Class 12 Maths, Worked in Full Past papersGet updates
Chapter 3 · Matrices

Example 5: Find a,b,c,da, b, c, d from a matrix equation

ExamplesEquality of matrices · Section 3.3.1
The question

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

[2a+ba−2b5c−d4c+3d]=[4−31124]\begin{bmatrix} 2a+b & a-2b \\ 5c-d & 4c+3d \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ 11 & 24 \end{bmatrix}
Exam answer

Equating the corresponding entries,

2a+b=4…(i)a−2b=−3…(ii)5c−d=11…(iii)4c+3d=24…(iv)\begin{aligned} 2a + b &= 4 \quad\ldots(\text{i}) \\ a - 2b &= -3 \quad\ldots(\text{ii}) \\ 5c - d &= 11 \quad\ldots(\text{iii}) \\ 4c + 3d &= 24 \quad\ldots(\text{iv}) \end{aligned}

2×2\times(i) ++ (ii): 5a=55a = 5⁠, so a=1a = 1⁠. From (i), b=4−2=2b = 4 - 2 = 2⁠.

From (iii), d=5c−11d = 5c - 11⁠. Putting this in (iv):

4c+3(5c−11)=2419c=57⇒c=3\begin{aligned} 4c + 3(5c - 11) &= 24 \\ 19c &= 57 \Rightarrow c = 3 \end{aligned}

Then d=5(3)−11=4d = 5(3) - 11 = 4⁠.

∴ a=1, b=2, c=3, d=4\therefore\ a = 1,\ b = 2,\ c = 3,\ d = 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 in every position, so this one matrix equation is four ordinary equations. Unlike Example 4, each unknown now appears in two entries, so the equations come in pairs: row 1 is a pair of simultaneous equations in a,ba, b⁠, and row 2 is a pair in c,dc, d⁠. Each pair is solved by the usual school methods.

Given, to find, and the plan
Given
The equality of two 2×22\times2 matrices.
To find
a,b,c,da, b, c, d⁠.
Plan
Check the orders, write the four entry equations, solve the (a,b)(a, b) pair by elimination and the (c,d)(c, d) pair by substitution, then substitute back into all four entries.

Basics you need

1Equality of matrices

Same order, and the entries in each position are equal. Lesson: equality of matrices

2Two equations in two unknowns

Elimination: multiply one equation so that adding (or subtracting) the two removes one unknown. Substitution: make one unknown the subject of one equation and put that into the other. Either way you are left with one equation in one unknown. Basics: solving equations, one unknown at a time

Solution

  1. Check the orders and write the four equations

    Both matrices are 2×22\times2⁠. Equating the entries in each position:

    (1,1):2a+b=4(1,2):a−2b=−3(2,1):5c−d=11(2,2):4c+3d=24\begin{aligned} (1,1):&\quad 2a + b = 4\\ (1,2):&\quad a - 2b = -3\\ (2,1):&\quad 5c - d = 11\\ (2,2):&\quad 4c + 3d = 24 \end{aligned}

    The first two involve only a,ba, b⁠; the last two involve only c,dc, d⁠. So we can solve two separate small systems.

    Equality of matrices is defined entry by entry, so the matrix equation says exactly these four things and nothing more. Noticing which unknowns appear where splits a four-unknown problem into two two-unknown problems.
    Definition 2, Section 3.3.1. Lesson: equality of matrices

  2. Solve for aa and bb by elimination

    Multiply 2a+b=42a + b = 4 by 2, so that bb gets the coefficient 2, then add a−2b=−3a - 2b = -3⁠:

    4a+2b=8a−2b=−35a=5\begin{array}{rcr} 4a + 2b &=& 8\\ a - 2b &=& -3\\ \hline 5a &=& 5 \end{array}

    So a=1a = 1⁠. Put this into 2a+b=42a + b = 4⁠: 2+b=42 + b = 4⁠, so b=2b = 2⁠.

    The coefficients of bb are +1+1 and −2-2⁠. Doubling the first makes them +2+2 and −2-2⁠, and adding the equations then cancels bb completely, leaving one equation in aa alone. Multiplying both sides of an equation by 2, and adding two true equations, both give true equations.
    Elimination: make the coefficients of one unknown opposite, then add. Basics: solving equations

  3. Solve for cc and dd by substitution

    From 5c−d=115c - d = 11⁠, make dd the subject: d=5c−11d = 5c - 11⁠. Substitute into 4c+3d=244c + 3d = 24⁠:

    4c+3(5c−11)=244c+15c−33=2419c=57c=3.\begin{aligned} 4c + 3(5c - 11) &= 24\\ 4c + 15c - 33 &= 24\\ 19c &= 57\\ c &= 3. \end{aligned}

    Then d=5⋅3−11=4d = 5\cdot3 - 11 = 4⁠.

    In 5c−d=115c - d = 11 the unknown dd has coefficient −1-1⁠, so it can be isolated without fractions. Substituting turns the second equation into one with only cc⁠. We expand the bracket (3⋅5c=15c3\cdot5c = 15c⁠, 3⋅(−11)=−333\cdot(-11) = -33⁠), collect like terms, add 33 to both sides and divide by 19.
    Substitution: replace an unknown by an expression equal to it. Basics: solving equations

  4. Substitute back into all four entries

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

    2a+b=2+2=4,a−2b=1−4=−3,5c−d=15−4=11,4c+3d=12+12=24.\begin{aligned} 2a + b &= 2 + 2 = 4,\\ a - 2b &= 1 - 4 = -3,\\ 5c - d &= 15 - 4 = 11,\\ 4c + 3d &= 12 + 12 = 24. \end{aligned}

    So the left matrix becomes

    [4−31124],\begin{bmatrix} 4 & -3 \\ 11 & 24 \end{bmatrix},

    exactly the right-hand matrix.

    We used only one equation of each pair to find the second unknown, so checking all four confirms nothing went wrong. Each pair has exactly one solution, because its coefficients are not in proportion (21≠1−2\frac21 \ne \frac{1}{-2} for the a,ba, b pair, 54≠−13\frac54 \ne \frac{-1}{3} for the c,dc, d pair), so these are the only values.
Result

a=1, b=2, c=3, d=4a = 1,\ b = 2,\ c = 3,\ d = 4. This agrees with the textbook's solution.

Check it with numbers

Solve the (a,b)(a, b) pair the other way, by substitution: b=4−2ab = 4 - 2a⁠, so a−2(4−2a)=−3a - 2(4 - 2a) = -3⁠, giving 5a−8=−35a - 8 = -3⁠, a=1a = 1⁠, b=2b = 2⁠. The same answer by a different route.

Common mistakes
  1. Mixing the pairs. Each pair of equations comes from one row. 2a+b=42a + b = 4 belongs with a−2b=−3a - 2b = -3⁠, not with 5c−d=115c - d = 11⁠.
  2. Sign error in the bracket. 3(5c−11)=15c−333(5c - 11) = 15c - 33⁠, not 15c−1115c - 11⁠.
  3. Stopping after one unknown. Once aa is found, substitute to get bb⁠; likewise dd from cc⁠.
What to take away
  1. Matrix equality turns into ordinary equations, one per entry; look for which unknowns share equations.
  2. When an unknown appears in two entries, solve the resulting system by elimination or substitution.
  3. Substitute back into every entry to confirm.

Every fact used, and where it comes from

FactUsed inSource
Equal matrices: same order and equal corresponding entriesStep 1Definition 2, Section 3.3.1 (Lesson: equality of matrices)
Elimination and substitution for two linear equationsSteps 2, 3Linear equations in two variables (Basics: solving equations)
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →