Matrix equations / solving for X

MATMatrix Operations

These problems ask you to find an unknown matrix $X$ from equations such as $2X+3A=B$ or $AX=B$ , treating matrices algebraically while respecting non-commutativity. Simple equations are rearranged entrywise; multiplicative equations $AX=B$ are solved by left-multiplying by $A^{-1}$ .

A special favourite uses a matrix's own characteristic relation (e.g. $A^2-5A+7I=O$ ) to derive $A^{-1}$ without inverting directly.

Linear matrix equation

2X + 3A = B \Rightarrow X = \tfrac{1}{2}(B - 3A)

A,B

same order; rearrange like ordinary algebra (no inverses needed) since only addition and scalar multiplication are involved.

Solving AX = B

AX = B \Rightarrow X = A^{-1}B \quad (|A| \ne 0)

Left-multiply by

A^{-1}

; keep

A^{-1}

on the LEFT. For

XA=B

instead,

X=BA^{-1}

(inverse on the right).

Inverse from a matrix polynomial

A^2 - 5A + 7I = O \Rightarrow A^{-1} = \tfrac{1}{7}(5I - A)

Multiply the relation by

A^{-1}

and rearrange; the constant term must be nonzero so

A^{-1}

exists. General: from

A^2=pA-qI

get

A^{-1}=\frac{1}{q}(pI-A)

2x2 inverse formula

A = \begin{pmatrix} a & b \\ c & d \end{pmatrix},\quad A^{-1} = \dfrac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

Used to compute

A^{-1}

when solving

AX=B

; requires

ad-bc \ne 0

For additive equations like $2X+3A=B$ , isolate $X$ exactly as in scalar algebra, then compute entrywise.
For $AX=B$ multiply on the LEFT by $A^{-1}$ ; for $XA=B$ multiply on the RIGHT — side matters because matrices don't commute.
Always confirm $|A|\ne 0$ before using $A^{-1}$ ; a singular $A$ may give no solution or infinitely many.
To get $A^{-1}$ from $A^2-5A+7I=O$ , multiply throughout by $A^{-1}$ : $A-5I+7A^{-1}=O$ , hence $A^{-1}=\frac{1}{7}(5I-A)$ .
Carry the identity $I$ explicitly; the constant in a matrix equation is a scalar times $I$ , not a bare number.
The order of $X$ is forced by conformability: in $AX=B$ , $X$ has rows = columns of $A$ and columns = columns of $B$ .
Verify the final $X$ by substituting back into the original equation.

Where the marks go

Solving $AX=B$ as $X=BA^{-1}$ (inverse on the wrong side) — must be $X=A^{-1}B$ .
Treating the constant in $A^2-5A+7I=O$ as a plain number $7$ instead of $7I$ .
Dividing by a matrix (matrices have no division) instead of multiplying by the inverse.
Using $A^{-1}$ without first checking $|A|\ne 0$ , or assuming a solution exists when $A$ is singular.

How the board asks it

Numericaladditive equation $2X+3A=B$ isolated entrywise
Given $A=\begin{pmatrix}1&2\\3&4\end{pmatrix}$ and $B=\begin{pmatrix}5&8\\11&14\end{pmatrix}$ , find the matrix $X$ such that $2X+3A=B$ .
Numerical $AX=B$ solved by left-multiplying by $A^{-1}$
If $A=\begin{pmatrix}2&3\\1&2\end{pmatrix}$ and $B=\begin{pmatrix}4\\3\end{pmatrix}$ , solve the matrix equation $AX=B$ for the matrix $X$ .
Numerical $XA=B$ solved by right-multiplying by $A^{-1}$
Find the matrix $X$ satisfying $XA=B$ , where $A=\begin{pmatrix}1&-1\\2&3\end{pmatrix}$ and $B=\begin{pmatrix}1&1\\3&7\end{pmatrix}$ , stating clearly the order of $X$ .
Numericalinverse from the characteristic relation $A^2-5A+7I=O$
If $A=\begin{pmatrix}3&1\\-1&2\end{pmatrix}$ satisfies $A^2-5A+7I=O$ , use this relation to obtain $A^{-1}$ and hence solve $AX=I$ .
Give reasonschecking $|A|\ne 0$ before using $A^{-1}$
For the equation $AX=B$ with $A=\begin{pmatrix}2&4\\1&2\end{pmatrix}$ , state with reasons whether $A^{-1}$ can be used to obtain a unique matrix $X$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.