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- Question 2, Exercise 2.3
- 1 & 2 \\ 0 & 0 & 1 \end{matrix} \right. \right]\text{ by }R_1+3R_3\text{ and }R_2+2R_3 \\ \underset{\sim}{R}&\left[\begin{matrix} 1 & 4 & 14 \\ 0 & 5 & 6 ... 1 & 2 \\ 1 & 0 & 4 \end{matrix} \right. \right]\text{ by }R_3+R_1\\ \underset{\sim}{R}&\left[\begin{ma... -1 & 2 \\ 1 & 0 & 4 \end{matrix} \right. \right]\text{by}-R_2+R_3\\ \underset{\sim}{R}&\left[\begin{ma
- Question 1, Exercise 2.3
- -1 \\ 0 & -5 & 6 \\ 0 & -5 & -2 \end{bmatrix} \text{ by } R_2-2R_1 \text{ and } R_3-3R_1 \\ \underset{\sim}{R}&\begin{bmatrix} 1 & 3 & -1 \\ 0 & -5 & 6 \\ 0 & -5 & -2 \end{bmatrix} \text{ by } R_3-R_2 \\ \underset{\sim}{R}&\begin{bmatri... 1 & 9 \\ 3 & \quad 1 & 3 & \quad 2 \end{bmatrix}\text{ by } R_1\leftrightarrow R_2\\ \underset{\sim}{R}
- Question 5, Exercise 2.2
- & 2-3b+3b & 3-3c+3c \\ 4 & 5 & 6 \end{vmatrix} \text{ by } R_2-3R_1 \\ &=\begin{vmatrix} a & b & c \\... a & b & c\\ a+b+c & b+c+a & c+a+b \end{vmatrix} \text{ by } R_3+R_2 \\ &=(a+b+c)\begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ 1 & 1 & 1 \end{vmatrix} \text{taking common from }R_3 \\ &=(a+b+c)0 \text{ by } R_1\simeq R_3 \\ &=0=R.H.S.\end{align} =====Questi
- Question 3, Exercise 2.3
- & -2 \\ 0 & 2 & 5 \\ 0 & 2 & 1 \end{bmatrix} \text{ by }R_2-2R_1 \text{ and } R_1-2R_3\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 0 & -2 \\ 0 & 0 & 4 \\ 0 & 2 & 1 \end{bmatrix}\text{ by }R_2-R_3\end{align} The last matrix is the ec... & -2 \\ 0 & 2 & 1 \\ 3 & 1 & -4 \end{bmatrix} \text{ by }R_1\leftrightarrow R_3\\ \underset{\sim}{R}&
- Question 4, Exercise 2.3
- 4 & 5 & 6 & 7 \\ 9 & 10 & 11 & 12 \end{bmatrix} \text{by}R_2-R_1\\ \underset{\sim}{R}&\begin{bmatrix} 1... 4 & 5 & 6 & 7 \\ 9 & 10 & 11 & 12 \end{bmatrix} \text{by}R_1\leftrightarrow R_2\\ \underset{\sim}{R}&\b... \\ 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 \end{bmatrix} \text{by}R_2-2R_1\text{,}R_3-4R_1 \text{and} R_4-9R_1\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 1 & 1 & 1 \\
- Question 5 & 6, Exercise 2.1
- 1 \\ \end{matrix} \right]$$ This gives $$ 3a=-2 \text{ and } 2b=3,$$ $$\implies a=-\dfrac{2}{3} \text{ and } b=\dfrac{3}{2}.$$ =====Question 6(i)===== Solv... end{matrix} \right].\end{align} ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit02:ex2-1-p4 |< Question 4]]</btn></text> <text align="right"><btn type="success">[[math-1
- Question 2, Exercise 2.1
- & 1 & 11 \end{bmatrix}\end{align} ====Go To==== <text align="left"><btn type="primary">[[ex2-1-p1 |< Question 1]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit02:ex2-1-p3|Question 3 >]]</btn></text>
- Question 3, Exercise 2.1
- (2), we have $$A(B-C)=AB-BC.$$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit02:ex2-1-p2 |< Question 2]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit02:ex2-1-p4|Question 4 >]]</btn></text>
- Question 4, Exercise 2.1
- frac{1}{3}A^2-2A-9I=0 \end{align} ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit02:ex2-1-p3 |< Question 3]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit02:ex2-1-p5|Question 5 & 6 >]]</btn></text>
- Question 7, Exercise 2.1
- (2), we have $(A+B)^t=A^t+B^t.$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit02:ex2-1-p5 |< Question 5 & 6]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit02:ex2-1-p7|Question 8 >]]</btn></text>
- Question 8, Exercise 2.1
- trix} \right]$$ $$AA^t\ne A^tA$$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit02:ex2-1-p6 |< Question 7]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit02:ex2-1-p8|Question 9 >]]</btn></text>
- Question 9, Exercise 2.1
- ght)}^{t}}={{B}^{t}}{{A}^{t}}$$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit02:ex2-1-p7 |< Question 8]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit02:ex2-1-p9|Question 10 >]]</btn></text>
- Question 10, Exercise 2.1
- \right]$$ $$( A+B )^t=A^t+B^t$$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit02:ex2-1-p8 |< Question 9]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit02:ex2-1-p10|Question 11 >]]</btn></text>
- Question 11, Exercise 2.1
- \right]$$ $$( A+B )^t=-( A+B )$$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit02:ex2-1-p9 |< Question 10]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit02:ex2-1-p11|Question 12 >]]</btn></text>
- Question 12, Exercise 2.1
- ight]$$ $$( A-A^t )^t=-( A-A^t)$$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit02:ex2-1-p10 |< Question 11]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit02:ex2-1-p12|Question 13 >]]</btn></text>