Q.Find the values of $a$, $b$, $c$ and $d$, if $3\begin{bmatrix}a & b\\ c & d\end{bmatrix} = \begin{bmatrix}a & 6\\ -1 & 2d\end{bmatrix} + \begin{bmatrix}4 & a+b\\ c+d & 3\end{bmatrix}$.

We are given a row vector $A = \begin{bmatrix} x & 1 & -2 \end{bmatrix}$ and a $3\times3$ matrix $B$. The product $A B A^t$ is a $1\times1$ matrix (a scalar), and we are told it equals $[-20]$. Our job is to find $x$.

Concept & Intuition

The expression $A B A^t$ is a quadratic form in the entries of $A$. Since $A$ is $1\times3$ and $B$ is $3\times3$, $A B$ is a $1\times3$ row vector; multiplying that by $A^t$ (a $3\times1$ column) yields a single number. We can compute step by step: first multiply $A$ by $B$, then dot the result with $A^t$. This will give a quadratic equation in $x$.

Step-by-step solution

1. Compute $A B$ $A$ is $1\times3$, $B$ is $3\times3$, so $AB$ is $1\times3$.

Multiply:

• First column: $x\cdot1 + 1\cdot4 + (-2)\cdot7 = x + 4 - 14 = x - 10$
• Second column: $x\cdot2 + 1\cdot5 + (-2)\cdot8 = 2x + 5 - 16 = 2x - 11$
• Third column: $x\cdot3 + 1\cdot6 + (-2)\cdot9 = 3x + 6 - 18 = 3x - 12$ So

2. Multiply $(AB)$ by $A^t$ $A^t$ is the column vector $\begin{bmatrix} x \\ 1 \\ -2 \end{bmatrix}$.

This is a dot product:

Simplify term by term:

• First term: $x(x-10) = x^2 - 10x$
• Second term: $2x - 11$
• Third term: $-2(3x-12) = -6x + 24$ Sum:

3. Set equal to $-20$ and solve We are given $A B A^t = [-20]$, so

Bring all terms to one side: …

Step 1 — Test for a unique solution

For $x+2y=3$ and $2x+3y=3$, the coefficient determinant is

Since this determinant is non-zero, the system has a unique solution.

Step 2 — Solve to confirm

From the first equation, $x = 3-2y$. Substituting into the second: …

Step 1 — Why we may expand binomially.

$I$ commutes with every matrix ($AI = IA = A$), so $I$ and $A$ commute and the ordinary binomial expansion is valid:

Step 2 — Use idempotency to reduce the powers.

Given $A^{2} = A$. Then

So both $A^{2}$ and $A^{3}$ equal $A$.

Step 3 — Substitute and simplify.

Step 4 — A neat cross-check. …

Step 1 — The concept.

By definition, $A^{-1}$ is the unique matrix with $A\,A^{-1} = I$. So if we can manipulate the given equation into the shape

that something must be $A^{-1}$. (This also silently proves $A$ is invertible, which is why the trick works.)

Step 2 — Rearrange the given relation.

Step 3 — Factor out $A$ on the right.

Matrix multiplication distributes over subtraction, and $A\cdot I = A$, so

Hence

Step 4 — Divide by the scalar and read off the inverse.

Comparing with $A A^{-1} = I$:

Step 5 — Verify (always check on an inverse question). …

Step 1 — The key property.

From the fundamental identity

take determinants of both sides:

(the last step uses $|kI_n| = k^n$). Dividing by $|A| \ne 0$:

Step 2 — Specialise to $n = 2$.

Here $B = \operatorname{adj}A$ is $2\times2$, so $A$ is $2\times2$ too, giving $n = 2$:

(A nice special fact: for a $2\times2$ matrix, the adjoint has the same determinant as the matrix.)

Step 3 — Compute $|B|$ directly from its entries.

Step 4 — Equate and solve.

Step 5 — Sanity check. …

Step 1 — Square the matrix.

Step 2 — Why this makes all powers easy.

Because $A^2 = 2A$ (a scalar times $A$), multiplying again by $A$ just pulls out another factor of $2$:

Step 3 — Induction.

Claim: $A^{n} = 2^{\,n-1}A$ for $n \ge 1$. True for $n=1$ ($A^1 = 2^0A$). If $A^{k}=2^{\,k-1}A$, then …

We are given two matrix equations involving $A$ and $B$ (both are $2 \times 3$ matrices, as we can infer from the dimensions of the given matrices). The trick is to treat these as simultaneous equations in the unknown matrices, but we must handle the transpose operation carefully.

1. Understand the dimensions and the unknowns

The first equation:

The right-hand side is $2 \times 3$, so $3A$ and $4B^t$ must also be $2 \times 3$. Hence $A$ is $2 \times 3$ and $B^t$ is $2 \times 3$, meaning $B$ is $3 \times 2$.

The second equation:

The right-hand side is $3 \times 2$, so $2B$ and $3A^t$ are $3 \times 2$. Since $A$ is $2 \times 3$, $A^t$ is $3 \times 2$, consistent.

So we have two matrix equations:

2. Eliminate $A$ by taking a transpose

If we transpose the second equation, we get:

Since $(B)^t = B^t$ and $(A^t)^t = A$, this becomes:

Now $N^t$ is a $2 \times 3$ matrix (transpose of the given $3 \times 2$ matrix):

So we now have two equations in the unknowns $A$ and $B^t$ (both $2 \times 3$):

3. Solve the linear system for $B^t$

Add the two equations to eliminate $A$:

Now compute $M + N^t$:

We are given $A M = B$, where

Since $A M = B$, isolate $A$ by right-multiplying with $M^{-1}$:

Find $M^{-1}$. The determinant is

So

Multiply.

Entry by entry: …

We are given a product of three matrices that equals the $1 \times 1$ matrix $[0]$. That means the entire expression is just a number — zero. The trick is to multiply step by step, keeping careful track of dimensions: the first is $1 \times 3$, the second is $3 \times 3$, the third is $3 \times 1$, so the result is $1 \times 1$.

Let’s denote:

We have $A (B C) = 0$ or $(A B) C = 0$ — matrix multiplication is associative, so we can choose whichever order is simpler.

1. First multiply $B$ and $C$ (the $3 \times 3$ times the $3 \times 1$):

2. Now multiply $A$ by that result (a $1 \times 3$ times a $3 \times 1$):

3. Simplify the expression:

4. Set equal to zero (since the product is $[0]$):

5. Solve the quadratic:

Step 1 — Given.

Step 2 — Show $A^2=A$.

Since $BA=A$, replace the second $A$:

Matrix multiplication is associative, so

Hence $A^2=A$.

Step 3 — Show $B^2=B$ (same trick).

Since $AB=B$, replace the second $B$:

$B^2=B\cdot B=B(AB)=(BA)B=A\,B=B$ …

We have

Multiply the second equation by 2:

Subtract the first equation:

$(2A+4B)-(2A+3B)=B=\begin{bmatrix}10-2&0-(-1)&6-4\\2-3&12-2&4-5\end{bmatrix}=\begin{bmatrix}8&1&2\\-1&10&-1\end{bmatrix}.$ …

Step 1 — Check (A): $\operatorname{adj}A = |A|\,A^{-1}$.

The defining property of the adjoint is $A\,(\operatorname{adj}A) = |A|\,I$. Pre-multiplying by $A^{-1}$ (which exists since $A$ is invertible):

Step 2 — Check (B): $\det(A^{-1}) = [\det A]^{-1}$.

From $AA^{-1} = I$ and the multiplicative property of determinants,

Step 3 — Check (C): $(AB)^{-1} = B^{-1}A^{-1}$ (the reversal law).

and likewise on the other side. So $B^{-1}A^{-1}$ is indeed the inverse of $AB$. TRUE

Step 4 — Check (D): $(A+B)^{-1} = B^{-1}+A^{-1}$.

This is false in general. A concrete counterexample: take $A = B = I_2$. Then …