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- Question 1, Exercise 2.2
- h is $a_{i j}=\dfrac{i+3 j}{2}$ ** Solution. ** Given \( a_{ij} = \dfrac{i + 3j}{2} \). For \( i = ... 4 \end{array}\right] \] **Alternative Method:** Given \( a_{ij}=\dfrac{i+3j}{2} \). So we have \begin... $a_{i j}=\dfrac{i \times j}{2}$ ** Solution. ** Given \( a_{ij}=\dfrac{i \times j}{2} \). So we have \b... s $a_{i j}=\dfrac{2 i-3 j}{3}$ ** Solution. ** Given \( a_{ij} = \frac{2i - 3j}{3} \), we need to find
- Question 4, Exercise 2.3
- (i)===== Find the value of $\lambda$, so that the given matrix is singular $\left[\begin{array}{lll}\lamb... ii)===== Find the value of $\lambda$, so that the given matrix is singular $\left[\begin{array}{lll}\lamb... ii)===== Find the value of $\lambda$, so that the given matrix is singular $\left[\begin{array}{lll}\lamb... iv)===== Find the value of $\lambda$, so that the given matrix is singular $\left[\begin{array}{ccc}2+i &
- Question 7, Exercise 2.2
- )}{x-1} & 1\end{array}\right]$. ** Solution. ** Given: $$A = \begin{bmatrix} x & 0 \\ y & 1 \end{bmatrix}.$$ We use mathematical induction to prove the given fact. For C-1, put $n = 1$ \begin{align}A^1 =\beg... \end{align} C-1 is satisfied. \\ For C-2, suppose given statement is true for $n=k$. \begin{align} A^k =
- Question 3, Exercise 2.6
- \ $2 x+y+z=5$\\ $3 x-2 y+z=-3$\\ ** Solution. ** Given the system of equations: \begin{align*} \begin{al... x+2 y+6 z=1$\\ $3 x-4 y-5 z=3$\\ ** Solution. ** Given the system of equations: \begin{align*} 5x - 2y +... x+z=2$\\ $2 y-z=3$\\ $x+3 y=5$\\ ** Solution. ** Given the system of equations: \begin{align*} 2x + z &=
- Question 3, Exercise 2.2
- matrix $C$ such that: $A+B+C=0$ ** Solution. ** Given $$A+B+C=0,$$ this given $$C=-A-B.$$ Thus \begin{align*} C&=-\begin{bmatrix}3 & -1 & 2 \\ 0 & 6 & 1 \\ -
- Question 3, Exercise 2.2
- matrix $C$ such that: $A+B+C=0$ ** Solution. ** Given $$A+B+C=0,$$ this given $$C=-A-B.$$ Thus \begin{align*} C&=-\begin{bmatrix}3 & -1 & 2 \\ 0 & 6 & 1 \\ -
- Question 4, Exercise 2.2
- array}{cc} a & b \\ c & d \end{array}\right] $ is given by: \begin{align*} \left[\begin{array}{cc} a & ... es of $z, t$ and $x^{2}+y^{2}$. ** Solution. ** Given $\begin{bmatrix}x y & 4 \\ 0 & x+y\end{bmatrix}=\
- Question 10, Exercise 2.2
- d $B A=A$. Find $A^{2}+B^{2}$ ** Solution. ** Given $$AB = B$$ and $$BA = A$$ \begin{align*} A^2 &=... {align*} Now, $$A^2 + B^2 = A + B$$ Therefore, given the conditions $AB = B$ and $BA = A$, we find tha
- Question 11, Exercise 2.2
- =a+{ji}$ and skew-symmetric if $a_{ij}=-a_{ji}$. Given $a_{i j}=i^{2}-j^{2}$, then \begin{align} a_{ji} ... j} \end{align} This gives $a_{ij}=-a_{ji}$, hence given matrix is skew-symmetric. ====Go to ==== <text
- Question 5, Exercise 2.3
- 1 & -2 & -1\end{array}\right]$. ** Solution. ** Given \begin{align*} A &= \left[\begin{array}{ccc} 1 &... 1 & 0 & 4 i\end{array}\right]$. ** Solution. ** Given \begin{align*} B &= \left[\begin{array}{ccc} i &
- Question 7, Exercise 2.3
- 2 \\ 0 & 2\end{array}\right]$. ** Solution. ** Given: \begin{align*} A &= \left[\begin{array}{ll}2 & 1... e, $(AB)^{-1} = B^{-1}A^{-1}$ is verified for the given matrices \(A\) and \(B\). =====Question 7(ii)==
- Question 7 and 8, Exercise 2.6
- x-y+3 z=4 ; \quad x+2 y-3 z=0$. ** Solution. ** Given \begin{align*} A &= \begin{bmatrix} 3 & 2 & 1 \\ ... rac{-5}{62} & \dfrac{-11}{62} \end{bmatrix}$$ Now given the system of equations: \begin{align*} 3x + 4y +
- Question 5, Exercise 2.2
- n prove that $X^{2}-4 X-5 I=0$. ** Solution. ** Given the matrix \begin{align}L.H.S. & =X^{2}-4 X-5 I
- Question 6, Exercise 2.2
- that, $A^{2}+\alpha I=\beta A$. ** Solution. ** Given the matrix \begin{align} & A^{2}+\alpha I=\beta
- Question 13, Exercise 2.2
- 1 & -3 & 0\end{array}\right]$. ** Solution. ** Given: \begin{align*} 2X - Y = \begin{pmatrix} 1 & 6 &