Operations on Matrices such as adding, subtracting, and multiplying by a scalar, along with the equality of matrices, are presented using clear examples and questions with fully worked-out solutions.
Only matrices of the same order (same number of rows and columns) can be added or subtracted. We add or subtract two matrices by adding or subtracting their corresponding entries.
Rewrite, if possible, each of the following expressions as a single matrix.
a) \[ \begin{bmatrix} 5 & -7 \\ 4 & 6 \end{bmatrix} + \begin{bmatrix} -8 & -3 \\ 4 & 0 \end{bmatrix} \]
b) \[ \begin{bmatrix} 5 & -7 & 5 \\ -2 & 3 & -4 \end{bmatrix} + \begin{bmatrix} -1 & -3 \\ 9 & -3 \\ 10 & -9 \end{bmatrix} \]
c) \[ \begin{bmatrix} x+y & -y & x \\ 3x & -9 & -2 \\ -3x & 4 & 4y-x \end{bmatrix} - \begin{bmatrix} 2x & y & 3x \\ -3 & 12 & -5 \\ y & x & 2+3y \end{bmatrix} \]
a) The matrices have the same order (\( 2 \times 2 \)), and we can therefore add them by adding their corresponding entries:
\[ \begin{bmatrix} 5 + (-8) & -7 + (-3) \\ 4 + 4 & 6 + 0 \end{bmatrix} = \begin{bmatrix} -3 & -10 \\ 8 & 6 \end{bmatrix} \]b) The matrices do not have the same order (a \( 2 \times 3 \) matrix and a \( 3 \times 2 \) matrix), and therefore we cannot add them.
c) The matrices have the same order, and we can therefore subtract them by subtracting their corresponding entries and simplifying the algebraic expressions:
$$ \begin{aligned} & \begin{bmatrix} x + y & -y & x \\ 3x & -9 & -2 \\ -3x & 4 & 4y - x \end{bmatrix} - \begin{bmatrix} 2x & y & 3x \\ -3 & 12 & -5 \\ y & x & 2 + 3y \end{bmatrix} \\[15pt] =\; & \begin{bmatrix} x + y - (2x) & -y - (y) & x - (3x) \\ 3x - (-3) & -9 - (12) & -2 - (-5) \\ -3x - (y) & 4 - (x) & 4y - x - (2 + 3y) \end{bmatrix} \\[15pt] =\; & \begin{bmatrix} -x + y & -2y & -2x \\ 3x + 3 & -21 & 3 \\ -3x - y & 4 - x & -x + y - 2 \end{bmatrix} \end{aligned} $$To multiply a matrix by a scalar (a real number), you multiply all entries of that matrix by the scalar. Note that scalar multiplication does not change the order of the matrix.
Express as a single matrix.
a) \[ -5 \begin{bmatrix} 5 & -7 \\ 4 & 6 \end{bmatrix} \]
b) \[ 2 \begin{bmatrix} x & x + y & 6 \\ -2 & -x & 1 \end{bmatrix} - 3 \begin{bmatrix} -x & -2y & 0 \\ 3x & x & -1 \end{bmatrix} \]
a) Multiply each entry by the scalar \( -5 \):
\[ = \begin{bmatrix} -5(5) & -5(-7) \\ -5(4) & -5(6) \end{bmatrix} = \begin{bmatrix} -25 & 35 \\ -20 & -30 \end{bmatrix} \]b) First, multiply the first matrix by \( 2 \) and the second matrix by \( 3 \):
\[ = \begin{bmatrix} 2(x) & 2(x + y) & 2(6) \\ 2(-2) & 2(-x) & 2(1) \end{bmatrix} - \begin{bmatrix} 3(-x) & 3(-2y) & 3(0) \\ 3(3x) & 3(x) & 3(-1) \end{bmatrix} \]Subtract, group, and simplify the algebraic expressions:
\[ = \begin{bmatrix} 2x - (-3x) & (2x + 2y) - (-6y) & 12 - 0 \\ -4 - 9x & -2x - 3x & 2 - (-3) \end{bmatrix} = \begin{bmatrix} 5x & 2x + 8y & 12 \\ -4 - 9x & -5x & 5 \end{bmatrix} \]Two matrices are equal if they have the same order (equal number of rows and equal number of columns) and all their corresponding entries are equal.
Which of the following matrices are equal?
\[ \text{a)} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad \text{b)} \begin{bmatrix} x & 2x \\ 2 & 1 \end{bmatrix}, \quad \text{c)} \begin{bmatrix} 1 & -x \\ 3x & 5 \end{bmatrix}, \quad \text{d)} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad \text{e)} \begin{bmatrix} 1 & 2 & 5 \\ 3 & 4 & 8 \end{bmatrix} \]Matrices a), b), c), and d) have the same order (\( 2 \times 2 \)). However, only matrices a) and d) are equal because all of their corresponding entries are identical. Matrix e) has an order of \( 2 \times 3 \), which is different from the rest, so it is not equal to any of the others.
Find \( x \) and \( y \) so that:
\[ \begin{bmatrix} 2x + y & -7 \\ 4 & 6 \end{bmatrix} = \begin{bmatrix} 2 & -7 \\ 4 & 4x + y \end{bmatrix} \]The two matrices have the same order. For these matrices to be equal, all their corresponding entries must be equal. The constants \(-7\) and \(4\) already match. Setting the unknown entries equal gives us a system of equations:
\[ 2x + y = 2 \quad \text{and} \quad 6 = 4x + y \]Solving this system of equations (by subtracting the first from the second), we find:
\[ x = 2, \quad y = -2 \]Check: Substitute \( x = 2 \) and \( y = -2 \) back into the matrices to verify that they are identical.
Find \( x \) and \( y \), if possible, such that:
A) \[ \begin{bmatrix} 2x + y & 3 & 10 \\ y + 1 & -2 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 10 \\ 3 & -2 & 0 \end{bmatrix} \]
The two matrices have the same order and to be equal, they need to have equal corresponding entries. Setting them equal yields the simultaneous equations:
\[ 2x + y = 2 \quad \text{and} \quad y + 1 = 3 \]Solve the second equation first to obtain \( y = 2 \). Substitute this into the first equation to find \( 2x + 2 = 2 \implies x = 0 \).
B) \[ \begin{bmatrix} 6 & -4 & -6 & x - y \end{bmatrix} = -2 \begin{bmatrix} -3 & 2 & 2x + 2y & 13 \end{bmatrix} \]
Multiply the second matrix by the scalar \(-2\) and rewrite the equation:
\[ \begin{bmatrix} 6 & -4 & -6 & x - y \end{bmatrix} = \begin{bmatrix} 6 & -4 & -2(2x + 2y) & -26 \end{bmatrix} \]For the matrices to be equal, we equate the unknown corresponding entries:
\[ -2(2x + 2y) = -6 \quad \text{and} \quad x - y = -26 \]Simplify the first equation by dividing by \(-4\), which gives \( x + y = 1.5 \) (or \( \frac{3}{2} \)). Solving this system yields:
\[ x = -\frac{49}{4}, \quad y = \frac{55}{4} \]Write as a single matrix:
A) \[ \begin{bmatrix} -2 & 4 & 3 \\ -6 & 3 & -4 \\ -1 & 0 & 9 \end{bmatrix} + \begin{bmatrix} -1 & 0 & 23 \\ -4 & -3 & 9 \\ 2 & -5 & 0 \end{bmatrix} \]
B) \[ -2 \begin{bmatrix} -2 & 3 \\ 1 & -5 \end{bmatrix} + 4 \begin{bmatrix} 3 & 4 \\ -2 & 9 \end{bmatrix} \]
C) \[ -3 \begin{bmatrix} -2x & y + 2 & 6 \\ 4 & -x & -4 \end{bmatrix} + 5 \begin{bmatrix} 3x & 5y & -5 \\ 3y & x + y & -1 \end{bmatrix} \]
Multiply each matrix by its scalar first:
\[ \begin{bmatrix} 6x & -3y - 6 & -18 \\ -12 & 3x & 12 \end{bmatrix} + \begin{bmatrix} 15x & 25y & -25 \\ 15y & 5x + 5y & -5 \end{bmatrix} \]Then add corresponding entries together:
\[ = \begin{bmatrix} 21x & 22y - 6 & -43 \\ -12 + 15y & 8x + 5y & 7 \end{bmatrix} \]