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- Question 3, Exercise 2.1
- prove that $$(AB)C=A(BC)$$ First, we take \begin{align} AB&=\begin{bmatrix}x & y & z\end{bmatrix}\begin{... {{y}^{2}}+2fyz+c{{z}^{2}} \right] \ldots (1) \end{align} Now we take \begin{align}BC&=\begin{bmatrix}a & h & g\\h & b & f\\g & f & c \end{bmatrix}\begin{bmatrix... ix}ax+hy+gz\\hx+by+fz\\gx+fy+cz \end{bmatrix}\end{align} \begin{align} R.H.S &= A(BC)\\ &=\begin{bmatrix}
- Question 2, Exercise 2.3
- 1 & 0 \\ -1 & 2 & 3 \end{bmatrix}.$$ Then \begin{align}|A|&=\begin{vmatrix}4 & -2 & 5 \\ 2 & 1 & 0 \\ ... ix}\\ &=4(3)+2(6)+5(4+1) \\ & =49\neq 0. \end{align} This gives, $A$ is non-singular and $A^{-1}$ exists. Now \begin{align} & \left[\begin{matrix} 4 & -2 & 5 \\ 2 & 1 & 0 ... ht]\text{ by }R_1+30R_3\text{ and } R_2-11R_3\end{align} Thus we have \begin{align} A^{-1}&=\begin{bmatri
- Question 1, Exercise 2.3
- 3 & 4 & -5\end{bmatrix}$. ====Solution==== \begin{align}&\begin{bmatrix} 1 & 3 & -1 \\ 2 & 1 & 4 \\ 3 &... \\ 0 & -5 & 6 \\ 0 & 0 & -8 \end{bmatrix}\end{align} =====Question 1(ii)===== Reduce the matrices to the reduce echelon form: $\begin{align} \begin{bmatrix}2 & 3 & -1 & 9 \\1 & -1 & 2 & -3 \\3 & \quad 1 & 3 & \quad 2\end{bmatrix} \end{align}$. ====Solution==== \begin{align} &\begin{bmatri
- Question 4, Exercise 2.2
- 1 \end{matrix} \right|.$ ====Solution==== \begin{align}&\left| \begin{matrix} 0 & 1 & 3 \\ -1 & 2... \right)+3\left( -1-4 \right)\\ =&0+3-15=-12 \end{align} =====Question 4(ii)===== Evaluate the determina... 0 \end{matrix} \right|.$ ====Solution==== \begin{align}&\left| \begin{matrix} 3 & 4 & -2 \\ 2 & 4... right)-2\left( 4+16 \right) \\ =&36+96-80=52 \end{align} =====Question 4(iii)===== Evaluate the determi
- Question 5, Exercise 2.2
- \c & n & z \end{vmatrix}$ ====Solution==== \begin{align}L.H.S.&=\begin{vmatrix} a & b & c \\ l & m & n ... & m & y \\ c & n & z \end{vmatrix} =R.H.S. \end{align} =====Question 5(ii)===== Show that $\begin{vm... 4 & 5 & 6 \end{vmatrix}.$ ====Solution==== \begin{align}L.H.S.&=\begin{vmatrix} a & b & c\\1-3a & 2-3b & ... \1 & 2 & 3 \\4 & 5 & 6 \end{vmatrix}=R.H.S. \end{align} =====Question 5(iii)===== Show that $\left| \b
- Question 1, Exercise 2.1
- \\ \end{matrix} \right]$$ ====Solution==== \begin{align}&\left[ \begin{matrix} 1 & 2 & 4 \\ \end{matr... \right]\\ &=\left[ 14+20+64 \right]\\ &=[98].\end{align} =====Question 1(ii)===== Express as a singl... \ \end{matrix} \right]$$ ====Solution==== \begin{align}&\left[ \begin{matrix} 1 & -2 & 3 \\ \end{mat... } -21 & 15 & -10 \\ \end{matrix} \right] \end{align} =====Question 1(iii)===== Express as a single m
- Question 4, Exercise 2.1
- 1 & 4 \\ 4 & 4 & 1 \end{bmatrix}$.\\ Now \begin{align}\frac{1}{3}A^2&=\frac{1}{3}\left[ \begin{matrix} ... 8 \\ 8 & 8 & 11 \\ \end{matrix} \right]\end{align} Now we take $$2A=\left[ \begin{matrix} 2 & 8 ... 8 & 8 & 2 \\ \end{matrix} \right]$$ and \begin{align}9I&=9\left[ \begin{matrix} 1 & 0 & 0 \\ 0 ... & 0 \\ 0 & 0 & 9 \\ \end{matrix} \right]\end{align} Now, we have \begin{align}&\dfrac{1}{3}A^2-2A-9I
- Question 7, Exercise 2.1
- \,4 & -1 \\ \end{matrix} \right].$$ Now \begin{align}A+B=\left[ \begin{matrix}1 & 0 & -1 & 2 \\ 3 & ... 9 \\3 & -1 & 3 & 5 \\\end{matrix} \right]\end{align} Now \begin{align}(A+B)^t=\left[ \begin{matrix}3 & 4 & 3 \\-1 & 4 & -1\\ 2 & 1 & 3 \\ 3 & 9 & 5 \\ \end{matrix} \right] ... (1)\end{align} Also \begin{align}A^t+B^t&=\left[ \begin{matrix
- Question 5 & 6, Exercise 2.1
- & -2 & 8 \\ \end{matrix} \right].$$ Thus \begin{align}X&=\left[ \begin{matrix}3 & 0 & 9 \\6 & 6 & 3 \... & 11 \\0 & 4 & 11 \\ \end{matrix} \right]\end{align} =====Question 6(ii)===== Solve the matrix equa... \0 & -2 & 1 \\ \end{matrix} \right]$$ So \begin{align}X&=\left[\begin{matrix}1 & 2 & 2 \\3 & -1 & 2 \... & 3 \\3 & -3 & 3 \\ \end{matrix} \right].\end{align} ====Go To==== <text align="left"><btn type="pr
- Question 3, Exercise 2.2
- & a_{32} & a_{33} \\ \end{bmatrix}$$ Then \begin{align}|A|&=a_{11} \left( a_{22} a_{33}-a_{23} a_{32} \r... 2}a_{21}a_{33}-a_{13}a_{22}a_{31} \ldots (1) \end{align} Now $$ {{A}^{t}}=\left[ \begin{matrix} {{a}_{... }_{33}} \\ \end{matrix} \right]\\ $$ Then \begin{align} |A^t|&=a_{11}\left( a_{22}a_{33}-a_{32}a_{23} \r... 2}a_{21}a_{33}-a_{13}a_{22}a_{31} \ldots (2) \end{align} Now comparing (1) and (2), we have $$|A|=|{{A}^{
- Question 6, Exercise 2.2
- nd{matrix} \right|=0$ ====Solution==== Let \begin{align} L.H.S&=\left| \begin{matrix} a-b & b-c & c-a \\... -R_1 \\ &=0 \quad R_1 \cong R_3 \\ &=R.H.S. \end{align} =====Question 6(ii)===== Prov that $\left| \be... trix} \right|=4a^2b^2c^2$ ====Solution==== \begin{align}L.H.S.&=\left| \begin{matrix} -a^2 & ab & ac ... +2a^2b^2c^2+2a^2b^2c^2\\ &=4a^2b^2c^2=R.H.S. \end{align} =====Question 6(v)===== Prov that $\left| \
- Question 3, Exercise 2.3
- \ \end{matrix} \right]$$ ====Solution==== \begin{align}&\begin{bmatrix} 1 & 0 & -2 \\ 2 & 2 & 1 \\ -1 ... \\ 0 & 2 & 1 \end{bmatrix}\text{ by }R_2-R_3\end{align} The last matrix is the echelon form of given mat... \\ \end{matrix} \right]$$ ====Solution==== \begin{align}&\begin{bmatrix} 3 & 1 & -4 \\ 0 & 2 & 1 \\ 1 &... \ 0 & 0 & 0 \end{bmatrix} \text{ by }R_3-2R_2\end{align} The last matrix is the echelon form of given mat
- Question 2, Exercise 2.1
- rix}0 & 4 & -8\\0 & -4 & -4\end{bmatrix}$. \begin{align}&2A+3B-4C\\&=\begin{bmatrix} 4 & -10 & 2\\ 6 & 0... bmatrix}7 & -20 & 1\\6 & 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:uni
- Question 8, Exercise 2.1
- & 4 \\ \end{matrix} \right]$$ This gives \begin{align}( A^t )^t&=\left[ \begin{matrix}1 & 2 & 0 \\3 & ... \end{matrix} \right]\\ \implies( A^t)^t&=A. \end{align} =====Question 8(ii)===== If $A=\begin{bmatrix}1... \right]$$ $$AA^t\ne A^tA$$ ====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:uni
- Question 18, Exercise 2.2
- 22}b_{21})(a_{11}b_{12}+a_{12}b_{22} )$$ $$\begin{align} & =a_{11}b_{11}a_{21}b_{12}+a_{11}b_{11}a_{22}b_... 1}a_{11}b_{12}+a_{22}b_{21}a_{12}b_{22}) \\ \end{align}$$ $$=a_{11}b_{11}a_{22}b_{22}+a_{12}b_{21}a_{21}... $$(AB)^{-1}=B^{-1}A^{-1}$$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit... x2-2-p13 |< Question 16 & 17]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:uni