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- Question 2, Exercise 2.2
- & 1 & 0 \\-1 & 2 & 0 \end{matrix}\right|=0$. ====Solution==== Given $$\left| \begin{matrix} 1 & 2 & 0 \... & -12 \\2 & -1 & 3 \end{matrix} \right|=0$. ====Solution==== Given $$\left| \begin{matrix} 1 & 2 & 3 \... & -1 & 1 \\-2 & 1 & 4 \end{matrix} \right|$. ====Solution==== Given $$\left| \begin{matrix} 1 & 3 & -2 ... 1 & 1 & 1 \\2 & 4 & 2 \end{matrix} \right|$. ====Solution==== Given $$\left| \begin{matrix}3 & 2 & 0 \\1 &
- Question 6, Exercise 2.2
- b \\c-a & a-b & b-c \end{matrix} \right|=0$ ====Solution==== Let \begin{align} L.H.S&=\left| \begin{matrix... trix} \right|=( a-b )( b-c )( c-a )( a+b+c )$ ====Solution==== Let $$L.H.S.=\left| \begin{matrix} 1 & a &... \end{matrix} \right|=( a-b )( b-c )( c-a )$ ====Solution==== Let $$L.H.S.=\left| \begin{matrix} 1 & a &... & bc & -c^2 \end{matrix} \right|=4a^2b^2c^2$ ====Solution==== \begin{align}L.H.S.&=\left| \begin{matrix}
- Question 1, Exercise 2.1
- rix} 2 \\ 4 \\ 6 \\ \end{matrix} \right]$$ ====Solution==== \begin{align}&\left[ \begin{matrix} 1 & 2 ... ix} 2 & -5 & 7 \\ \end{matrix} \right]$$ ====Solution==== \begin{align}&\left[ \begin{matrix} 1 & -2... ix} 4 \\ 2 \\ \end{matrix} \right]$$ ====Solution==== \begin{align}&\left[ \begin{matrix} 7 & 1 ... 5 \\ 2 & 4 & 6 \\ \end{matrix} \right]$$ ====Solution==== \begin{align}&\left\{ \begin{bmatrix} 1 & 3 \
- Question 4, Exercise 2.2
- 1 & 2 & 1 \\2 & 1 & 1 \end{matrix} \right|.$ ====Solution==== \begin{align}&\left| \begin{matrix} 0 & 1 ... & 4 & -6 \\4 & 2 & 0 \end{matrix} \right|.$ ====Solution==== \begin{align}&\left| \begin{matrix} 3 & 4 ... -5 & 4 \\-9 & 8 & -7 \end{matrix} \right|.$ ====Solution==== \begin{align}&\left| \begin{matrix} 3 & 1 ... 1 & 1 & 0 \\-2 & 3 & 4\end{matrix} \right|.$ ====Solution==== \begin{align}&\left| \begin{matrix} 2 & 1
- Question 5, Exercise 2.2
- & l & x\\b & m & y\\c & n & z \end{vmatrix}$ ====Solution==== \begin{align}L.H.S.&=\begin{vmatrix} a & b & ... & b & c\\1 & 2 & 3\\4 & 5 & 6 \end{vmatrix}.$ ====Solution==== \begin{align}L.H.S.&=\begin{vmatrix} a & b & ... c \\b+c & c+a & a+b \end{matrix} \right|=0$ ====Solution==== \begin{align}L.H.S.&=\begin{vmatrix} 1 & 1 & ... c^2 \\a^3 & b^3 & c^3 \end{matrix} \right|$ ====Solution==== \begin{align}L.H.S.&=\begin{vmatrix} bc & ca
- Question 1, Exercise 2.3
- -1 \\2 & 1 & 4 \\3 & 4 & -5\end{bmatrix}$. ====Solution==== \begin{align}&\begin{bmatrix} 1 & 3 & -1 \\ ... d 1 & 3 & \quad 2\end{bmatrix} \end{align}$. ====Solution==== \begin{align} &\begin{bmatrix} 2 & 3 & -1 & 9... & 1 \\1 & 1 & 2 \\4 & 1 & 7\end{bmatrix}$. ====Solution==== \begin{align}&\begin{bmatrix} 2 & -3 & 1 \\ ... 1 & 1 \\3 & 2 & 3\end{bmatrix}\end{align}$. ====Solution==== \begin{align} &\begin{bmatrix} 1 & 0 & -2 \\
- Question 2, Exercise 2.3
- \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{bmatrix}$$ ====Solution==== Let $$A=\begin{bmatrix} 4 & -2 & 5 \\ 2 & 1 ... \\ -1 & 5 & 1 \\ \end{matrix} \right]$$ ====Solution==== Let $$A=\begin{bmatrix} 3 & -1 & 6 \\ 1 & 3 ... \\ -2 & -2 & 2 \\ \end{matrix} \right]$$ ====Solution==== Let $$A=\begin{bmatrix} 1 & 2 & -3 \\ 0 & -2... \\ 1 & 0 & 2 \\ \end{matrix} \right]$$ ====Solution==== Let $$A=\begin{bmatrix} 1 & 2 & -1 \\ 0 & -1
- Question 3, Exercise 2.1
- that $\left( AB \right)C=A\left( BC \right)$. ====Solution==== Given: $A=\begin{bmatrix}x & y & z\end{bmatri... \end{bmatrix}$. Verify that $A(B+C)=AB+AC$. ====Solution==== Given: $A=\begin{bmatrix} 1 & 3 \\ -1 & 4 ... \end{bmatrix}.$ Verify that $A( B-C )=AB-AC$. ====Solution==== Given: $ A=\begin{bmatrix}1 & 3 \\-1 & 4 \en
- Question 5 & 6, Exercise 2.1
- be symmetric. Find the value of $a$ and $b$. ====Solution==== Given: $A=\begin{bmatrix} 0 & 2b & -2 \\ 3 &... trix}2 & 1 & 1 \\ 3 & -1 & 4 \end{bmatrix}$. ====Solution==== Given $A=\left[ \begin{matrix} 1 & 0 & 3 \\... trix} 4 & 6 & 2 \\ 0 & -4 & 2 \end{bmatrix}$. ====Solution==== Given $A=\left[ \begin{matrix}1 & 2 & 2 \\3
- Question 8,9 & 10, Exercise 2.2
- \\x & y & 1+z \end{matrix} \right|=1+x+y+z$ ====Solution==== Let $$L.H.S.=\left| \begin{matrix} 1+x & y... \end{matrix} \right|=( x-p )( x-q )( x+p+q )$ ====Solution==== Let $$L.H.S.=\left| \begin{matrix} x & p &... ( 1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} )$ ====Solution==== Let $$L.H.S.=\left| \begin{matrix} 1+a & 1
- Question 11, Exercise 2.2
- \6 & 2 & -2 \\5 & 1 & 1\end{matrix} \right]$ ====Solution==== Let $$A=\left[ \begin{matrix} 7 & 1 & 3 \... & -2 & 1 \\-2 & -3 & 2 \end{matrix} \right]$ ====Solution==== Let $$A=\left[ \begin{matrix} 1 & -1 & 1 ... & 6 & -3 \\-1 & 0 & 1 \end{matrix} \right]$ ====Solution==== Let $$A=\left[ \begin{matrix} 3 & 2 & -3
- Question 13, Exercise 2.2
- & -1 & 1 \\0 & 4 & 5 \end{matrix} \right|=9$ ====Solution==== Given $$\left| \begin{matrix} x & 2 & 3 \... & 1 & x \\2 & 3 & 4 \end{matrix} \right|=-6$ ====Solution==== Given $$\left| \begin{matrix} -1 & 0 & 1 ... x+3 & 4 \\2 & 3 & x+4\end{matrix} \right|=0$ ====Solution==== Given $$\left| \begin{matrix} x+2 & 3 & 4
- Question 8, Exercise 2.1
- & 4 \end{bmatrix}$, show that $( A^t )^t=A$. ====Solution==== Given $$A=\left[ \begin{matrix} 1 & 2 & 0 ... & 4\end{bmatrix}$, show that $AA^t\ne A^tA$. ====Solution==== $$A=\left[ \begin{matrix} 1 & 2 & 0 \\
- Question 9, Exercise 2.1
- \end{bmatrix}$, show that $( AB )^t=B^tA^t$. ====Solution==== $$A=\left[ \begin{matrix} 2 & -1 & 3 \\ ... 2 \end{bmatrix}$, show that $( AB)^t=B^tA^t$. ====Solution==== $$A=\left[ \begin{matrix} \quad 1 & 2 & 0
- Question 12, Exercise 2.1
- d{bmatrix}$. Verify that$A+A^t$ is symmetric. ====Solution==== $$A=\left[ \begin{matrix} 3 & 2 & 1 \\ ... trix}$. Verify that$A-A^t$ is skew symmetric. ====Solution==== $$A=\left[ \begin{matrix} 3 & 2 & 1 \\