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Question 2, Exercise 2.2
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& 2 & 0 \end{matrix}\right|=0$. ====Solution==== Given $$\left| \begin{matrix} 1 & 2 & 0 \\ 3 & 1... -1 & 3 \end{matrix} \right|=0$. ====Solution==== Given $$\left| \begin{matrix} 1 & 2 & 3 \\ -8 & ... 2 & 1 & 4 \end{matrix} \right|$. ====Solution==== Given $$\left| \begin{matrix} 1 & 3 & -2 \\ 3 & ... 2 & 4 & 2 \end{matrix} \right|$. ====Solution==== Given $$\left| \begin{matrix}3 & 2 & 0 \\1 & 1 & -3 \
Question 5 & 6, Exercise 2.1
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2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix}$ is given to be symmetric. Find the value of $a$ and $b$. ====Solution==== Given: $A=\begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\... & 3 & -1 \\ \end{matrix} \right]$$ Since $A$ is given to be symmetirc, $A^t=A$, implies $$\left[ \begin... 1 \\ 3 & -1 & 4 \end{bmatrix}$. ====Solution==== Given $A=\left[ \begin{matrix} 1 & 0 & 3 \\ -2 & 2 &
Question 3, Exercise 2.3
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end{align} The last matrix is the echelon form of given matrix having $3$ non-zero rows. Hence rank of given matrix is $3$. =====Question 3(ii)===== Find the... end{align} The last matrix is the echelon form of given matrix having $2$ non-zero rows. Hence rank of the given matrix is $2$. ====Go To==== <text align="left"
Question 3, Exercise 2.1
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AB \right)C=A\left( BC \right)$. ====Solution==== Given: $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B=\b... }$. Verify that $A(B+C)=AB+AC$. ====Solution==== Given: $A=\begin{bmatrix} 1 & 3 \\ -1 & 4 \end{bmatr... .$ Verify that $A( B-C )=AB-AC$. ====Solution==== Given: $ A=\begin{bmatrix}1 & 3 \\-1 & 4 \end{bmatrix}
Question 13, Exercise 2.2
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& 4 & 5 \end{matrix} \right|=9$ ====Solution==== Given $$\left| \begin{matrix} x & 2 & 3 \\ 0 & -... & 3 & 4 \end{matrix} \right|=-6$ ====Solution==== Given $$\left| \begin{matrix} -1 & 0 & 1 \\ x^2 ... & 3 & x+4\end{matrix} \right|=0$ ====Solution==== Given $$\left| \begin{matrix} x+2 & 3 & 4 \\ 2 &
Question 7, Exercise 2.2
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862 & 3863 \end{matrix} \right|$ ====Solution==== Given $$\left| \begin{matrix} 3860 & 3861 \\ 386... 7 & 88 & 89\end{matrix} \right|$ ====Solution==== Given $$\left| \begin{matrix} 81 & 82 & 83 \\ 84
Question 16 & 17, Exercise 2.2
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that $|A^{-1}|=\dfrac{1}{|A|}$. ====Solution==== Given $$A=\left[ \begin{matrix} 3 & -1 \\ 4 & 2 ... -1 & 1 \\2 & 3\end{bmatrix}$. ====Solution==== Given $$A=\left[ \begin{matrix} 2 & 3 \\ 1 & 0
Question 4, Exercise 2.3
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end{align} The last matrix is the echelon form of given matrix having $2$ non-zero rows. Hence rank of the given matrix is $2$. ====Go To==== <text align="left">
Question 2, Exercise 2.1
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end{bmatrix}$. Find $2A+3B-4C.$ ====Solution==== Given: $A=\begin{bmatrix}2 & -5 & 1\\ 3 & 0 & -4\end{bm
Question 4, Exercise 2.1
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that $\dfrac{1}{3}A^2-2A-9I=0$. ====Solution==== Given: $A=\begin{bmatrix} 1 & 4 & 4 \\ 4 & 1 & 4 \\ 4
Question 7, Exercise 2.1
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n show that $( A+B )^t=A^t+B^t$. ====Solution==== Given $A=\left[ \begin{matrix}1 & 0 & -1 & 2 \\3 & 1
Question 8, Exercise 2.1
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trix}$, show that $( A^t )^t=A$. ====Solution==== Given $$A=\left[ \begin{matrix} 1 & 2 & 0 \\ 3 &
Question 1, Exercise 2.2
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32},A_{33}.$ Also find $|A|.$ =====Solution===== Given $$A=\left[ \begin{matrix} 1 & 3 & 1 \\ -1
Question 12, Exercise 2.2
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\0 & 1 & -\lambda \end{bmatrix}$ ====Solution==== Given $$A=\left[ \begin{matrix} -\lambda & 1 & 0 \
Question 14 & 15, Exercise 2.2
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5\end{bmatrix}$. Find $A^{-1}$. ====Solution==== Given $$A=\left[ \begin{matrix} 0 & 2 & 2 \\ -1
Question 19, Exercise 2.2
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