Question 3, Exercise 2.1

Solutions of Question 3 of Exercise 2.1 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

If $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B=\begin{bmatrix}a & h & g\\h & b & f\\g & f & c\end{bmatrix}$ and $C=\begin{bmatrix}x\\y\\z\end{bmatrix}$. Verify that $\left( AB \right)C=A\left( BC \right)$.

Given: $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B=\begin{bmatrix}a & h & g\\h & b & f\\g & f & c\end{bmatrix}$ and $C=\begin{bmatrix}x\\y\\z\end{bmatrix}$.
We have to prove that $$(AB)C=A(BC)$$ First, we take \begin{align} AB&=\begin{bmatrix}x & y & z\end{bmatrix}\begin{bmatrix}a & h & g\\h & b & f\\g & f & c\end{bmatrix}\\ &=\begin{bmatrix}ax+hy+gz & hx+by+fz & gx+fy+cz \end{bmatrix} \\ &=\begin{bmatrix}ax+hy+gz & hx+by+fz & gx+fy+cz \end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}\\ &=\left[ x( ax+hy+gz )+y( hx+by+fz )+z( gx+fy+cz ) \right] \\ &=\left[ a{{x}^{2}}+hxy+gxz+hxy+b{{y}^{2}}+fyz+gxz+fyz+c{{z}^{2}} \right] \\ &=\left[ a{{x}^{2}}+2hxy+2gxz+b{{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}x\\ y\\z\end{bmatrix}\\ &=\begin{bmatrix}ax+hy+gz\\hx+by+fz\\gx+fy+cz \end{bmatrix}\end{align} \begin{align} R.H.S &= A(BC)\\ &=\begin{bmatrix}x & y & z\end{bmatrix}\begin{bmatrix}ax+hy+gz\\hx+by+fz\\gx+fy+cz\end{bmatrix}\\ &=\left[ x( ax+hy+gz )+y( hx+by+fz )+z( gx+fy+cz ) \right]\\ &=\left[ a{{x}^{2}}+hxy+gxz+hxy+b{{y}^{2}}+fyz+gxz+fyz+c{{z}^{2}} \right]\\ &=\left[ a{{x}^{2}}+2hxy+2gxz+b{{y}^{2}}+2fyz+c{{z}^{2}} \right] \ldots (2)\end{align} From (1) and (2), we have $$(AB)C=A(BC).$$

If $A=\begin{bmatrix}1 & 3 \\ -1 & 4 \end{bmatrix}$, $B=\begin{bmatrix} 2 & 1 & -3 \\ 0 & 4 & 2 \end{bmatrix}$ and $C=\begin{bmatrix}3 & -1 & 5 \\ 2 & \quad1 & 0 \end{bmatrix}$. Verify that $A(B+C)=AB+AC$.

Given: $A=\begin{bmatrix} 1 & 3
-1 & 4 \end{bmatrix}$, $B=\begin{bmatrix} 2 & 1 & -3
0 & 4 & 2 \end{bmatrix}$ and $C=\begin{bmatrix}3 & -1 & 5 \\2 & 1 & 0 \end{bmatrix}$.
Now \begin{align}B+C&=\left[ \begin{matrix} 2 & 1 & -3 \\0 & 4 & 2 \\\end{matrix} \right]+\left[ \begin{matrix}3 & -1 & 5 \\2 & 1 & 0 \\\end{matrix} \right]\\ &=\left[ \begin{matrix}2+3 & 1-1 & -3+5 \\0+2 & 4+1 & 2+0 \\\end{matrix} \right]\\ &=\left[ \begin{matrix}5 & 0 & 2 \\2 & 5 & 2 \\\end{matrix} \right]\end{align} This gives \begin{align}A( B+C )&=\left[ \begin{matrix}1 & 3 \\-1 & 4 \\ \end{matrix} \right]\left[ \begin{matrix}5 & 0 & 2 \\2 & 5 & 2 \\\end{matrix} \right]\\ &=\left[ \begin{matrix}5+6 & 0+15 & 2+6 \\-5+8 & 0+20 & -2+8 \\\end{matrix} \right]\\ \implies A( B+C )&=\left[ \begin{matrix}11 & 15 & 8 \\3 & \,\,20 & \,6 \\\end{matrix}\right] ... (1)\end{align} Now, we take \begin{align}AB &=\left[ \begin{matrix}1 & 3 \\-1 & 4 \\\end{matrix} \right]\left[ \begin{matrix}2 & 1 & -3 \\0 & 4 & 2 \\ \end{matrix} \right]\\ &=\left[ \begin{matrix}2-0 & 1+12 & -3+6 \\-2+0 & -1+16 & 3+8 \\\end{matrix} \right]\\ &=\left[ \begin{matrix}\quad 2 & 13 & 3 \\-2 & 15 & 11 \\\end{matrix} \right].\end{align} and \begin{align}AC& =\left[ \begin{matrix}1 & 3 \\-1 & 4 \\\end{matrix} \right]\left[ \begin{matrix}3 & -1 & 5\\2 & \quad 1 & 0 \\ \end{matrix}\right]\\ &=\left[ \begin{matrix}3+6 & -1+3 & 5+0 \\-3+8 & 1+4 & -5+0 \\\end{matrix} \right]\\ &=\left[ \begin{matrix}9 & 2 & \quad5 \\5 & 5 & -5 \\ \end{matrix} \right]\end{align} Now we take \begin{align}AB+AC &=\left[ \begin{matrix}2 & 13 & 3 \\-2 & 15 & 11 \\ \end{matrix} \right]+\left[ \begin{matrix}9 & 2 & \quad 5 \\5 & 5 & -5 \\ \end{matrix} \right]\\ &=\left[ \begin{matrix}\quad2+9 & 13+2 & 3+5\\-2+5 & 15+5 & 11-5\\ \end{matrix} \right] \\ \implies AB+AC&=\left[ \begin{matrix} 11 & 15 & 8 \\ 3 & 20 & 6 \\ \end{matrix} \right] ... (2) \end{align} From (1) and (2), we have $$A(B+C)=AB+AC.$$

If $A=\begin{bmatrix} 1 & 3 \\ -1 & 4 \end{bmatrix},$ $B=\begin{bmatrix} 2 & 1 & -3 \\ 0 & 4 & 2 \end{bmatrix}$ and $C=\begin{bmatrix} 3 & -1 & 5 \\ 2 & \quad1 & 0 \end{bmatrix}.$ Verify that $A( B-C )=AB-AC$.

Given: $ A=\begin{bmatrix}1 & 3 \\-1 & 4 \end{bmatrix}$, $B=\begin{bmatrix}2 & 1 & -3 \\ 0 & 4 & 2 \end{bmatrix}$ and $C=\begin{bmatrix}3 & -1 & 5\\ 2 & 1 & 0 \end{bmatrix}$.
Now \begin{align}B-C&=\left[ \begin{matrix}2 & 1 & -3\\0 & 4 & 2 \\\end{matrix} \right]-\left[\begin{matrix}3 & -1 & 5 \\2 & \quad 1 & 0 \\ \end{matrix}\right]\\ &=\left[ \begin{matrix}2-3 & 1+1 & -3-5 \\ 0-2 & 4-1 & 2-0 \\ \end{matrix} \right]\\ \implies B-C&=\left[ \begin{matrix}-1 & 2 & -8 \\ -2 & 3 & \quad 2 \\ \end{matrix}\right]\end{align} Now \begin{align}A(B-C)&=\left[ \begin{matrix}1 & 3\\-1 & 4 \\ \end{matrix} \right] \left[ \begin{matrix}-1 & 2 & -8\\-2 & 3 & 2 \\ \end{matrix} \right]\\ &=\left[ \begin{matrix}-1-6 & 2+9 & -8+6 \\ 1-8 & -2+12 & 8+8\\ \end{matrix} \right]\\ \implies A(B-C)&=\left[ \begin{matrix} -7 & 11 & -2 \\ -7 & 10 & 16 \\ \end{matrix} \right] ... (1) \end{align} Now we take \begin{align}AB& =\left[ \begin{matrix}1 & 3\\-1 & 4\\ \end{matrix}\right] \left[\begin{matrix}2 & 1 & -3\\0 & 4 & 2\\ \end{matrix}\right] \\ &=\left[ \begin{matrix}2+0 & 1+12 & -3+6\\ -2+0 & -1+16 & 3+8\\ \end{matrix}\right] \\ &=\left[ \begin{matrix} 2 & 13 & 3 \\ -2 & 15 & 11 \\ \end{matrix}\right] \end{align} Now \begin{align} AC &=\left[ \begin{matrix} 1 & 3 \\ -1 & 4 \\ \end{matrix} \right] \left[ \begin{matrix}3 & -1 & 5\\2 & 1 & 0 \\ \end{matrix}\right]\\ &=\left[ \begin{matrix} 3+6 & -1+3 & 5+0 \\ 3+8 & 1+4 & -5+0 \\ \end{matrix} \right]\\ &=\left[ \begin{matrix} 9 & 2 & \quad 5\\ 5 & 5 & -5\\ \end{matrix}\right]\end{align} \begin{align} AB-AC&=\left[\begin{matrix} 2 & 13 & 3\\-2 & 15 & 11 \\ \end{matrix}\right]-\left[\begin{matrix} 9 & 2 & 5 \\ 5 & 5 & -5 \\ \end{matrix} \right] \\ &=\left[ \begin{matrix} 2-9 & 13-2 & 3-5 \\ -2-5 & 15-5 & 11+5 \\ \end{matrix}\right]\\ \implies AB-AC&=\left[\begin{matrix} -7 & 11 & -2 \\ -7 & 10 & 16 \\ \end{matrix}\right]... (2) \end{align} From (1) and (2), we have $$A(B-C)=AB-BC.$$

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