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- Question 8, Exercise 2.2
- \begin{align*} A &= \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix}\\ B &= \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31} & b_{32} \end{bmatrix}\\ AB &= \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix} \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31} & b_{32} \end{bma
- Question 2, Exercise 2.3
- The elements of \(R_1\) are \(a_{11} = 3\), \(a_{12} = 2\), and \(a_{13} = 3\). Now we find their cor... } (5 \cdot 0 - 1 \cdot 1) = (1) (-1) = -1 \\ & A_{12} = (-1)^{1+2} \left|\begin{array}{cc} 4 & 1 \\ 2 ... ant: \begin{align*} \det(A) &= a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} \\ &= 3(-1) + 2(2) + 3(-6) \\ &= -3 + 4 - 18 \\ &= -17 \end{align*} Thus, the
- Question 9, Exercise 2.2
- \begin{align*} A &= \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31}... end{pmatrix} \\ B &= \begin{pmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31}... } A + B &= \begin{pmatrix} a_{11} + b_{11} & a_{12} + b_{12} & a_{13} + b_{13} \\ a_{21} + b_{21} & a_{22} + b_{22} & a_{23} + b_{23} \\ a_{31} + b_{3
- Question 6, Exercise 2.6
- 3 \\ 2 & 4 \end{array} \right| = 4 - 6 = -2\\ A_{12} &= (-1)^{1+2} \left| \begin{array}{cc} 2 & 3 \\ ... array}{cc} 3 & 1 \\ 2 & 4 \end{array} \right| = -(12 - 2) = -10\\ A_{22} &= (-1)^{2+2} \left| \begin{a... t| = 5 - 6 = -1\\ A&= \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} &... \\ 2 & -5 \end{array} \right| = 15 - 4 = 11\\ A_{12} &= (-1)^{1+2} \left| \begin{array}{cc} 2 & 2 \\
- Question 2, Exercise 2.5
- gin{array}{ccc} 1 & 2 & -3 \\ 0 & 4 & -4 \\ 0 & 12 & -12 \end{array} \right] \quad R3 + 5R1 \\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & 2 & -3 \\ 0 & 1 & -1 \\ 0 & 12 & -12 \end{array} \right] \quad \frac{1}{4}R2 \\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & 2 & -3
- Question 1, Exercise 2.2
- frac{4}{2} = 2 \] For \( i = 1, j = 2 \): \[ a_{12} = \dfrac{1 + 3 \cdot 2}{2} = \dfrac{1 + 6}{2} = ... ve \begin{align*} A&=\begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \\ &=\begin{bma... ve \begin{align*} A &= \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \\ &= \begin{bm... ): \begin{align*} A &= \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \\ &= \begin{bm
- Question 5, Exercise 2.6
- &=\frac{ 0-2 ( 4 + 2) + 1(1 + 5)}{42}\\ &=\frac{ -12 + 6}{42}\\ &=-\frac{6}{42}\\ &=-\frac{1}{7}\end{a... &=\frac{2 ( 4 + 2 )-0 + 1(2 - 8)}{42}\\ &=\frac{ 12 - 6}{42}\\ &=\frac{6}{42}\\ & = \frac{1}{7} \end{... frac{2 ( -5 - 1 ) - 2 ( 2 - 8 )+0}{42}\\ &=\frac{-12 + 12}{42}\\ &= \frac{0}{42} = 0 \end{align*} The solution set for the given system of equations using
- Question 12, Exercise 2.2
- ====== Question 12, Exercise 2.2 ====== Solutions of Question 12 of Exercise 2.2 of Unit 02: Matrices and Determinants. ... tbook Board, Islamabad, Pakistan. =====Question 12===== For any square matrix $A$; prove that $\left
- Question 7 and 8, Exercise 2.6
- \\ 3 & -3 \end{array} \right| = 3 - 6 = -3 \\ A_{12} &= (-1)^{1+2} \left| \begin{array}{cc} 4 & 2 \\ 7 & -3 \end{array} \right| = -(-12 - 14) = 26 \\ A_{13} &= (-1)^{1+3} \left| \begin{array}{cc} 4 & -1 \\ 7 & 3 \end{array} \right| = 12 + 7 = 19 \\ A_{21} &= (-1)^{2+1} \left| \begin{ar
- Question 11, Exercise 2.2
- text align="right"><btn type="success">[[math-11-nbf:sol:unit02:ex2-2-p12|Question 12 >]]</btn></text>
- Question 13, Exercise 2.2
- text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:ex2-2-p12|< Question 12]]</btn></text>
- Question 5, Exercise 2.3
- (1) [(1)(-1) - (-1)(-2)] \\ &= -1 - 2 = -3 \\ A_{12} &= (-1)^{1+2} \left|\begin{array}{cc} 2 & -1 \\ ... \right|\\ &= (1) [(-1)(4i) - (-i)(0)] = -4i \\ B_{12} &= (-1)^{1+2} \left|\begin{array}{cc} 2i & -i \\
- Question 7, Exercise 2.3
- y}{ll}2 & 1 \\ 8 & 6\end{array}\right] \\ |A|& = 12 - 8 = 4\\ A^{-1} &= \dfrac{1}{4} \left[\begin{ar... frac{1}{2} + \frac{2}{3}\right) & \left(-\frac{1}{12} - \frac{1}{6}\right) \\ -1 & \frac{1}{4} \end{ar
- Question 2, Exercise 2.6
- 3 & -1 \\ 3 & -2 & 4 \end{array} \right|=0\\ &2(12-2)+\lambda(8+3)+1(-4-9)=0\\ &20+11\lambda-13=0\\ ... bda \right) = 0\\ &\lambda^2 + 2 + 8\lambda - 4 - 12-3\lambda = 0\\ &\lambda^2 + 5\lambda - 14 = 0\\ &
- Question 6, Exercise 2.2
- \implies &\begin{bmatrix} 7 + \alpha & -1 \\ -3 & 12 + \alpha \end{bmatrix} = \begin{bmatrix} 2\beta &