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Chapter 3 · Matrices

Exercise 3.1 · Q7: Find a,b,c,da, b, c, d from equal matrices

Exercise 3.1Equality of matrices · Section 3.3.1
The question

Find the value of aa⁠, bb⁠, cc and dd from the equation:

[a−b2a+c2a−b3c+d]=[−15013]\begin{bmatrix} a - b & 2a + c \\ 2a - b & 3c + d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}
Exam answer

Equating corresponding entries,

a−b=−1…(1)2a+c=5…(2)2a−b=0…(3)3c+d=13…(4)\begin{aligned} a - b &= -1 \quad\ldots(1) \\ 2a + c &= 5 \quad\ldots(2) \\ 2a - b &= 0 \quad\ldots(3) \\ 3c + d &= 13 \quad\ldots(4) \end{aligned}

(3) − (1): a=1a = 1⁠.

From (3): b=2a=2b = 2a = 2⁠.

From (2): c=5−2a=3c = 5 - 2a = 3⁠.

From (4): d=13−3c=4d = 13 - 3c = 4⁠.

∴ a=1\therefore\ a = 1⁠, b=2b = 2⁠, c=3c = 3⁠, d=4d = 4⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Equal matrices have equal entries at every address. That turns this one matrix equation into four ordinary equations. They can be solved as a chain: the first column involves only aa and bb⁠; once aa is known, the top-right entry gives cc⁠; once cc is known, the bottom-right entry gives dd⁠.

Given, to find, and the plan
Given
An equation between two 2×22\times2 matrices.
To find
aa⁠, bb⁠, cc⁠, dd⁠.
Plan
(1) Check the orders match and write the four entry equations. (2) Solve the first-column pair for aa and bb⁠. (3) Use aa to get cc⁠, then cc to get dd⁠. (4) Check all four entries.

Basics you need

1Equality of matrices

Same order, and aij=bija_{ij} = b_{ij} for every ii⁠, jj (Definition 2). Lesson: equality of matrices

2Elimination

If two equations contain the same term (here −b-b⁠), subtracting one from the other removes it, leaving one unknown. Basics: solving equations, one unknown at a time

3Substitution

Once an unknown is known, put its value into another equation to find the next one.

Solution

  1. Same order, then equate entries

    Both matrices are 2×22\times2⁠, so equality is possible. Matching the four addresses:

    (1,1):a−b=−1(1)(1,2):2a+c=5(2)(2,1):2a−b=0(3)(2,2):3c+d=13(4)\begin{aligned} (1,1):&\quad a - b = -1 &\text{(1)} \\ (1,2):&\quad 2a + c = 5 &\text{(2)} \\ (2,1):&\quad 2a - b = 0 &\text{(3)} \\ (2,2):&\quad 3c + d = 13 &\text{(4)} \end{aligned}
    By Definition 2, the two matrices are equal exactly when all four of these hold together. So solving the system (1)–(4) is the same problem as solving the matrix equation.
    Equal matrices: entry (i,j)(i,j) on the left equals entry (i,j)(i,j) on the right. Lesson: equality of matrices

  2. Find aa by eliminating bb

    Equations (1) and (3) both contain −b-b⁠. Subtract (1) from (3):

    (2a−b)−(a−b)=0−(−1)  ⇒  a=1.(2a - b) - (a - b) = 0 - (-1) \;\Rightarrow\; a = 1.
    On the left, 2a−a=a2a - a = a and −b−(−b)=0-b - (-b) = 0⁠, so bb disappears. On the right, 0−(−1)=10 - (-1) = 1⁠. Subtracting an equation from another keeps the result true, because we subtract equal amounts from equal amounts.
    Subtracting a negative: 0−(−1)=0+1=10 - (-1) = 0 + 1 = 1⁠. Basics: elimination

  3. Find bb

    Put a=1a = 1 into (3): 2(1)−b=02(1) - b = 0⁠, so b=2b = 2⁠.

    Equation (3) now has only one unknown. Adding bb to both sides of 2−b=02 - b = 0 gives 2=b2 = b⁠.

  4. Find cc

    Put a=1a = 1 into (2): 2(1)+c=52(1) + c = 5⁠, so c=5−2=3c = 5 - 2 = 3⁠.

    Equation (2) involves only aa and cc⁠, and aa is known now, so it gives cc directly.

  5. Find dd

    Put c=3c = 3 into (4): 3(3)+d=133(3) + d = 13⁠, so d=13−9=4d = 13 - 9 = 4⁠.

    Equation (4) involves only cc and dd⁠. Every step of the chain was forced, with no choices, so the solution is unique.

  6. Check all four entries

    With a=1a = 1⁠, b=2b = 2⁠, c=3c = 3⁠, d=4d = 4⁠:

    a−b=1−2=−12a+c=2+3=52a−b=2−2=03c+d=9+4=13\begin{aligned} a - b &= 1 - 2 = -1 \\ 2a + c &= 2 + 3 = 5 \\ 2a - b &= 2 - 2 = 0 \\ 3c + d &= 9 + 4 = 13 \end{aligned}

    so the left matrix becomes

    [−15013],\begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix},

    exactly the right side ✓. Hence a=1a = 1⁠, b=2b = 2⁠, c=3c = 3⁠, d=4d = 4.

    Checking in the original equation, entry by entry, confirms the values make the matrices equal, not just that some of the equations hold. Equation (1) was used only through the subtraction, so checking it directly catches any slip made there.
Result

a=1a = 1⁠, b=2b = 2⁠, c=3c = 3⁠, d=4d = 4⁠. (Textbook answer: a=1a = 1⁠, b=2b = 2⁠, c=3c = 3⁠, d=4d = 4⁠.)

Check it with numbers

Another route to aa and bb⁠: from (1), b=a+1b = a + 1⁠; put it into (3): 2a−(a+1)=02a - (a + 1) = 0⁠, so a=1a = 1 and b=2b = 2 ✓. The four values also make the sum of the entries on the left (−1)+5+0+13=17(-1) + 5 + 0 + 13 = 17⁠, equal to the sum on the right ✓.

Common mistakes
  1. Sign slip in the elimination: 0−(−1)0 - (-1) is +1+1⁠, not −1-1⁠.
  2. Pairing the wrong equations, for instance trying to solve (1) and (2) as a pair. Together they contain three unknowns, aa⁠, bb and cc⁠, so two equations cannot fix all three. Equations (1) and (3) share the same two unknowns, which is why they are the pair to start with.
  3. Skipping the check. A slip in Step 2 would carry through to bb⁠, cc and dd⁠.
What to take away
  1. Write one equation per entry, number them, and look for the pair that involves the fewest unknowns.
  2. Solve in a chain, each new equation bringing in only one new unknown.
  3. Check by rebuilding the matrix from your values.

Every fact used, and where it comes from

FactUsed inSource
Definition of equal matricesStep 1Definition 2, Section 3.3.1 (Lesson: equality of matrices)
Equating entries of a 2×22\times2 equation to find four unknownsSteps 1–5Example 5
Elimination and substitutionSteps 2–5Linear equations (Basics: solving equations)
Solution a=1,b=2,c=3,d=4a = 1, b = 2, c = 3, d = 4Step 6Recomputed in the site's verification script
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