Matrix Equality

Matching Matrix Dimensions and Corresponding Entries

Matrix equality compares both structure and content. Matching values are not enough if they occupy different positions or if the matrices have different dimensions.

Matrices $A$ and $B$ are equal, written $A=B$, if and only if both conditions below hold:

1. Same dimensions: Matrix $A$ and matrix $B$ must have the same number of rows and columns. If $A$ has dimensions $m \times n$, then $B$ must also have dimensions $m \times n$.
2. Corresponding Elements are Equal: Every corresponding element (located in the same row and column position) in matrix $A$ and matrix $B$ must have the same value. If $A = [a_{ij}]$ and $B = [b_{ij}]$, then $a_{ij} = b_{ij}$ for all values of $i$ (row index) and $j$ (column index).

If one of these two conditions is not met, then matrix $A$ is not equal to matrix $B$ (written as $A \neq B$).

Examples of Matrix Equality

Matrix equality is decided by two checks. The two matrices must have the same format, and then every entry at the same position must match. The examples below show one pair that passes both checks and two pairs that fail exactly one of them, followed by a case where the entries contain unknown variables.

Equal Matrices

Given two matrices:

Matrix $P$ and matrix $Q$ are equal ($P=Q$) because:

• Both have an order of $2 \times 2$.

• Corresponding elements have the same value:

  $p_{11} = q_{11} = 1$, $p_{12} = q_{12} = 2$ , $p_{21} = q_{21} = 3$, $p_{22} = q_{22} = 4$ .

Unequal Matrices with Different Dimensions

Given two matrices:

Matrix $R$ is not equal to matrix $C$ ($R \neq C$) because $R$ has dimensions $2 \times 2$, while $C$ has dimensions $3 \times 2$.

Unequal Matrices with Different Corresponding Elements

Given two matrices:

Although matrix $S$ and matrix $T$ have the same dimensions, $2 \times 2$, they are not equal ($S \neq T$). Their entries in row $2$, column $1$ differ: $s_{21} = -2$, whereas $t_{21} = 2$.

Determining Variable Values from Matrix Equality

Given the following matrices:

If matrix $A$ is equal to matrix $B$ ($A=B$), determine the values of $x$, $y$, and $z$.

Solution:

Since $A=B$, the corresponding elements must be equal:

1. $a_{11} = b_{11} \implies -x = -1 \implies x = 1$
2. $a_{12} = b_{12} \implies 2 = 2$ (already equal)
3. $a_{21} = b_{21} \implies -3y = 6 \implies y = \frac{6}{-3} \implies y = -2$
4. $a_{22} = b_{22} \implies z^2 = 9 \implies z = \pm\sqrt{9} \implies z = 3 \text{ or } z = -3$

The values are $x=1$, $y=-2$, and $z = \pm 3$.

Exercises

Decide whether the first three statements are True or False.

1. Having the same dimensions is one condition that two equal matrices must satisfy.

2. Two unequal matrices always have different dimensions.

3. The two matrices below are equal.

   $$K = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$$

   $$L = \begin{bmatrix} 2 & 0 \\ 0 & 2 \\ 0 & 0 \end{bmatrix}$$

4. For the following matrices, $P=Q$. Determine $x$ and $y$.

   $$P = \begin{bmatrix} 2x & 4 \\ -1 & y+3 \end{bmatrix}$$

   $$Q = \begin{bmatrix} -10 & 4 \\ -1 & 2 \end{bmatrix}$$

5. Matrix $I$ is the $2 \times 2$ identity matrix. For the following matrix $A$, $A = I$. Determine $x+y$.

   $$A = \begin{bmatrix} -x-3y & 0 \\ (x-2y)^2 & 1 \end{bmatrix}$$

6. Calculate the value of $a+b+c+d$ that satisfies the following matrix equality:

   $$\begin{bmatrix} a+2b & 2a+b \\ c+d & 2c+d \end{bmatrix} = \begin{bmatrix} 3 & -3 \\ 7 & 1 \end{bmatrix}$$

Worked Solutions

1. True. Equal matrices must have the same dimensions, although that condition alone is not sufficient.

2. False. Two unequal matrices can have the same dimensions while differing in one or more corresponding entries, as in Example 3.

3. False. Matrix $K$ has dimensions $2 \times 2$, while matrix $L$ has dimensions $3 \times 2$. Matrices with different dimensions cannot be equal.

4. Given $P=Q$:

   $$\begin{bmatrix} 2x & 4 \\ -1 & y+3 \end{bmatrix} = \begin{bmatrix} -10 & 4 \\ -1 & 2 \end{bmatrix}$$

   From the equality of corresponding elements:

   • $2x = -10 \implies x = -5$
   • $y+3 = 2 \implies y = 2-3 \implies y = -1$

     Thus, $x=-5$ and $y=-1$.

5. The $2 \times 2$ identity matrix is:

   $$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

   Given $A=I$:

   $$\begin{bmatrix} -x-3y & 0 \\ (x-2y)^2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

   From the equality of corresponding elements, we obtain a system of equations:

   1. $-x-3y = 1$ (Equation 1)
   2. $(x-2y)^2 = 0$ (Equation 2)

   Solve Equation 2 first:

   $$(x-2y)^2 = 0$$

   $$x-2y = \sqrt{0}$$

   $$x-2y = 0$$

   $$x = 2y \quad \text{(Equation 2')}$$

   Substitute Equation 2' into Equation 1:

   $$-(2y)-3y = 1$$

   $$-2y-3y = 1$$

   $$-5y = 1$$

   $$y = -\frac{1}{5}$$

   Substitute the value of $y = -\frac{1}{5}$ into Equation 2':

   $$x = 2\left(-\frac{1}{5}\right)$$

   $$x = -\frac{2}{5}$$

   Then, the value of $x+y$ is:

   $$x+y = \left(-\frac{2}{5}\right) + \left(-\frac{1}{5}\right)$$

   $$= -\frac{2}{5} - \frac{1}{5}$$

   $$= \frac{-2-1}{5}$$

   $$= -\frac{3}{5}$$

6. Given the matrix equality:

   $$\begin{bmatrix} a+2b & 2a+b \\ c+d & 2c+d \end{bmatrix} = \begin{bmatrix} 3 & -3 \\ 7 & 1 \end{bmatrix}$$

   From the equality of corresponding elements, we obtain a system of equations:

   1. $a+2b = 3$
   2. $2a+b = -3$
   3. $c+d = 7$
   4. $2c+d = 1$

   Solve the system of equations for $a$ and $b$ (equations 1 and 2):

   Initial equations:

   $$a+2b = 3 \quad \text{(1)}$$

   $$2a+b = -3 \quad \text{(2)}$$

   To eliminate $b$, multiply equation (2) by $2$:

   $$2(2a+b) = 2(-3)$$

   $$4a+2b = -6 \quad \text{(2')}$$

   Subtract equation (1) from equation (2'):

   $$(4a+2b) - (a+2b) = -6 - 3$$

   $$4a - a + 2b - 2b = -9$$

   $$3a = -9$$

   $$a = \frac{-9}{3}$$

   $$a = -3$$

   Substitute the value of $a=-3$ into equation (1):

   $$-3+2b=3$$

   $$2b=3+3$$

   $$2b=6$$

   $$b = \frac{6}{2}$$

   $$b = 3$$

   Solve the system of equations for $c$ and $d$ (equations 3 and 4):

   Initial equations:

   $$c+d = 7 \quad \text{(3)}$$

   $$2c+d = 1 \quad \text{(4)}$$

   Subtract equation (3) from equation (4) to eliminate $d$:

   $$(2c+d) - (c+d) = 1 - 7$$

   $$2c - c + d - d = -6$$

   $$c = -6$$

   Substitute the value of $c=-6$ into equation (3):

   $$-6+d=7$$

   $$d=7+6$$

   $$d=13$$

   Then, the value of $a+b+c+d$ is:

   $$\begin{aligned} &a+b+c +d \\ &= (-3) + 3 + (-6) + 13 \end{aligned}$$

   $$= 0 - 6 + 13$$

   $$= 7$$