Add, Subtract and Scalar Multiply Matrices
Operations on Matrices such as adding, subtracting and multiplying by a scalar along with the equality of matrices are presented using examples and questions with solutions.
Add and Subtract MatricesOnly matrices of the same order can be added or subtracted. We add (or subtract) two matrices by adding (or subtracting) their corresponding entries.Example 1Rewrite, if possible, the following pairs of matrices as a single matrix.
Solutiona) The matrices in part a) have the same order and we therefore can add them by adding their corresponding entries. Hence\( \begin{bmatrix} 5 & - 7 \\ 4 & 6 \end{bmatrix} + \begin{bmatrix} - 8 & - 3 \\ 4 & 0 \end{bmatrix} = \begin{bmatrix} 5 + (-8) & - 7 +(-3) \\ 4 + 4 & 6 + 0 \end{bmatrix} = \begin{bmatrix} -3 & -10 \\ 8 & 6 \end{bmatrix}\) b) The matrices in part b) do not have the same order and we therefore cannot add them. c) The matrices in part c) have the same order and we therefore can add them by adding their corresponding entries and simplifying the algebraic expressions. \( \begin{bmatrix} x + y & - y & x\\ 3x & - 9 & -2 \\ -3x & 4 & 4y - x \end{bmatrix} - \begin{bmatrix} 2 x & y & 3 x \\ -3 & 12 & -5\\ y & x & 2 + 3y \end{bmatrix} \) \( = \begin{bmatrix} x + y - (2x) & - y - (y) & x - (3x)\\ 3x - (-3) & - 9 - ( 12) & -2 - (-5)\\ -3x - (y) & 4 - (x) & 4y - x - (2 + 3y) \end{bmatrix} \) \( = \begin{bmatrix} -x + y & - 2 y & -2x\\ 3x + 3 & -21 & 3\\ -3x - y & 4 - x & - x + y- 2 \end{bmatrix}\) Multiply a Matrix by a ScalarTo multiply a matrix by a scalar, you multiply all entries of that matrix by the scalar. Note that scalar multiplication does not change the order of the matrix.Example 1Express as a single matrix.a) - 5 \( \begin{bmatrix} 5 & - 7 \\ 4 & 6 \end{bmatrix} \) b) \( 2 \begin{bmatrix} x & x + y & 6\\ -2 & - x & 1 \end{bmatrix} - 3 \begin{bmatrix} - x & -2y & 0\\ 3x & x & -1 \end{bmatrix} \) Solutiona) - 5 \( \begin{bmatrix} 5 & - 7 \\ 4 & 6 \end{bmatrix} = \begin{bmatrix} -5(5) & -5(- 7) \\ -5(4) & -5(6) \end{bmatrix}= \begin{bmatrix} -25 & 35 \\ -20 & - 30 \end{bmatrix}\)b) \( 2 \begin{bmatrix} x & x + y & 6\\ -2 & - x & 1 \end{bmatrix} - 3 \begin{bmatrix} - x & -2y & 0\\ 3x & x & -1 \end{bmatrix} \) 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 \( = \begin{bmatrix} 5x & 2x+8y & 12\\ -4 - 9x & -5x & 5 \end{bmatrix}\) Equality of two MatricesTwo matrices are equal if they have the same order and their corresponding entries are equal.Example 1Which of the following matrices are equal? SolutionMatrices a), b), c) and d) have the same order 2 by 2. But only matrices a) and c) are equal as they have equal corresponding entries. Matrix e) has the order 2 by 3 which is different to the order of all other matrices and is therefore not equal to any of the matrices.Example 2Find x and y so that \[ \begin{bmatrix} 2x + y & - 7 \\ 4 & 6 \end{bmatrix} = \begin{bmatrix} 2 & - 7 \\ 4 & 4x + y \end{bmatrix} \].SolutionThe two matrices have the same order. For these matrices to be equal, all their corresponding entries have to be equal. Corresponding entries - 7 and 4 are equal. The entries with the unknowns x and y has to be equal. Hence2 x + y = 2 and 6 = 4 x + y Solve the above system of equations to find x = 2 and y = - 2 As an exercise, substitute x by 2 and y by - 2 into the matrices and verify that they are equal.
QuestionsPart 1Find x and y, if possible, such thatA) \( \begin{bmatrix} 2x + y & 3 & 10 \\ y + 1 & -2 & 0 \\ \end{bmatrix} = \begin{bmatrix} 2 & 3 & 10 \\ 3 & -2 & 0 \\ \end{bmatrix} \) B) \( \begin{bmatrix} 6 & -4 & -6 & x - y \end{bmatrix} = - 2 \begin{bmatrix} -3 & 2 & 2x + 2y & 13 \end{bmatrix} \) Part 2Write as one matrixA) \( \begin{bmatrix} -2 & 4 & 3 \\ -6 & 3 & -4 \\ -1 & 0 & 9 \\ \end{bmatrix} + \begin{bmatrix} -1 & 0 & 23 \\ -4 & -3 & 9 \\ 2 & -5 & 0 \\ \end{bmatrix} \) B) \( - 2 \begin{bmatrix} -2 & 3 \\ 1 & -5 \end{bmatrix} + 4 \begin{bmatrix} 3 & 4 \\ -2 & 9 \end{bmatrix} \) C) \( -3 \begin{bmatrix} -2x & y + 2 & 6\\ 4 & - x & -4 \\ \end{bmatrix} + 5 \begin{bmatrix} 3x & 5y & -5\\ 3y & x+y & -1 \end{bmatrix} \)
Solutions to the QuestionsPart 1A) The two matrices have the same order and to be equal, they need to have equal corresponding entries. Hence the simultaneous equations. 2x + y = 2 and y + 1 = 3Solve the above equations to obtain: x = 0 and y = 2. B) Multiply the second equation by -2 and rewrite the given equation as follows: \( \begin{bmatrix} 6 & -4 & -6 & x - y \end{bmatrix} = \begin{bmatrix} 6 & -4 & -2(2x + 2y) & -26 \end{bmatrix}\) For the two equations to be equal, we need to have -2(2x + 2y) = - 6 and x - y = - 26 Solve the above system to obtain: x = -49/4 and y = 55/4 Part 2A) \( \begin{bmatrix} -2 & 4 & 3 \\ -6 & 3 & -4 \\ -1 & 0 & 9 \end{bmatrix} + \begin{bmatrix} -1 & 0 & 23 \\ -4 & -3 & 9 \\ 2 & -5 & 0 \end{bmatrix} = \begin{bmatrix}-3&4&26\\ -10&0&5\\ 1&-5&9\end{bmatrix} \)B) \( - 2 \begin{bmatrix} -2 & 3 \\ 1 & -5 \end{bmatrix} + 4 \begin{bmatrix} 3 & 4 \\ -2 & 9 \end{bmatrix} = \begin{bmatrix}16&10\\ -10&46\end{bmatrix}\) C) \( -3 \begin{bmatrix} -2x & y + 2 & 6\\ 4 & - x & -4 \end{bmatrix} + 5 \begin{bmatrix} 3x & 5y & -5\\ 3y & x+y & -1 \end{bmatrix} = \begin{bmatrix}21x&22y-6&-43\\ -12+15y&8x+5y&7\end{bmatrix}\) More References and linksMatrices with Examples and Questions with SolutionsLinear Algebra Matrix (mathematics) Matrices Applied to Electric Circuits The Inverse of a Square Matrix |
