array}{l}2 \\ 5\end{array}\right]-\left[\begin{array}{l}1 \\ 4\end{array}\right]-\left[\begin{array}{c}6 \\ -2\end{array}\right]$
$=\left[\begin{array}{l}2+1-6 \\ 5+4+2\end{array}\right]$
$=\left[\begin{array}{c}-3 \\ 11\end{array}\right]$
View full question & answer→Question 332 Marks
Given $A=\left[\begin{array}{cc}-1 & 0 \\ 2 & -4\end{array}\right]$ and $B=\left[\begin{array}{cc}3 & -3 \\ -2 & 0\end{array}\right]$ Find the matrix $X$ in following
X - B = A
AnswerX - B = A
X = A + B
$X=\left[\begin{array}{cc}-1 & 0 \\ 2 & -4\end{array}\right]+\left[\begin{array}{cc}3 & -3 \\ -2 & 0\end{array}\right]=\left[\begin{array}{cc}-1+3 & 0-3 \\ 2-2 & -4+0\end{array}\right]=\left[\begin{array}{cc}2 & -3 \\ 0 & -4\end{array}\right]$
View full question & answer→Question 342 Marks
Given $A=\left[\begin{array}{cc}-1 & 0 \\ 2 & 0\end{array}\right]$ and $B=\left[\begin{array}{cc}3 & -3 \\ -2 & 0\end{array}\right]$ find the matrix $X$ in of the following
A- X = B
AnswerA - X = B
X = A - B
$X=\left[\begin{array}{cc}-1 & 0 \\ 2 & -4\end{array}\right]-\left[\begin{array}{cc}3 & -3 \\ -2 & 0\end{array}\right]=\left[\begin{array}{cc}-1-3 & 0+3 \\ 2+2 & -4-0\end{array}\right]=\left[\begin{array}{cc}-4 & 3 \\ 4 & -4\end{array}\right]$
View full question & answer→Question 352 Marks
Given $A=\left[\begin{array}{cc}-1 & 0 \\ 2 & -4\end{array}\right]$ and $B=\left[\begin{array}{cc}3 & -3 \\ -2 & 0\end{array}\right]$ Find the following
$A+X=B$
AnswerA + X = B
X = B - A
$X=\left[\begin{array}{cc}3 & -3 \\ -2 & 0\end{array}\right]-\left[\begin{array}{cc}-1 & 0 \\ 2 & -4\end{array}\right]=\left[\begin{array}{cc}3+1 & -3-0 \\ -2-2 & 0+4\end{array}\right]=\left[\begin{array}{cc}4 & -3 \\ -4 & 4\end{array}\right]$
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Given A = [2 -3], B = [0 2] and C = [-1 4]; Find the matrx X in the following
A - X = B + C
AnswerA - X = B + C
[2 -3] - C = [0 2] + [-1 4]
[2 -3] - X = [0 - 1 2 + 4]
[2 -3] - X = [-1 6]
[2 -3] - [-1 6] = x
X = [2 + 1 -3-6] = [3 -9]
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Given A = [2 -3], B = [0 2] and C = [-1 4]; Find the matrix X in the following:
X + B = C - A
AnswerX + B = C - A
X + [0 2] = [-1 4] - [2 -3]
X + [0 2] = [-1 -2 4+ 3] = [-3 7]
X = [-3 7] - [0 2] = [-3 -0 7 - 2] = [-3 5]
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