Question 4, Exercise 2.2

Solutions of Question 4 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Find $A$ if $$\begin{align}\left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}\right]A\left[\begin{array}{cc} 1 & 3 \\ 2 & 4 \end{array}\right]&=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\end{align}$$

Solution.

Let $B = \left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}\right]$ and $C = \left[\begin{array}{cc} 1 & 3 \\ 2 & 4 \end{array}\right]$

$$\begin{align*}BAC &= I,\\ A &= B^{-1} I C^{-1} = B^{-1} C^{-1}. \end{align*}$$

The inverse of a 2×2 matrix $\left[\begin{array}{cc} a & b \\ c & d \end{array}\right]$ is given by:

$$\begin{align*} \left[\begin{array}{cc} a & b \\ c & d \end{array}\right]^{-1} &= \frac{1}{ad - bc} \left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]. \end{align*}$$

For $$\begin{align*} B &= \left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}\right] \\ \det(B) &= 2 \cdot 2 - 1 \cdot 3 = 4 - 3 = 1.\\ B^{-1} &= \left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}\right]^{-1}\\ & = \left[\begin{array}{cc} 2 & -1 \\ -3 & 2 \end{array}\right]. \end{align*}$$ $$\begin{align*} C &= \left[\begin{array}{cc} 1 & 3 \\ 2 & 4 \end{array}\right] \\ C^{-1} &= \dfrac{1}{-2} \left[\begin{array}{cc} 4 & -3 \\ -2 & 1 \end{array}\right]\\ & = \left[\begin{array}{cc} -2 & \dfrac{3}{2} \\ 1 & -\dfrac{1}{2} \end{array}\right]\end{align*}$$ $$\begin{align*} A &= B^{-1} C^{-1} \\ &= \left[\begin{array}{cc} 2 & -1 \\ -3 & 2 \end{array}\right] \left[\begin{array}{cc} -2 & \dfrac{3}{2} \\ 1 & -\dfrac{1}{2} \end{array}\right]\\ A &= \left[\begin{array}{cc} 2 & -1 \\ -3 & 2 \end{array}\right] \left[\begin{array}{cc} -2 & \dfrac{3}{2} \\ 1 & -\dfrac{1}{2} \end{array}\right] \\ &= \left[\begin{array}{cc} 2(-2) + (-1)(1) & 2\left(\dfrac{3}{2}\right) + (-1)\left(-\dfrac{1}{2}\right) \\ -3(-2) + 2(1) & -3\left(\dfrac{3}{2}\right) + 2\left(-\frac{1}{2}\right) \end{array}\right] \\ &= \left[\begin{array}{cc} -4 - 1 & 3 + \dfrac{1}{2} \\ 6 + 2 & -\dfrac{9}{2} - 1 \end{array}\right] \\ &= \left[\begin{array}{cc} -5 & \dfrac{7}{2} \\ 8 & -\dfrac{11}{2} \end{array}\right]. \end{align*}$$

Therefore, the matrix $A$ is:

$$\begin{align*} A &= \left[\begin{array}{cc} -5 & \dfrac{7}{2} \\ 8 & -\dfrac{11}{2} \end{array}\right]. \end{align*}$$

Find $X$ if $$\begin{bmatrix}3 & 2 \\ 0 & 1 \\ 2 & 0\end{bmatrix} X=\begin{bmatrix}7 / 2 & 11 & 2 \\ 2 & 4 & 1 \\ 1 & 2 & 0\end{bmatrix}.$$

Solution.

If $A=\left[\begin{array}{ll}3 & 7\end{array}\right]$ and $B=\left[\begin{array}{ll}2 & 14\end{array}\right]$ then find a non-zero matrix $C$ such that $A C=B C$.

Solution.

$\left[\begin{array}{cc}x y & 4 \\ 0 & x+y\end{array}\right]=\left[\begin{array}{ll}8 & z \\ t & 6\end{array}\right]$ then find the values of $z, t$ and $x^{2}+y^{2}$.

Solution.

Given $\begin{bmatrix}x y & 4 \\ 0 & x+y\end{bmatrix}=\begin{bmatrix}8 & z \\ t & 6\end{bmatrix}$.

Equating the elements, we get $$xy=8, z=4, t=0. x+y=6$$

As we know $$\begin{align*} &(x+y)^2=x^2+y^2+2xy \\ \implies & x^2+y^2 = (x+y)^2-2xy \\ \implies & x^2+y^2 = 6^2-2(8) =20 \end{align*}$$ Hence, we conclude $z=4$, $t=0$ and $x^2+y^2=20$.

If $A=\left[\begin{array}{ll}3 & 4 \\ 7 & 6\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ then find $\alpha$ and $\beta$ so that $A^{2}+\alpha I=\beta A$.

Solution.

Find the values of $x$ if $\left[\begin{array}{lll}x & -4 & 2\end{array}\right]\left[\begin{array}{lll}1 & 0 & 3 \\ 0 & 1 & 0 \\ 2 & 0 & 4\end{array}\right]\left[\begin{array}{c}x \\ 1 \\ -1\end{array}\right]=0$.

Solution.

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